Chapter 3 Matrices

Given:

Information regarding the number of men and women workers in three factories I, II, and III is provided in a table format.

To Represent and Find:

Represent the given information in the form of a $3 \times 2$ matrix and interpret the entry in the third row and second column.

Solution:

A matrix is a rectangular arrangement of numbers or functions.

The information can be represented as a matrix where the rows represent the factories (I, II, III) and the columns represent the categories of workers (Men, Women).

Factory I has 30 men and 25 women.

Factory II has 25 men and 31 women.

Factory III has 27 men and 26 women.

Arranging these numbers in a $3 \times 2$ matrix, where the first column is Men Workers and the second column is Women Workers, and the rows are Factory I, Factory II, and Factory III, respectively, we get:

$\begin{bmatrix} 30 & 25 \\ 25 & 31 \\ 27 & 26 \end{bmatrix}$

This is a $3 \times 2$ matrix as it has 3 rows and 2 columns.

Now, we need to interpret the entry in the third row and second column.

The third row corresponds to Factory III.

The second column corresponds to Women Workers.

Therefore, the entry in the third row and second column represents the number of women workers in Factory III.

Looking at the matrix, the entry in the third row and second column is 26.

So, the entry in the third row and second column represents the number of women workers in Factory III, which is 26.

Given:

A matrix has 8 elements.

To Find:

The possible orders the matrix can have.

Solution:

The order of a matrix is given by $m \times n$, where $m$ is the number of rows and $n$ is the number of columns.

The total number of elements in a matrix of order $m \times n$ is the product of the number of rows and the number of columns, i.e., $m \times n$.

In this case, the matrix has 8 elements. Therefore, the product of the number of rows ($m$) and the number of columns ($n$) must be equal to 8.

$m \times n = 8$

We need to find all possible pairs of positive integers $(m, n)$ whose product is 8.

The pairs of positive integers whose product is 8 are the pairs of factors of 8.

The factors of 8 are 1, 2, 4, and 8.

The possible pairs $(m, n)$ such that $m \times n = 8$ are:

• $m=1$, $n=8$. The order is $1 \times 8$.
• $m=8$, $n=1$. The order is $8 \times 1$.
• $m=2$, $n=4$. The order is $2 \times 4$.
• $m=4$, $n=2$. The order is $4 \times 2$.

These are the only possible pairs of positive integers whose product is 8.

The possible orders a matrix with 8 elements can have are $1 \times 8$, $8 \times 1$, $2 \times 4$, and $4 \times 2$.

Given:

The rule for the elements of a matrix $A = [a_{ij}]$ is given by $a_{ij} = \frac{1}{2} |i - 3j|$.

To Construct:

A $3 \times 2$ matrix $A$.

Solution:

A $3 \times 2$ matrix has 3 rows and 2 columns. Let the matrix be denoted by $A$. Its general form is:

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}$

We need to calculate each element $a_{ij}$ using the given formula $a_{ij} = \frac{1}{2} |i - 3j|$, where $i$ represents the row number ($i = 1, 2, 3$) and $j$ represents the column number ($j = 1, 2$).

For the first row ($i=1$):

• For $j=1$: $a_{11} = \frac{1}{2} |1 - 3(1)| = \frac{1}{2} |1 - 3| = \frac{1}{2} |-2| = \frac{1}{2}(2) = 1$
• For $j=2$: $a_{12} = \frac{1}{2} |1 - 3(2)| = \frac{1}{2} |1 - 6| = \frac{1}{2} |-5| = \frac{1}{2}(5) = \frac{5}{2}$

For the second row ($i=2$):

• For $j=1$: $a_{21} = \frac{1}{2} |2 - 3(1)| = \frac{1}{2} |2 - 3| = \frac{1}{2} |-1| = \frac{1}{2}(1) = \frac{1}{2}$
• For $j=2$: $a_{22} = \frac{1}{2} |2 - 3(2)| = \frac{1}{2} |2 - 6| = \frac{1}{2} |-4| = \frac{1}{2}(4) = 2$

For the third row ($i=3$):

• For $j=1$: $a_{31} = \frac{1}{2} |3 - 3(1)| = \frac{1}{2} |3 - 3| = \frac{1}{2} |0| = \frac{1}{2}(0) = 0$
• For $j=2$: $a_{32} = \frac{1}{2} |3 - 3(2)| = \frac{1}{2} |3 - 6| = \frac{1}{2} |-3| = \frac{1}{2}(3) = \frac{3}{2}$

Now, we arrange these calculated elements into the $3 \times 2$ matrix:

$A = \begin{bmatrix} 1 & \frac{5}{2} \\ \frac{1}{2} & 2 \\ 0 & \frac{3}{2} \end{bmatrix}$

The constructed $3 \times 2$ matrix is $\begin{bmatrix} 1 & \frac{5}{2} \\ \frac{1}{2} & 2 \\ 0 & \frac{3}{2} \end{bmatrix}$.

Given:

The matrix equation is $\begin{bmatrix}x+3 & z+4 & 2y-7 \\ -6 & a-1 & 0 \\ b-3 & -21 & 0 \end{bmatrix} = \begin{bmatrix}0 & 6 & 3y − 2 \\ −6 & −3 & 2c + 2 \\ 2b + 4 & −21 & 0 \end{bmatrix}$.

To Find:

The values of $a, b, c, x, y,$ and $z$.

Solution:

Two matrices are equal if and only if they have the same order and their corresponding elements are equal.

The given matrices have the same order, which is $3 \times 3$.

Now, we equate the corresponding elements:

Equating the elements in the first row:

• $(1,1)$: $x+3 = 0 \implies x = -3$
• $(1,2)$: $z+4 = 6 \implies z = 6 - 4 \implies z = 2$
• $(1,3)$: $2y-7 = 3y - 2 \implies 2y - 3y = -2 + 7 \implies -y = 5 \implies y = -5$

Equating the elements in the second row:

• $(2,1)$: $-6 = -6$ (This is consistent and doesn't give a value for the variables.)
• $(2,2)$: $a-1 = -3 \implies a = -3 + 1 \implies a = -2$
• $(2,3)$: $0 = 2c + 2 \implies -2 = 2c \implies c = \frac{-2}{2} \implies c = -1$

Equating the elements in the third row:

• $(3,1)$: $b-3 = 2b + 4 \implies b - 2b = 4 + 3 \implies -b = 7 \implies b = -7$
• $(3,2)$: $-21 = -21$ (Consistent.)
• $(3,3)$: $0 = 0$ (Consistent.)

We have found the values for $a, b, c, x, y,$ and $z$ by equating the corresponding elements.

The values are $a = -2$, $b = -7$, $c = -1$, $x = -3$, $y = -5$, and $z = 2$.

Given:

The matrix equation is $\begin{bmatrix}2a + b& a − 2b \\ 5c − d& 4c + 3d \end{bmatrix} = \begin{bmatrix}4& −3 \\ 11& 24 \end{bmatrix}$.

To Find:

The values of $a, b, c,$ and $d$.

Solution:

Two matrices are equal if and only if they have the same order and their corresponding elements are equal.

The given matrices have the same order, which is $2 \times 2$.

Equating the corresponding elements, we get the following system of linear equations:

We have two independent systems of equations: one for $a$ and $b$, and one for $c$ and $d$.

Solving for $a$ and $b$:

From equation (2), we can express $a$ in terms of $b$: $a = 2b - 3$.

Substitute this expression for $a$ into equation (1):

$2(2b - 3) + b = 4$

$4b - 6 + b = 4$

$5b - 6 = 4$

$5b = 4 + 6$

$5b = 10$

$b = \frac{10}{5}$

Now substitute the value of $b$ from equation (5) back into the expression for $a$ ($a = 2b - 3$):

$a = 2(2) - 3$

$a = 4 - 3$

So, $a=1$ and $b=2$.

Solving for $c$ and $d$:

From equation (3), we can express $d$ in terms of $c$: $d = 5c - 11$.

Substitute this expression for $d$ into equation (4):

$4c + 3(5c - 11) = 24$

$4c + 15c - 33 = 24$

$19c - 33 = 24$

$19c = 24 + 33$

$19c = 57$

$c = \frac{57}{19}$

Now substitute the value of $c$ from equation (7) back into the expression for $d$ ($d = 5c - 11$):

$d = 5(3) - 11$

$d = 15 - 11$

So, $c=3$ and $d=4$.

The values are $a = 1$, $b = 2$, $c = 3$, and $d = 4$.

Given:

The matrix A is $A = \begin{bmatrix}2& 5& 19& -7 \\ 35& -2& \frac{5}{2} & 12 \\ \sqrt{3}& 1& -5& 17 \end{bmatrix}$.

To Find:

(i) The order of the matrix.

(ii) The number of elements in the matrix.

(iii) The elements $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$.

Solution:

The given matrix is $A = \begin{bmatrix}2& 5& 19& -7 \\ 35& -2& \frac{5}{2} & 12 \\ \sqrt{3}& 1& -5& 17 \end{bmatrix}$.

(i) The order of the matrix:

The order of a matrix is given by the number of rows by the number of columns ($m \times n$).

In matrix A, there are 3 rows and 4 columns.

So, the order of matrix A is $3 \times 4$.

(ii) The number of elements:

The number of elements in a matrix is the product of its number of rows and columns.

Number of elements = Number of rows $\times$ Number of columns $= 3 \times 4 = 12$.

So, there are 12 elements in matrix A.

(iii) The elements $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$:

The element $a_{ij}$ refers to the element in the $i$-th row and $j$-th column.

$a_{13}$: element in the 1st row and 3rd column is 19.

$a_{21}$: element in the 2nd row and 1st column is 35.

$a_{33}$: element in the 3rd row and 3rd column is -5.

$a_{24}$: element in the 2nd row and 4th column is 12.

$a_{23}$: element in the 2nd row and 3rd column is $\frac{5}{2}$.

Given:

A matrix has 24 elements in the first case and 13 elements in the second case.

To Find:

The possible orders the matrix can have in both cases.

Solution:

The order of a matrix is given by $m \times n$, where $m$ is the number of rows and $n$ is the number of columns.

The total number of elements in a matrix of order $m \times n$ is the product of the number of rows and the number of columns, i.e., $m \times n$.

Case 1: The matrix has 24 elements.

We need to find all possible pairs of positive integers $(m, n)$ whose product is 24.

$m \times n = 24$

The pairs of positive integers whose product is 24 are the pairs of factors of 24.

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

The possible pairs $(m, n)$ such that $m \times n = 24$ are:

• $m=1$, $n=24$. The order is $1 \times 24$.
• $m=24$, $n=1$. The order is $24 \times 1$.
• $m=2$, $n=12$. The order is $2 \times 12$.
• $m=12$, $n=2$. The order is $12 \times 2$.
• $m=3$, $n=8$. The order is $3 \times 8$.
• $m=8$, $n=3$. The order is $8 \times 3$.
• $m=4$, $n=6$. The order is $4 \times 6$.
• $m=6$, $n=4$. The order is $6 \times 4$.

These are the 8 possible orders for a matrix with 24 elements.

Case 2: The matrix has 13 elements.

We need to find all possible pairs of positive integers $(m, n)$ whose product is 13.

$m \times n = 13$

The number 13 is a prime number. The only positive integer factors of a prime number are 1 and the number itself.

The factors of 13 are 1 and 13.

The possible pairs $(m, n)$ such that $m \times n = 13$ are:

• $m=1$, $n=13$. The order is $1 \times 13$.
• $m=13$, $n=1$. The order is $13 \times 1$.

These are the only 2 possible orders for a matrix with 13 elements.

If a matrix has 24 elements, the possible orders are $1 \times 24, 24 \times 1, 2 \times 12, 12 \times 2, 3 \times 8, 8 \times 3, 4 \times 6,$ and $6 \times 4$.

If a matrix has 13 elements, the possible orders are $1 \times 13$ and $13 \times 1$.

Given:

A matrix has 18 elements in the first case and 5 elements in the second case.

To Find:

The possible orders the matrix can have in both cases.

Solution:

The order of a matrix is given by $m \times n$, where $m$ is the number of rows and $n$ is the number of columns.

The total number of elements in a matrix of order $m \times n$ is the product of the number of rows and the number of columns, i.e., $m \times n$.

Case 1: The matrix has 18 elements.

We need to find all possible pairs of positive integers $(m, n)$ whose product is 18.

$m \times n = 18$

The pairs of positive integers whose product is 18 are the pairs of factors of 18.

The factors of 18 are 1, 2, 3, 6, 9, and 18.

The possible pairs $(m, n)$ such that $m \times n = 18$ are:

• $m=1$, $n=18$. The order is $1 \times 18$.
• $m=18$, $n=1$. The order is $18 \times 1$.
• $m=2$, $n=9$. The order is $2 \times 9$.
• $m=9$, $n=2$. The order is $9 \times 2$.
• $m=3$, $n=6$. The order is $3 \times 6$.
• $m=6$, $n=3$. The order is $6 \times 3$.

These are the 6 possible orders for a matrix with 18 elements.

Case 2: The matrix has 5 elements.

We need to find all possible pairs of positive integers $(m, n)$ whose product is 5.

$m \times n = 5$

The number 5 is a prime number. The only positive integer factors of a prime number are 1 and the number itself.

The factors of 5 are 1 and 5.

The possible pairs $(m, n)$ such that $m \times n = 5$ are:

• $m=1$, $n=5$. The order is $1 \times 5$.
• $m=5$, $n=1$. The order is $5 \times 1$.

These are the only 2 possible orders for a matrix with 5 elements.

If a matrix has 18 elements, the possible orders are $1 \times 18, 18 \times 1, 2 \times 9, 9 \times 2, 3 \times 6,$ and $6 \times 3$.

If a matrix has 5 elements, the possible orders are $1 \times 5$ and $5 \times 1$.

Given:

The general element of a $2 \times 2$ matrix $A = [a_{ij}]$ is given by different formulas.

To Construct:

A $2 \times 2$ matrix for each of the given rules for $a_{ij}$.

Solution:

A $2 \times 2$ matrix has 2 rows and 2 columns. Its general form is:

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$

We need to calculate each element $a_{ij}$ using the given formula, where $i$ represents the row number ($i = 1, 2$) and $j$ represents the column number ($j = 1, 2$).

(i) a_ij = $\frac{(i + j)^2}{2}$

• For $i=1, j=1$: $a_{11} = \frac{(1 + 1)^2}{2} = \frac{(2)^2}{2} = \frac{4}{2} = 2$
• For $i=1, j=2$: $a_{12} = \frac{(1 + 2)^2}{2} = \frac{(3)^2}{2} = \frac{9}{2}$
• For $i=2, j=1$: $a_{21} = \frac{(2 + 1)^2}{2} = \frac{(3)^2}{2} = \frac{9}{2}$
• For $i=2, j=2$: $a_{22} = \frac{(2 + 2)^2}{2} = \frac{(4)^2}{2} = \frac{16}{2} = 8$

The matrix is $\begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}$.

(ii) a_ij = $\frac{i}{j}$

• For $i=1, j=1$: $a_{11} = \frac{1}{1} = 1$
• For $i=1, j=2$: $a_{12} = \frac{1}{2}$
• For $i=2, j=1$: $a_{21} = \frac{2}{1} = 2$
• For $i=2, j=2$: $a_{22} = \frac{2}{2} = 1$

The matrix is $\begin{bmatrix} 1 & \frac{1}{2} \\ 2 & 1 \end{bmatrix}$.

(iii) a_ij = $\frac{(i + 2j)^2}{2}$

• For $i=1, j=1$: $a_{11} = \frac{(1 + 2(1))^2}{2} = \frac{(1 + 2)^2}{2} = \frac{3^2}{2} = \frac{9}{2}$
• For $i=1, j=2$: $a_{12} = \frac{(1 + 2(2))^2}{2} = \frac{(1 + 4)^2}{2} = \frac{5^2}{2} = \frac{25}{2}$
• For $i=2, j=1$: $a_{21} = \frac{(2 + 2(1))^2}{2} = \frac{(2 + 2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8$
• For $i=2, j=2$: $a_{22} = \frac{(2 + 2(2))^2}{2} = \frac{(2 + 4)^2}{2} = \frac{6^2}{2} = \frac{36}{2} = 18$

The matrix is $\begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}$.

Given:

The general element of a $3 \times 4$ matrix $A = [a_{ij}]$ is given by different formulas.

To Construct:

A $3 \times 4$ matrix for each of the given rules for $a_{ij}$.

Solution:

A $3 \times 4$ matrix has 3 rows and 4 columns. Its general form is:

$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$

We need to calculate each element $a_{ij}$ using the given formula, where $i$ represents the row number ($i = 1, 2, 3$) and $j$ represents the column number ($j = 1, 2, 3, 4$).

(i) a_ij = $\frac{1}{2}$ |-3i + j|

For the first row ($i=1$):

• $j=1$: $a_{11} = \frac{1}{2} |-3(1) + 1| = \frac{1}{2} |-3 + 1| = \frac{1}{2} |-2| = \frac{1}{2}(2) = 1$
• $j=2$: $a_{12} = \frac{1}{2} |-3(1) + 2| = \frac{1}{2} |-3 + 2| = \frac{1}{2} |-1| = \frac{1}{2}(1) = \frac{1}{2}$
• $j=3$: $a_{13} = \frac{1}{2} |-3(1) + 3| = \frac{1}{2} |-3 + 3| = \frac{1}{2} |0| = \frac{1}{2}(0) = 0$
• $j=4$: $a_{14} = \frac{1}{2} |-3(1) + 4| = \frac{1}{2} |-3 + 4| = \frac{1}{2} |1| = \frac{1}{2}(1) = \frac{1}{2}$

For the second row ($i=2$):

• $j=1$: $a_{21} = \frac{1}{2} |-3(2) + 1| = \frac{1}{2} |-6 + 1| = \frac{1}{2} |-5| = \frac{1}{2}(5) = \frac{5}{2}$
• $j=2$: $a_{22} = \frac{1}{2} |-3(2) + 2| = \frac{1}{2} |-6 + 2| = \frac{1}{2} |-4| = \frac{1}{2}(4) = 2$
• $j=3$: $a_{23} = \frac{1}{2} |-3(2) + 3| = \frac{1}{2} |-6 + 3| = \frac{1}{2} |-3| = \frac{1}{2}(3) = \frac{3}{2}$
• $j=4$: $a_{24} = \frac{1}{2} |-3(2) + 4| = \frac{1}{2} |-6 + 4| = \frac{1}{2} |-2| = \frac{1}{2}(2) = 1$

For the third row ($i=3$):

• $j=1$: $a_{31} = \frac{1}{2} |-3(3) + 1| = \frac{1}{2} |-9 + 1| = \frac{1}{2} |-8| = \frac{1}{2}(8) = 4$
• $j=2$: $a_{32} = \frac{1}{2} |-3(3) + 2| = \frac{1}{2} |-9 + 2| = \frac{1}{2} |-7| = \frac{1}{2}(7) = \frac{7}{2}$
• $j=3$: $a_{33} = \frac{1}{2} |-3(3) + 3| = \frac{1}{2} |-9 + 3| = \frac{1}{2} |-6| = \frac{1}{2}(6) = 3$
• $j=4$: $a_{34} = \frac{1}{2} |-3(3) + 4| = \frac{1}{2} |-9 + 4| = \frac{1}{2} |-5| = \frac{1}{2}(5) = \frac{5}{2}$

The matrix is $\begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix}$.

(ii) a_ij = 2i - j

For the first row ($i=1$):

• $j=1$: $a_{11} = 2(1) - 1 = 2 - 1 = 1$
• $j=2$: $a_{12} = 2(1) - 2 = 2 - 2 = 0$
• $j=3$: $a_{13} = 2(1) - 3 = 2 - 3 = -1$
• $j=4$: $a_{14} = 2(1) - 4 = 2 - 4 = -2$

For the second row ($i=2$):

• $j=1$: $a_{21} = 2(2) - 1 = 4 - 1 = 3$
• $j=2$: $a_{22} = 2(2) - 2 = 4 - 2 = 2$
• $j=3$: $a_{23} = 2(2) - 3 = 4 - 3 = 1$
• $j=4$: $a_{24} = 2(2) - 4 = 4 - 4 = 0$

For the third row ($i=3$):

• $j=1$: $a_{31} = 2(3) - 1 = 6 - 1 = 5$
• $j=2$: $a_{32} = 2(3) - 2 = 6 - 2 = 4$
• $j=3$: $a_{33} = 2(3) - 3 = 6 - 3 = 3$
• $j=4$: $a_{34} = 2(3) - 4 = 6 - 4 = 2$

The matrix is $\begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}$.

Given:

Matrix equations are provided in three parts.

To Find:

The values of $x, y,$ and $z$ from the given equations.

Solution:

In each part, we use the property that two matrices are equal if and only if they have the same order and their corresponding elements are equal.

(i) $\begin{bmatrix}4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix}y & z \\ 1 & 5 \end{bmatrix}$

Both matrices are $2 \times 2$. Equating the corresponding elements:

• $(1,1)$: $4 = y$
• $(1,2)$: $3 = z$
• $(2,1)$: $x = 1$
• $(2,2)$: $5 = 5$ (Consistent)

From these equalities, we directly get the values of $x, y,$ and $z$.

$x = 1$

$y = 4$

$z = 3$

(ii) $\begin{bmatrix}x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix}6& 2\\5& 8 \end{bmatrix}$

Both matrices are $2 \times 2$. Equating the corresponding elements:

$(1,2)$: $2 = 2$ (Consistent)

From equation (2): $5 + z = 5 \implies z = 5 - 5 \implies z = 0$.

From equation (1), we can express $y$ in terms of $x$: $y = 6 - x$.

Substitute this into equation (3):

$x(6 - x) = 8$

$6x - x^2 = 8$

Rearrange into a quadratic equation:

$x^2 - 6x + 8 = 0$

Factor the quadratic equation:

$(x - 2)(x - 4) = 0$

This gives two possible values for $x$: $x = 2$ or $x = 4$.

If $x = 2$, substitute into $y = 6 - x$ to find $y$: $y = 6 - 2 = 4$. Check with equation (3): $xy = (2)(4) = 8$, which is correct.

If $x = 4$, substitute into $y = 6 - x$ to find $y$: $y = 6 - 4 = 2$. Check with equation (3): $xy = (4)(2) = 8$, which is correct.

So, there are two possible sets of values for $(x, y)$: $(2, 4)$ and $(4, 2)$. The value of $z$ is 0 in both cases.

The possible values are $x=2, y=4, z=0$ or $x=4, y=2, z=0$.

(iii) $\begin{bmatrix}x + y + z\\x + z\\y + z \end{bmatrix} = \begin{bmatrix}9\\5\\7 \end{bmatrix}$

Both matrices are $3 \times 1$. Equating the corresponding elements:

Substitute equation (2) into equation (1):

$(x + z) + y = 9$

$5 + y = 9$

$y = 9 - 5$

Substitute equation (3) into equation (1):

$x + (y + z) = 9$

$x + 7 = 9$

$x = 9 - 7$

Substitute the value of $x$ from equation (5) into equation (2):

$2 + z = 5$

$z = 5 - 2$

We can check these values by substituting $y=4$ into equation (3): $4+z=7 \implies z=3$, which is consistent.

Also, check with equation (1): $x+y+z = 2+4+3 = 9$, which is correct.

So, the values are $x=2$, $y=4$, and $z=3$.

Given:

The matrix equation is $\begin{bmatrix}a− b& 2a + c \\ 2a − b& 3c + d \end{bmatrix} = \begin{bmatrix}−1& 5\\0& 13 \end{bmatrix}$.

To Find:

The values of $a, b, c,$ and $d$.

Solution:

Two matrices are equal if and only if they have the same order and their corresponding elements are equal.

The given matrices have the same order, which is $2 \times 2$.

Equating the corresponding elements, we get the following system of linear equations:

We can solve the system of equations involving $a$ and $b$ first, using equations (1) and (3).

Subtract equation (1) from equation (3):

$(2a - b) - (a - b) = 0 - (-1)$

$2a - b - a + b = 0 + 1$

$a = 1$

Substitute the value of $a$ from equation (5) into equation (1):

$1 - b = -1$

$-b = -1 - 1$

$-b = -2$

Now, we use the value of $a$ from equation (5) to find $c$ from equation (2):

$2(1) + c = 5$

$2 + c = 5$

$c = 5 - 2$

Finally, use the value of $c$ from equation (7) to find $d$ from equation (4):

$3(3) + d = 13$

$9 + d = 13$

$d = 13 - 9$

We can verify the values by substituting $a=1$ and $b=2$ into equation (3): $2(1) - 2 = 2 - 2 = 0$, which is correct.

The values are $a = 1$, $b = 2$, $c = 3$, and $d = 4$.

Given:

A matrix $A = [a_{ij}]_{m \times n}$ is a square matrix.

To Determine:

The condition for a matrix to be a square matrix.

Solution:

A square matrix is defined as a matrix in which the number of rows is equal to the number of columns.

For the matrix $A = [a_{ij}]_{m \times n}$, $m$ represents the number of rows and $n$ represents the number of columns.

According to the definition of a square matrix, the number of rows must be equal to the number of columns.

Substituting the given notation, we have:

$m = n$

Comparing this condition with the given options:

(A) $m < n$ (Number of rows is less than the number of columns - This is a rectangular matrix)

(B) $m > n$ (Number of rows is greater than the number of columns - This is a rectangular matrix)

(C) $m = n$ (Number of rows is equal to the number of columns - This is a square matrix)

(D) None of these

The condition for a matrix to be a square matrix is $m = n$.

The correct option is (C).

Given:

Two matrices are given as $\begin{bmatrix}3x + 7& 5 \\ y + 1& 2-3x \end{bmatrix}$ and $\begin{bmatrix}0& y − 2 \\ 8& 4 \end{bmatrix}$. We are asked to find the values of $x$ and $y$ that make these matrices equal.

To Find:

The values of $x$ and $y$ for which the given pair of matrices are equal.

Solution:

Two matrices are equal if and only if they have the same order and their corresponding elements are equal.

Both given matrices are $2 \times 2$. Equating the corresponding elements, we get the following equations:

Solve equation (1) for $x$:

$3x = -7$

Solve equation (4) for $x$:

$-3x = 4 - 2$

$-3x = 2$

We have obtained two different values for $x$ from two different equations derived from the equality of corresponding elements. For the matrices to be equal, the value of $x$ must be the same in both cases.

Since $-\frac{7}{3} \ne -\frac{2}{3}$, there is no single value of $x$ that satisfies both equations (1) and (4) simultaneously.

Let's also solve for $y$ from equations (2) and (3).

From equation (2):

$y = 5 + 2$

From equation (3):

$y = 8 - 1$

The value of $y$ obtained from equations (2) and (3) is consistent, $y=7$.

However, since the values of $x$ are not consistent across all corresponding equations, there are no values of $x$ and $y$ that can make the given pair of matrices equal.

Let's examine the given options based on our findings:

(A) x = $-\frac{1}{3}$, y = 7: This would make the $(1,1)$ elements $3(-\frac{1}{3})+7 = -1+7 = 6$ and $0$. Not equal.

(B) not possible to find: This aligns with our finding that no consistent values exist.

(C) y = 7, x = $-\frac{2}{3}$: This would make the $(1,1)$ elements $3(-\frac{2}{3})+7 = -2+7 = 5$ and $0$. Not equal. This would make the $(1,2)$ elements $5$ and $7-2=5$. Equal. This would make the $(2,1)$ elements $7+1=8$ and $8$. Equal. This would make the $(2,2)$ elements $2-3(-\frac{2}{3}) = 2+2=4$ and $4$. Equal. However, the $(1,1)$ elements are not equal.

(D) x = $-\frac{1}{3}$, y = $-\frac{2}{3}$: This would make the $(1,1)$ elements $3(-\frac{1}{3})+7 = -1+7 = 6$ and $0$. Not equal.

Since equating the corresponding elements leads to contradictory values for $x$, the given pair of matrices cannot be equal for any values of $x$ and $y$.

The correct option is (B).

Given:

We are considering matrices of order $3 \times 3$, where each entry can be either 0 or 1.

To Find:

The total number of all possible matrices of order $3 \times 3$ with each entry being 0 or 1.

Solution:

A matrix of order $3 \times 3$ has 3 rows and 3 columns. The total number of elements in a $3 \times 3$ matrix is the product of the number of rows and the number of columns.

Number of elements = $3 \times 3 = 9$.

Each of these 9 elements can be chosen independently from the set of possible entries, which is $\{0, 1\}$.

For each of the 9 positions in the matrix, there are 2 choices (either 0 or 1).

Since the choice for each element is independent, the total number of possible matrices is the product of the number of choices for each element.

Total number of possible matrices = (Number of choices for the first element) $\times$ (Number of choices for the second element) $\times \dots \times$ (Number of choices for the ninth element)

Total number of possible matrices = $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

This can be written in exponential form as $2^9$.

Calculate $2^9$:

$2^1 = 2$

$2^2 = 4$

$2^3 = 8$

$2^4 = 16$

$2^5 = 32$

$2^6 = 64$

$2^7 = 128$

$2^8 = 256$

$2^9 = 512$

So, there are 512 possible matrices of order $3 \times 3$ with each entry 0 or 1.

Comparing this result with the given options:

(A) 27

(B) 18

(C) 81

(D) 512

The result matches option (D).

The correct option is (D).

Given:

Matrix A = $\begin{bmatrix}\sqrt{3}& 1& −1 \\ 2& 3& 0 \end{bmatrix}$ and Matrix B = $\begin{bmatrix}2& \sqrt{5}& 1\\−2& 3& \frac{1}{2} \end{bmatrix}$.

To Find:

The sum of matrices A and B, i.e., A + B.

Solution:

To add two matrices, they must have the same order. Matrix A has 2 rows and 3 columns, so its order is $2 \times 3$. Matrix B also has 2 rows and 3 columns, so its order is $2 \times 3$. Since they have the same order, we can add them.

Matrix addition is performed by adding the corresponding elements of the matrices.

Let $A = [a_{ij}]$ and $B = [b_{ij}]$. Then $A + B = [a_{ij} + b_{ij}]$.

$A + B = \begin{bmatrix}\sqrt{3}& 1& −1 \\ 2& 3& 0 \end{bmatrix} + \begin{bmatrix}2& \sqrt{5}& 1\\−2& 3& \frac{1}{2} \end{bmatrix}$

Add the corresponding elements:

• Element (1,1): $\sqrt{3} + 2$
• Element (1,2): $1 + \sqrt{5}$
• Element (1,3): $-1 + 1 = 0$
• Element (2,1): $2 + (-2) = 2 - 2 = 0$
• Element (2,2): $3 + 3 = 6$
• Element (2,3): $0 + \frac{1}{2} = \frac{1}{2}$

So, the resulting matrix $A + B$ is:

$A + B = \begin{bmatrix}\sqrt{3} + 2& 1 + \sqrt{5}& 0 \\ 0& 6& \frac{1}{2} \end{bmatrix}$

The sum of matrices A and B is $\begin{bmatrix}\sqrt{3} + 2& 1 + \sqrt{5}& 0 \\ 0& 6& \frac{1}{2} \end{bmatrix}$.

Given:

Matrix A = $\begin{bmatrix}1& 2& 3\\2& 3& 1 \end{bmatrix}$ and Matrix B = $\begin{bmatrix}3& −1& 3\\−1& 0& 2 \end{bmatrix}$.

To Find:

The result of $2A - B$.

Solution:

First, we need to calculate the scalar multiple of matrix A, which is $2A$. Scalar multiplication is performed by multiplying each element of the matrix by the scalar.

$2A = 2 \begin{bmatrix}1& 2& 3\\2& 3& 1 \end{bmatrix} = \begin{bmatrix}2 \times 1& 2 \times 2& 2 \times 3\\2 \times 2& 2 \times 3& 2 \times 1 \end{bmatrix} = \begin{bmatrix}2& 4& 6\\4& 6& 2 \end{bmatrix}$

Now we need to find $2A - B$. Matrix subtraction is performed by subtracting the corresponding elements of the matrices. For subtraction to be possible, the matrices must have the same order. Both $2A$ (which is the same order as A) and B are $2 \times 3$ matrices, so we can subtract them.

$2A - B = \begin{bmatrix}2& 4& 6\\4& 6& 2 \end{bmatrix} - \begin{bmatrix}3& −1& 3\\−1& 0& 2 \end{bmatrix}$

Subtract the corresponding elements:

• Element (1,1): $2 - 3 = -1$
• Element (1,2): $4 - (-1) = 4 + 1 = 5$
• Element (1,3): $6 - 3 = 3$
• Element (2,1): $4 - (-1) = 4 + 1 = 5$
• Element (2,2): $6 - 0 = 6$
• Element (2,3): $2 - 2 = 0$

So, the resulting matrix $2A - B$ is:

$2A - B = \begin{bmatrix}-1& 5& 3\\5& 6& 0 \end{bmatrix}$

The result of $2A - B$ is $\begin{bmatrix}-1& 5& 3\\5& 6& 0 \end{bmatrix}$.

Given:

Matrix A = $\begin{bmatrix}8& 0\\4& −2\\3& 6 \end{bmatrix}$ and Matrix B = $\begin{bmatrix}2& −2\\4& 2\\−5& 1 \end{bmatrix}$.

The equation is $2A + 3X = 5B$.

To Find:

The matrix X that satisfies the given equation.

Solution:

The given equation is $2A + 3X = 5B$.

We need to solve for matrix X. First, isolate the term with X:

$3X = 5B - 2A$

To find X, we can multiply both sides by $\frac{1}{3}$:

$X = \frac{1}{3}(5B - 2A)$

Now, we calculate the matrices $5B$ and $2A$. Scalar multiplication is performed by multiplying each element of the matrix by the scalar.

$5B = 5 \begin{bmatrix}2& −2\\4& 2\\−5& 1 \end{bmatrix} = \begin{bmatrix}5 \times 2& 5 \times (−2)\\5 \times 4& 5 \times 2\\5 \times (−5)& 5 \times 1 \end{bmatrix} = \begin{bmatrix}10& −10\\20& 10\\−25& 5 \end{bmatrix}$

$2A = 2 \begin{bmatrix}8& 0\\4& −2\\3& 6 \end{bmatrix} = \begin{bmatrix}2 \times 8& 2 \times 0\\2 \times 4& 2 \times (−2)\\2 \times 3& 2 \times 6 \end{bmatrix} = \begin{bmatrix}16& 0\\8& −4\\6& 12 \end{bmatrix}$

Now, we calculate $5B - 2A$. Matrix subtraction is performed by subtracting the corresponding elements. Both matrices $5B$ and $2A$ have the same order ($3 \times 2$), so subtraction is possible.

$5B - 2A = \begin{bmatrix}10& −10\\20& 10\\−25& 5 \end{bmatrix} - \begin{bmatrix}16& 0\\8& −4\\6& 12 \end{bmatrix}$

Subtract the corresponding elements:

• Element (1,1): $10 - 16 = -6$
• Element (1,2): $-10 - 0 = -10$
• Element (2,1): $20 - 8 = 12$
• Element (2,2): $10 - (-4) = 10 + 4 = 14$
• Element (3,1): $-25 - 6 = -31$
• Element (3,2): $5 - 12 = -7$

So, $5B - 2A = \begin{bmatrix}-6& −10\\12& 14\\−31& −7 \end{bmatrix}$.

Finally, we find X by multiplying this matrix by $\frac{1}{3}$:

$X = \frac{1}{3} \begin{bmatrix}-6& −10\\12& 14\\−31& −7 \end{bmatrix} = \begin{bmatrix}\frac{1}{3} \times (-6)& \frac{1}{3} \times (−10)\\\frac{1}{3} \times 12& \frac{1}{3} \times 14\\\frac{1}{3} \times (−31)& \frac{1}{3} \times (−7) \end{bmatrix} = \begin{bmatrix}-2& −\frac{10}{3}\\4& \frac{14}{3}\\−\frac{31}{3}& −\frac{7}{3} \end{bmatrix}$

The matrix X is $\begin{bmatrix}-2& −\frac{10}{3}\\4& \frac{14}{3}\\−\frac{31}{3}& −\frac{7}{3} \end{bmatrix}$.

Given:

Two matrix equations are provided:

To Find:

The matrices X and Y.

Solution:

We have a system of two linear equations involving matrices X and Y. We can solve this system using methods similar to solving systems of linear equations with variables.

Method 1: Addition and Subtraction of equations

Add equation (1) and equation (2):

$(X + Y) + (X - Y) = \begin{bmatrix}5& 2\\0& 9 \end{bmatrix} + \begin{bmatrix}3& 6\\0& −1 \end{bmatrix}$

$X + Y + X - Y = \begin{bmatrix}5+3& 2+6\\0+0& 9+(-1) \end{bmatrix}$

$2X = \begin{bmatrix}8& 8\\0& 8 \end{bmatrix}$

Now, multiply by $\frac{1}{2}$ to find X:

$X = \frac{1}{2} \begin{bmatrix}8& 8\\0& 8 \end{bmatrix} = \begin{bmatrix}\frac{1}{2} \times 8& \frac{1}{2} \times 8\\\frac{1}{2} \times 0& \frac{1}{2} \times 8 \end{bmatrix} = \begin{bmatrix}4& 4\\0& 4 \end{bmatrix}$

Now, subtract equation (2) from equation (1):

$(X + Y) - (X - Y) = \begin{bmatrix}5& 2\\0& 9 \end{bmatrix} - \begin{bmatrix}3& 6\\0& −1 \end{bmatrix}$

$X + Y - X + Y = \begin{bmatrix}5-3& 2-6\\0-0& 9-(-1) \end{bmatrix}$

$2Y = \begin{bmatrix}2& -4\\0& 10 \end{bmatrix}$

Now, multiply by $\frac{1}{2}$ to find Y:

$Y = \frac{1}{2} \begin{bmatrix}2& -4\\0& 10 \end{bmatrix} = \begin{bmatrix}\frac{1}{2} \times 2& \frac{1}{2} \times (-4)\\\frac{1}{2} \times 0& \frac{1}{2} \times 10 \end{bmatrix} = \begin{bmatrix}1& -2\\0& 5 \end{bmatrix}$

We can verify these results by substituting X and Y back into the original equations.

Check equation (1): $X + Y = \begin{bmatrix}4& 4\\0& 4 \end{bmatrix} + \begin{bmatrix}1& -2\\0& 5 \end{bmatrix} = \begin{bmatrix}4+1& 4+(-2)\\0+0& 4+5 \end{bmatrix} = \begin{bmatrix}5& 2\\0& 9 \end{bmatrix}$, which is correct.

Check equation (2): $X - Y = \begin{bmatrix}4& 4\\0& 4 \end{bmatrix} - \begin{bmatrix}1& -2\\0& 5 \end{bmatrix} = \begin{bmatrix}4-1& 4-(-2)\\0-0& 4-5 \end{bmatrix} = \begin{bmatrix}3& 6\\0& -1 \end{bmatrix}$, which is correct.

The matrices are $X = \begin{bmatrix}4& 4\\0& 4 \end{bmatrix}$ and $Y = \begin{bmatrix}1& -2\\0& 5 \end{bmatrix}$.

Given:

The matrix equation is $2\begin{bmatrix}x& 5\\7& y−3 \end{bmatrix} + \begin{bmatrix}3& −4\\1& 2 \end{bmatrix} = \begin{bmatrix}7& 6\\15& 14 \end{bmatrix}$.

To Find:

The values of $x$ and $y$ that satisfy the given equation.

Solution:

First, perform the scalar multiplication on the left-hand side ($L.H.S.$) of the equation:

$2\begin{bmatrix}x& 5\\7& y−3 \end{bmatrix} = \begin{bmatrix}2 \times x& 2 \times 5\\2 \times 7& 2 \times (y−3) \end{bmatrix} = \begin{bmatrix}2x& 10\\14& 2y−6 \end{bmatrix}$

Now, add the resulting matrix to the second matrix on the $L.H.S.$:

$\begin{bmatrix}2x& 10\\14& 2y−6 \end{bmatrix} + \begin{bmatrix}3& −4\\1& 2 \end{bmatrix} = \begin{bmatrix}2x + 3& 10 + (−4)\\14 + 1& (2y−6) + 2 \end{bmatrix} = \begin{bmatrix}2x + 3& 6\\15& 2y−4 \end{bmatrix}$

So, the given equation becomes:

$\begin{bmatrix}2x + 3& 6\\15& 2y−4 \end{bmatrix} = \begin{bmatrix}7& 6\\15& 14 \end{bmatrix}$

For these two matrices to be equal, their corresponding elements must be equal.

Equating the elements in the first row, first column:

Equating the elements in the first row, second column: $6 = 6$ (Consistent)

Equating the elements in the second row, first column: $15 = 15$ (Consistent)

Equating the elements in the second row, second column:

Now, solve the linear equations for $x$ and $y$.

From equation (1):

$2x + 3 = 7$

$2x = 7 - 3$

$2x = 4$

$x = \frac{4}{2}$

From equation (2):

$2y - 4 = 14$

$2y = 14 + 4$

$2y = 18$

$y = \frac{18}{2}$

The values of $x$ and $y$ are 2 and 9, respectively.

The values are $x = 2$ and $y = 9$.

Given:

September Sales matrix A = $\begin{bmatrix}10000 & 20000 & 30000 \\ 50000 & 30000 & 10000 \end{bmatrix}$

October Sales matrix B = $\begin{bmatrix}5000 & 10000 & 6000 \\ 20000 & 10000 & 10000 \end{bmatrix}$

In these matrices, the rows represent the farmers (Row 1: Ramkishan, Row 2: Gurcharan Singh) and the columns represent the varieties of rice (Column 1: Basmati, Column 2: Permal, Column 3: Naura).

To Find:

(i) Combined sales in September and October (A + B).

(ii) Decrease in sales from September to October (A - B).

(iii) 2% profit on October sales for each farmer and each variety.

Solution:

(i) Combined sales in September and October:

To find the combined sales, we need to add the corresponding elements of matrix A and matrix B.

Combined Sales Matrix = A + B

$A + B = \begin{bmatrix}10000 & 20000 & 30000 \\ 50000 & 30000 & 10000 \end{bmatrix} + \begin{bmatrix}5000 & 10000 & 6000 \\ 20000 & 10000 & 10000 \end{bmatrix}$

$A + B = \begin{bmatrix}10000 + 5000 & 20000 + 10000 & 30000 + 6000 \\ 50000 + 20000 & 30000 + 10000 & 10000 + 10000 \end{bmatrix}$

$A + B = \begin{bmatrix}15000 & 30000 & 36000 \\ 70000 & 40000 & 20000 \end{bmatrix}$

The combined sales in September and October are:

• Ramkishan: Basmati - $\textsf{₹}15000$, Permal - $\textsf{₹}30000$, Naura - $\textsf{₹}36000$
• Gurcharan Singh: Basmati - $\textsf{₹}70000$, Permal - $\textsf{₹}40000$, Naura - $\textsf{₹}20000$

(ii) Decrease in sales from September to October:

To find the decrease in sales, we need to subtract the October sales (Matrix B) from the September sales (Matrix A).

Decrease in Sales Matrix = A - B

$A - B = \begin{bmatrix}10000 & 20000 & 30000 \\ 50000 & 30000 & 10000 \end{bmatrix} - \begin{bmatrix}5000 & 10000 & 6000 \\ 20000 & 10000 & 10000 \end{bmatrix}$

$A - B = \begin{bmatrix}10000 - 5000 & 20000 - 10000 & 30000 - 6000 \\ 50000 - 20000 & 30000 - 10000 & 10000 - 10000 \end{bmatrix}$

$A - B = \begin{bmatrix}5000 & 10000 & 24000 \\ 30000 & 20000 & 0 \end{bmatrix}$

The decrease in sales from September to October is:

• Ramkishan: Basmati - $\textsf{₹}5000$, Permal - $\textsf{₹}10000$, Naura - $\textsf{₹}24000$
• Gurcharan Singh: Basmati - $\textsf{₹}30000$, Permal - $\textsf{₹}20000$, Naura - $\textsf{₹}0$

(iii) 2% profit on gross sales in October:

To compute the profit, we need to find 2% of each entry in the October Sales matrix B.

Profit Matrix = 2% of B = $0.02 \times B$

$0.02 \times B = 0.02 \begin{bmatrix}5000 & 10000 & 6000 \\ 20000 & 10000 & 10000 \end{bmatrix}$

$0.02 \times B = \begin{bmatrix}0.02 \times 5000 & 0.02 \times 10000 & 0.02 \times 6000 \\ 0.02 \times 20000 & 0.02 \times 10000 & 0.02 \times 10000 \end{bmatrix}$

$0.02 \times B = \begin{bmatrix}100 & 200 & 120 \\ 400 & 200 & 200 \end{bmatrix}$

The profit for each farmer and each variety sold in October is:

• Ramkishan: Basmati - $\textsf{₹}100$, Permal - $\textsf{₹}200$, Naura - $\textsf{₹}120$
• Gurcharan Singh: Basmati - $\textsf{₹}400$, Permal - $\textsf{₹}200$, Naura - $\textsf{₹}200$

Given:

Matrix A = $\begin{bmatrix}6& 9\\2& 3 \end{bmatrix}$ and Matrix B = $\begin{bmatrix}2& 6& 0\\7& 9& 8 \end{bmatrix}$.

To Find:

The product of matrices A and B, i.e., AB.

Solution:

To multiply two matrices A and B to get the product AB, the number of columns in matrix A must be equal to the number of rows in matrix B.

The order of matrix A is $2 \times 2$ (2 rows, 2 columns).

The order of matrix B is $2 \times 3$ (2 rows, 3 columns).

The number of columns in A is 2, and the number of rows in B is 2. Since these numbers are equal, the multiplication AB is possible.

The order of the resulting matrix AB will be (number of rows in A) $\times$ (number of columns in B), which is $2 \times 3$.

Let AB = C = $[c_{ij}]$, where $c_{ij}$ is the element in the $i$-th row of A and the $j$-th column of B. The formula for $c_{ij}$ is $c_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj}$, where $n$ is the number of columns in A (which is equal to the number of rows in B). Here, $n=2$.

$c_{ij} = a_{i1} b_{1j} + a_{i2} b_{2j}$

Let's calculate each element of the resulting $2 \times 3$ matrix AB:

For the first row of AB ($i=1$):

• $j=1$: $c_{11} = a_{11} b_{11} + a_{12} b_{21} = (6)(2) + (9)(7) = 12 + 63 = 75$
• $j=2$: $c_{12} = a_{11} b_{12} + a_{12} b_{22} = (6)(6) + (9)(9) = 36 + 81 = 117$
• $j=3$: $c_{13} = a_{11} b_{13} + a_{12} b_{23} = (6)(0) + (9)(8) = 0 + 72 = 72$

For the second row of AB ($i=2$):

• $j=1$: $c_{21} = a_{21} b_{11} + a_{22} b_{21} = (2)(2) + (3)(7) = 4 + 21 = 25$
• $j=2$: $c_{22} = a_{21} b_{12} + a_{22} b_{22} = (2)(6) + (3)(9) = 12 + 27 = 39$
• $j=3$: $c_{23} = a_{21} b_{13} + a_{22} b_{23} = (2)(0) + (3)(8) = 0 + 24 = 24$

Arranging these elements into the $2 \times 3$ matrix AB:

$AB = \begin{bmatrix}75& 117& 72\\25& 39& 24 \end{bmatrix}$

The product AB is $\begin{bmatrix}75& 117& 72\\25& 39& 24 \end{bmatrix}$.

Given:

Matrix A = $\begin{bmatrix}1& −2& 3\\−4& 2& 5 \end{bmatrix}$ and Matrix B = $\begin{bmatrix}2& 3\\4& 5\\2& 1 \end{bmatrix}$.

To Find:

The products AB and BA, and to show that AB ≠ BA.

Solution:

First, let's find the product AB.

The order of matrix A is $2 \times 3$ (2 rows, 3 columns).

The order of matrix B is $3 \times 2$ (3 rows, 2 columns).

The number of columns in A (3) is equal to the number of rows in B (3). So, the product AB is possible. The order of the resulting matrix AB will be $2 \times 2$.

Let AB = C = $[c_{ij}]$, where $c_{ij} = \sum_{k=1}^{3} a_{ik} b_{kj} = a_{i1} b_{1j} + a_{i2} b_{2j} + a_{i3} b_{3j}$.

For the first row of AB ($i=1$):

• $j=1$: $c_{11} = (1)(2) + (-2)(4) + (3)(2) = 2 - 8 + 6 = 0$
• $j=2$: $c_{12} = (1)(3) + (-2)(5) + (3)(1) = 3 - 10 + 3 = -4$

For the second row of AB ($i=2$):

• $j=1$: $c_{21} = (-4)(2) + (2)(4) + (5)(2) = -8 + 8 + 10 = 10$
• $j=2$: $c_{22} = (-4)(3) + (2)(5) + (5)(1) = -12 + 10 + 5 = 3$

So, $AB = \begin{bmatrix}0& -4\\10& 3 \end{bmatrix}$.

Next, let's find the product BA.

The order of matrix B is $3 \times 2$ (3 rows, 2 columns).

The order of matrix A is $2 \times 3$ (2 rows, 3 columns).

The number of columns in B (2) is equal to the number of rows in A (2). So, the product BA is possible. The order of the resulting matrix BA will be $3 \times 3$.

Let BA = D = $[d_{ij}]$, where $d_{ij} = \sum_{k=1}^{2} b_{ik} a_{kj} = b_{i1} a_{1j} + b_{i2} a_{2j}$.

For the first row of BA ($i=1$):

• $j=1$: $d_{11} = (2)(1) + (3)(-4) = 2 - 12 = -10$
• $j=2$: $d_{12} = (2)(-2) + (3)(2) = -4 + 6 = 2$
• $j=3$: $d_{13} = (2)(3) + (3)(5) = 6 + 15 = 21$

For the second row of BA ($i=2$):

• $j=1$: $d_{21} = (4)(1) + (5)(-4) = 4 - 20 = -16$
• $j=2$: $d_{22} = (4)(-2) + (5)(2) = -8 + 10 = 2$
• $j=3$: $d_{23} = (4)(3) + (5)(5) = 12 + 25 = 37$

For the third row of BA ($i=3$):

• $j=1$: $d_{31} = (2)(1) + (1)(-4) = 2 - 4 = -2$
• $j=2$: $d_{32} = (2)(-2) + (1)(2) = -4 + 2 = -2$
• $j=3$: $d_{33} = (2)(3) + (1)(5) = 6 + 5 = 11$

So, $BA = \begin{bmatrix}-10& 2& 21\\-16& 2& 37\\-2& -2& 11 \end{bmatrix}$.

Showing that AB ≠ BA:

We found $AB = \begin{bmatrix}0& -4\\10& 3 \end{bmatrix}$ and $BA = \begin{bmatrix}-10& 2& 21\\-16& 2& 37\\-2& -2& 11 \end{bmatrix}$.

The order of matrix AB is $2 \times 2$, and the order of matrix BA is $3 \times 3$.

Since the orders of the matrices AB and BA are different, they are not equal.

Therefore, $AB \ne BA$.

Given:

Matrix A = $\begin{bmatrix}1& 0\\0& −1 \end{bmatrix}$ and Matrix B = $\begin{bmatrix}0& 1\\1& 0 \end{bmatrix}$.

The products AB and BA are also stated: AB = $\begin{bmatrix}0& 1\\−1& 0 \end{bmatrix}$ and BA = $\begin{bmatrix}0& -1\\1& 0 \end{bmatrix}$.

The statement that AB ≠ BA and the conclusion that matrix multiplication is not commutative are provided.

Verification:

We will verify the given matrix products AB and BA and check if AB ≠ BA.

Calculation of AB:

Matrix A is $2 \times 2$ and Matrix B is $2 \times 2$. The product AB is possible and will be a $2 \times 2$ matrix.

$AB = \begin{bmatrix}1& 0\\0& −1 \end{bmatrix} \begin{bmatrix}0& 1\\1& 0 \end{bmatrix}$

• Element (1,1): $(1)(0) + (0)(1) = 0 + 0 = 0$
• Element (1,2): $(1)(1) + (0)(0) = 1 + 0 = 1$
• Element (2,1): $(0)(0) + (-1)(1) = 0 - 1 = -1$
• Element (2,2): $(0)(1) + (-1)(0) = 0 + 0 = 0$

So, $AB = \begin{bmatrix}0& 1\\−1& 0 \end{bmatrix}$. This matches the given result for AB.

Calculation of BA:

Matrix B is $2 \times 2$ and Matrix A is $2 \times 2$. The product BA is possible and will be a $2 \times 2$ matrix.

$BA = \begin{bmatrix}0& 1\\1& 0 \end{bmatrix} \begin{bmatrix}1& 0\\0& −1 \end{bmatrix}$

• Element (1,1): $(0)(1) + (1)(0) = 0 + 0 = 0$
• Element (1,2): $(0)(0) + (1)(-1) = 0 - 1 = -1$
• Element (2,1): $(1)(1) + (0)(0) = 1 + 0 = 1$
• Element (2,2): $(1)(0) + (0)(-1) = 0 + 0 = 0$

So, $BA = \begin{bmatrix}0& -1\\1& 0 \end{bmatrix}$. This matches the given result for BA.

Comparison of AB and BA:

$AB = \begin{bmatrix}0& 1\\−1& 0 \end{bmatrix}$ and $BA = \begin{bmatrix}0& -1\\1& 0 \end{bmatrix}$.

Comparing the corresponding elements, we see that the element in the first row, second column of AB is 1, while the element in the first row, second column of BA is -1. These are not equal.

Also, the element in the second row, first column of AB is -1, while the element in the second row, first column of BA is 1. These are not equal.

Since not all corresponding elements are equal, the matrices AB and BA are not equal.

Thus, $AB \ne BA$.

Conclusion:

The verification confirms that $AB \ne BA$ for the given matrices A and B.

This example illustrates that in general, matrix multiplication is not commutative, meaning the order of multiplication matters (AB is not always equal to BA).

Given:

Matrix A = $\begin{bmatrix}0& −1\\0& 2 \end{bmatrix}$ and Matrix B = $\begin{bmatrix}3& 5\\0& 0 \end{bmatrix}$.

To Find:

The product of matrices A and B, i.e., AB.

Solution:

The order of matrix A is $2 \times 2$ (2 rows, 2 columns).

The order of matrix B is $2 \times 2$ (2 rows, 2 columns).

The number of columns in A (2) is equal to the number of rows in B (2). So, the product AB is possible. The order of the resulting matrix AB will be $2 \times 2$.

Let AB = C = $[c_{ij}]$, where $c_{ij} = \sum_{k=1}^{2} a_{ik} b_{kj} = a_{i1} b_{1j} + a_{i2} b_{2j}$.

For the first row of AB ($i=1$):

• $j=1$: $c_{11} = a_{11} b_{11} + a_{12} b_{21} = (0)(3) + (-1)(0) = 0 + 0 = 0$
• $j=2$: $c_{12} = a_{11} b_{12} + a_{12} b_{22} = (0)(5) + (-1)(0) = 0 + 0 = 0$

For the second row of AB ($i=2$):

• $j=1$: $c_{21} = a_{21} b_{11} + a_{22} b_{21} = (0)(3) + (2)(0) = 0 + 0 = 0$
• $j=2$: $c_{22} = a_{21} b_{12} + a_{22} b_{22} = (0)(5) + (2)(0) = 0 + 0 = 0$

Arranging these elements into the $2 \times 2$ matrix AB:

$AB = \begin{bmatrix}0& 0\\0& 0 \end{bmatrix}$

The product AB is the zero matrix of order $2 \times 2$, which is $\begin{bmatrix}0& 0\\0& 0 \end{bmatrix}$. This example shows that the product of two non-zero matrices can be a zero matrix.

Given:

Matrix A = $\begin{bmatrix}1& 1& −1\\2& 0& 3\\3& −1& 2 \end{bmatrix}$ (order $3 \times 3$)

Matrix B = $\begin{bmatrix}1& 3\\0& 2\\−1& 4 \end{bmatrix}$ (order $3 \times 2$)

Matrix C = $\begin{bmatrix}1& 2& 3& -4\\2& 0& −2& 1 \end{bmatrix}$ (order $2 \times 4$)

To Find and Show:

Find A(BC) and (AB)C, and show that (AB)C = A(BC).

Solution:

We need to calculate A(BC) and (AB)C separately.

Calculating A(BC):

First, calculate BC. The order of B is $3 \times 2$ and the order of C is $2 \times 4$. The number of columns in B (2) is equal to the number of rows in C (2), so BC is possible. The order of BC will be $3 \times 4$.

$BC = \begin{bmatrix}1& 3\\0& 2\\−1& 4 \end{bmatrix} \begin{bmatrix}1& 2& 3& -4\\2& 0& −2& 1 \end{bmatrix}$

• Element (1,1): $(1)(1) + (3)(2) = 1 + 6 = 7$
• Element (1,2): $(1)(2) + (3)(0) = 2 + 0 = 2$
• Element (1,3): $(1)(3) + (3)(-2) = 3 - 6 = -3$
• Element (1,4): $(1)(-4) + (3)(1) = -4 + 3 = -1$
• Element (2,1): $(0)(1) + (2)(2) = 0 + 4 = 4$
• Element (2,2): $(0)(2) + (2)(0) = 0 + 0 = 0$
• Element (2,3): $(0)(3) + (2)(-2) = 0 - 4 = -4$
• Element (2,4): $(0)(-4) + (2)(1) = 0 + 2 = 2$
• Element (3,1): $(-1)(1) + (4)(2) = -1 + 8 = 7$
• Element (3,2): $(-1)(2) + (4)(0) = -2 + 0 = -2$
• Element (3,3): $(-1)(3) + (4)(-2) = -3 - 8 = -11$
• Element (3,4): $(-1)(-4) + (4)(1) = 4 + 4 = 8$

So, $BC = \begin{bmatrix}7& 2& -3& -1\\4& 0& -4& 2\\7& -2& -11& 8 \end{bmatrix}$ (order $3 \times 4$).

Now, calculate A(BC). The order of A is $3 \times 3$ and the order of BC is $3 \times 4$. The number of columns in A (3) is equal to the number of rows in BC (3), so A(BC) is possible. The order of A(BC) will be $3 \times 4$.

$A(BC) = \begin{bmatrix}1& 1& −1\\2& 0& 3\\3& −1& 2 \end{bmatrix} \begin{bmatrix}7& 2& -3& -1\\4& 0& -4& 2\\7& -2& -11& 8 \end{bmatrix}$

• Element (1,1): $(1)(7) + (1)(4) + (-1)(7) = 7 + 4 - 7 = 4$
• Element (1,2): $(1)(2) + (1)(0) + (-1)(-2) = 2 + 0 + 2 = 4$
• Element (1,3): $(1)(-3) + (1)(-4) + (-1)(-11) = -3 - 4 + 11 = 4$
• Element (1,4): $(1)(-1) + (1)(2) + (-1)(8) = -1 + 2 - 8 = -7$
• Element (2,1): $(2)(7) + (0)(4) + (3)(7) = 14 + 0 + 21 = 35$
• Element (2,2): $(2)(2) + (0)(0) + (3)(-2) = 4 + 0 - 6 = -2$
• Element (2,3): $(2)(-3) + (0)(-4) + (3)(-11) = -6 + 0 - 33 = -39$
• Element (2,4): $(2)(-1) + (0)(2) + (3)(8) = -2 + 0 + 24 = 22$
• Element (3,1): $(3)(7) + (-1)(4) + (2)(7) = 21 - 4 + 14 = 31$
• Element (3,2): $(3)(2) + (-1)(0) + (2)(-2) = 6 + 0 - 4 = 2$
• Element (3,3): $(3)(-3) + (-1)(-4) + (2)(-11) = -9 + 4 - 22 = -27$
• Element (3,4): $(3)(-1) + (-1)(2) + (2)(8) = -3 - 2 + 16 = 11$

So, $A(BC) = \begin{bmatrix}4& 4& 4& -7\\35& -2& -39& 22\\31& 2& -27& 11 \end{bmatrix}$.

Calculating (AB)C:

First, calculate AB. The order of A is $3 \times 3$ and the order of B is $3 \times 2$. The number of columns in A (3) is equal to the number of rows in B (3), so AB is possible. The order of AB will be $3 \times 2$.

$AB = \begin{bmatrix}1& 1& −1\\2& 0& 3\\3& −1& 2 \end{bmatrix} \begin{bmatrix}1& 3\\0& 2\\−1& 4 \end{bmatrix}$

• Element (1,1): $(1)(1) + (1)(0) + (-1)(-1) = 1 + 0 + 1 = 2$
• Element (1,2): $(1)(3) + (1)(2) + (-1)(4) = 3 + 2 - 4 = 1$
• Element (2,1): $(2)(1) + (0)(0) + (3)(-1) = 2 + 0 - 3 = -1$
• Element (2,2): $(2)(3) + (0)(2) + (3)(4) = 6 + 0 + 12 = 18$
• Element (3,1): $(3)(1) + (-1)(0) + (2)(-1) = 3 + 0 - 2 = 1$
• Element (3,2): $(3)(3) + (-1)(2) + (2)(4) = 9 - 2 + 8 = 15$

So, $AB = \begin{bmatrix}2& 1\\-1& 18\\1& 15 \end{bmatrix}$ (order $3 \times 2$).

Now, calculate (AB)C. The order of AB is $3 \times 2$ and the order of C is $2 \times 4$. The number of columns in AB (2) is equal to the number of rows in C (2), so (AB)C is possible. The order of (AB)C will be $3 \times 4$.

$(AB)C = \begin{bmatrix}2& 1\\-1& 18\\1& 15 \end{bmatrix} \begin{bmatrix}1& 2& 3& -4\\2& 0& −2& 1 \end{bmatrix}$

• Element (1,1): $(2)(1) + (1)(2) = 2 + 2 = 4$
• Element (1,2): $(2)(2) + (1)(0) = 4 + 0 = 4$
• Element (1,3): $(2)(3) + (1)(-2) = 6 - 2 = 4$
• Element (1,4): $(2)(-4) + (1)(1) = -8 + 1 = -7$
• Element (2,1): $(-1)(1) + (18)(2) = -1 + 36 = 35$
• Element (2,2): $(-1)(2) + (18)(0) = -2 + 0 = -2$
• Element (2,3): $(-1)(3) + (18)(-2) = -3 - 36 = -39$
• Element (2,4): $(-1)(-4) + (18)(1) = 4 + 18 = 22$
• Element (3,1): $(1)(1) + (15)(2) = 1 + 30 = 31$
• Element (3,2): $(1)(2) + (15)(0) = 2 + 0 = 2$
• Element (3,3): $(1)(3) + (15)(-2) = 3 - 30 = -27$
• Element (3,4): $(1)(-4) + (15)(1) = -4 + 15 = 11$

So, $(AB)C = \begin{bmatrix}4& 4& 4& -7\\35& -2& -39& 22\\31& 2& -27& 11 \end{bmatrix}$.

Showing that (AB)C = A(BC):

We found $A(BC) = \begin{bmatrix}4& 4& 4& -7\\35& -2& -39& 22\\31& 2& -27& 11 \end{bmatrix}$ and $(AB)C = \begin{bmatrix}4& 4& 4& -7\\35& -2& -39& 22\\31& 2& -27& 11 \end{bmatrix}$.

The matrices A(BC) and (AB)C have the same order ($3 \times 4$) and their corresponding elements are equal.

Therefore, $(AB)C = A(BC)$.

This example verifies the associative property of matrix multiplication.

Given:

Matrix A = $\begin{bmatrix}0& 6& 7\\−6& 0& 8\\7& −8& 0 \end{bmatrix}$ (order $3 \times 3$)

Matrix B = $\begin{bmatrix}0& 1& 1\\1& 0& 2\\1& 2& 0 \end{bmatrix}$ (order $3 \times 3$)

Matrix C = $\begin{bmatrix}2\\−2\\3 \end{bmatrix}$ (order $3 \times 1$)

To Calculate and Verify:

Calculate AC, BC, and (A + B)C. Verify that (A + B)C = AC + BC.

Solution:

Calculate AC:

The order of A is $3 \times 3$ and the order of C is $3 \times 1$. The number of columns in A (3) is equal to the number of rows in C (3), so AC is possible. The order of AC will be $3 \times 1$.

$AC = \begin{bmatrix}0& 6& 7\\−6& 0& 8\\7& −8& 0 \end{bmatrix} \begin{bmatrix}2\\−2\\3 \end{bmatrix}$

• Element (1,1): $(0)(2) + (6)(-2) + (7)(3) = 0 - 12 + 21 = 9$
• Element (2,1): $(-6)(2) + (0)(-2) + (8)(3) = -12 + 0 + 24 = 12$
• Element (3,1): $(7)(2) + (-8)(-2) + (0)(3) = 14 + 16 + 0 = 30$

So, $AC = \begin{bmatrix}9\\12\\30 \end{bmatrix}$.

Calculate BC:

The order of B is $3 \times 3$ and the order of C is $3 \times 1$. The number of columns in B (3) is equal to the number of rows in C (3), so BC is possible. The order of BC will be $3 \times 1$.

$BC = \begin{bmatrix}0& 1& 1\\1& 0& 2\\1& 2& 0 \end{bmatrix} \begin{bmatrix}2\\−2\\3 \end{bmatrix}$

• Element (1,1): $(0)(2) + (1)(-2) + (1)(3) = 0 - 2 + 3 = 1$
• Element (2,1): $(1)(2) + (0)(-2) + (2)(3) = 2 + 0 + 6 = 8$
• Element (3,1): $(1)(2) + (2)(-2) + (0)(3) = 2 - 4 + 0 = -2$

So, $BC = \begin{bmatrix}1\\8\\-2 \end{bmatrix}$.

Calculate (A + B)C:

First, calculate A + B. The order of A is $3 \times 3$ and the order of B is $3 \times 3$. Addition is possible and the order of A + B will be $3 \times 3$.

$A + B = \begin{bmatrix}0& 6& 7\\−6& 0& 8\\7& −8& 0 \end{bmatrix} + \begin{bmatrix}0& 1& 1\\1& 0& 2\\1& 2& 0 \end{bmatrix} = \begin{bmatrix}0+0& 6+1& 7+1\\-6+1& 0+0& 8+2\\7+1& -8+2& 0+0 \end{bmatrix} = \begin{bmatrix}0& 7& 8\\-5& 0& 10\\8& -6& 0 \end{bmatrix}$

Now, calculate (A + B)C. The order of (A + B) is $3 \times 3$ and the order of C is $3 \times 1$. The number of columns in (A + B) (3) is equal to the number of rows in C (3), so (A + B)C is possible. The order of (A + B)C will be $3 \times 1$.

$(A + B)C = \begin{bmatrix}0& 7& 8\\-5& 0& 10\\8& -6& 0 \end{bmatrix} \begin{bmatrix}2\\−2\\3 \end{bmatrix}$

• Element (1,1): $(0)(2) + (7)(-2) + (8)(3) = 0 - 14 + 24 = 10$
• Element (2,1): $(-5)(2) + (0)(-2) + (10)(3) = -10 + 0 + 30 = 20$
• Element (3,1): $(8)(2) + (-6)(-2) + (0)(3) = 16 + 12 + 0 = 28$

So, $(A + B)C = \begin{bmatrix}10\\20\\28 \end{bmatrix}$.

Verify that (A + B)C = AC + BC:

First, calculate AC + BC. The order of AC is $3 \times 1$ and the order of BC is $3 \times 1$. Addition is possible and the order of AC + BC will be $3 \times 1$.

$AC + BC = \begin{bmatrix}9\\12\\30 \end{bmatrix} + \begin{bmatrix}1\\8\\-2 \end{bmatrix} = \begin{bmatrix}9+1\\12+8\\30+(-2) \end{bmatrix} = \begin{bmatrix}10\\20\\28 \end{bmatrix}$

We found $(A + B)C = \begin{bmatrix}10\\20\\28 \end{bmatrix}$ and $AC + BC = \begin{bmatrix}10\\20\\28 \end{bmatrix}$.

The matrices (A + B)C and AC + BC have the same order ($3 \times 1$) and their corresponding elements are equal.

Therefore, $(A + B)C = AC + BC$.

This example verifies the distributive property of matrix multiplication over matrix addition (right distributive law).

Given:

Matrix A = $\begin{bmatrix}1& 2& 3\\3& −2& 1\\4& 2& 1 \end{bmatrix}$.

The identity matrix I of the same order as A is $I = \begin{bmatrix}1& 0& 0\\0& 1& 0\\0& 0& 1 \end{bmatrix}$.

The zero matrix O of the same order is $O = \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$.

To Show:

$A^3 – 23A – 40 I = O$

Solution:

We need to calculate $A^2$ and $A^3$, and then substitute them into the given expression.

First, calculate $A^2 = A \times A$. The order of A is $3 \times 3$, so $A^2$ will also be $3 \times 3$.

$A^2 = \begin{bmatrix}1& 2& 3\\3& −2& 1\\4& 2& 1 \end{bmatrix} \begin{bmatrix}1& 2& 3\\3& −2& 1\\4& 2& 1 \end{bmatrix}$

• Element (1,1): $(1)(1) + (2)(3) + (3)(4) = 1 + 6 + 12 = 19$
• Element (1,2): $(1)(2) + (2)(-2) + (3)(2) = 2 - 4 + 6 = 4$
• Element (1,3): $(1)(3) + (2)(1) + (3)(1) = 3 + 2 + 3 = 8$
• Element (2,1): $(3)(1) + (-2)(3) + (1)(4) = 3 - 6 + 4 = 1$
• Element (2,2): $(3)(2) + (-2)(-2) + (1)(2) = 6 + 4 + 2 = 12$
• Element (2,3): $(3)(3) + (-2)(1) + (1)(1) = 9 - 2 + 1 = 8$
• Element (3,1): $(4)(1) + (2)(3) + (1)(4) = 4 + 6 + 4 = 14$
• Element (3,2): $(4)(2) + (2)(-2) + (1)(2) = 8 - 4 + 2 = 6$
• Element (3,3): $(4)(3) + (2)(1) + (1)(1) = 12 + 2 + 1 = 15$

So, $A^2 = \begin{bmatrix}19& 4& 8\\1& 12& 8\\14& 6& 15 \end{bmatrix}$.

Now, calculate $A^3 = A^2 \times A$. The order of $A^2$ is $3 \times 3$ and the order of A is $3 \times 3$, so $A^3$ will be $3 \times 3$.

$A^3 = \begin{bmatrix}19& 4& 8\\1& 12& 8\\14& 6& 15 \end{bmatrix} \begin{bmatrix}1& 2& 3\\3& −2& 1\\4& 2& 1 \end{bmatrix}$

• Element (1,1): $(19)(1) + (4)(3) + (8)(4) = 19 + 12 + 32 = 63$
• Element (1,2): $(19)(2) + (4)(-2) + (8)(2) = 38 - 8 + 16 = 46$
• Element (1,3): $(19)(3) + (4)(1) + (8)(1) = 57 + 4 + 8 = 69$
• Element (2,1): $(1)(1) + (12)(3) + (8)(4) = 1 + 36 + 32 = 69$
• Element (2,2): $(1)(2) + (12)(-2) + (8)(2) = 2 - 24 + 16 = -6$
• Element (2,3): $(1)(3) + (12)(1) + (8)(1) = 3 + 12 + 8 = 23$
• Element (3,1): $(14)(1) + (6)(3) + (15)(4) = 14 + 18 + 60 = 92$
• Element (3,2): $(14)(2) + (6)(-2) + (15)(2) = 28 - 12 + 30 = 46$
• Element (3,3): $(14)(3) + (6)(1) + (15)(1) = 42 + 6 + 15 = 63$

So, $A^3 = \begin{bmatrix}63& 46& 69\\69& -6& 23\\92& 46& 63 \end{bmatrix}$.

Now, calculate $23A$ and $40I$.

$23A = 23 \begin{bmatrix}1& 2& 3\\3& −2& 1\\4& 2& 1 \end{bmatrix} = \begin{bmatrix}23 \times 1& 23 \times 2& 23 \times 3\\23 \times 3& 23 \times (−2)& 23 \times 1\\23 \times 4& 23 \times 2& 23 \times 1 \end{bmatrix} = \begin{bmatrix}23& 46& 69\\69& -46& 23\\92& 46& 23 \end{bmatrix}$

$40I = 40 \begin{bmatrix}1& 0& 0\\0& 1& 0\\0& 0& 1 \end{bmatrix} = \begin{bmatrix}40 \times 1& 40 \times 0& 40 \times 0\\40 \times 0& 40 \times 1& 40 \times 0\\40 \times 0& 40 \times 0& 40 \times 1 \end{bmatrix} = \begin{bmatrix}40& 0& 0\\0& 40& 0\\0& 0& 40 \end{bmatrix}$

Now, calculate $A^3 – 23A – 40 I$. This involves matrix subtraction.

$A^3 – 23A – 40 I = \begin{bmatrix}63& 46& 69\\69& -6& 23\\92& 46& 63 \end{bmatrix} - \begin{bmatrix}23& 46& 69\\69& -46& 23\\92& 46& 23 \end{bmatrix} - \begin{bmatrix}40& 0& 0\\0& 40& 0\\0& 0& 40 \end{bmatrix}$

$A^3 – 23A – 40 I = \begin{bmatrix}63 - 23 - 40 & 46 - 46 - 0 & 69 - 69 - 0 \\ 69 - 69 - 0 & -6 - (-46) - 40 & 23 - 23 - 0 \\ 92 - 92 - 0 & 46 - 46 - 0 & 63 - 23 - 40 \end{bmatrix}$

$A^3 – 23A – 40 I = \begin{bmatrix}40 - 40 & 0 & 0 \\ 0 & -6 + 46 - 40 & 0 \\ 0 & 0 & 40 - 40 \end{bmatrix}$

$A^3 – 23A – 40 I = \begin{bmatrix}0 & 0 & 0 \\ 0 & 40 - 40 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

$A^3 – 23A – 40 I = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

The resulting matrix is the zero matrix O of order $3 \times 3$.

Therefore, $A^3 – 23A – 40 I = O$.

The identity $A^3 – 23A – 40 I = O$ is shown to be true for the given matrix A.

Given:

Cost per contact matrix A = $\begin{bmatrix}40 \\ 100 \\ 50 \end{bmatrix}$ (order $3 \times 1$). The rows represent the contact type (Telephone, Housecall, Letter) and the single column represents the cost in paise.

Number of contacts matrix B = $\begin{bmatrix}1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix}$ (order $2 \times 3$). The rows represent the cities (Row 1: X, Row 2: Y) and the columns represent the contact types (Column 1: Telephone, Column 2: Housecall, Column 3: Letter).

Note: The second given matrix should be denoted as B, not A, to avoid confusion with the cost matrix.

To Find:

The total amount spent by the group in the two cities X and Y.

Solution:

To find the total cost in each city, we need to perform matrix multiplication. We want the result to be a matrix where each entry represents the total cost for a city.

The number of contacts matrix (let's call it B) has cities as rows and contact types as columns (order $2 \times 3$). The cost per contact matrix (A) has contact types as rows and cost as a column (order $3 \times 1$).

To get a matrix with cities as rows and total cost as a column (order $2 \times 1$), we should multiply the number of contacts matrix by the cost per contact matrix in the order BA.

Order of B is $2 \times 3$. Order of A is $3 \times 1$. The number of columns in B (3) is equal to the number of rows in A (3). So, the product BA is possible and its order will be $2 \times 1$.

Total Cost Matrix = BA

$BA = \begin{bmatrix}1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{bmatrix}40 \\ 100 \\ 50 \end{bmatrix}$

• Element (1,1) (Total cost in City X): $(1000)(40) + (500)(100) + (5000)(50) = 40000 + 50000 + 250000 = 340000$
• Element (2,1) (Total cost in City Y): $(3000)(40) + (1000)(100) + (10000)(50) = 120000 + 100000 + 500000 = 720000$

The resulting matrix is $\begin{bmatrix}340000 \\ 720000 \end{bmatrix}$.

The entries in this matrix represent the total cost in paise for City X and City Y, respectively.

Total cost in City X = 340000 paise.

Total cost in City Y = 720000 paise.

To convert paise to Rupees, divide by 100.

Total cost in City X = $\frac{340000}{100}$ Rupees = $\textsf{₹}3400$.

Total cost in City Y = $\frac{720000}{100}$ Rupees = $\textsf{₹}7200$.

The total amount spent by the group in City X is $\textsf{₹}3400$, and in City Y is $\textsf{₹}7200$.

Given:

Matrix A = $\begin{bmatrix}2& 4\\3& 2 \end{bmatrix}$

Matrix B = $\begin{bmatrix}1& 3\\−2& 5 \end{bmatrix}$

Matrix C = $\begin{bmatrix}−2& 5\\3& 4 \end{bmatrix}$

To Find:

(i) A + B

(ii) A – B

(iii) 3A – C

(iv) AB

(v) BA

Solution:

(i) A + B:

Since both matrices A and B are of order $2 \times 2$, addition is possible. Add the corresponding elements.

$A + B = \begin{bmatrix}2& 4\\3& 2 \end{bmatrix} + \begin{bmatrix}1& 3\\−2& 5 \end{bmatrix} = \begin{bmatrix}2+1& 4+3\\3+(-2)& 2+5 \end{bmatrix} = \begin{bmatrix}3& 7\\3-2& 7 \end{bmatrix} = \begin{bmatrix}3& 7\\1& 7 \end{bmatrix}$

(ii) A – B:

Since both matrices A and B are of order $2 \times 2$, subtraction is possible. Subtract the corresponding elements.

$A – B = \begin{bmatrix}2& 4\\3& 2 \end{bmatrix} - \begin{bmatrix}1& 3\\−2& 5 \end{bmatrix} = \begin{bmatrix}2-1& 4-3\\3-(-2)& 2-5 \end{bmatrix} = \begin{bmatrix}1& 1\\3+2& -3 \end{bmatrix} = \begin{bmatrix}1& 1\\5& -3 \end{bmatrix}$

(iii) 3A – C:

First, calculate 3A. Scalar multiplication is performed by multiplying each element of A by 3.

$3A = 3 \begin{bmatrix}2& 4\\3& 2 \end{bmatrix} = \begin{bmatrix}3 \times 2& 3 \times 4\\3 \times 3& 3 \times 2 \end{bmatrix} = \begin{bmatrix}6& 12\\9& 6 \end{bmatrix}$

Now, subtract C from 3A. Both matrices are of order $2 \times 2$, so subtraction is possible.

$3A – C = \begin{bmatrix}6& 12\\9& 6 \end{bmatrix} - \begin{bmatrix}−2& 5\\3& 4 \end{bmatrix} = \begin{bmatrix}6-(-2)& 12-5\\9-3& 6-4 \end{bmatrix} = \begin{bmatrix}6+2& 7\\6& 2 \end{bmatrix} = \begin{bmatrix}8& 7\\6& 2 \end{bmatrix}$

(iv) AB:

The order of A is $2 \times 2$. The order of B is $2 \times 2$. The number of columns in A (2) is equal to the number of rows in B (2). So, the product AB is possible and will be a $2 \times 2$ matrix.

$AB = \begin{bmatrix}2& 4\\3& 2 \end{bmatrix} \begin{bmatrix}1& 3\\−2& 5 \end{bmatrix}$

• Element (1,1): $(2)(1) + (4)(-2) = 2 - 8 = -6$
• Element (1,2): $(2)(3) + (4)(5) = 6 + 20 = 26$
• Element (2,1): $(3)(1) + (2)(-2) = 3 - 4 = -1$
• Element (2,2): $(3)(3) + (2)(5) = 9 + 10 = 19$

So, $AB = \begin{bmatrix}-6& 26\\-1& 19 \end{bmatrix}$.

(v) BA:

The order of B is $2 \times 2$. The order of A is $2 \times 2$. The number of columns in B (2) is equal to the number of rows in A (2). So, the product BA is possible and will be a $2 \times 2$ matrix.

$BA = \begin{bmatrix}1& 3\\−2& 5 \end{bmatrix} \begin{bmatrix}2& 4\\3& 2 \end{bmatrix}$

• Element (1,1): $(1)(2) + (3)(3) = 2 + 9 = 11$
• Element (1,2): $(1)(4) + (3)(2) = 4 + 6 = 10$
• Element (2,1): $(-2)(2) + (5)(3) = -4 + 15 = 11$
• Element (2,2): $(-2)(4) + (5)(2) = -8 + 10 = 2$

So, $BA = \begin{bmatrix}11& 10\\11& 2 \end{bmatrix}$.

Given:

Different matrix addition problems.

To Compute:

The sum of the given matrices in each part.

Solution:

Matrix addition is performed by adding the corresponding elements of the matrices. This is possible only if the matrices have the same order.

(i) $\begin{bmatrix}a& b\\−b& a \end{bmatrix}$ + $\begin{bmatrix}a& b\\b& a \end{bmatrix}$

Both matrices are $2 \times 2$.

$\begin{bmatrix}a& b\\−b& a \end{bmatrix} + \begin{bmatrix}a& b\\b& a \end{bmatrix} = \begin{bmatrix}a+a& b+b\\−b+b& a+a \end{bmatrix} = \begin{bmatrix}2a& 2b\\0& 2a \end{bmatrix}$

(ii) $\begin{bmatrix}a^2 + b^2& b^2 + c^2\\a^2 + c^2& a^2 + b^2 \end{bmatrix}$ + $\begin{bmatrix}2ab& 2bc\\−2ac& −2ab \end{bmatrix}$

Both matrices are $2 \times 2$.

$\begin{bmatrix}a^2 + b^2& b^2 + c^2\\a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix}2ab& 2bc\\−2ac& −2ab \end{bmatrix} = \begin{bmatrix}(a^2 + b^2) + 2ab& (b^2 + c^2) + 2bc\\(a^2 + c^2) + (−2ac)& (a^2 + b^2) + (−2ab) \end{bmatrix}$

Using the algebraic identities $(x+y)^2 = x^2 + y^2 + 2xy$ and $(x-y)^2 = x^2 + y^2 - 2xy$:

= $\begin{bmatrix}(a + b)^2& (b + c)^2\\a^2 + c^2 - 2ac& a^2 + b^2 - 2ab \end{bmatrix}$

= $\begin{bmatrix}(a + b)^2& (b + c)^2\\(a - c)^2& (a - b)^2 \end{bmatrix}$

(iii) $\begin{bmatrix}−1& 4& −6\\8& 5& 16\\2& 8& 5 \end{bmatrix}$ + $\begin{bmatrix}12& 7& 6\\8& 0& 5\\3& 2& 4 \end{bmatrix}$

Both matrices are $3 \times 3$.

$\begin{bmatrix}−1& 4& −6\\8& 5& 16\\2& 8& 5 \end{bmatrix} + \begin{bmatrix}12& 7& 6\\8& 0& 5\\3& 2& 4 \end{bmatrix} = \begin{bmatrix}-1+12& 4+7& -6+6\\8+8& 5+0& 16+5\\2+3& 8+2& 5+4 \end{bmatrix} = \begin{bmatrix}11& 11& 0\\16& 5& 21\\5& 10& 9 \end{bmatrix}$

(iv) $\begin{bmatrix}\cos^2x& \sin^2x\\\sin^2x& \cos^2x \end{bmatrix}$ + $\begin{bmatrix}\sin^2x& \cos^2x \\ \cos^2x& \sin^2x \end{bmatrix}$

Both matrices are $2 \times 2$.

$\begin{bmatrix}\cos^2x& \sin^2x\\\sin^2x& \cos^2x \end{bmatrix}$ + $\begin{bmatrix}\sin^2x& \cos^2x \\ \cos^2x& \sin^2x \end{bmatrix} = \begin{bmatrix}\cos^2x + \sin^2x& \sin^2x + \cos^2x\\\sin^2x + \cos^2x& \cos^2x + \sin^2x \end{bmatrix}$

Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$:

= $\begin{bmatrix}1& 1\\1& 1 \end{bmatrix}$

Given:

Different matrix multiplication problems.

To Compute:

The product of the given matrices in each part.

Solution:

To multiply two matrices A and B (to get AB), the number of columns in A must be equal to the number of rows in B. The order of the resulting matrix AB is (number of rows in A) $\times$ (number of columns in B).

(i) $\begin{bmatrix}a& b\\−b& a \end{bmatrix} \begin{bmatrix}a& −b\\b& a \end{bmatrix}$

Order of first matrix is $2 \times 2$. Order of second matrix is $2 \times 2$. Product is possible and will be $2 \times 2$.

= $\begin{bmatrix}(a)(a) + (b)(b)& (a)(−b) + (b)(a)\\(−b)(a) + (a)(b)& (−b)(−b) + (a)(a) \end{bmatrix}$

= $\begin{bmatrix}a^2 + b^2& -ab + ab\\-ab + ab& b^2 + a^2 \end{bmatrix} = \begin{bmatrix}a^2 + b^2& 0\\0& a^2 + b^2 \end{bmatrix}$

(ii) $\begin{bmatrix}1\\2\\3 \end{bmatrix} \begin{bmatrix}2& 3& 4 \end{bmatrix}$

Order of first matrix is $3 \times 1$. Order of second matrix is $1 \times 3$. Product is possible and will be $3 \times 3$.

= $\begin{bmatrix}(1)(2)& (1)(3)& (1)(4)\\(2)(2)& (2)(3)& (2)(4)\\(3)(2)& (3)(3)& (3)(4) \end{bmatrix} = \begin{bmatrix}2& 3& 4\\4& 6& 8\\6& 9& 12 \end{bmatrix}$

(iii) $\begin{bmatrix}1& −2\\2& 3 \end{bmatrix} \begin{bmatrix}1& 2& 3\\2& 3& 1 \end{bmatrix}$

Order of first matrix is $2 \times 2$. Order of second matrix is $2 \times 3$. Product is possible and will be $2 \times 3$.

= $\begin{bmatrix}(1)(1) + (−2)(2)& (1)(2) + (−2)(3)& (1)(3) + (−2)(1)\\(2)(1) + (3)(2)& (2)(2) + (3)(3)& (2)(3) + (3)(1) \end{bmatrix}$

= $\begin{bmatrix}1 - 4& 2 - 6& 3 - 2\\2 + 6& 4 + 9& 6 + 3 \end{bmatrix} = \begin{bmatrix}-3& -4& 1\\8& 13& 9 \end{bmatrix}$

(iv) $\begin{bmatrix}2& 3& 4\\3& 4& 5\\4& 5& 6 \end{bmatrix} \begin{bmatrix}1& −3& 5\\0& 2& 4\\3& 0& 5 \end{bmatrix}$

Order of first matrix is $3 \times 3$. Order of second matrix is $3 \times 3$. Product is possible and will be $3 \times 3$.

= $\begin{bmatrix}(2)(1)+(3)(0)+(4)(3)& (2)(−3)+(3)(2)+(4)(0)& (2)(5)+(3)(4)+(4)(5)\\(3)(1)+(4)(0)+(5)(3)& (3)(−3)+(4)(2)+(5)(0)& (3)(5)+(4)(4)+(5)(5)\\(4)(1)+(5)(0)+(6)(3)& (4)(−3)+(5)(2)+(6)(0)& (4)(5)+(5)(4)+(6)(5) \end{bmatrix}$

= $\begin{bmatrix}2+0+12& -6+6+0& 10+12+20\\3+0+15& -9+8+0& 15+16+25\\4+0+18& -12+10+0& 20+20+30 \end{bmatrix}$

= $\begin{bmatrix}14& 0& 42\\18& -1& 56\\22& -2& 70 \end{bmatrix}$

(v) $\begin{bmatrix}2& 1\\3& 2\\−1& 1 \end{bmatrix} \begin{bmatrix}1& 0& 1\\−1& 2& 1 \end{bmatrix}$

Order of first matrix is $3 \times 2$. Order of second matrix is $2 \times 3$. Product is possible and will be $3 \times 3$.

= $\begin{bmatrix}(2)(1)+(1)(−1)& (2)(0)+(1)(2)& (2)(1)+(1)(1)\\(3)(1)+(2)(−1)& (3)(0)+(2)(2)& (3)(1)+(2)(1)\\(−1)(1)+(1)(−1)& (−1)(0)+(1)(2)& (−1)(1)+(1)(1) \end{bmatrix}$

= $\begin{bmatrix}2-1& 0+2& 2+1\\3-2& 0+4& 3+2\\-1-1& 0+2& -1+1 \end{bmatrix} = \begin{bmatrix}1& 2& 3\\1& 4& 5\\-2& 2& 0 \end{bmatrix}$

(vi) $\begin{bmatrix}3& −1& 3\\−1& 0& 2 \end{bmatrix} \begin{bmatrix}2& −3\\1& 0\\3& 1 \end{bmatrix}$

Order of first matrix is $2 \times 3$. Order of second matrix is $3 \times 2$. Product is possible and will be $2 \times 2$.

= $\begin{bmatrix}(3)(2)+(−1)(1)+(3)(3)& (3)(−3)+(−1)(0)+(3)(1)\\(−1)(2)+(0)(1)+(2)(3)& (−1)(−3)+(0)(0)+(2)(1) \end{bmatrix}$

= $\begin{bmatrix}6-1+9& -9+0+3\\-2+0+6& 3+0+2 \end{bmatrix} = \begin{bmatrix}14& -6\\4& 5 \end{bmatrix}$

Given:

Matrix A = $\begin{bmatrix}1& 2& −3\\5& 0& 2\\1& −1& 1 \end{bmatrix}$

Matrix B = $\begin{bmatrix}3& −1& 2\\4& 2& 5\\2& 0& 3 \end{bmatrix}$

Matrix C = $\begin{bmatrix}4& 1& 2\\0& 3& 2\\1& −2& 3 \end{bmatrix}$

All matrices are of order $3 \times 3$.

To Compute and Verify:

Compute (A + B) and (B – C). Verify that A + (B – C) = (A + B) – C.

Solution:

Compute (A + B):

Add the corresponding elements of A and B.

$A + B = \begin{bmatrix}1& 2& −3\\5& 0& 2\\1& −1& 1 \end{bmatrix} + \begin{bmatrix}3& −1& 2\\4& 2& 5\\2& 0& 3 \end{bmatrix} = \begin{bmatrix}1+3& 2+(-1)& -3+2\\5+4& 0+2& 2+5\\1+2& -1+0& 1+3 \end{bmatrix} = \begin{bmatrix}4& 1& -1\\9& 2& 7\\3& -1& 4 \end{bmatrix}$

Compute (B – C):

Subtract the corresponding elements of C from B.

$B – C = \begin{bmatrix}3& −1& 2\\4& 2& 5\\2& 0& 3 \end{bmatrix} - \begin{bmatrix}4& 1& 2\\0& 3& 2\\1& −2& 3 \end{bmatrix} = \begin{bmatrix}3-4& -1-1& 2-2\\4-0& 2-3& 5-2\\2-1& 0-(-2)& 3-3 \end{bmatrix} = \begin{bmatrix}-1& -2& 0\\4& -1& 3\\1& 0+2& 0 \end{bmatrix} = \begin{bmatrix}-1& -2& 0\\4& -1& 3\\1& 2& 0 \end{bmatrix}$

Verify that A + (B – C) = (A + B) – C:

Calculate the left-hand side ($L.H.S.$), A + (B – C):

$A + (B – C) = \begin{bmatrix}1& 2& −3\\5& 0& 2\\1& −1& 1 \end{bmatrix} + \begin{bmatrix}-1& -2& 0\\4& -1& 3\\1& 2& 0 \end{bmatrix} = \begin{bmatrix}1+(-1)& 2+(-2)& -3+0\\5+4& 0+(-1)& 2+3\\1+1& -1+2& 1+0 \end{bmatrix} = \begin{bmatrix}0& 0& -3\\9& -1& 5\\2& 1& 1 \end{bmatrix}$

Calculate the right-hand side ($R.H.S.$), (A + B) – C:

$(A + B) – C = \begin{bmatrix}4& 1& -1\\9& 2& 7\\3& -1& 4 \end{bmatrix} - \begin{bmatrix}4& 1& 2\\0& 3& 2\\1& −2& 3 \end{bmatrix} = \begin{bmatrix}4-4& 1-1& -1-2\\9-0& 2-3& 7-2\\3-1& -1-(-2)& 4-3 \end{bmatrix} = \begin{bmatrix}0& 0& -3\\9& -1& 5\\2& -1+2& 1 \end{bmatrix} = \begin{bmatrix}0& 0& -3\\9& -1& 5\\2& 1& 1 \end{bmatrix}$

Comparing the $L.H.S.$ and $R.H.S.$, we see that they are equal.

$A + (B – C) = \begin{bmatrix}0& 0& -3\\9& -1& 5\\2& 1& 1 \end{bmatrix}$

$(A + B) – C = \begin{bmatrix}0& 0& -3\\9& -1& 5\\2& 1& 1 \end{bmatrix}$

Thus, A + (B – C) = (A + B) – C is verified.

This demonstrates the associative property of matrix addition: (A + B) + C = A + (B + C). Subtracting C is equivalent to adding -C, so (A+B) - C = (A+B) + (-C) and A + (B-C) = A + (B + (-C)). The property holds.

Given:

Matrix A = $\begin{bmatrix}\frac{2}{3}& 1& \frac{5}{3}\\\frac{1}{3}& \frac{2}{3}& \frac{4}{3}\\ \frac{7}{3}& 2& \frac{2}{3} \end{bmatrix}$

Matrix B = $\begin{bmatrix}\frac{2}{5}& \frac{3}{5}& 1\\\frac{1}{5}& \frac{2}{5}& \frac{4}{5}\\\frac{7}{5}& \frac{6}{5}& \frac{2}{5} \end{bmatrix}$

Both matrices are of order $3 \times 3$.

To Compute:

The expression 3A – 5B.

Solution:

First, calculate the scalar multiple 3A by multiplying each element of A by 3.

$3A = 3 \begin{bmatrix}\frac{2}{3}& 1& \frac{5}{3}\\\frac{1}{3}& \frac{2}{3}& \frac{4}{3}\\ \frac{7}{3}& 2& \frac{2}{3} \end{bmatrix} = \begin{bmatrix}3 \times \frac{2}{3}& 3 \times 1& 3 \times \frac{5}{3}\\3 \times \frac{1}{3}& 3 \times \frac{2}{3}& 3 \times \frac{4}{3}\\3 \times \frac{7}{3}& 3 \times 2& 3 \times \frac{2}{3} \end{bmatrix} = \begin{bmatrix}\frac{6}{3}& 3& \frac{15}{3}\\\frac{3}{3}& \frac{6}{3}& \frac{12}{3}\\\frac{21}{3}& 6& \frac{6}{3} \end{bmatrix} = \begin{bmatrix}2& 3& 5\\1& 2& 4\\7& 6& 2 \end{bmatrix}$

Next, calculate the scalar multiple 5B by multiplying each element of B by 5.

$5B = 5 \begin{bmatrix}\frac{2}{5}& \frac{3}{5}& 1\\\frac{1}{5}& \frac{2}{5}& \frac{4}{5}\\\frac{7}{5}& \frac{6}{5}& \frac{2}{5} \end{bmatrix} = \begin{bmatrix}5 \times \frac{2}{5}& 5 \times \frac{3}{5}& 5 \times 1\\5 \times \frac{1}{5}& 5 \times \frac{2}{5}& 5 \times \frac{4}{5}\\5 \times \frac{7}{5}& 5 \times \frac{6}{5}& 5 \times \frac{2}{5} \end{bmatrix} = \begin{bmatrix}\frac{10}{5}& \frac{15}{5}& 5\\\frac{5}{5}& \frac{10}{5}& \frac{20}{5}\\\frac{35}{5}& \frac{30}{5}& \frac{10}{5} \end{bmatrix} = \begin{bmatrix}2& 3& 5\\1& 2& 4\\7& 6& 2 \end{bmatrix}$

Now, compute 3A – 5B by subtracting the corresponding elements of 5B from 3A.

$3A – 5B = \begin{bmatrix}2& 3& 5\\1& 2& 4\\7& 6& 2 \end{bmatrix} - \begin{bmatrix}2& 3& 5\\1& 2& 4\\7& 6& 2 \end{bmatrix}$

$3A – 5B = \begin{bmatrix}2-2& 3-3& 5-5\\1-1& 2-2& 4-4\\7-7& 6-6& 2-2 \end{bmatrix} = \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$

The resulting matrix is the zero matrix of order $3 \times 3$.

The result of 3A – 5B is $\begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$.

Given:

The expression is $\cos \;θ\begin{bmatrix}\cos\;θ& \sin\;θ\\− \sin\;θ& \cos\;θ \end{bmatrix} + \sin \;θ \begin{bmatrix}\sin\;θ& −\cos\;θ\\\cos\;θ& \sin\;θ \end{bmatrix}$.

To Simplify:

Simplify the given expression.

Solution:

First, perform the scalar multiplication for each term.

Multiply the first matrix by $\cos \theta$:

$\cos \;θ\begin{bmatrix}\cos\;θ& \sin\;θ\\− \sin\;θ& \cos\;θ \end{bmatrix} = \begin{bmatrix}\cos \theta \times \cos \theta& \cos \theta \times \sin \theta\\ \cos \theta \times (-\sin \theta)& \cos \theta \times \cos \theta \end{bmatrix} = \begin{bmatrix}\cos^2 \theta& \sin \theta \cos \theta\\ −\sin \theta \cos \theta& \cos^2 \theta \end{bmatrix}$

Multiply the second matrix by $\sin \theta$:

$\sin \;θ \begin{bmatrix}\sin\;θ& −\cos\;θ\\\cos\;θ& \sin\;θ \end{bmatrix} = \begin{bmatrix}\sin \theta \times \sin \theta& \sin \theta \times (-\cos \theta)\\ \sin \theta \times \cos \theta& \sin \theta \times \sin \theta \end{bmatrix} = \begin{bmatrix}\sin^2 \theta& −\sin \theta \cos \theta\\ \sin \theta \cos \theta& \sin^2 \theta \end{bmatrix}$

Now, add the resulting matrices. Addition is possible since both matrices are of order $2 \times 2$.

$\begin{bmatrix}\cos^2 \theta& \sin \theta \cos \theta\\ −\sin \theta \cos \theta& \cos^2 \theta \end{bmatrix} + \begin{bmatrix}\sin^2 \theta& −\sin \theta \cos \theta\\ \sin \theta \cos \theta& \sin^2 \theta \end{bmatrix}$

Add the corresponding elements:

• Element (1,1): $\cos^2 \theta + \sin^2 \theta$
• Element (1,2): $\sin \theta \cos \theta + (-\sin \theta \cos \theta) = \sin \theta \cos \theta - \sin \theta \cos \theta = 0$
• Element (2,1): $-\sin \theta \cos \theta + \sin \theta \cos \theta = 0$
• Element (2,2): $\cos^2 \theta + \sin^2 \theta$

Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, the resulting matrix is:

= $\begin{bmatrix}1& 0\\0& 1 \end{bmatrix}$

This is the identity matrix of order $2 \times 2$, denoted by $I_2$ or just $I$.

The simplified expression is $\begin{bmatrix}1& 0\\0& 1 \end{bmatrix}$.

Given:

Systems of matrix equations involving matrices X and Y.

To Find:

The matrices X and Y that satisfy the given equations in each part.

Solution:

We can solve these systems of matrix equations using methods similar to solving systems of linear equations with variables.

(i) X + Y = $\begin{bmatrix}7& 0\\2& 5 \end{bmatrix}$ and X - Y = $\begin{bmatrix}3& 0\\0& 3 \end{bmatrix}$

Let the given equations be:

Add equation (1) and equation (2):

$(X + Y) + (X - Y) = \begin{bmatrix}7& 0\\2& 5 \end{bmatrix} + \begin{bmatrix}3& 0\\0& 3 \end{bmatrix}$

$2X = \begin{bmatrix}7+3& 0+0\\2+0& 5+3 \end{bmatrix} = \begin{bmatrix}10& 0\\2& 8 \end{bmatrix}$

$X = \frac{1}{2} \begin{bmatrix}10& 0\\2& 8 \end{bmatrix} = \begin{bmatrix}\frac{1}{2} \times 10& \frac{1}{2} \times 0\\\frac{1}{2} \times 2& \frac{1}{2} \times 8 \end{bmatrix} = \begin{bmatrix}5& 0\\1& 4 \end{bmatrix}$

Subtract equation (2) from equation (1):

$(X + Y) - (X - Y) = \begin{bmatrix}7& 0\\2& 5 \end{bmatrix} - \begin{bmatrix}3& 0\\0& 3 \end{bmatrix}$

$2Y = \begin{bmatrix}7-3& 0-0\\2-0& 5-3 \end{bmatrix} = \begin{bmatrix}4& 0\\2& 2 \end{bmatrix}$

$Y = \frac{1}{2} \begin{bmatrix}4& 0\\2& 2 \end{bmatrix} = \begin{bmatrix}\frac{1}{2} \times 4& \frac{1}{2} \times 0\\\frac{1}{2} \times 2& \frac{1}{2} \times 2 \end{bmatrix} = \begin{bmatrix}2& 0\\1& 1 \end{bmatrix}$

The matrices are $X = \begin{bmatrix}5& 0\\1& 4 \end{bmatrix}$ and $Y = \begin{bmatrix}2& 0\\1& 1 \end{bmatrix}$.

(ii) 2X + 3Y = $\begin{bmatrix}2& 3\\4& 0 \end{bmatrix}$ and 3X + 2Y = $\begin{bmatrix}2& −2\\−1& 5 \end{bmatrix}$

Let the given equations be:

Multiply equation (3) by 3 and equation (4) by 2 to eliminate X:

$3 \times (2X + 3Y) = 3 \times \begin{bmatrix}2& 3\\4& 0 \end{bmatrix} \implies 6X + 9Y = \begin{bmatrix}6& 9\\12& 0 \end{bmatrix}$

$2 \times (3X + 2Y) = 2 \times \begin{bmatrix}2& −2\\−1& 5 \end{bmatrix} \implies 6X + 4Y = \begin{bmatrix}4& −4\\−2& 10 \end{bmatrix}$

Subtract equation (6) from equation (5):

$(6X + 9Y) - (6X + 4Y) = \begin{bmatrix}6& 9\\12& 0 \end{bmatrix} - \begin{bmatrix}4& −4\\−2& 10 \end{bmatrix}$

$5Y = \begin{bmatrix}6-4& 9-(-4)\\12-(-2)& 0-10 \end{bmatrix} = \begin{bmatrix}2& 9+4\\12+2& -10 \end{bmatrix} = \begin{bmatrix}2& 13\\14& -10 \end{bmatrix}$

$Y = \frac{1}{5} \begin{bmatrix}2& 13\\14& -10 \end{bmatrix} = \begin{bmatrix}\frac{2}{5}& \frac{13}{5}\\\frac{14}{5}& \frac{-10}{5} \end{bmatrix} = \begin{bmatrix}\frac{2}{5}& \frac{13}{5}\\\frac{14}{5}& -2 \end{bmatrix}$

Now, we can substitute the matrix Y into one of the original equations (say, equation 3) to find X.

$2X + 3Y = \begin{bmatrix}2& 3\\4& 0 \end{bmatrix}$

$2X = \begin{bmatrix}2& 3\\4& 0 \end{bmatrix} - 3Y$

$3Y = 3 \begin{bmatrix}\frac{2}{5}& \frac{13}{5}\\\frac{14}{5}& -2 \end{bmatrix} = \begin{bmatrix}3 \times \frac{2}{5}& 3 \times \frac{13}{5}\\3 \times \frac{14}{5}& 3 \times -2 \end{bmatrix} = \begin{bmatrix}\frac{6}{5}& \frac{39}{5}\\\frac{42}{5}& -6 \end{bmatrix}$

$2X = \begin{bmatrix}2& 3\\4& 0 \end{bmatrix} - \begin{bmatrix}\frac{6}{5}& \frac{39}{5}\\\frac{42}{5}& -6 \end{bmatrix} = \begin{bmatrix}2 - \frac{6}{5}& 3 - \frac{39}{5}\\4 - \frac{42}{5}& 0 - (-6) \end{bmatrix}$

$2 - \frac{6}{5} = \frac{10}{5} - \frac{6}{5} = \frac{4}{5}$

$3 - \frac{39}{5} = \frac{15}{5} - \frac{39}{5} = \frac{-24}{5}$

$4 - \frac{42}{5} = \frac{20}{5} - \frac{42}{5} = \frac{-22}{5}$

$0 - (-6) = 6$

$2X = \begin{bmatrix}\frac{4}{5}& \frac{-24}{5}\\\frac{-22}{5}& 6 \end{bmatrix}$

$X = \frac{1}{2} \begin{bmatrix}\frac{4}{5}& \frac{-24}{5}\\\frac{-22}{5}& 6 \end{bmatrix} = \begin{bmatrix}\frac{1}{2} \times \frac{4}{5}& \frac{1}{2} \times \frac{-24}{5}\\\frac{1}{2} \times \frac{-22}{5}& \frac{1}{2} \times 6 \end{bmatrix} = \begin{bmatrix}\frac{2}{5}& \frac{-12}{5}\\\frac{-11}{5}& 3 \end{bmatrix}$

The matrices are $X = \begin{bmatrix}\frac{2}{5}& -\frac{12}{5}\\-\frac{11}{5}& 3 \end{bmatrix}$ and $Y = \begin{bmatrix}\frac{2}{5}& \frac{13}{5}\\\frac{14}{5}& -2 \end{bmatrix}$.

Given:

Matrix Y = $\begin{bmatrix}3& 2\\1& 4 \end{bmatrix}$ and the equation $2X + Y = \begin{bmatrix}1& 0\\−3& 2 \end{bmatrix}$.

To Find:

The matrix X that satisfies the given equation.

Solution:

The given equation is:

We want to solve for matrix X. First, isolate the term with X by subtracting Y from both sides:

$2X = \begin{bmatrix}1& 0\\−3& 2 \end{bmatrix} - Y$

Substitute the given matrix Y into the equation:

$2X = \begin{bmatrix}1& 0\\−3& 2 \end{bmatrix} - \begin{bmatrix}3& 2\\1& 4 \end{bmatrix}$

Perform the matrix subtraction:

$2X = \begin{bmatrix}1-3& 0-2\\-3-1& 2-4 \end{bmatrix} = \begin{bmatrix}-2& -2\\-4& -2 \end{bmatrix}$

Now, to find X, multiply the resulting matrix by the scalar $\frac{1}{2}$:

$X = \frac{1}{2} \begin{bmatrix}-2& -2\\-4& -2 \end{bmatrix}$

Multiply each element of the matrix by $\frac{1}{2}$:

$X = \begin{bmatrix}\frac{1}{2} \times (-2)& \frac{1}{2} \times (-2)\\\frac{1}{2} \times (-4)& \frac{1}{2} \times (-2) \end{bmatrix} = \begin{bmatrix}-1& -1\\-2& -1 \end{bmatrix}$

We can verify the result by substituting X and Y back into the original equation (1).

$2X + Y = 2\begin{bmatrix}-1& -1\\-2& -1 \end{bmatrix} + \begin{bmatrix}3& 2\\1& 4 \end{bmatrix} = \begin{bmatrix}2 \times (-1)& 2 \times (-1)\\2 \times (-2)& 2 \times (-1) \end{bmatrix} + \begin{bmatrix}3& 2\\1& 4 \end{bmatrix}$

= $\begin{bmatrix}-2& -2\\-4& -2 \end{bmatrix} + \begin{bmatrix}3& 2\\1& 4 \end{bmatrix} = \begin{bmatrix}-2+3& -2+2\\-4+1& -2+4 \end{bmatrix} = \begin{bmatrix}1& 0\\-3& 2 \end{bmatrix}$

This matches the right-hand side of the original equation, so our value for X is correct.

The matrix X is $\begin{bmatrix}-1& -1\\-2& -1 \end{bmatrix}$.

Given:

The matrix equation $2\begin{bmatrix}1& 3\\0& x \end{bmatrix} + \begin{bmatrix}y& 0\\1& 2 \end{bmatrix} = \begin{bmatrix}5& 6\\1& 8 \end{bmatrix}$.

To Find:

The values of $x$ and $y$ that satisfy the given equation.

Solution:

First, perform the scalar multiplication on the left-hand side ($L.H.S.$) of the equation:

$2\begin{bmatrix}1& 3\\0& x \end{bmatrix} = \begin{bmatrix}2 \times 1& 2 \times 3\\2 \times 0& 2 \times x \end{bmatrix} = \begin{bmatrix}2& 6\\0& 2x \end{bmatrix}$

Now, add the resulting matrix to the second matrix on the $L.H.S.$:

$\begin{bmatrix}2& 6\\0& 2x \end{bmatrix} + \begin{bmatrix}y& 0\\1& 2 \end{bmatrix} = \begin{bmatrix}2 + y& 6 + 0\\0 + 1& 2x + 2 \end{bmatrix} = \begin{bmatrix}2 + y& 6\\1& 2x + 2 \end{bmatrix}$

So, the given equation becomes:

$\begin{bmatrix}2 + y& 6\\1& 2x + 2 \end{bmatrix} = \begin{bmatrix}5& 6\\1& 8 \end{bmatrix}$

For these two matrices to be equal, their corresponding elements must be equal. Both matrices are of order $2 \times 2$.

Equating the elements in the first row, first column:

Equating the elements in the first row, second column: $6 = 6$ (Consistent)

Equating the elements in the second row, first column: $1 = 1$ (Consistent)

Equating the elements in the second row, second column:

Now, solve the linear equations for $x$ and $y$.

From equation (1):

$2 + y = 5$

$y = 5 - 2$

From equation (2):

$2x + 2 = 8$

$2x = 8 - 2$

$2x = 6$

$x = \frac{6}{2}$

The values of $x$ and $y$ are 3 and 3, respectively.

The values are $x = 3$ and $y = 3$.

Given:

The matrix equation $2\begin{bmatrix}x& z\\y& t \end{bmatrix} + 3\begin{bmatrix}1& −1\\0& 2 \end{bmatrix} = 3\begin{bmatrix}3& 5\\4& 6 \end{bmatrix}$.

To Solve:

Find the values of $x, y, z,$ and $t$ that satisfy the given equation.

Solution:

Perform the scalar multiplications on both sides of the equation.

Left-hand side ($L.H.S.$):

$2\begin{bmatrix}x& z\\y& t \end{bmatrix} = \begin{bmatrix}2x& 2z\\2y& 2t \end{bmatrix}$

$3\begin{bmatrix}1& −1\\0& 2 \end{bmatrix} = \begin{bmatrix}3 \times 1& 3 \times (−1)\\3 \times 0& 3 \times 2 \end{bmatrix} = \begin{bmatrix}3& −3\\0& 6 \end{bmatrix}$

$L.H.S. = \begin{bmatrix}2x& 2z\\2y& 2t \end{bmatrix} + \begin{bmatrix}3& −3\\0& 6 \end{bmatrix} = \begin{bmatrix}2x + 3& 2z - 3\\2y + 0& 2t + 6 \end{bmatrix} = \begin{bmatrix}2x + 3& 2z - 3\\2y& 2t + 6 \end{bmatrix}$

Right-hand side ($R.H.S.$):

$3\begin{bmatrix}3& 5\\4& 6 \end{bmatrix} = \begin{bmatrix}3 \times 3& 3 \times 5\\3 \times 4& 3 \times 6 \end{bmatrix} = \begin{bmatrix}9& 15\\12& 18 \end{bmatrix}$

Equating the $L.H.S.$ and $R.H.S.$, the matrix equation becomes:

$\begin{bmatrix}2x + 3& 2z - 3\\2y& 2t + 6 \end{bmatrix} = \begin{bmatrix}9& 15\\12& 18 \end{bmatrix}$

For these two matrices to be equal, their corresponding elements must be equal. Both matrices are of order $2 \times 2$.

Equating the elements in the first row, first column:

Equating the elements in the first row, second column:

Equating the elements in the second row, first column:

Equating the elements in the second row, second column:

Now, solve the linear equations for $x, y, z,$ and $t$.

From equation (1):

$2x + 3 = 9$

$2x = 9 - 3$

$2x = 6$

$x = \frac{6}{2}$

From equation (2):

$2z - 3 = 15$

$2z = 15 + 3$

$2z = 18$

$z = \frac{18}{2}$

From equation (3):

$2y = 12$

$y = \frac{12}{2}$

From equation (4):

$2t + 6 = 18$

$2t = 18 - 6$

$2t = 12$

$t = \frac{12}{2}$

The values are $x=3, y=6, z=9,$ and $t=6$.

The values are $x = 3$, $y = 6$, $z = 9$, and $t = 6$.

Given:

The vector equation $x\begin{bmatrix}2\\3 \end{bmatrix} + y\begin{bmatrix}−1\\1 \end{bmatrix} = \begin{bmatrix}10\\5 \end{bmatrix}$.

To Find:

The values of $x$ and $y$ that satisfy the given equation.

Solution:

Perform the scalar multiplication on the left-hand side ($L.H.S.$) of the equation:

$x\begin{bmatrix}2\\3 \end{bmatrix} = \begin{bmatrix}x \times 2\\x \times 3 \end{bmatrix} = \begin{bmatrix}2x\\3x \end{bmatrix}$

$y\begin{bmatrix}−1\\1 \end{bmatrix} = \begin{bmatrix}y \times (−1)\\y \times 1 \end{bmatrix} = \begin{bmatrix}-y\\y \end{bmatrix}$

Now, add the resulting column matrices:

$\begin{bmatrix}2x\\3x \end{bmatrix} + \begin{bmatrix}-y\\y \end{bmatrix} = \begin{bmatrix}2x + (-y)\\3x + y \end{bmatrix} = \begin{bmatrix}2x - y\\3x + y \end{bmatrix}$

So, the given equation becomes:

$\begin{bmatrix}2x - y\\3x + y \end{bmatrix} = \begin{bmatrix}10\\5 \end{bmatrix}$

For these two column matrices to be equal, their corresponding elements must be equal. Both matrices are of order $2 \times 1$.

Equating the elements in the first row:

Equating the elements in the second row:

We have a system of two linear equations with two variables $x$ and $y$. We can solve this system using elimination or substitution.

Using the elimination method, add equation (1) and equation (2):

$(2x - y) + (3x + y) = 10 + 5$

$2x - y + 3x + y = 15$

$5x = 15$

$x = \frac{15}{5}$

Now substitute the value of $x$ from equation (3) into equation (2):

$3(3) + y = 5$

$9 + y = 5$

$y = 5 - 9$

We can check these values by substituting them into equation (1): $2(3) - (-4) = 6 + 4 = 10$, which is correct.

The values of $x$ and $y$ are 3 and -4, respectively.

The values are $x = 3$ and $y = -4$.

Given:

The matrix equation is $3\begin{bmatrix}x& y\\z& w \end{bmatrix} = \begin{bmatrix}x& 6\\−1& 2w \end{bmatrix} + \begin{bmatrix}4& x+y\\z+w& 3 \end{bmatrix}$.

To Find:

The values of $x, y, z,$ and $w$ that satisfy the given equation.

Solution:

Perform the scalar multiplication on the left-hand side ($L.H.S.$) and the matrix addition on the right-hand side ($R.H.S.$) of the equation.

$L.H.S. = 3\begin{bmatrix}x& y\\z& w \end{bmatrix} = \begin{bmatrix}3x& 3y\\3z& 3w \end{bmatrix}$

$R.H.S. = \begin{bmatrix}x& 6\\−1& 2w \end{bmatrix} + \begin{bmatrix}4& x+y\\z+w& 3 \end{bmatrix} = \begin{bmatrix}x+4& 6+(x+y)\\-1+(z+w)& 2w+3 \end{bmatrix} = \begin{bmatrix}x+4& 6+x+y\\-1+z+w& 2w+3 \end{bmatrix}$

Equating the $L.H.S.$ and $R.H.S.$, the matrix equation becomes:

$\begin{bmatrix}3x& 3y\\3z& 3w \end{bmatrix} = \begin{bmatrix}x+4& 6+x+y\\-1+z+w& 2w+3 \end{bmatrix}$

For these two matrices to be equal, their corresponding elements must be equal. Both matrices are of order $2 \times 2$.

Equating the elements in the first row, first column:

Equating the elements in the first row, second column:

Equating the elements in the second row, first column:

Equating the elements in the second row, second column:

Now, solve the system of linear equations for $x, y, z,$ and $w$.

From equation (1):

$3x - x = 4$

$2x = 4$

$x = \frac{4}{2}$

From equation (4):

$3w - 2w = 3$

Substitute the value of $x$ from equation (5) into equation (2):

$3y = 6 + 2 + y$

$3y = 8 + y$

$3y - y = 8$

$2y = 8$

$y = \frac{8}{2}$

Substitute the value of $w$ from equation (6) into equation (3):

$3z = -1 + z + 3$

$3z = z + 2$

$3z - z = 2$

$2z = 2$

$z = \frac{2}{2}$

The values are $x=2, y=4, z=1,$ and $w=3$.

The values are $x = 2$, $y = 4$, $z = 1$, and $w = 3$.

Given:

The matrix function $F(x) = \begin{bmatrix}\cos\;x& −\sin\;x& 0\\\sin\;x& \cos\;x& 0\\0& 0& 1 \end{bmatrix}$.

To Show:

$F(x) F(y) = F(x + y)$.

Solution:

First, let's find $F(y)$. Replace $x$ with $y$ in the definition of $F(x)$:

$F(y) = \begin{bmatrix}\cos\;y& −\sin\;y& 0\\\sin\;y& \cos\;y& 0\\0& 0& 1 \end{bmatrix}$

Now, calculate the product $F(x) F(y)$. The order of $F(x)$ is $3 \times 3$ and the order of $F(y)$ is $3 \times 3$. The product is possible and will be a $3 \times 3$ matrix.

$F(x) F(y) = \begin{bmatrix}\cos\;x& −\sin\;x& 0\\\sin\;x& \cos\;x& 0\\0& 0& 1 \end{bmatrix} \begin{bmatrix}\cos\;y& −\sin\;y& 0\\\sin\;y& \cos\;y& 0\\0& 0& 1 \end{bmatrix}$

Perform the matrix multiplication:

• Element (1,1): $(\cos x)(\cos y) + (-\sin x)(\sin y) + (0)(0) = \cos x \cos y - \sin x \sin y$
• Element (1,2): $(\cos x)(-\sin y) + (-\sin x)(\cos y) + (0)(0) = -\cos x \sin y - \sin x \cos y = -(\cos x \sin y + \sin x \cos y)$
• Element (1,3): $(\cos x)(0) + (-\sin x)(0) + (0)(1) = 0 + 0 + 0 = 0$
• Element (2,1): $(\sin x)(\cos y) + (\cos x)(\sin y) + (0)(0) = \sin x \cos y + \cos x \sin y$
• Element (2,2): $(\sin x)(-\sin y) + (\cos x)(\cos y) + (0)(0) = -\sin x \sin y + \cos x \cos y = \cos x \cos y - \sin x \sin y$
• Element (2,3): $(\sin x)(0) + (\cos x)(0) + (0)(1) = 0 + 0 + 0 = 0$
• Element (3,1): $(0)(\cos y) + (0)(\sin y) + (1)(0) = 0 + 0 + 0 = 0$
• Element (3,2): $(0)(-\sin y) + (0)(\cos y) + (1)(0) = 0 + 0 + 0 = 0$
• Element (3,3): $(0)(0) + (0)(0) + (1)(1) = 0 + 0 + 1 = 1$

Using the trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$, the resulting matrix is:

$F(x) F(y) = \begin{bmatrix}\cos(x+y)& -(\sin(x+y))& 0\\\sin(x+y)& \cos(x+y)& 0\\0& 0& 1 \end{bmatrix}$

Now, let's find $F(x + y)$. Replace $x$ with $(x + y)$ in the definition of $F(x)$:

$F(x + y) = \begin{bmatrix}\cos(x+y)& −\sin(x+y)& 0\\\sin(x+y)& \cos(x+y)& 0\\0& 0& 1 \end{bmatrix}$

Comparing the matrices in equation (1) and equation (2), we see that they are equal.

Therefore, $F(x) F(y) = F(x + y)$.

The identity $F(x) F(y) = F(x + y)$ is shown.

Given:

Pairs of matrices for which we need to show that matrix multiplication is not commutative.

To Show:

(i) $\begin{bmatrix}5& −1\\6& 7 \end{bmatrix} \begin{bmatrix}2& 1\\3& 4 \end{bmatrix} ≠ \begin{bmatrix}2& 1\\3& 4 \end{bmatrix} \begin{bmatrix}5& −1\\6& 7 \end{bmatrix}$

(ii) $\begin{bmatrix}1& 2& 3\\0& 1& 0\\1& 1& 0 \end{bmatrix} \begin{bmatrix}−1& 1& 0\\0& −1& 1\\2& 3& 4 \end{bmatrix} ≠ \begin{bmatrix}−1& 1& 0\\0& −1& 1\\2& 3& 4 \end{bmatrix} \begin{bmatrix}1& 2& 3\\0& 1& 0\\1& 1& 0 \end{bmatrix}$

Solution:

To show that AB ≠ BA, we need to compute both products and demonstrate that the resulting matrices are not equal.

(i) Let A = $\begin{bmatrix}5& −1\\6& 7 \end{bmatrix}$ and B = $\begin{bmatrix}2& 1\\3& 4 \end{bmatrix}$. Both matrices are $2 \times 2$. The products AB and BA are possible and will be $2 \times 2$ matrices.

Calculate AB:

$AB = \begin{bmatrix}5& −1\\6& 7 \end{bmatrix} \begin{bmatrix}2& 1\\3& 4 \end{bmatrix} = \begin{bmatrix}(5)(2) + (−1)(3)& (5)(1) + (−1)(4)\\(6)(2) + (7)(3)& (6)(1) + (7)(4) \end{bmatrix} = \begin{bmatrix}10 - 3& 5 - 4\\12 + 21& 6 + 28 \end{bmatrix} = \begin{bmatrix}7& 1\\33& 34 \end{bmatrix}$

Calculate BA:

$BA = \begin{bmatrix}2& 1\\3& 4 \end{bmatrix} \begin{bmatrix}5& −1\\6& 7 \end{bmatrix} = \begin{bmatrix}(2)(5) + (1)(6)& (2)(−1) + (1)(7)\\(3)(5) + (4)(6)& (3)(−1) + (4)(7) \end{bmatrix} = \begin{bmatrix}10 + 6& -2 + 7\\15 + 24& -3 + 28 \end{bmatrix} = \begin{bmatrix}16& 5\\39& 25 \end{bmatrix}$

Comparing AB = $\begin{bmatrix}7& 1\\33& 34 \end{bmatrix}$ and BA = $\begin{bmatrix}16& 5\\39& 25 \end{bmatrix}$, we see that their corresponding elements are not equal.

For example, the element in the first row, first column of AB is 7, while in BA it is 16. Since the matrices are not identical, AB ≠ BA.

(ii) Let A = $\begin{bmatrix}1& 2& 3\\0& 1& 0\\1& 1& 0 \end{bmatrix}$ and B = $\begin{bmatrix}−1& 1& 0\\0& −1& 1\\2& 3& 4 \end{bmatrix}$. Both matrices are $3 \times 3$. The products AB and BA are possible and will be $3 \times 3$ matrices.

Calculate AB:

$AB = \begin{bmatrix}1& 2& 3\\0& 1& 0\\1& 1& 0 \end{bmatrix} \begin{bmatrix}−1& 1& 0\\0& −1& 1\\2& 3& 4 \end{bmatrix}$

• Element (1,1): $(1)(−1) + (2)(0) + (3)(2) = -1 + 0 + 6 = 5$
• Element (1,2): $(1)(1) + (2)(−1) + (3)(3) = 1 - 2 + 9 = 8$
• Element (1,3): $(1)(0) + (2)(1) + (3)(4) = 0 + 2 + 12 = 14$
• Element (2,1): $(0)(−1) + (1)(0) + (0)(2) = 0 + 0 + 0 = 0$
• Element (2,2): $(0)(1) + (1)(−1) + (0)(3) = 0 - 1 + 0 = -1$
• Element (2,3): $(0)(0) + (1)(1) + (0)(4) = 0 + 1 + 0 = 1$
• Element (3,1): $(1)(−1) + (1)(0) + (0)(2) = -1 + 0 + 0 = -1$
• Element (3,2): $(1)(1) + (1)(−1) + (0)(3) = 1 - 1 + 0 = 0$
• Element (3,3): $(1)(0) + (1)(1) + (0)(4) = 0 + 1 + 0 = 1$

So, $AB = \begin{bmatrix}5& 8& 14\\0& -1& 1\\-1& 0& 1 \end{bmatrix}$.

Calculate BA:

$BA = \begin{bmatrix}−1& 1& 0\\0& −1& 1\\2& 3& 4 \end{bmatrix} \begin{bmatrix}1& 2& 3\\0& 1& 0\\1& 1& 0 \end{bmatrix}$

• Element (1,1): $(-1)(1) + (1)(0) + (0)(1) = -1 + 0 + 0 = -1$
• Element (1,2): $(-1)(2) + (1)(1) + (0)(1) = -2 + 1 + 0 = -1$
• Element (1,3): $(-1)(3) + (1)(0) + (0)(0) = -3 + 0 + 0 = -3$
• Element (2,1): $(0)(1) + (-1)(0) + (1)(1) = 0 + 0 + 1 = 1$
• Element (2,2): $(0)(2) + (-1)(1) + (1)(1) = 0 - 1 + 1 = 0$
• Element (2,3): $(0)(3) + (-1)(0) + (1)(0) = 0 + 0 + 0 = 0$
• Element (3,1): $(2)(1) + (3)(0) + (4)(1) = 2 + 0 + 4 = 6$
• Element (3,2): $(2)(2) + (3)(1) + (4)(1) = 4 + 3 + 4 = 11$
• Element (3,3): $(2)(3) + (3)(0) + (4)(0) = 6 + 0 + 0 = 6$

So, $BA = \begin{bmatrix}-1& -1& -3\\1& 0& 0\\6& 11& 6 \end{bmatrix}$.

Comparing AB = $\begin{bmatrix}5& 8& 14\\0& -1& 1\\-1& 0& 1 \end{bmatrix}$ and BA = $\begin{bmatrix}-1& -1& -3\\1& 0& 0\\6& 11& 6 \end{bmatrix}$, we see that their corresponding elements are not equal.

For example, the element in the first row, first column of AB is 5, while in BA it is -1. Since the matrices are not identical, AB ≠ BA.

In both cases, we have shown that AB ≠ BA, which demonstrates that matrix multiplication is generally not commutative.

Given:

Matrix A = $\begin{bmatrix}2& 0& 1\\2& 1& 3\\1& −1& 0 \end{bmatrix}$.

We need to compute the expression $A^2 – 5A + 6I$. The identity matrix I of the same order as A is $I = \begin{bmatrix}1& 0& 0\\0& 1& 0\\0& 0& 1 \end{bmatrix}$.

To Compute:

$A^2 – 5A + 6I$.

Solution:

First, calculate $A^2 = A \times A$. The order of A is $3 \times 3$, so $A^2$ will also be $3 \times 3$.

$A^2 = \begin{bmatrix}2& 0& 1\\2& 1& 3\\1& −1& 0 \end{bmatrix} \begin{bmatrix}2& 0& 1\\2& 1& 3\\1& −1& 0 \end{bmatrix}$

• Element (1,1): $(2)(2) + (0)(2) + (1)(1) = 4 + 0 + 1 = 5$
• Element (1,2): $(2)(0) + (0)(1) + (1)(-1) = 0 + 0 - 1 = -1$
• Element (1,3): $(2)(1) + (0)(3) + (1)(0) = 2 + 0 + 0 = 2$
• Element (2,1): $(2)(2) + (1)(2) + (3)(1) = 4 + 2 + 3 = 9$
• Element (2,2): $(2)(0) + (1)(1) + (3)(-1) = 0 + 1 - 3 = -2$
• Element (2,3): $(2)(1) + (1)(3) + (3)(0) = 2 + 3 + 0 = 5$
• Element (3,1): $(1)(2) + (-1)(2) + (0)(1) = 2 - 2 + 0 = 0$
• Element (3,2): $(1)(0) + (-1)(1) + (0)(-1) = 0 - 1 + 0 = -1$
• Element (3,3): $(1)(1) + (-1)(3) + (0)(0) = 1 - 3 + 0 = -2$

So, $A^2 = \begin{bmatrix}5& -1& 2\\9& -2& 5\\0& -1& -2 \end{bmatrix}$.

Next, calculate $5A$ and $6I$.

$5A = 5 \begin{bmatrix}2& 0& 1\\2& 1& 3\\1& −1& 0 \end{bmatrix} = \begin{bmatrix}5 \times 2& 5 \times 0& 5 \times 1\\5 \times 2& 5 \times 1& 5 \times 3\\5 \times 1& 5 \times (-1)& 5 \times 0 \end{bmatrix} = \begin{bmatrix}10& 0& 5\\10& 5& 15\\5& -5& 0 \end{bmatrix}$

$6I = 6 \begin{bmatrix}1& 0& 0\\0& 1& 0\\0& 0& 1 \end{bmatrix} = \begin{bmatrix}6 \times 1& 6 \times 0& 6 \times 0\\6 \times 0& 6 \times 1& 6 \times 0\\6 \times 0& 6 \times 0& 6 \times 1 \end{bmatrix} = \begin{bmatrix}6& 0& 0\\0& 6& 0\\0& 0& 6 \end{bmatrix}$

Now, compute $A^2 – 5A + 6I$. This involves matrix subtraction and addition.

$A^2 – 5A + 6I = \begin{bmatrix}5& -1& 2\\9& -2& 5\\0& -1& -2 \end{bmatrix} - \begin{bmatrix}10& 0& 5\\10& 5& 15\\5& -5& 0 \end{bmatrix} + \begin{bmatrix}6& 0& 0\\0& 6& 0\\0& 0& 6 \end{bmatrix}$

$A^2 – 5A + 6I = \begin{bmatrix}5 - 10 + 6& -1 - 0 + 0& 2 - 5 + 0\\9 - 10 + 0& -2 - 5 + 6& 5 - 15 + 0\\0 - 5 + 0& -1 - (-5) + 0& -2 - 0 + 6 \end{bmatrix}$

$A^2 – 5A + 6I = \begin{bmatrix}1 & -1 & -3 \\ -1 & -7 + 6 & -10 \\ -5 & -1 + 5 & 4 \end{bmatrix}$

$A^2 – 5A + 6I = \begin{bmatrix}1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$

The result of $A^2 – 5A + 6I$ is $\begin{bmatrix}1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$.

Given:

Matrix A = $\begin{bmatrix}1& 0& 2\\0& 2& 1\\2& 0& 3 \end{bmatrix}$.

The identity matrix of order $3 \times 3$ is $I = \begin{bmatrix}1& 0& 0\\0& 1& 0\\0& 0& 1 \end{bmatrix}$.

The zero matrix of order $3 \times 3$ is $O = \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$.

To Prove:

$A^3 – 6A^2 + 7A + 2I = O$

Solution:

First, we need to calculate $A^2 = A \times A$. The order of A is $3 \times 3$, so the order of $A^2$ will also be $3 \times 3$.

$A^2 = \begin{bmatrix}1& 0& 2\\0& 2& 1\\2& 0& 3 \end{bmatrix} \begin{bmatrix}1& 0& 2\\0& 2& 1\\2& 0& 3 \end{bmatrix}$

• Element (1,1): $(1)(1) + (0)(0) + (2)(2) = 1 + 0 + 4 = 5$
• Element (1,2): $(1)(0) + (0)(2) + (2)(0) = 0 + 0 + 0 = 0$
• Element (1,3): $(1)(2) + (0)(1) + (2)(3) = 2 + 0 + 6 = 8$
• Element (2,1): $(0)(1) + (2)(0) + (1)(2) = 0 + 0 + 2 = 2$
• Element (2,2): $(0)(0) + (2)(2) + (1)(0) = 0 + 4 + 0 = 4$
• Element (2,3): $(0)(2) + (2)(1) + (1)(3) = 0 + 2 + 3 = 5$
• Element (3,1): $(2)(1) + (0)(0) + (3)(2) = 2 + 0 + 6 = 8$
• Element (3,2): $(2)(0) + (0)(2) + (3)(0) = 0 + 0 + 0 = 0$
• Element (3,3): $(2)(2) + (0)(1) + (3)(3) = 4 + 0 + 9 = 13$

$A^2 = \begin{bmatrix}5& 0& 8\\2& 4& 5\\8& 0& 13 \end{bmatrix}$

Next, calculate $A^3 = A^2 \times A$. The order of $A^2$ is $3 \times 3$ and the order of A is $3 \times 3$, so the order of $A^3$ will be $3 \times 3$.

$A^3 = \begin{bmatrix}5& 0& 8\\2& 4& 5\\8& 0& 13 \end{bmatrix} \begin{bmatrix}1& 0& 2\\0& 2& 1\\2& 0& 3 \end{bmatrix}$

• Element (1,1): $(5)(1) + (0)(0) + (8)(2) = 5 + 0 + 16 = 21$
• Element (1,2): $(5)(0) + (0)(2) + (8)(0) = 0 + 0 + 0 = 0$
• Element (1,3): $(5)(2) + (0)(1) + (8)(3) = 10 + 0 + 24 = 34$
• Element (2,1): $(2)(1) + (4)(0) + (5)(2) = 2 + 0 + 10 = 12$
• Element (2,2): $(2)(0) + (4)(2) + (5)(0) = 0 + 8 + 0 = 8$
• Element (2,3): $(2)(2) + (4)(1) + (5)(3) = 4 + 4 + 15 = 23$
• Element (3,1): $(8)(1) + (0)(0) + (13)(2) = 8 + 0 + 26 = 34$
• Element (3,2): $(8)(0) + (0)(2) + (13)(0) = 0 + 0 + 0 = 0$
• Element (3,3): $(8)(2) + (0)(1) + (13)(3) = 16 + 0 + 39 = 55$

$A^3 = \begin{bmatrix}21& 0& 34\\12& 8& 23\\34& 0& 55 \end{bmatrix}$

Now, calculate $6A^2$, $7A$, and $2I$ by scalar multiplication.

$6A^2 = 6 \begin{bmatrix}5& 0& 8\\2& 4& 5\\8& 0& 13 \end{bmatrix} = \begin{bmatrix}30& 0& 48\\12& 24& 30\\48& 0& 78 \end{bmatrix}$

$7A = 7 \begin{bmatrix}1& 0& 2\\0& 2& 1\\2& 0& 3 \end{bmatrix} = \begin{bmatrix}7& 0& 14\\0& 14& 7\\14& 0& 21 \end{bmatrix}$

$2I = 2 \begin{bmatrix}1& 0& 0\\0& 1& 0\\0& 0& 1 \end{bmatrix} = \begin{bmatrix}2& 0& 0\\0& 2& 0\\0& 0& 2 \end{bmatrix}$

Now, compute the expression $A^3 – 6A^2 + 7A + 2I$:

$A^3 – 6A^2 + 7A + 2I = \begin{bmatrix}21& 0& 34\\12& 8& 23\\34& 0& 55 \end{bmatrix} - \begin{bmatrix}30& 0& 48\\12& 24& 30\\48& 0& 78 \end{bmatrix} + \begin{bmatrix}7& 0& 14\\0& 14& 7\\14& 0& 21 \end{bmatrix} + \begin{bmatrix}2& 0& 0\\0& 2& 0\\0& 0& 2 \end{bmatrix}$

Perform the addition and subtraction of corresponding elements:

• Element (1,1): $21 - 30 + 7 + 2 = -9 + 9 = 0$
• Element (1,2): $0 - 0 + 0 + 0 = 0$
• Element (1,3): $34 - 48 + 14 + 0 = -14 + 14 = 0$
• Element (2,1): $12 - 12 + 0 + 0 = 0$
• Element (2,2): $8 - 24 + 14 + 2 = -16 + 16 = 0$
• Element (2,3): $23 - 30 + 7 + 0 = -7 + 7 = 0$
• Element (3,1): $34 - 48 + 14 + 0 = -14 + 14 = 0$
• Element (3,2): $0 - 0 + 0 + 0 = 0$
• Element (3,3): $55 - 78 + 21 + 2 = -23 + 23 = 0$

$A^3 – 6A^2 + 7A + 2I = \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$

The resulting matrix is the zero matrix O.

Therefore, $A^3 – 6A^2 + 7A + 2I = O$ is proven.

Given:

Matrix A = $\begin{bmatrix}3& −2\\4& −2 \end{bmatrix}$ and Identity Matrix I = $\begin{bmatrix}1& 0\\0& 1 \end{bmatrix}$.

The equation is $A^2 = kA - 2I$, where $k$ is a scalar.

To Find:

The value of the scalar $k$ that satisfies the given equation.

Solution:

First, calculate $A^2 = A \times A$. The order of A is $2 \times 2$, so the order of $A^2$ will also be $2 \times 2$.

$A^2 = \begin{bmatrix}3& −2\\4& −2 \end{bmatrix} \begin{bmatrix}3& −2\\4& −2 \end{bmatrix}$

• Element (1,1): $(3)(3) + (−2)(4) = 9 - 8 = 1$
• Element (1,2): $(3)(−2) + (−2)(−2) = -6 + 4 = -2$
• Element (2,1): $(4)(3) + (−2)(4) = 12 - 8 = 4$
• Element (2,2): $(4)(−2) + (−2)(−2) = -8 + 4 = -4$

So, $A^2 = \begin{bmatrix}1& -2\\4& -4 \end{bmatrix}$.

Next, calculate $kA$ and $2I$ by scalar multiplication.

$kA = k \begin{bmatrix}3& −2\\4& −2 \end{bmatrix} = \begin{bmatrix}3k& -2k\\4k& -2k \end{bmatrix}$

$2I = 2 \begin{bmatrix}1& 0\\0& 1 \end{bmatrix} = \begin{bmatrix}2& 0\\0& 2 \end{bmatrix}$

Substitute these matrices into the given equation $A^2 = kA - 2I$:

$\begin{bmatrix}1& -2\\4& -4 \end{bmatrix} = \begin{bmatrix}3k& -2k\\4k& -2k \end{bmatrix} - \begin{bmatrix}2& 0\\0& 2 \end{bmatrix}$

Perform the matrix subtraction on the right-hand side ($R.H.S.$):

$R.H.S. = \begin{bmatrix}3k - 2& -2k - 0\\4k - 0& -2k - 2 \end{bmatrix} = \begin{bmatrix}3k - 2& -2k\\4k& -2k - 2 \end{bmatrix}$

Equating the left-hand side ($L.H.S.$) and $R.H.S.$:

$\begin{bmatrix}1& -2\\4& -4 \end{bmatrix} = \begin{bmatrix}3k - 2& -2k\\4k& -2k - 2 \end{bmatrix}$

For these two matrices to be equal, their corresponding elements must be equal. We can equate any corresponding elements to find the value of $k$. Let's use the elements in the first row, first column:

From equation (1):

$1 + 2 = 3k$

$3 = 3k$

$k = \frac{3}{3}$

$k = 1$

Let's check if this value of $k$ is consistent with the other corresponding elements.

Equating the elements in the first row, second column:

$-2 = -2k$

$k = \frac{-2}{-2} = 1$. (Consistent)

Equating the elements in the second row, first column:

$4 = 4k$

$k = \frac{4}{4} = 1$. (Consistent)

Equating the elements in the second row, second column:

$-4 = -2k - 2$

$-4 + 2 = -2k$

$-2 = -2k$

$k = \frac{-2}{-2} = 1$. (Consistent)

Since the value of $k=1$ is consistent across all equations, it is the correct value.

The value of $k$ is 1.

Given:

Matrix A = $\begin{bmatrix}0& −\tan \frac{\alpha}{2}\\\tan \frac{\alpha}{2}& 0 \end{bmatrix}$.

Identity matrix of order 2, $I = \begin{bmatrix}1& 0\\0& 1 \end{bmatrix}$.

To Show:

$I + A = (I – A) \begin{bmatrix}\cos\;\alpha& −\sin\;\alpha \\ \sin\;\alpha& \cos\;\alpha \end{bmatrix}$.

Solution:

Let $t = \tan \frac{\alpha}{2}$. Then matrix A can be written as $A = \begin{bmatrix}0& -t\\t& 0 \end{bmatrix}$.

We will compute both the left-hand side ($L.H.S.$) and the right-hand side ($R.H.S.$) of the equation and show that they are equal.

Calculate L.H.S. = I + A:

$I + A = \begin{bmatrix}1& 0\\0& 1 \end{bmatrix} + \begin{bmatrix}0& -t\\t& 0 \end{bmatrix} = \begin{bmatrix}1+0& 0+(-t)\\0+t& 1+0 \end{bmatrix} = \begin{bmatrix}1& -t\\t& 1 \end{bmatrix}$

Calculate R.H.S. = (I – A) $\begin{bmatrix}\cos\;\alpha& −\sin\; α \\ \sin\;α& \cos\;\alpha \end{bmatrix}$:

First, calculate $I - A$:

$I - A = \begin{bmatrix}1& 0\\0& 1 \end{bmatrix} - \begin{bmatrix}0& -t\\t& 0 \end{bmatrix} = \begin{bmatrix}1-0& 0-(-t)\\0-t& 1-0 \end{bmatrix} = \begin{bmatrix}1& t\\-t& 1 \end{bmatrix}$

Now, we need to express $\cos \alpha$ and $\sin \alpha$ in terms of $t = \tan \frac{\alpha}{2}$. We use the double angle formulas:

$\cos \alpha = \frac{1 - \tan^2 \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} = \frac{1 - t^2}{1 + t^2}$

$\sin \alpha = \frac{2 \tan \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} = \frac{2t}{1 + t^2}$

Substitute these into the matrix $\begin{bmatrix}\cos\;\alpha& −\sin\; α \\ \sin\;α& \cos\;\alpha \end{bmatrix}$:

$\begin{bmatrix}\frac{1 - t^2}{1 + t^2}& -\frac{2t}{1 + t^2}\\\frac{2t}{1 + t^2}& \frac{1 - t^2}{1 + t^2} \end{bmatrix} = \frac{1}{1 + t^2} \begin{bmatrix}1 - t^2& -2t\\2t& 1 - t^2 \end{bmatrix}$

Now, compute the product $(I – A) \begin{bmatrix}\cos\;\alpha& −\sin\; α \\ \sin\;α& \cos\;\alpha \end{bmatrix}$:

$R.H.S. = \begin{bmatrix}1& t\\-t& 1 \end{bmatrix} \begin{bmatrix}\frac{1 - t^2}{1 + t^2}& -\frac{2t}{1 + t^2}\\\frac{2t}{1 + t^2}& \frac{1 - t^2}{1 + t^2} \end{bmatrix}$

$R.H.S. = \begin{bmatrix}1& t\\-t& 1 \end{bmatrix} \frac{1}{1 + t^2} \begin{bmatrix}1 - t^2& -2t\\2t& 1 - t^2 \end{bmatrix}$

$R.H.S. = \frac{1}{1 + t^2} \begin{bmatrix}1& t\\-t& 1 \end{bmatrix} \begin{bmatrix}1 - t^2& -2t\\2t& 1 - t^2 \end{bmatrix}$

Perform the matrix multiplication:

• Element (1,1): $(1)(1 - t^2) + (t)(2t) = 1 - t^2 + 2t^2 = 1 + t^2$
• Element (1,2): $(1)(-2t) + (t)(1 - t^2) = -2t + t - t^3 = -t - t^3 = -t(1 + t^2)$
• Element (2,1): $(-t)(1 - t^2) + (1)(2t) = -t + t^3 + 2t = t + t^3 = t(1 + t^2)$
• Element (2,2): $(-t)(-2t) + (1)(1 - t^2) = 2t^2 + 1 - t^2 = 1 + t^2$

The resulting matrix is $\begin{bmatrix}1 + t^2& -t(1 + t^2)\\t(1 + t^2)& 1 + t^2 \end{bmatrix}$.

Now, multiply by the scalar $\frac{1}{1 + t^2}$:

$R.H.S. = \frac{1}{1 + t^2} \begin{bmatrix}1 + t^2& -t(1 + t^2)\\t(1 + t^2)& 1 + t^2 \end{bmatrix} = \begin{bmatrix}\frac{1 + t^2}{1 + t^2}& \frac{-t(1 + t^2)}{1 + t^2}\\\frac{t(1 + t^2)}{1 + t^2}& \frac{1 + t^2}{1 + t^2} \end{bmatrix} = \begin{bmatrix}1& -t\\t& 1 \end{bmatrix}$

Comparing the matrices in equation (1) and equation (2), we see that they are equal.

Therefore, $I + A = (I – A) \begin{bmatrix}\cos\;\alpha& −\sin\;\alpha \\ \sin\;\alpha& \cos\;\alpha \end{bmatrix}$ is shown.

Given:

Total amount to be invested = $\textsf{₹}30,000$.

Interest rate for the first bond = 5% per year.

Interest rate for the second bond = 7% per year.

We need to use matrix multiplication to determine the amount invested in each bond for given total annual interests.

To Find:

The amount invested in each bond for a total annual interest of (a) $\textsf{₹}1800$ and (b) $\textsf{₹}2000$.

Solution:

Let the amount invested in the first bond be $x$ Rupees.

Since the total amount is $\textsf{₹}30,000$, the amount invested in the second bond will be $(30000 - x)$ Rupees.

The interest earned from the first bond is 5% of $x = 0.05x$.

The interest earned from the second bond is 7% of $(30000 - x) = 0.07(30000 - x)$.

The total annual interest is the sum of the interests from both bonds:

Total Interest = $0.05x + 0.07(30000 - x)$

We can represent the amounts invested as a row matrix (or a column matrix) and the interest rates as a column matrix (or a row matrix) such that their product gives the total interest using matrix multiplication.

Let the investment amounts be represented by the row matrix $X = \begin{bmatrix}x & 30000 - x \end{bmatrix}$.

Let the interest rates be represented by the column matrix $R = \begin{bmatrix}0.05 \\ 0.07 \end{bmatrix}$.

The product $XR$ will give the total interest:

$XR = \begin{bmatrix}x & 30000 - x \end{bmatrix} \begin{bmatrix}0.05 \\ 0.07 \end{bmatrix} = [(x)(0.05) + (30000 - x)(0.07)] = [0.05x + 0.07(30000 - x)]$

This is a $1 \times 1$ matrix containing the total interest.

The equation based on the total annual interest is:

$0.05x + 0.07(30000 - x) = \text{Total Interest}$

$0.05x + 2100 - 0.07x = \text{Total Interest}$

$2100 - 0.02x = \text{Total Interest}$

We need to solve for $x$ in two different cases.

(a) Total annual interest = ₹ 1800

$2100 - 0.02x = 1800$

$-0.02x = 1800 - 2100$

$-0.02x = -300$

$x = \frac{-300}{-0.02} = \frac{300}{0.02}$

$x = \frac{300}{\frac{2}{100}} = 300 \times \frac{100}{2} = 150 \times 100 = 15000$

Amount invested in the first bond ($x$) = $\textsf{₹}15000$.

Amount invested in the second bond ($30000 - x$) = $30000 - 15000 = \textsf{₹}15000$.

(b) Total annual interest = ₹ 2000

$2100 - 0.02x = 2000$

$-0.02x = 2000 - 2100$

$-0.02x = -100$

$x = \frac{-100}{-0.02} = \frac{100}{0.02}$

$x = \frac{100}{\frac{2}{100}} = 100 \times \frac{100}{2} = 50 \times 100 = 5000$

Amount invested in the first bond ($x$) = $\textsf{₹}5000$.

Amount invested in the second bond ($30000 - x$) = $30000 - 5000 = \textsf{₹}25000$.

For an annual total interest of $\textsf{₹}1800$, the trust fund should invest $\textsf{₹}15000$ in the first bond and $\textsf{₹}15000$ in the second bond.

For an annual total interest of $\textsf{₹}2000$, the trust fund should invest $\textsf{₹}5000$ in the first bond and $\textsf{₹}25000$ in the second bond.

Alternatively, we could represent the amounts invested as a column matrix and the interest rates as a row matrix: Let $X' = \begin{bmatrix}x \\ 30000 - x \end{bmatrix}$ and $R' = \begin{bmatrix}0.05 & 0.07 \end{bmatrix}$. The product $R'X'$ would also give the total interest.

$R'X' = \begin{bmatrix}0.05 & 0.07 \end{bmatrix} \begin{bmatrix}x \\ 30000 - x \end{bmatrix} = [(0.05)(x) + (0.07)(30000 - x)] = [0.05x + 2100 - 0.07x] = [2100 - 0.02x]$.

This leads to the same equation as before.

Given:

• Quantity of Chemistry books = 10 dozen = $10 \times 12 = 120$ books.
• Quantity of Physics books = 8 dozen = $8 \times 12 = 96$ books.
• Quantity of Economics books = 10 dozen = $10 \times 12 = 120$ books.
• Selling price per Chemistry book = $\textsf{₹}80$.
• Selling price per Physics book = $\textsf{₹}60$.
• Selling price per Economics book = $\textsf{₹}40$.

To Find:

The total amount the bookshop will receive from selling all the books using matrix algebra.

Solution:

We can represent the number of books of each type as a row matrix (or a column matrix) and the selling price of each book as a column matrix (or a row matrix) such that their product gives the total amount using matrix multiplication.

Let the quantities of the books be represented by the row matrix $Q = \begin{bmatrix}120 & 96 & 120 \end{bmatrix}$. The entries represent the number of Chemistry, Physics, and Economics books, respectively.

Let the selling prices per book be represented by the column matrix $P = \begin{bmatrix}80 \\ 60 \\ 40 \end{bmatrix}$. The entries represent the price per Chemistry, Physics, and Economics book, respectively.

The product QP will give the total amount received:

Total Amount = QP

$QP = \begin{bmatrix}120& 96& 120 \end{bmatrix} \begin{bmatrix}80 \\ 60 \\ 40 \end{bmatrix}$

The order of Q is $1 \times 3$. The order of P is $3 \times 1$. The number of columns in Q (3) is equal to the number of rows in P (3). So, the product QP is possible and its order will be $1 \times 1$.

$QP = [(120)(80) + (96)(60) + (120)(40)]$

• $(120)(80) = 9600$
• $(96)(60) = 5760$
• $(120)(40) = 4800$

Total Amount = $[9600 + 5760 + 4800]$

Total Amount = $[20160]$

The total amount received is $\textsf{₹}20160$.

Alternatively, we could represent the quantities as a column matrix and the prices as a row matrix: Let $Q' = \begin{bmatrix}120 \\ 96 \\ 120 \end{bmatrix}$ and $P' = \begin{bmatrix}80& 60& 40 \end{bmatrix}$. The product $P'Q'$ would also give the total amount.

$P'Q' = \begin{bmatrix}80& 60& 40 \end{bmatrix} \begin{bmatrix}120 \\ 96 \\ 120 \end{bmatrix} = [(80)(120) + (60)(96) + (40)(120)] = [9600 + 5760 + 4800] = [20160]$.

This also gives the total amount as $\textsf{₹}20160$.

The total amount the bookshop will receive from selling all the books is $\textsf{₹}20160$.

Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

Given:

Orders of the matrices X, Y, Z, W, and P:

• Order of X: $2 \times n$
• Order of Y: $3 \times k$
• Order of Z: $2 \times p$
• Order of W: $n \times 3$
• Order of P: $p \times k$

The expression is PY + WY.

To Find:

The restrictions on $n, k,$ and $p$ so that the expression PY + WY is defined.

Solution:

For the sum of two matrices to be defined, they must have the same order. So, the order of PY must be equal to the order of WY.

First, consider the product PY:

Order of P is $p \times k$.

Order of Y is $3 \times k$.

For the product PY to be defined, the number of columns in P must be equal to the number of rows in Y.

$k = 3$

The order of the product PY is (number of rows in P) $\times$ (number of columns in Y), which is $p \times k$. Since $k=3$, the order of PY is $p \times 3$.

Next, consider the product WY:

Order of W is $n \times 3$.

Order of Y is $3 \times k$.

For the product WY to be defined, the number of columns in W must be equal to the number of rows in Y.

$3 = 3$

This condition is always satisfied, which means the product WY is defined for any value of $k$. The order of the product WY is (number of rows in W) $\times$ (number of columns in Y), which is $n \times k$. Since $k=3$, the order of WY is $n \times 3$.

For the sum PY + WY to be defined, the orders of PY and WY must be the same.

Order of PY is $p \times 3$.

Order of WY is $n \times 3$.

Equating the orders:

$p \times 3 = n \times 3$

This implies that the number of rows must be equal and the number of columns must be equal.

$p = n$

$3 = 3$ (Consistent)

So, the restrictions on $n, k,$ and $p$ are $k = 3$ and $p = n$.

Let's check the given options:

(A) $k = 3, p = n$: This matches our findings.

(B) $k$ is arbitrary, $p = 2$: Our finding requires $k=3$.

(C) $p$ is arbitrary, $k = 3$: Our finding requires $p=n$.

(D) $k = 2, p = 3$: Our finding requires $k=3$ and $p=n$.

The restrictions are $k=3$ and $p=n$.

The correct option is (A).

Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

Given:

Orders of the matrices X, Y, Z, W, and P:

• Order of X: $2 \times n$
• Order of Z: $2 \times p$

The condition $n = p$ is given.

The expression is $7X - 5Z$.

To Find:

The order of the matrix $7X - 5Z$, given that $n = p$.

Solution:

For the subtraction of two matrices to be defined, they must have the same order. The scalar multiplication of a matrix does not change its order.

The order of $7X$ is the same as the order of X, which is $2 \times n$.

The order of $5Z$ is the same as the order of Z, which is $2 \times p$.

For the subtraction $7X - 5Z$ to be defined, the order of $7X$ must be equal to the order of $5Z$.

$2 \times n = 2 \times p$

This requires the number of rows to be equal (which is 2) and the number of columns to be equal, i.e., $n = p$.

We are given that $n = p$. This condition ensures that the subtraction is possible.

The order of the resulting matrix $7X - 5Z$ is the same as the order of $7X$ (or $5Z$), which is $2 \times n$.

Since $n = p$, the order can also be written as $2 \times p$.

We need to find the order among the given options:

(A) $p \times 2$: Incorrect (rows $\times$ columns should be $2 \times n$ or $2 \times p$)

(B) $2 \times n$: Correct (matches the order we found)

(C) $n \times 3$: Incorrect

(D) $p \times n$: Incorrect (unless $p=2$ and $n=2$, which is not a general restriction)

Given $n=p$, the order is $2 \times n$ or $2 \times p$. Option (B) is $2 \times n$, which is a valid representation of the order when $n=p$.

The correct option is (B).

Given:

Matrix A = $\begin{bmatrix}3& \sqrt{3}& 2\\4& 2& 0 \end{bmatrix}$ (order $2 \times 3$)

Matrix B = $\begin{bmatrix}2& −1& 2\\1& 2& 4 \end{bmatrix}$ (order $2 \times 3$)

$k$ is any constant.

To Verify:

(i) $(A')' = A$

(ii) $(A + B)' = A' + B'$

(iii) $(kB)' = kB'$

Solution:

The transpose of a matrix is obtained by interchanging its rows and columns.

(i) Verify $(A')' = A$:

First, find the transpose of A, denoted by A'. The rows of A become the columns of A'. The order of A' will be $3 \times 2$.

$A' = \begin{bmatrix}3& 4\\\sqrt{3}& 2\\2& 0 \end{bmatrix}$

Now, find the transpose of A', denoted by $(A')'$. The rows of A' become the columns of $(A')'$. The order of $(A')'$ will be $2 \times 3$.

$(A')' = \begin{bmatrix}3& \sqrt{3}& 2\\4& 2& 0 \end{bmatrix}$

Comparing $(A')'$ with A, we see that they are identical.

Thus, $(A')' = A$ is verified.

(ii) Verify $(A + B)' = A' + B'$:

First, calculate A + B. Since both A and B are $2 \times 3$ matrices, addition is possible. The order of A + B is $2 \times 3$.

$A + B = \begin{bmatrix}3& \sqrt{3}& 2\\4& 2& 0 \end{bmatrix} + \begin{bmatrix}2& −1& 2\\1& 2& 4 \end{bmatrix} = \begin{bmatrix}3+2& \sqrt{3}+(-1)& 2+2\\4+1& 2+2& 0+4 \end{bmatrix} = \begin{bmatrix}5& \sqrt{3}-1& 4\\5& 4& 4 \end{bmatrix}$

Now, find the transpose of (A + B), denoted by $(A + B)'$. The order of $(A + B)'$ will be $3 \times 2$.

$(A + B)' = \begin{bmatrix}5& 5\\\sqrt{3}-1& 4\\4& 4 \end{bmatrix}$

Next, find the transpose of B, denoted by B'. The order of B' will be $3 \times 2$.

$B' = \begin{bmatrix}2& 1\\−1& 2\\2& 4 \end{bmatrix}$

Now, calculate A' + B'. Since both A' and B' are $3 \times 2$ matrices, addition is possible. The order of A' + B' is $3 \times 2$.

$A' + B' = \begin{bmatrix}3& 4\\\sqrt{3}& 2\\2& 0 \end{bmatrix} + \begin{bmatrix}2& 1\\−1& 2\\2& 4 \end{bmatrix} = \begin{bmatrix}3+2& 4+1\\\sqrt{3}+(-1)& 2+2\\2+2& 0+4 \end{bmatrix} = \begin{bmatrix}5& 5\\\sqrt{3}-1& 4\\4& 4 \end{bmatrix}$

Comparing $(A + B)'$ and $A' + B'$, we see that they are identical.

Thus, $(A + B)' = A' + B'$ is verified.

(iii) Verify $(kB)' = kB'$:

First, calculate the scalar multiple kB. Multiply each element of B by $k$. The order of kB is $2 \times 3$.

$kB = k \begin{bmatrix}2& −1& 2\\1& 2& 4 \end{bmatrix} = \begin{bmatrix}2k& -k& 2k\\k& 2k& 4k \end{bmatrix}$

Now, find the transpose of kB, denoted by $(kB)'$. The order of $(kB)'$ will be $3 \times 2$.

$(kB)' = \begin{bmatrix}2k& k\\-k& 2k\\2k& 4k \end{bmatrix}$

Next, calculate the scalar multiple $kB'$. We already found $B' = \begin{bmatrix}2& 1\\−1& 2\\2& 4 \end{bmatrix}$. Multiply each element of B' by $k$. The order of $kB'$ is $3 \times 2$.

$kB' = k \begin{bmatrix}2& 1\\−1& 2\\2& 4 \end{bmatrix} = \begin{bmatrix}k \times 2& k \times 1\\k \times (−1)& k \times 2\\k \times 2& k \times 4 \end{bmatrix} = \begin{bmatrix}2k& k\\-k& 2k\\2k& 4k \end{bmatrix}$

Comparing $(kB)'$ and $kB'$, we see that they are identical.

Thus, $(kB)' = kB'$ is verified.

Given:

Matrix A = $\begin{bmatrix}−2\\4\\5 \end{bmatrix}$ (order $3 \times 1$)

Matrix B = $\begin{bmatrix}1& 3& −6 \end{bmatrix}$ (order $1 \times 3$)

To Verify:

$(AB)' = B'A'$.

Solution:

We need to compute $(AB)'$ and $B'A'$ separately and show that they are equal.

Calculate (AB)':

First, calculate the product AB. The order of A is $3 \times 1$ and the order of B is $1 \times 3$. The number of columns in A (1) is equal to the number of rows in B (1), so the product AB is possible. The order of AB will be $3 \times 3$.

$AB = \begin{bmatrix}−2\\4\\5 \end{bmatrix} \begin{bmatrix}1& 3& −6 \end{bmatrix}$

• Element (1,1): $(-2)(1) = -2$
• Element (1,2): $(-2)(3) = -6$
• Element (1,3): $(-2)(-6) = 12$
• Element (2,1): $(4)(1) = 4$
• Element (2,2): $(4)(3) = 12$
• Element (2,3): $(4)(-6) = -24$
• Element (3,1): $(5)(1) = 5$
• Element (3,2): $(5)(3) = 15$
• Element (3,3): $(5)(-6) = -30$

So, $AB = \begin{bmatrix}−2& −6& 12\\4& 12& −24\\5& 15& −30 \end{bmatrix}$.

Now, find the transpose of AB, denoted by $(AB)'$. The rows of AB become the columns of $(AB)'$. The order of $(AB)'$ will be $3 \times 3$.

$(AB)' = \begin{bmatrix}−2& 4& 5\\−6& 12& 15\\12& −24& −30 \end{bmatrix}$

Calculate B'A':

First, find the transpose of B, denoted by B'. The rows of B become the columns of B'. The order of B' will be $3 \times 1$.

$B' = \begin{bmatrix}1\\3\\−6 \end{bmatrix}$

Next, find the transpose of A, denoted by A'. The rows of A become the columns of A'. The order of A' will be $1 \times 3$.

$A' = \begin{bmatrix}−2& 4& 5 \end{bmatrix}$

Now, calculate the product B'A'. The order of B' is $3 \times 1$ and the order of A' is $1 \times 3$. The number of columns in B' (1) is equal to the number of rows in A' (1), so the product B'A' is possible. The order of B'A' will be $3 \times 3$.

$B'A' = \begin{bmatrix}1\\3\\−6 \end{bmatrix} \begin{bmatrix}−2& 4& 5 \end{bmatrix}$

• Element (1,1): $(1)(-2) = -2$
• Element (1,2): $(1)(4) = 4$
• Element (1,3): $(1)(5) = 5$
• Element (2,1): $(3)(-2) = -6$
• Element (2,2): $(3)(4) = 12$
• Element (2,3): $(3)(5) = 15$
• Element (3,1): $(-6)(-2) = 12$
• Element (3,2): $(-6)(4) = -24$
• Element (3,3): $(-6)(5) = -30$

So, $B'A' = \begin{bmatrix}−2& 4& 5\\−6& 12& 15\\12& −24& −30 \end{bmatrix}$.

Comparison:

We found $(AB)' = \begin{bmatrix}−2& 4& 5\\−6& 12& 15\\12& −24& −30 \end{bmatrix}$ and $B'A' = \begin{bmatrix}−2& 4& 5\\−6& 12& 15\\12& −24& −30 \end{bmatrix}$.

The matrices $(AB)'$ and $B'A'$ have the same order ($3 \times 3$) and their corresponding elements are equal.

Therefore, $(AB)' = B'A'$ is verified.

Given:

Matrix B = $\begin{bmatrix}2& −2& −4\\−1& 3& 4\\1& −2& −3 \end{bmatrix}$ (order $3 \times 3$).

To Express:

Express B as the sum of a symmetric and a skew-symmetric matrix.

Solution:

Any square matrix B can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix. The symmetric part is given by $P = \frac{1}{2}(B + B')$ and the skew-symmetric part is given by $Q = \frac{1}{2}(B - B')$. Then $B = P + Q$.

First, find the transpose of B, denoted by B'. The rows of B become the columns of B'. The order of B' is $3 \times 3$.

$B' = \begin{bmatrix}2& −1& 1\\−2& 3& −2\\−4& 4& −3 \end{bmatrix}$

Now, calculate $B + B'$:

$B + B' = \begin{bmatrix}2& −2& −4\\−1& 3& 4\\1& −2& −3 \end{bmatrix} + \begin{bmatrix}2& −1& 1\\−2& 3& −2\\−4& 4& −3 \end{bmatrix} = \begin{bmatrix}2+2& -2+(-1)& -4+1\\-1+(-2)& 3+3& 4+(-2)\\1+(-4)& -2+4& -3+(-3) \end{bmatrix} = \begin{bmatrix}4& -3& -3\\-3& 6& 2\\-3& 2& -6 \end{bmatrix}$

Calculate the symmetric part $P = \frac{1}{2}(B + B')$:

$P = \frac{1}{2} \begin{bmatrix}4& -3& -3\\-3& 6& 2\\-3& 2& -6 \end{bmatrix} = \begin{bmatrix}\frac{4}{2}& \frac{-3}{2}& \frac{-3}{2}\\\frac{-3}{2}& \frac{6}{2}& \frac{2}{2}\\\frac{-3}{2}& \frac{2}{2}& \frac{-6}{2} \end{bmatrix} = \begin{bmatrix}2& -\frac{3}{2}& -\frac{3}{2}\\-\frac{3}{2}& 3& 1\\-\frac{3}{2}& 1& -3 \end{bmatrix}$

To verify that P is symmetric, check if $P' = P$.

$P' = \begin{bmatrix}2& -\frac{3}{2}& -\frac{3}{2}\\-\frac{3}{2}& 3& 1\\-\frac{3}{2}& 1& -3 \end{bmatrix}' = \begin{bmatrix}2& -\frac{3}{2}& -\frac{3}{2}\\-\frac{3}{2}& 3& 1\\-\frac{3}{2}& 1& -3 \end{bmatrix}$

Since $P' = P$, P is a symmetric matrix.

Now, calculate $B - B'$:

$B - B' = \begin{bmatrix}2& −2& −4\\−1& 3& 4\\1& −2& −3 \end{bmatrix} - \begin{bmatrix}2& −1& 1\\−2& 3& −2\\−4& 4& −3 \end{bmatrix} = \begin{bmatrix}2-2& -2-(-1)& -4-1\\-1-(-2)& 3-3& 4-(-2)\\1-(-4)& -2-4& -3-(-3) \end{bmatrix} = \begin{bmatrix}0& -1& -5\\1& 0& 6\\5& -6& 0 \end{bmatrix}$

Calculate the skew-symmetric part $Q = \frac{1}{2}(B - B')$:

$Q = \frac{1}{2} \begin{bmatrix}0& -1& -5\\1& 0& 6\\5& -6& 0 \end{bmatrix} = \begin{bmatrix}\frac{0}{2}& \frac{-1}{2}& \frac{-5}{2}\\\frac{1}{2}& \frac{0}{2}& \frac{6}{2}\\\frac{5}{2}& \frac{-6}{2}& \frac{0}{2} \end{bmatrix} = \begin{bmatrix}0& -\frac{1}{2}& -\frac{5}{2}\\\frac{1}{2}& 0& 3\\\frac{5}{2}& -3& 0 \end{bmatrix}$

To verify that Q is skew-symmetric, check if $Q' = -Q$.

$Q' = \begin{bmatrix}0& -\frac{1}{2}& -\frac{5}{2}\\\frac{1}{2}& 0& 3\\\frac{5}{2}& -3& 0 \end{bmatrix}' = \begin{bmatrix}0& \frac{1}{2}& \frac{5}{2}\\-\frac{1}{2}& 0& -3\\-\frac{5}{2}& 3& 0 \end{bmatrix}$

$-Q = -\begin{bmatrix}0& -\frac{1}{2}& -\frac{5}{2}\\\frac{1}{2}& 0& 3\\\frac{5}{2}& -3& 0 \end{bmatrix} = \begin{bmatrix}-0& -(-\frac{1}{2})& -(-\frac{5}{2})\\-(\frac{1}{2})& -0& -3\\-(\frac{5}{2})& -(-3)& -0 \end{bmatrix} = \begin{bmatrix}0& \frac{1}{2}& \frac{5}{2}\\-\frac{1}{2}& 0& -3\\-\frac{5}{2}& 3& 0 \end{bmatrix}$

Since $Q' = -Q$, Q is a skew-symmetric matrix.

Finally, express B as the sum of P and Q:

$P + Q = \begin{bmatrix}2& -\frac{3}{2}& -\frac{3}{2}\\-\frac{3}{2}& 3& 1\\-\frac{3}{2}& 1& -3 \end{bmatrix} + \begin{bmatrix}0& -\frac{1}{2}& -\frac{5}{2}\\\frac{1}{2}& 0& 3\\\frac{5}{2}& -3& 0 \end{bmatrix}$

$P + Q = \begin{bmatrix}2+0& -\frac{3}{2}+(-\frac{1}{2})& -\frac{3}{2}+(-\frac{5}{2})\\-\frac{3}{2}+\frac{1}{2}& 3+0& 1+3\\-\frac{3}{2}+\frac{5}{2}& 1-3& -3+0 \end{bmatrix} = \begin{bmatrix}2& -\frac{4}{2}& -\frac{8}{2}\\-\frac{2}{2}& 3& 4\\\frac{2}{2}& -2& -3 \end{bmatrix} = \begin{bmatrix}2& -2& -4\\-1& 3& 4\\1& -2& -3 \end{bmatrix}$

This is equal to the original matrix B.

Therefore, the matrix B is expressed as the sum of the symmetric matrix P and the skew-symmetric matrix Q.

$B = P + Q = \underbrace{\begin{bmatrix}2& -\frac{3}{2}& -\frac{3}{2}\\-\frac{3}{2}& 3& 1\\-\frac{3}{2}& 1& -3 \end{bmatrix}}_{\text{Symmetric}} + \underbrace{\begin{bmatrix}0& -\frac{1}{2}& -\frac{5}{2}\\\frac{1}{2}& 0& 3\\\frac{5}{2}& -3& 0 \end{bmatrix}}_{\text{Skew-Symmetric}}$

Given:

Three matrices are provided.

To Find:

The transpose of each of the given matrices.

Solution:

The transpose of a matrix is obtained by interchanging its rows and columns.

(i) Let the given matrix be $A = \begin{bmatrix}5\\\frac{1}{2}\\−1 \end{bmatrix}$. This is a column matrix of order $3 \times 1$.

The transpose of A, denoted by $A'$, is obtained by making the column the row. The order of $A'$ will be $1 \times 3$.

$A' = \begin{bmatrix}5& \frac{1}{2}& −1 \end{bmatrix}$

(ii) Let the given matrix be $B = \begin{bmatrix}1& −1\\2& 3 \end{bmatrix}$. This is a square matrix of order $2 \times 2$.

The transpose of B, denoted by $B'$, is obtained by interchanging rows and columns.

$B' = \begin{bmatrix}1& 2\\−1& 3 \end{bmatrix}$

(iii) Let the given matrix be $C = \begin{bmatrix}−1& 5& 6\\\sqrt{3}& 5& 6\\2& 3& −1 \end{bmatrix}$. This is a square matrix of order $3 \times 3$.

The transpose of C, denoted by $C'$, is obtained by interchanging rows and columns.

$C' = \begin{bmatrix}−1& \sqrt{3}& 2\\5& 5& 3\\6& 6& −1 \end{bmatrix}$

Given:

Matrix A = $\begin{bmatrix}−1& 2& 3\\5& 7& 9\\−2& 1& 1 \end{bmatrix}$ (order $3 \times 3$)

Matrix B = $\begin{bmatrix}−4& 1& −5\\1& 2& 0\\1& 3& 1 \end{bmatrix}$ (order $3 \times 3$)

To Verify:

(i) $(A + B)' = A' + B'$

(ii) $(A – B)' = A' – B'$

Solution:

First, find the transposes of A and B.

$A' = \begin{bmatrix}−1& 5& −2\\2& 7& 1\\3& 9& 1 \end{bmatrix}$ (order $3 \times 3$)

$B' = \begin{bmatrix}−4& 1& 1\\1& 2& 3\\−5& 0& 1 \end{bmatrix}$ (order $3 \times 3$)

(i) Verify $(A + B)' = A' + B'$:

Calculate $A + B$. Since both A and B are $3 \times 3$, addition is possible. The order of A + B is $3 \times 3$.

$A + B = \begin{bmatrix}−1& 2& 3\\5& 7& 9\\−2& 1& 1 \end{bmatrix} + \begin{bmatrix}−4& 1& −5\\1& 2& 0\\1& 3& 1 \end{bmatrix} = \begin{bmatrix}-1+(-4)& 2+1& 3+(-5)\\5+1& 7+2& 9+0\\-2+1& 1+3& 1+1 \end{bmatrix} = \begin{bmatrix}-5& 3& -2\\6& 9& 9\\-1& 4& 2 \end{bmatrix}$

Now, find the transpose of (A + B), denoted by $(A + B)'$. The order of $(A + B)'$ will be $3 \times 3$.

$(A + B)' = \begin{bmatrix}-5& 6& -1\\3& 9& 4\\-2& 9& 2 \end{bmatrix}$

Calculate $A' + B'$. Since both A' and B' are $3 \times 3$, addition is possible. The order of A' + B' is $3 \times 3$.

$A' + B' = \begin{bmatrix}−1& 5& −2\\2& 7& 1\\3& 9& 1 \end{bmatrix} + \begin{bmatrix}−4& 1& 1\\1& 2& 3\\−5& 0& 1 \end{bmatrix} = \begin{bmatrix}-1+(-4)& 5+1& -2+1\\2+1& 7+2& 1+3\\3+(-5)& 9+0& 1+1 \end{bmatrix} = \begin{bmatrix}-5& 6& -1\\3& 9& 4\\-2& 9& 2 \end{bmatrix}$

Comparing $(A + B)'$ and $A' + B'$, we see that they are identical.

Thus, $(A + B)' = A' + B'$ is verified.

(ii) Verify $(A – B)' = A' – B'$:

Calculate $A – B$. Since both A and B are $3 \times 3$, subtraction is possible. The order of A – B is $3 \times 3$.

$A – B = \begin{bmatrix}−1& 2& 3\\5& 7& 9\\−2& 1& 1 \end{bmatrix} - \begin{bmatrix}−4& 1& −5\\1& 2& 0\\1& 3& 1 \end{bmatrix} = \begin{bmatrix}-1-(-4)& 2-1& 3-(-5)\\5-1& 7-2& 9-0\\-2-1& 1-3& 1-1 \end{bmatrix} = \begin{bmatrix}-1+4& 1& 3+5\\4& 5& 9\\-3& -2& 0 \end{bmatrix} = \begin{bmatrix}3& 1& 8\\4& 5& 9\\-3& -2& 0 \end{bmatrix}$

Now, find the transpose of (A – B), denoted by $(A – B)'$. The order of $(A – B)'$ will be $3 \times 3$.

$(A – B)' = \begin{bmatrix}3& 4& -3\\1& 5& -2\\8& 9& 0 \end{bmatrix}$

Calculate $A' – B'$. Since both A' and B' are $3 \times 3$, subtraction is possible. The order of A' – B' is $3 \times 3$.

$A' – B' = \begin{bmatrix}−1& 5& −2\\2& 7& 1\\3& 9& 1 \end{bmatrix} - \begin{bmatrix}−4& 1& 1\\1& 2& 3\\−5& 0& 1 \end{bmatrix} = \begin{bmatrix}-1-(-4)& 5-1& -2-1\\2-1& 7-2& 1-3\\3-(-5)& 9-0& 1-1 \end{bmatrix} = \begin{bmatrix}-1+4& 4& -3\\1& 5& -2\\3+5& 9& 0 \end{bmatrix} = \begin{bmatrix}3& 4& -3\\1& 5& -2\\8& 9& 0 \end{bmatrix}$

Comparing $(A – B)'$ and $A' – B'$, we see that they are identical.

Thus, $(A – B)' = A' – B'$ is verified.

Given:

Transpose of matrix A, A’ = $\begin{bmatrix}3& 4\\−1& 2\\0& 1 \end{bmatrix}$ (order $3 \times 2$)

Matrix B = $\begin{bmatrix}−1& 2& 1\\1& 2& 3 \end{bmatrix}$ (order $2 \times 3$)

To Verify:

(i) $(A + B)' = A' + B'$

(ii) $(A – B)' = A' – B'$

Solution:

First, we need to find matrix A from A'. The transpose of A' is A.

$A = (A')' = \begin{bmatrix}3& 4\\−1& 2\\0& 1 \end{bmatrix}' = \begin{bmatrix}3& −1& 0\\4& 2& 1 \end{bmatrix}$ (order $2 \times 3$)

Also, find the transpose of B, B'.

$B' = \begin{bmatrix}−1& 2& 1\\1& 2& 3 \end{bmatrix}' = \begin{bmatrix}−1& 1\\2& 2\\1& 3 \end{bmatrix}$ (order $3 \times 2$)

(i) Verify $(A + B)' = A' + B'$:

Calculate $A + B$. The order of A is $2 \times 3$ and the order of B is $2 \times 3$. Addition is possible. The order of A + B is $2 \times 3$.

$A + B = \begin{bmatrix}3& −1& 0\\4& 2& 1 \end{bmatrix} + \begin{bmatrix}−1& 2& 1\\1& 2& 3 \end{bmatrix} = \begin{bmatrix}3+(-1)& -1+2& 0+1\\4+1& 2+2& 1+3 \end{bmatrix} = \begin{bmatrix}2& 1& 1\\5& 4& 4 \end{bmatrix}$

Now, find the transpose of (A + B), denoted by $(A + B)'$. The order of $(A + B)'$ will be $3 \times 2$.

$(A + B)' = \begin{bmatrix}2& 5\\1& 4\\1& 4 \end{bmatrix}$

Calculate $A' + B'$. The order of A' is $3 \times 2$ and the order of B' is $3 \times 2$. Addition is possible. The order of A' + B' is $3 \times 2$.

$A' + B' = \begin{bmatrix}3& 4\\−1& 2\\0& 1 \end{bmatrix} + \begin{bmatrix}−1& 1\\2& 2\\1& 3 \end{bmatrix} = \begin{bmatrix}3+(-1)& 4+1\\-1+2& 2+2\\0+1& 1+3 \end{bmatrix} = \begin{bmatrix}2& 5\\1& 4\\1& 4 \end{bmatrix}$

Comparing $(A + B)'$ and $A' + B'$, we see that they are identical.

Thus, $(A + B)' = A' + B'$ is verified.

(ii) Verify $(A – B)' = A' – B'$:

Calculate $A – B$. The order of A is $2 \times 3$ and the order of B is $2 \times 3$. Subtraction is possible. The order of A – B is $2 \times 3$.

$A – B = \begin{bmatrix}3& −1& 0\\4& 2& 1 \end{bmatrix} - \begin{bmatrix}−1& 2& 1\\1& 2& 3 \end{bmatrix} = \begin{bmatrix}3-(-1)& -1-2& 0-1\\4-1& 2-2& 1-3 \end{bmatrix} = \begin{bmatrix}3+1& -3& -1\\3& 0& -2 \end{bmatrix} = \begin{bmatrix}4& -3& -1\\3& 0& -2 \end{bmatrix}$

Now, find the transpose of (A – B), denoted by $(A – B)'$. The order of $(A – B)'$ will be $3 \times 2$.

$(A – B)' = \begin{bmatrix}4& 3\\-3& 0\\-1& -2 \end{bmatrix}$

Calculate $A' – B'$. The order of A' is $3 \times 2$ and the order of B' is $3 \times 2$. Subtraction is possible. The order of A' – B' is $3 \times 2$.

$A' – B' = \begin{bmatrix}3& 4\\−1& 2\\0& 1 \end{bmatrix} - \begin{bmatrix}−1& 1\\2& 2\\1& 3 \end{bmatrix} = \begin{bmatrix}3-(-1)& 4-1\\-1-2& 2-2\\0-1& 1-3 \end{bmatrix} = \begin{bmatrix}3+1& 3\\-3& 0\\-1& -2 \end{bmatrix} = \begin{bmatrix}4& 3\\-3& 0\\-1& -2 \end{bmatrix}$

Comparing $(A – B)'$ and $A' – B'$, we see that they are identical.

Thus, $(A – B)' = A' – B'$ is verified.

Given:

Transpose of matrix A, A’ = $\begin{bmatrix}−2& 3\\1& 2 \end{bmatrix}$ (order $2 \times 2$)

Matrix B = $\begin{bmatrix}−1& 0\\1& 2 \end{bmatrix}$ (order $2 \times 2$)

To Find:

(A + 2B)’.

Solution:

First, we need to find matrix A from A'. The transpose of A' is A.

$A = (A')' = \begin{bmatrix}−2& 3\\1& 2 \end{bmatrix}' = \begin{bmatrix}−2& 1\\3& 2 \end{bmatrix}$ (order $2 \times 2$)

Next, calculate the scalar multiple 2B. Multiply each element of B by 2.

$2B = 2 \begin{bmatrix}−1& 0\\1& 2 \end{bmatrix} = \begin{bmatrix}2 \times (-1)& 2 \times 0\\2 \times 1& 2 \times 2 \end{bmatrix} = \begin{bmatrix}−2& 0\\2& 4 \end{bmatrix}$

Now, calculate A + 2B. Since both A and 2B are $2 \times 2$ matrices, addition is possible. The order of A + 2B is $2 \times 2$.

$A + 2B = \begin{bmatrix}−2& 1\\3& 2 \end{bmatrix} + \begin{bmatrix}−2& 0\\2& 4 \end{bmatrix} = \begin{bmatrix}-2+(-2)& 1+0\\3+2& 2+4 \end{bmatrix} = \begin{bmatrix}-4& 1\\5& 6 \end{bmatrix}$

Finally, find the transpose of (A + 2B), denoted by $(A + 2B)'$. The rows of (A + 2B) become the columns of $(A + 2B)'$. The order of $(A + 2B)'$ will be $2 \times 2$.

$(A + 2B)' = \begin{bmatrix}-4& 5\\1& 6 \end{bmatrix}$

Alternatively, using the property $(A + B)' = A' + B'$ and $(kB)' = kB'$, we can write $(A + 2B)' = A' + (2B)' = A' + 2B'$.

$A' + 2B' = \begin{bmatrix}−2& 3\\1& 2 \end{bmatrix} + 2 \begin{bmatrix}−1& 0\\1& 2 \end{bmatrix}'$

First, find $B' = \begin{bmatrix}−1& 0\\1& 2 \end{bmatrix}' = \begin{bmatrix}−1& 1\\0& 2 \end{bmatrix}$.

$A' + 2B' = \begin{bmatrix}−2& 3\\1& 2 \end{bmatrix} + 2 \begin{bmatrix}−1& 1\\0& 2 \end{bmatrix} = \begin{bmatrix}−2& 3\\1& 2 \end{bmatrix} + \begin{bmatrix}-2& 2\\0& 4 \end{bmatrix} = \begin{bmatrix}-2+(-2)& 3+2\\1+0& 2+4 \end{bmatrix} = \begin{bmatrix}-4& 5\\1& 6 \end{bmatrix}$

Both methods yield the same result.

The matrix (A + 2B)’ is $\begin{bmatrix}-4& 5\\1& 6 \end{bmatrix}$.

Given:

Matrices A and B in two different parts.

To Verify:

$(AB)' = B'A'$ for the given matrices.

Solution:

We need to compute $(AB)'$ and $B'A'$ separately for each pair of matrices and show that they are equal.

(i) A = $\begin{bmatrix}1\\−4\\3 \end{bmatrix}$ (order $3 \times 1$), B = $\begin{bmatrix}−1& 2& 1 \end{bmatrix}$ (order $1 \times 3$).

Calculate (AB)':

Product AB is possible (number of columns in A = 1, number of rows in B = 1). The order of AB will be $3 \times 3$.

$AB = \begin{bmatrix}1\\−4\\3 \end{bmatrix} \begin{bmatrix}−1& 2& 1 \end{bmatrix} = \begin{bmatrix}(1)(-1)& (1)(2)& (1)(1)\\(−4)(-1)& (−4)(2)& (−4)(1)\\(3)(-1)& (3)(2)& (3)(1) \end{bmatrix} = \begin{bmatrix}-1& 2& 1\\4& -8& -4\\-3& 6& 3 \end{bmatrix}$

The transpose of AB is $(AB)'$. The order of $(AB)'$ will be $3 \times 3$.

$(AB)' = \begin{bmatrix}-1& 4& -3\\2& -8& 6\\1& -4& 3 \end{bmatrix}$

Calculate B'A':

First, find the transposes of B and A.

$B' = \begin{bmatrix}−1& 2& 1 \end{bmatrix}' = \begin{bmatrix}−1\\2\\1 \end{bmatrix}$ (order $3 \times 1$)

$A' = \begin{bmatrix}1\\−4\\3 \end{bmatrix}' = \begin{bmatrix}1& −4& 3 \end{bmatrix}$ (order $1 \times 3$)

Product B'A' is possible (number of columns in B' = 1, number of rows in A' = 1). The order of B'A' will be $3 \times 3$.

$B'A' = \begin{bmatrix}−1\\2\\1 \end{bmatrix} \begin{bmatrix}1& −4& 3 \end{bmatrix} = \begin{bmatrix}(−1)(1)& (−1)(−4)& (−1)(3)\\(2)(1)& (2)(−4)& (2)(3)\\(1)(1)& (1)(−4)& (1)(3) \end{bmatrix} = \begin{bmatrix}-1& 4& -3\\2& -8& 6\\1& -4& 3 \end{bmatrix}$

Comparing $(AB)'$ and $B'A'$, we see that they are identical.

Thus, $(AB)' = B'A'$ is verified for this case.

(ii) A = $\begin{bmatrix}0\\1\\2 \end{bmatrix}$ (order $3 \times 1$), B = $\begin{bmatrix}1& 5& 7 \end{bmatrix}$ (order $1 \times 3$).

Calculate (AB)':

Product AB is possible (number of columns in A = 1, number of rows in B = 1). The order of AB will be $3 \times 3$.

$AB = \begin{bmatrix}0\\1\\2 \end{bmatrix} \begin{bmatrix}1& 5& 7 \end{bmatrix} = \begin{bmatrix}(0)(1)& (0)(5)& (0)(7)\\(1)(1)& (1)(5)& (1)(7)\\(2)(1)& (2)(5)& (2)(7) \end{bmatrix} = \begin{bmatrix}0& 0& 0\\1& 5& 7\\2& 10& 14 \end{bmatrix}$

The transpose of AB is $(AB)'$. The order of $(AB)'$ will be $3 \times 3$.

$(AB)' = \begin{bmatrix}0& 1& 2\\0& 5& 10\\0& 7& 14 \end{bmatrix}$

Calculate B'A':

First, find the transposes of B and A.

$B' = \begin{bmatrix}1& 5& 7 \end{bmatrix}' = \begin{bmatrix}1\\5\\7 \end{bmatrix}$ (order $3 \times 1$)

$A' = \begin{bmatrix}0\\1\\2 \end{bmatrix}' = \begin{bmatrix}0& 1& 2 \end{bmatrix}$ (order $1 \times 3$)

Product B'A' is possible (number of columns in B' = 1, number of rows in A' = 1). The order of B'A' will be $3 \times 3$.

$B'A' = \begin{bmatrix}1\\5\\7 \end{bmatrix} \begin{bmatrix}0& 1& 2 \end{bmatrix} = \begin{bmatrix}(1)(0)& (1)(1)& (1)(2)\\(5)(0)& (5)(1)& (5)(2)\\(7)(0)& (7)(1)& (7)(2) \end{bmatrix} = \begin{bmatrix}0& 1& 2\\0& 5& 10\\0& 7& 14 \end{bmatrix}$

Comparing $(AB)'$ and $B'A'$, we see that they are identical.

Thus, $(AB)' = B'A'$ is verified for this case.

Given:

Two matrices A involving trigonometric functions.

Identity matrix $I = \begin{bmatrix}1& 0\\0& 1 \end{bmatrix}$ of order 2.

To Verify:

(i) $A' A = I$ for $A = \begin{bmatrix}\cosα& \sinα\\−\sinα& \cosα \end{bmatrix}$.

(ii) $A' A = I$ for $A = \begin{bmatrix}\sinα& \cosα \\ −\cosα& \sinα \end{bmatrix}$.

Solution:

(i) Let A = $\begin{bmatrix}\cosα& \sinα\\−\sinα& \cosα \end{bmatrix}$.

First, find the transpose of A, A'.

$A' = \begin{bmatrix}\cosα& −\sinα\\\sinα& \cosα \end{bmatrix}$

Now, calculate the product A'A. The order of A' is $2 \times 2$ and the order of A is $2 \times 2$. The product is possible and will be $2 \times 2$.

$A'A = \begin{bmatrix}\cosα& −\sinα\\\sinα& \cosα \end{bmatrix} \begin{bmatrix}\cosα& \sinα\\−\sinα& \cosα \end{bmatrix}$

• Element (1,1): $(\cosα)(\cosα) + (−\sinα)(−\sinα) = \cos^2α + \sin^2α$
• Element (1,2): $(\cosα)(\sinα) + (−\sinα)(\cosα) = \sinα \cosα - \sinα \cosα = 0$
• Element (2,1): $(\sinα)(\cosα) + (\cosα)(−\sinα) = \sinα \cosα - \sinα \cosα = 0$
• Element (2,2): $(\sinα)(\sinα) + (\cosα)(\cosα) = \sin^2α + \cos^2α$

Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we get:

$A'A = \begin{bmatrix}1& 0\\0& 1 \end{bmatrix}$

This is the identity matrix I.

Thus, $A' A = I$ is verified for this case.

(ii) Let A = $\begin{bmatrix}\sinα& \cosα \\ −\cosα& \sinα \end{bmatrix}$.

First, find the transpose of A, A'.

$A' = \begin{bmatrix}\sinα& −\cosα \\ \cosα& \sinα \end{bmatrix}$

Now, calculate the product A'A. The order of A' is $2 \times 2$ and the order of A is $2 \times 2$. The product is possible and will be $2 \times 2$.

$A'A = \begin{bmatrix}\sinα& −\cosα \\ \cosα& \sinα \end{bmatrix} \begin{bmatrix}\sinα& \cosα \\ −\cosα& \sinα \end{bmatrix}$

• Element (1,1): $(\sinα)(\sinα) + (−\cosα)(−\cosα) = \sin^2α + \cos^2α$
• Element (1,2): $(\sinα)(\cosα) + (−\cosα)(\sinα) = \sinα \cosα - \sinα \cosα = 0$
• Element (2,1): $(\cosα)(\sinα) + (\sinα)(−\cosα) = \sinα \cosα - \sinα \cosα = 0$
• Element (2,2): $(\cosα)(\cosα) + (\sinα)(\sinα) = \cos^2α + \sin^2α$

Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we get:

$A'A = \begin{bmatrix}1& 0\\0& 1 \end{bmatrix}$

This is the identity matrix I.

Thus, $A' A = I$ is verified for this case.

Given:

Two matrices A are provided.

To Show:

(i) The first matrix is symmetric.

(ii) The second matrix is skew-symmetric.

Solution:

(i) Let the given matrix be A = $\begin{bmatrix}1& −1& 5\\−1& 2& 1\\5& 1& 3 \end{bmatrix}$.

A square matrix A is called symmetric if its transpose is equal to the matrix itself, i.e., $A' = A$.

Find the transpose of A, A'. The rows of A become the columns of A'.

$A' = \begin{bmatrix}1& −1& 5\\−1& 2& 1\\5& 1& 3 \end{bmatrix}' = \begin{bmatrix}1& −1& 5\\−1& 2& 1\\5& 1& 3 \end{bmatrix}$

Comparing A' with A, we see that they are identical.

Thus, $A' = A$.

Therefore, the matrix A is a symmetric matrix.

(ii) Let the given matrix be A = $\begin{bmatrix}0& 1& −1\\−1& 0& 1\\1& −1& 0 \end{bmatrix}$.

A square matrix A is called skew-symmetric if its transpose is equal to the negative of the matrix itself, i.e., $A' = -A$. In a skew-symmetric matrix, the diagonal elements are always zero.

Find the transpose of A, A'. The rows of A become the columns of A'.

$A' = \begin{bmatrix}0& 1& −1\\−1& 0& 1\\1& −1& 0 \end{bmatrix}' = \begin{bmatrix}0& −1& 1\\1& 0& −1\\−1& 1& 0 \end{bmatrix}$

Now, find the negative of the matrix A, -A.

$-A = -\begin{bmatrix}0& 1& −1\\−1& 0& 1\\1& −1& 0 \end{bmatrix} = \begin{bmatrix}-0& -1& -(-1)\\-(-1)& -0& -1\\-1& -(-1)& -0 \end{bmatrix} = \begin{bmatrix}0& -1& 1\\1& 0& -1\\-1& 1& 0 \end{bmatrix}$

Comparing A' with -A, we see that they are identical.

Thus, $A' = -A$.

Also, the diagonal elements of A are all zero, which is consistent with the property of a skew-symmetric matrix.

Therefore, the matrix A is a skew-symmetric matrix.

Given:

Matrix A = $\begin{bmatrix}1& 5\\6& 7 \end{bmatrix}$ (order $2 \times 2$).

To Verify:

(i) $(A + A')$ is a symmetric matrix.

(ii) $(A – A')$ is a skew-symmetric matrix.

Solution:

First, find the transpose of A, A'.

$A' = \begin{bmatrix}1& 5\\6& 7 \end{bmatrix}' = \begin{bmatrix}1& 6\\5& 7 \end{bmatrix}$

(i) Verify (A + A′) is a symmetric matrix:

Calculate $A + A'$:

$A + A' = \begin{bmatrix}1& 5\\6& 7 \end{bmatrix} + \begin{bmatrix}1& 6\\5& 7 \end{bmatrix} = \begin{bmatrix}1+1& 5+6\\6+5& 7+7 \end{bmatrix} = \begin{bmatrix}2& 11\\11& 14 \end{bmatrix}$

Let $P = A + A' = \begin{bmatrix}2& 11\\11& 14 \end{bmatrix}$.

To verify that P is symmetric, we need to show that $P' = P$. Find the transpose of P.

$P' = \begin{bmatrix}2& 11\\11& 14 \end{bmatrix}' = \begin{bmatrix}2& 11\\11& 14 \end{bmatrix}$

Comparing P' and P, we see that they are identical.

Thus, $P' = P$. Therefore, $(A + A')$ is a symmetric matrix.

(ii) Verify (A – A′) is a skew-symmetric matrix:

Calculate $A – A'$:

$A – A' = \begin{bmatrix}1& 5\\6& 7 \end{bmatrix} - \begin{bmatrix}1& 6\\5& 7 \end{bmatrix} = \begin{bmatrix}1-1& 5-6\\6-5& 7-7 \end{bmatrix} = \begin{bmatrix}0& -1\\1& 0 \end{bmatrix}$

Let $Q = A – A' = \begin{bmatrix}0& -1\\1& 0 \end{bmatrix}$.

To verify that Q is skew-symmetric, we need to show that $Q' = -Q$. Find the transpose of Q.

$Q' = \begin{bmatrix}0& -1\\1& 0 \end{bmatrix}' = \begin{bmatrix}0& 1\\-1& 0 \end{bmatrix}$

Find the negative of Q.

$-Q = -\begin{bmatrix}0& -1\\1& 0 \end{bmatrix} = \begin{bmatrix}-0& -(-1)\\-1& -0 \end{bmatrix} = \begin{bmatrix}0& 1\\-1& 0 \end{bmatrix}$

Comparing Q' and -Q, we see that they are identical.

Thus, $Q' = -Q$. Therefore, $(A – A')$ is a skew-symmetric matrix.

Given:

Matrix A = $\begin{bmatrix}0& a& b\\−a& 0& c\\−b& −c& 0 \end{bmatrix}$ (order $3 \times 3$).

To Find:

$\frac{1}{2}(A + A')$ and $\frac{1}{2}(A - A')$.

Solution:

First, find the transpose of A, A'. The rows of A become the columns of A'.

$A' = \begin{bmatrix}0& a& b\\−a& 0& c\\−b& −c& 0 \end{bmatrix}' = \begin{bmatrix}0& −a& −b\\a& 0& −c\\b& c& 0 \end{bmatrix}$

Now, calculate $A + A'$:

$A + A' = \begin{bmatrix}0& a& b\\−a& 0& c\\−b& −c& 0 \end{bmatrix} + \begin{bmatrix}0& −a& −b\\a& 0& −c\\b& c& 0 \end{bmatrix} = \begin{bmatrix}0+0& a+(-a)& b+(-b)\\-a+a& 0+0& c+(-c)\\-b+b& -c+c& 0+0 \end{bmatrix} = \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$

Calculate $\frac{1}{2}(A + A')$ by multiplying each element of the resulting matrix by $\frac{1}{2}$:

$\frac{1}{2}(A + A') = \frac{1}{2} \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix} = \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$

Next, calculate $A - A'$:

$A - A' = \begin{bmatrix}0& a& b\\−a& 0& c\\−b& −c& 0 \end{bmatrix} - \begin{bmatrix}0& −a& −b\\a& 0& −c\\b& c& 0 \end{bmatrix} = \begin{bmatrix}0-0& a-(-a)& b-(-b)\\-a-a& 0-0& c-(-c)\\-b-b& -c-c& 0-0 \end{bmatrix} = \begin{bmatrix}0& 2a& 2b\\-2a& 0& 2c\\-2b& -2c& 0 \end{bmatrix}$

Calculate $\frac{1}{2}(A - A')$ by multiplying each element of the resulting matrix by $\frac{1}{2}$:

$\frac{1}{2}(A - A') = \frac{1}{2} \begin{bmatrix}0& 2a& 2b\\-2a& 0& 2c\\-2b& -2c& 0 \end{bmatrix} = \begin{bmatrix}\frac{1}{2}(0)& \frac{1}{2}(2a)& \frac{1}{2}(2b)\\\frac{1}{2}(-2a)& \frac{1}{2}(0)& \frac{1}{2}(2c)\\\frac{1}{2}(-2b)& \frac{1}{2}(-2c)& \frac{1}{2}(0) \end{bmatrix} = \begin{bmatrix}0& a& b\\-a& 0& c\\-b& -c& 0 \end{bmatrix}$

The results are $\frac{1}{2}(A + A') = \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$ and $\frac{1}{2}(A - A') = \begin{bmatrix}0& a& b\\-a& 0& c\\-b& -c& 0 \end{bmatrix}$.

Given:

Four matrices are provided.

To Express:

Express each matrix as the sum of a symmetric and a skew-symmetric matrix.

Solution:

Any square matrix A can be uniquely expressed as the sum of a symmetric matrix P and a skew-symmetric matrix Q, where $P = \frac{1}{2}(A + A')$ and $Q = \frac{1}{2}(A - A')$.

(i) Let A = $\begin{bmatrix}3& 5\\1& −1 \end{bmatrix}$.

Find A': $A' = \begin{bmatrix}3& 1\\5& −1 \end{bmatrix}$.

Calculate $A + A'$: $A + A' = \begin{bmatrix}3& 5\\1& −1 \end{bmatrix} + \begin{bmatrix}3& 1\\5& −1 \end{bmatrix} = \begin{bmatrix}3+3& 5+1\\1+5& -1+(-1) \end{bmatrix} = \begin{bmatrix}6& 6\\6& -2 \end{bmatrix}$.

Calculate the symmetric part $P = \frac{1}{2}(A + A')$: $P = \frac{1}{2} \begin{bmatrix}6& 6\\6& -2 \end{bmatrix} = \begin{bmatrix}3& 3\\3& -1 \end{bmatrix}$.

Calculate $A - A'$: $A - A' = \begin{bmatrix}3& 5\\1& −1 \end{bmatrix} - \begin{bmatrix}3& 1\\5& −1 \end{bmatrix} = \begin{bmatrix}3-3& 5-1\\1-5& -1-(-1) \end{bmatrix} = \begin{bmatrix}0& 4\\-4& 0 \end{bmatrix}$.

Calculate the skew-symmetric part $Q = \frac{1}{2}(A - A')$: $Q = \frac{1}{2} \begin{bmatrix}0& 4\\-4& 0 \end{bmatrix} = \begin{bmatrix}0& 2\\-2& 0 \end{bmatrix}$.

Express A as P + Q:

$A = \begin{bmatrix}3& 3\\3& -1 \end{bmatrix} + \begin{bmatrix}0& 2\\-2& 0 \end{bmatrix}$

(ii) Let A = $\begin{bmatrix}6& −2& 2\\−2& 3& −1\\2& −1& 3 \end{bmatrix}$.

Find A': $A' = \begin{bmatrix}6& −2& 2\\−2& 3& −1\\2& −1& 3 \end{bmatrix}' = \begin{bmatrix}6& −2& 2\\−2& 3& −1\\2& −1& 3 \end{bmatrix}$.

Calculate $A + A'$: $A + A' = \begin{bmatrix}6& −2& 2\\−2& 3& −1\\2& −1& 3 \end{bmatrix} + \begin{bmatrix}6& −2& 2\\−2& 3& −1\\2& −1& 3 \end{bmatrix} = \begin{bmatrix}12& -4& 4\\-4& 6& -2\\4& -2& 6 \end{bmatrix}$.

Calculate the symmetric part $P = \frac{1}{2}(A + A')$: $P = \frac{1}{2} \begin{bmatrix}12& -4& 4\\-4& 6& -2\\4& -2& 6 \end{bmatrix} = \begin{bmatrix}6& -2& 2\\-2& 3& -1\\2& -1& 3 \end{bmatrix}$.

Calculate $A - A'$: $A - A' = \begin{bmatrix}6& −2& 2\\−2& 3& −1\\2& −1& 3 \end{bmatrix} - \begin{bmatrix}6& −2& 2\\−2& 3& −1\\2& −1& 3 \end{bmatrix} = \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$.

Calculate the skew-symmetric part $Q = \frac{1}{2}(A - A')$: $Q = \frac{1}{2} \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix} = \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$.

Express A as P + Q:

$A = \begin{bmatrix}6& -2& 2\\-2& 3& -1\\2& -1& 3 \end{bmatrix} + \begin{bmatrix}0& 0& 0\\0& 0& 0\\0& 0& 0 \end{bmatrix}$

Note that the given matrix A in this part is already symmetric.

(iii) Let A = $\begin{bmatrix}3& 3& −1\\−2& −2& 1\\−4& −5& 2 \end{bmatrix}$.

Find A': $A' = \begin{bmatrix}3& −2& −4\\3& −2& −5\\−1& 1& 2 \end{bmatrix}$.

Calculate $A + A'$: $A + A' = \begin{bmatrix}3& 3& −1\\−2& −2& 1\\−4& −5& 2 \end{bmatrix} + \begin{bmatrix}3& −2& −4\\3& −2& −5\\−1& 1& 2 \end{bmatrix} = \begin{bmatrix}6& 1& -5\\1& -4& -4\\-5& -4& 4 \end{bmatrix}$.

Calculate the symmetric part $P = \frac{1}{2}(A + A')$: $P = \frac{1}{2} \begin{bmatrix}6& 1& -5\\1& -4& -4\\-5& -4& 4 \end{bmatrix} = \begin{bmatrix}3& \frac{1}{2}& -\frac{5}{2}\\\frac{1}{2}& -2& -2\\-\frac{5}{2}& -2& 2 \end{bmatrix}$.

Calculate $A - A'$: $A - A' = \begin{bmatrix}3& 3& −1\\−2& −2& 1\\−4& −5& 2 \end{bmatrix} - \begin{bmatrix}3& −2& −4\\3& −2& −5\\−1& 1& 2 \end{bmatrix} = \begin{bmatrix}0& 5& 3\\-5& 0& 6\\-3& -6& 0 \end{bmatrix}$.

Calculate the skew-symmetric part $Q = \frac{1}{2}(A - A')$: $Q = \frac{1}{2} \begin{bmatrix}0& 5& 3\\-5& 0& 6\\-3& -6& 0 \end{bmatrix} = \begin{bmatrix}0& \frac{5}{2}& \frac{3}{2}\\-\frac{5}{2}& 0& 3\\-\frac{3}{2}& -3& 0 \end{bmatrix}$.

Express A as P + Q:

$A = \begin{bmatrix}3& \frac{1}{2}& -\frac{5}{2}\\\frac{1}{2}& -2& -2\\-\frac{5}{2}& -2& 2 \end{bmatrix} + \begin{bmatrix}0& \frac{5}{2}& \frac{3}{2}\\-\frac{5}{2}& 0& 3\\-\frac{3}{2}& -3& 0 \end{bmatrix}$

(iv) Let A = $\begin{bmatrix}1& 5\\−1& 2 \end{bmatrix}$.

Find A': $A' = \begin{bmatrix}1& −1\\5& 2 \end{bmatrix}$.

Calculate $A + A'$: $A + A' = \begin{bmatrix}1& 5\\−1& 2 \end{bmatrix} + \begin{bmatrix}1& −1\\5& 2 \end{bmatrix} = \begin{bmatrix}1+1& 5+(-1)\\-1+5& 2+2 \end{bmatrix} = \begin{bmatrix}2& 4\\4& 4 \end{bmatrix}$.

Calculate the symmetric part $P = \frac{1}{2}(A + A')$: $P = \frac{1}{2} \begin{bmatrix}2& 4\\4& 4 \end{bmatrix} = \begin{bmatrix}1& 2\\2& 2 \end{bmatrix}$.

Calculate $A - A'$: $A - A' = \begin{bmatrix}1& 5\\−1& 2 \end{bmatrix} - \begin{bmatrix}1& −1\\5& 2 \end{bmatrix} = \begin{bmatrix}1-1& 5-(-1)\\-1-5& 2-2 \end{bmatrix} = \begin{bmatrix}0& 6\\-6& 0 \end{bmatrix}$.

Calculate the skew-symmetric part $Q = \frac{1}{2}(A - A')$: $Q = \frac{1}{2} \begin{bmatrix}0& 6\\-6& 0 \end{bmatrix} = \begin{bmatrix}0& 3\\-3& 0 \end{bmatrix}$.

Express A as P + Q:

$A = \begin{bmatrix}1& 2\\2& 2 \end{bmatrix} + \begin{bmatrix}0& 3\\-3& 0 \end{bmatrix}$

Given:

Matrices A and B are symmetric matrices of the same order.

To Determine:

The type of matrix that AB – BA is.

Solution:

Since A and B are symmetric matrices, by definition, their transposes are equal to the original matrices:

We need to determine if the matrix $C = AB - BA$ is symmetric, skew-symmetric, or something else. To do this, we will find the transpose of C, i.e., $C' = (AB - BA)'$.

Using the property of transpose of the difference of matrices, $(P - Q)' = P' - Q'$, we have:

$C' = (AB)' - (BA)'$

Using the property of transpose of the product of matrices, $(PQ)' = Q'P'$, we have:

$(AB)' = B'A'$

$(BA)' = A'B'$

Substitute these into the expression for C':

$C' = B'A' - A'B'$

Now, use the given conditions that A' = A and B' = B:

$C' = BA - AB$

We can rewrite the expression $BA - AB$ as $-(AB - BA)$.

$C' = -(AB - BA)$

Since $C = AB - BA$, we have:

$C' = -C$

By definition, a matrix C is skew-symmetric if $C' = -C$.

Therefore, the matrix AB – BA is a skew-symmetric matrix.

Let's examine the given options:

(A) Skew symmetric matrix: This matches our finding.

(B) Symmetric matrix: This would require $C' = C$, i.e., $BA - AB = AB - BA$, which implies $2BA = 2AB$, or $BA = AB$. This is not always true for symmetric matrices A and B.

(C) Zero matrix: This is a special case of a skew-symmetric matrix, but AB – BA is not always the zero matrix.

(D) Identity matrix: This is a special case of a symmetric matrix, but AB – BA is not always the identity matrix.

The property $AB - BA$ being skew-symmetric when A and B are symmetric is a general result.

The correct option is (A).

Given:

Matrix $A = \begin{bmatrix}\cos\alpha&−\sin\alpha\\\sin\alpha&\cos\alpha \end{bmatrix}$.

The equation $A + A′ = I$, where $A′$ is the transpose of matrix A and I is the $2 \times 2$ identity matrix.

To Find:

The value of $\alpha$ that satisfies the equation $A + A′ = I$.

Solution:

First, find the transpose of matrix A, denoted by $A′$. The transpose of a matrix is obtained by interchanging its rows and columns.

$A′ = \begin{bmatrix}\cos\alpha&\sin\alpha\\−\sin\alpha&\cos\alpha \end{bmatrix}$

The identity matrix of order $2 \times 2$ is $I = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$.

Now, substitute A, $A′$, and I into the given equation $A + A′ = I$:

$\begin{bmatrix}\cos\alpha&−\sin\alpha\\\sin\alpha&\cos\alpha \end{bmatrix} + \begin{bmatrix}\cos\alpha&\sin\alpha\\−\sin\alpha&\cos\alpha \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

Perform the matrix addition on the left side:

$\begin{bmatrix}\cos\alpha + \cos\alpha&−\sin\alpha + \sin\alpha\\\sin\alpha − \sin\alpha&\cos\alpha + \cos\alpha \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

$\begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

For two matrices to be equal, their corresponding elements must be equal. Equating the elements of the matrices on both sides:

And the off-diagonal elements match $0=0$.

From the equation $2\cos\alpha = 1$, we solve for $\cos\alpha$:

We need to find the value of $\alpha$ among the given options that satisfies this equation.

The general solution for $\cos\alpha = \frac{1}{2}$ is $\alpha = 2n\pi \pm \frac{\pi}{3}$, where $n$ is an integer.

Let's check the given options:

(A) $\alpha = \frac{\pi}{6}$: $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \neq \frac{1}{2}$.

(B) $\alpha = \frac{\pi}{3}$: $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$. This matches the required value. ($\frac{\pi}{3}$ is a specific solution when $n=0$ and the positive sign is taken).

(C) $\alpha = \pi$: $\cos(\pi) = -1 \neq \frac{1}{2}$.

(D) $\alpha = \frac{3\pi}{2}$: $\cos\left(\frac{3\pi}{2}\right) = 0 \neq \frac{1}{2}$.

Only option (B) satisfies the condition $\cos\alpha = \frac{1}{2}$.

The value of $\alpha$ is $\frac{\pi}{3}$.

The final answer is $\boxed{\frac{π}{3}}$.

Given:

Matrices A and B.

To Determine:

The condition for matrices A and B to be inverses of each other.

Solution:

Two square matrices A and B of the same order are said to be inverse of each other if their product is the identity matrix.

This means that if B is the inverse of A, then $AB = I$ and $BA = I$, where I is the identity matrix of the same order as A and B.

The condition is that both products, AB and BA, must be equal to the identity matrix.

Let's examine the given options:

(A) AB = BA: This means the matrices commute, but it does not guarantee that they are inverses.

(B) AB = BA = 0: This means both products are the zero matrix, not the identity matrix. This is not the condition for inverse matrices (unless the matrices themselves are zero matrices, which are not invertible).

(C) AB = 0, BA = I: This means the product in one order is the zero matrix and in the other order is the identity matrix. For invertible matrices, this cannot happen (unless A or B is the zero matrix, which is not invertible). If AB=0, and if A were invertible, then $A^{-1}(AB) = A^{-1}0 \implies (A^{-1}A)B = 0 \implies IB = 0 \implies B=0$, but then $BA=0$, not I (for a non-zero A). Similarly if B were invertible.

(D) AB = BA = I: This matches the definition of inverse matrices. When the product of two square matrices in both orders is the identity matrix, they are inverses of each other.

The correct option is (D).

Given:

Matrix A = $\begin{bmatrix}\cosθ& \sinθ\\−\sinθ& \cosθ \end{bmatrix}$.

To Prove:

$A^n = \begin{bmatrix}\cos nθ& \sin nθ\\−\sin nθ& \cos nθ \end{bmatrix}$ for all natural numbers $n \in \mathbb{N}$.

Solution:

We will prove this by the principle of mathematical induction on $n$.

Base Case: For $n = 1$

$A^1 = A = \begin{bmatrix}\cosθ& \sinθ\\−\sinθ& \cosθ \end{bmatrix}$

The formula gives $\begin{bmatrix}\cos(1)\theta& \sin(1)\theta\\−\sin(1)\theta& \cos(1)\theta \end{bmatrix} = \begin{bmatrix}\cosθ& \sinθ\\−\sinθ& \cosθ \end{bmatrix}$.

Thus, the formula is true for $n = 1$.

Inductive Hypothesis: Assume that the formula is true for some natural number $k \ge 1$.

That is, $A^k = \begin{bmatrix}\cos kθ& \sin kθ\\−\sin kθ& \cos k θ \end{bmatrix}$

Inductive Step: We need to prove that the formula is true for $n = k + 1$.

Consider $A^{k+1} = A^k \cdot A$.

$A^{k+1} = \begin{bmatrix}\cos kθ& \sin kθ\\−\sin kθ& \cos kθ \end{bmatrix} \begin{bmatrix}\cosθ& \sinθ\\−\sinθ& \cosθ \end{bmatrix}$

Perform the matrix multiplication:

• Element (1,1): $(\cos kθ)(\cosθ) + (\sin kθ)(−\sinθ) = \cos kθ \cosθ - \sin kθ \sinθ$
• Element (1,2): $(\cos kθ)(\sinθ) + (\sin kθ)(\cosθ) = \cos kθ \sinθ + \sin kθ \cosθ$
• Element (2,1): $(−\sin kθ)(\cosθ) + (\cos kθ)(−\sinθ) = -\sin kθ \cosθ - \cos kθ \sinθ = -(\sin kθ \cosθ + \cos kθ \sinθ)$
• Element (2,2): $(−\sin kθ)(\sinθ) + (\cos kθ)(\cosθ) = -\sin kθ \sinθ + \cos kθ \cosθ = \cos kθ \cosθ - \sin kθ \sinθ$

Using the trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$, we get:

$A^{k+1} = \begin{bmatrix}\cos(kθ + θ)& \sin(kθ + θ)\\−\sin(kθ + θ)& \cos(kθ + θ) \end{bmatrix} = \begin{bmatrix}\cos((k+1)θ)& \sin((k+1)θ)\\−\sin((k+1)θ)& \cos((k+1)θ) \end{bmatrix}$

This is the form of the formula for $n = k+1$.

Therefore, by the principle of mathematical induction, the formula is true for all natural numbers $n \in \mathbb{N}$.

The statement $A^n = \begin{bmatrix}\cos nθ& \sin nθ\\−\sin nθ& \cos nθ \end{bmatrix}$ for $n \in N$ is proven by mathematical induction.

Given:

A and B are symmetric matrices of the same order. This means $A' = A$ and $B' = B$.

To Show:

AB is symmetric if and only if A and B commute (AB = BA).

The statement "if and only if" requires proving two implications:

• If AB is symmetric, then AB = BA.
• If AB = BA, then AB is symmetric.

Solution:

Part 1: Assume AB is symmetric. We need to show that AB = BA.

If AB is symmetric, then its transpose is equal to itself:

Using the property of the transpose of a product, $(PQ)' = Q'P'$, we have:

$(AB)' = B'A'$

Since A and B are symmetric, $A' = A$ and $B' = B$. Substitute these into the expression for $(AB)'$:

$(AB)' = BA$

Now, substitute this result back into equation (1):

$BA = AB$

Thus, if AB is symmetric, then A and B commute (AB = BA).

Part 2: Assume A and B commute, i.e., AB = BA. We need to show that AB is symmetric.

We need to show that $(AB)' = AB$.

Consider the transpose of the product AB:

$(AB)' = B'A'$

Since A and B are symmetric, $A' = A$ and $B' = B$. Substitute these into the expression for $(AB)'$:

$(AB)' = BA$

We are given that A and B commute, which means $BA = AB$. Substitute this into the expression for $(AB)'$:

$(AB)' = AB$

Thus, if A and B commute (AB = BA), then AB is symmetric.

Combining both parts, we have shown that AB is symmetric if and only if A and B commute (AB = BA).

Given:

Matrix A = $\begin{bmatrix}2& −1\\3& 4 \end{bmatrix}$ (order $2 \times 2$)

Matrix B = $\begin{bmatrix}5& 2\\7& 4 \end{bmatrix}$ (order $2 \times 2$)

Matrix C = $\begin{bmatrix}2& 5\\3& 8 \end{bmatrix}$ (order $2 \times 2$)

The equation is $CD - AB = O$, where O is the zero matrix of the appropriate order.

To Find:

A matrix D such that $CD - AB = O$.

Solution:

The given equation is $CD - AB = O$. We can rewrite this as $CD = AB$.

We need to find the matrix D. First, let's calculate the product AB.

$AB = \begin{bmatrix}2& −1\\3& 4 \end{bmatrix} \begin{bmatrix}5& 2\\7& 4 \end{bmatrix}$

• Element (1,1): $(2)(5) + (−1)(7) = 10 - 7 = 3$
• Element (1,2): $(2)(2) + (−1)(4) = 4 - 4 = 0$
• Element (2,1): $(3)(5) + (4)(7) = 15 + 28 = 43$
• Element (2,2): $(3)(2) + (4)(4) = 6 + 16 = 22$

So, $AB = \begin{bmatrix}3& 0\\43& 22 \end{bmatrix}$.

The equation is now $CD = \begin{bmatrix}3& 0\\43& 22 \end{bmatrix}$.

Let the matrix D have order $m \times n$. The order of C is $2 \times 2$. For the product CD to be defined, the number of columns in C must be equal to the number of rows in D, so $m=2$. The order of the product CD is (number of rows in C) $\times$ (number of columns in D), which is $2 \times n$. The order of AB is $2 \times 2$. For $CD = AB$, their orders must be the same, so $2 \times n = 2 \times 2$, which implies $n=2$. Thus, the order of matrix D must be $2 \times 2$.

Let $D = \begin{bmatrix}a& b\\c& d \end{bmatrix}$.

The equation $CD = AB$ is:

$\begin{bmatrix}2& 5\\3& 8 \end{bmatrix} \begin{bmatrix}a& b\\c& d \end{bmatrix} = \begin{bmatrix}3& 0\\43& 22 \end{bmatrix}$

Perform the matrix multiplication on the left-hand side:

$\begin{bmatrix}(2)(a)+(5)(c)& (2)(b)+(5)(d)\\(3)(a)+(8)(c)& (3)(b)+(8)(d) \end{bmatrix} = \begin{bmatrix}2a+5c& 2b+5d\\3a+8c& 3b+8d \end{bmatrix}$

Equating the corresponding elements, we get a system of linear equations:

We have two systems of linear equations. Solve the first system for $a$ and $c$.

Multiply equation (1) by 3 and equation (2) by 2:

$3 \times (2a + 5c) = 3 \times 3 \implies 6a + 15c = 9$

$2 \times (3a + 8c) = 2 \times 43 \implies 6a + 16c = 86$

Subtract the first modified equation from the second modified equation:

$(6a + 16c) - (6a + 15c) = 86 - 9$

$c = 77$

Substitute $c=77$ into equation (1):

$2a + 5(77) = 3$

$2a + 385 = 3$

$2a = 3 - 385$

$2a = -382$

$a = \frac{-382}{2} = -191$

So, $a = -191$ and $c = 77$.

Now, solve the second system for $b$ and $d$.

Multiply equation (3) by 3 and equation (4) by 2:

$3 \times (2b + 5d) = 3 \times 0 \implies 6b + 15d = 0$

$2 \times (3b + 8d) = 2 \times 22 \implies 6b + 16d = 44$

Subtract the first modified equation from the second modified equation:

$(6b + 16d) - (6b + 15d) = 44 - 0$

$d = 44$

Substitute $d=44$ into equation (3):

$2b + 5(44) = 0$

$2b + 220 = 0$

$2b = -220$

$b = \frac{-220}{2} = -110$

So, $b = -110$ and $d = 44$.

Thus, the matrix D is:

$D = \begin{bmatrix}-191& -110\\77& 44 \end{bmatrix}$

Alternatively, we can solve for D using the inverse of C, if it exists. For a $2 \times 2$ matrix $C = \begin{bmatrix}c_{11}& c_{12}\\c_{21}& c_{22} \end{bmatrix}$, its inverse $C^{-1}$ exists if and only if the determinant $\det(C) = c_{11}c_{22} - c_{12}c_{21} \ne 0$. If the inverse exists, $C^{-1} = \frac{1}{\det(C)} \begin{bmatrix}c_{22}& -c_{12}\\-c_{21}& c_{11} \end{bmatrix}$.

For matrix C = $\begin{bmatrix}2& 5\\3& 8 \end{bmatrix}$, the determinant is $\det(C) = (2)(8) - (5)(3) = 16 - 15 = 1$.

Since $\det(C) = 1 \ne 0$, the inverse of C exists.

$C^{-1} = \frac{1}{1} \begin{bmatrix}8& -5\\-3& 2 \end{bmatrix} = \begin{bmatrix}8& -5\\-3& 2 \end{bmatrix}$.

From the equation $CD = AB$, multiply by $C^{-1}$ from the left:

$C^{-1}(CD) = C^{-1}(AB)$

$(C^{-1}C)D = C^{-1}(AB)$

$ID = C^{-1}(AB)$

$D = C^{-1}(AB)$

We already calculated $AB = \begin{bmatrix}3& 0\\43& 22 \end{bmatrix}$.

$D = \begin{bmatrix}8& -5\\-3& 2 \end{bmatrix} \begin{bmatrix}3& 0\\43& 22 \end{bmatrix}$

Perform the matrix multiplication:

• Element (1,1): $(8)(3) + (-5)(43) = 24 - 215 = -191$
• Element (1,2): $(8)(0) + (-5)(22) = 0 - 110 = -110$
• Element (2,1): $(-3)(3) + (2)(43) = -9 + 86 = 77$
• Element (2,2): $(-3)(0) + (2)(22) = 0 + 44 = 44$

$D = \begin{bmatrix}-191& -110\\77& 44 \end{bmatrix}$

Both methods give the same result for matrix D.

The matrix D is $\begin{bmatrix}-191& -110\\77& 44 \end{bmatrix}$.

Given:

Matrices A and B are symmetric matrices of the same order.

By definition, a matrix A is symmetric if its transpose is equal to itself, i.e., $A' = A$.

Similarly, for matrix B, $B' = B$.

To Prove:

The matrix AB – BA is a skew-symmetric matrix.

By definition, a matrix C is skew-symmetric if its transpose is equal to the negative of the matrix itself, i.e., $C' = -C$.

We need to show that $(AB - BA)' = -(AB - BA)$.

Proof:

Let $C = AB - BA$. We want to find the transpose of C, which is $C' = (AB - BA)'$.

Using the property of the transpose of the difference of two matrices, $(P - Q)' = P' - Q'$, we have:

Now, we use the property of the transpose of the product of two matrices, $(PQ)' = Q'P'$. Applying this to the terms on the right-hand side of equation (i):

Substitute equations (ii) and (iii) back into equation (i):

$(AB - BA)' = B'A' - A'B'$

We are given that A and B are symmetric matrices, which means $A' = A$ and $B' = B$. Substitute these conditions into the equation:

$(AB - BA)' = BA - AB$

We can rewrite the expression $BA - AB$ as the negative of $AB - BA$:

$BA - AB = -(AB - BA)$

So, we have:

$(AB - BA)' = -(AB - BA)$

Let $C = AB - BA$. Then the equation is $C' = -C$.

By the definition of a skew-symmetric matrix, since the transpose of $(AB - BA)$ is equal to the negative of $(AB - BA)$, the matrix $(AB - BA)$ is a skew-symmetric matrix.

Hence, the proof is complete.

Given:

Matrix A and matrix B.

To Show:

The matrix $C = B'AB$ is symmetric if A is symmetric, and $C = B'AB$ is skew-symmetric if A is skew-symmetric.

Proof:

We need to analyze the transpose of the matrix $C = B'AB$. Using the property of the transpose of a product $(PQR)' = R'Q'P'$, we have:

Using the property of the transpose of a transpose, $(P')' = P$, we have $(B')' = B$.

Substitute this into equation (i):

$(B'AB)' = B' A' B$

Now, we consider the two cases: A is symmetric and A is skew-symmetric.

Case 1: A is symmetric.

If A is symmetric, then by definition, $A' = A$.

Substitute $A' = A$ into the expression for $(B'AB)'$:

$(B'AB)' = B' (A) B = B'AB$

Let $C = B'AB$. We have shown that $C' = C$.

By the definition of a symmetric matrix, the matrix $B'AB$ is symmetric if A is symmetric.

Case 2: A is skew-symmetric.

If A is skew-symmetric, then by definition, $A' = -A$.

Substitute $A' = -A$ into the expression for $(B'AB)'$:

$(B'AB)' = B' (-A) B$

We can factor out the scalar -1:

$(B'AB)' = - (B' A B)$

Let $C = B'AB$. We have shown that $C' = -C$.

By the definition of a skew-symmetric matrix, the matrix $B'AB$ is skew-symmetric if A is skew-symmetric.

From both cases, we have shown that the matrix $B'AB$ is symmetric if A is symmetric, and $B'AB$ is skew-symmetric if A is skew-symmetric.

Hence, the statement is proven.

Given:

The matrix $A = \begin{bmatrix}0&2y&z\\x&y&−z\\x&−y&z \end{bmatrix}$.

The equation $A′A = I$, where $A′$ is the transpose of matrix A and I is the identity matrix.

To Find:

The values of x, y, and z.

Solution:

First, we find the transpose of matrix A, denoted by $A′$. The transpose is obtained by interchanging the rows and columns of A.

$A′ = \begin{bmatrix}0&x&x\\2y&y&−y\\z&−z&z \end{bmatrix}$

Next, we calculate the matrix product $A′A$:

$A′A = \begin{bmatrix}0&x&x\\2y&y&−y\\z&−z&z \end{bmatrix} \begin{bmatrix}0&2y&z\\x&y&−z\\x&−y&z \end{bmatrix}$

Multiplying the matrices, we get:

$A′A = \begin{bmatrix}(0)(0)+x(x)+x(x)&(0)(2y)+x(y)+x(−y)&(0)(z)+x(−z)+x(z)\\(2y)(0)+y(x)+(−y)(x)&(2y)(2y)+y(y)+(−y)(−y)&(2y)(z)+y(−z)+(−y)(z)\\(z)(0)+(−z)(x)+z(x)&(z)(2y)+(−z)(y)+z(−y)&(z)(z)+(−z)(−z)+z(z) \end{bmatrix}$

$A′A = \begin{bmatrix}0+x^2+x^2&0+xy−xy&0−xz+xz\\0+xy−xy&4y^2+y^2+y^2&2yz−yz−yz\\0−xz+xz&2yz−yz−yz&z^2+z^2+z^2 \end{bmatrix}$

$A′A = \begin{bmatrix}2x^2&0&0\\0&6y^2&0\\0&0&3z^2 \end{bmatrix}$

We are given that $A′A = I$. Since A is a $3 \times 3$ matrix, $A′A$ is also a $3 \times 3$ matrix. The identity matrix I of order $3 \times 3$ is:

$I = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$

So, we have the equation:

$\begin{bmatrix}2x^2&0&0\\0&6y^2&0\\0&0&3z^2 \end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$

For two matrices to be equal, their corresponding elements must be equal. Equating the diagonal elements, we get the following system of equations:

Solving the first equation for x:

$2x^2 = 1$

$x^2 = \frac{1}{2}$

$x = \pm\sqrt{\frac{1}{2}} = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2}$

Solving the second equation for y:

$6y^2 = 1$

$y^2 = \frac{1}{6}$

$y = \pm\sqrt{\frac{1}{6}} = \pm\frac{1}{\sqrt{6}} = \pm\frac{\sqrt{6}}{6}$

Solving the third equation for z:

$3z^2 = 1$

$z^2 = \frac{1}{3}$

$z = \pm\sqrt{\frac{1}{3}} = \pm\frac{1}{\sqrt{3}} = \pm\frac{\sqrt{3}}{3}$

The values of x, y, and z are $x = \pm \frac{1}{\sqrt{2}}$, $y = \pm \frac{1}{\sqrt{6}}$, and $z = \pm \frac{1}{\sqrt{3}}$.

Given:

The matrix equation $\begin{bmatrix}1& 2& 1 \end{bmatrix} \begin{bmatrix}1& 2& 0\\2& 0& 1\\1& 0& 2 \end{bmatrix} \begin{bmatrix}0\\2\\x \end{bmatrix} = O$, where O is the zero matrix.

To Find:

The values of $x$ that satisfy the given equation.

Solution:

We need to perform the matrix multiplications on the left-hand side ($L.H.S.$) of the equation and equate the result to the zero matrix O.

Let the matrices be A = $\begin{bmatrix}1& 2& 1 \end{bmatrix}$ (order $1 \times 3$), B = $\begin{bmatrix}1& 2& 0\\2& 0& 1\\1& 0& 2 \end{bmatrix}$ (order $3 \times 3$), and C = $\begin{bmatrix}0\\2\\x \end{bmatrix}$ (order $3 \times 1$).

The equation is $ABC = O$.

First, calculate the product AB. The number of columns in A (3) is equal to the number of rows in B (3). The order of AB will be $1 \times 3$.

$AB = \begin{bmatrix}1& 2& 1 \end{bmatrix} \begin{bmatrix}1& 2& 0\\2& 0& 1\\1& 0& 2 \end{bmatrix} = [(1)(1)+(2)(2)+(1)(1) \quad (1)(2)+(2)(0)+(1)(0) \quad (1)(0)+(2)(1)+(1)(2)]$

$AB = [1+4+1 \quad 2+0+0 \quad 0+2+2] = [6 \quad 2 \quad 4]$

Now, calculate the product $(AB)C$. The order of AB is $1 \times 3$ and the order of C is $3 \times 1$. The number of columns in AB (3) is equal to the number of rows in C (3). The order of (AB)C will be $1 \times 1$.

$(AB)C = \begin{bmatrix}6& 2& 4 \end{bmatrix} \begin{bmatrix}0\\2\\x \end{bmatrix} = [(6)(0) + (2)(2) + (4)(x)] = [0 + 4 + 4x] = [4 + 4x]$

The equation is $(AB)C = O$. Since $(AB)C$ is a $1 \times 1$ matrix, the zero matrix O must also be a $1 \times 1$ matrix, which is [0].

So, $[4 + 4x] = [0]$.

Equating the single elements of the matrices:

Solve the linear equation for $x$:

$4x = -4$

$x = \frac{-4}{4}$

The value of $x$ that satisfies the equation is -1.

The value of $x$ is -1.

Given:

Matrix A = $\begin{bmatrix}3& 1\\−1& 2 \end{bmatrix}$.

The identity matrix I of order $2 \times 2$ is $I = \begin{bmatrix}1& 0\\0& 1 \end{bmatrix}$.

The zero matrix O of order $2 \times 2$ is $O = \begin{bmatrix}0& 0\\0& 0 \end{bmatrix}$.

To Show:

$A^2 - 5A + 7I = O$

Solution:

First, calculate $A^2 = A \times A$. The order of A is $2 \times 2$, so the order of $A^2$ will also be $2 \times 2$.

$A^2 = \begin{bmatrix}3& 1\\−1& 2 \end{bmatrix} \begin{bmatrix}3& 1\\−1& 2 \end{bmatrix}$

• Element (1,1): $(3)(3) + (1)(−1) = 9 - 1 = 8$
• Element (1,2): $(3)(1) + (1)(2) = 3 + 2 = 5$
• Element (2,1): $(−1)(3) + (2)(−1) = -3 - 2 = -5$
• Element (2,2): $(−1)(1) + (2)(2) = -1 + 4 = 3$

$A^2 = \begin{bmatrix}8& 5\\-5& 3 \end{bmatrix}$

Next, calculate $5A$ and $7I$ by scalar multiplication.

$5A = 5 \begin{bmatrix}3& 1\\−1& 2 \end{bmatrix} = \begin{bmatrix}5 \times 3& 5 \times 1\\5 \times (−1)& 5 \times 2 \end{bmatrix} = \begin{bmatrix}15& 5\\-5& 10 \end{bmatrix}$

$7I = 7 \begin{bmatrix}1& 0\\0& 1 \end{bmatrix} = \begin{bmatrix}7 \times 1& 7 \times 0\\7 \times 0& 7 \times 1 \end{bmatrix} = \begin{bmatrix}7& 0\\0& 7 \end{bmatrix}$

Now, compute the expression $A^2 - 5A + 7I$. This involves matrix subtraction and addition.

$A^2 - 5A + 7I = \begin{bmatrix}8& 5\\-5& 3 \end{bmatrix} - \begin{bmatrix}15& 5\\-5& 10 \end{bmatrix} + \begin{bmatrix}7& 0\\0& 7 \end{bmatrix}$

$A^2 - 5A + 7I = \begin{bmatrix}8 - 15 + 7& 5 - 5 + 0\\-5 - (-5) + 0& 3 - 10 + 7 \end{bmatrix}$

$A^2 - 5A + 7I = \begin{bmatrix}-7 + 7& 0\\-5 + 5& -7 + 7 \end{bmatrix}$

$A^2 - 5A + 7I = \begin{bmatrix}0& 0\\0& 0 \end{bmatrix}$

The resulting matrix is the zero matrix O of order $2 \times 2$.

Therefore, $A^2 - 5A + 7I = O$ is shown.

Given:

The matrix equation $\begin{bmatrix}x& −5& −1 \end{bmatrix} \begin{bmatrix}1& 0& 2\\0& 2& 1\\2& 0& 3 \end{bmatrix} \begin{bmatrix}x\\4\\1 \end{bmatrix} = 0$. Note that the right-hand side is the scalar 0, not the zero matrix O.

To Find:

The value(s) of $x$ that satisfy the given equation.

Solution:

We need to perform the matrix multiplications on the left-hand side ($L.H.S.$) of the equation and equate the result to the scalar 0.

Let the matrices be A = $\begin{bmatrix}x& −5& −1 \end{bmatrix}$ (order $1 \times 3$), B = $\begin{bmatrix}1& 0& 2\\0& 2& 1\\2& 0& 3 \end{bmatrix}$ (order $3 \times 3$), and C = $\begin{bmatrix}x\\4\\1 \end{bmatrix}$ (order $3 \times 1$).

The equation is $ABC = 0$.

First, calculate the product AB. The number of columns in A (3) is equal to the number of rows in B (3). The order of AB will be $1 \times 3$.

$AB = \begin{bmatrix}x& −5& −1 \end{bmatrix} \begin{bmatrix}1& 0& 2\\0& 2& 1\\2& 0& 3 \end{bmatrix}$

$AB = [(x)(1)+(-5)(0)+(-1)(2) \quad (x)(0)+(-5)(2)+(-1)(0) \quad (x)(2)+(-5)(1)+(-1)(3)]$

$AB = [x+0-2 \quad 0-10-0 \quad 2x-5-3]$

$AB = [x-2 \quad -10 \quad 2x-8]$

Now, calculate the product $(AB)C$. The order of AB is $1 \times 3$ and the order of C is $3 \times 1$. The number of columns in AB (3) is equal to the number of rows in C (3). The order of (AB)C will be $1 \times 1$, which is a scalar.

$(AB)C = \begin{bmatrix}x-2& -10& 2x-8 \end{bmatrix} \begin{bmatrix}x\\4\\1 \end{bmatrix}$

$(AB)C = [(x-2)(x) + (-10)(4) + (2x-8)(1)]$

$(AB)C = [x(x-2) - 40 + (2x-8)]$

$(AB)C = [x^2 - 2x - 40 + 2x - 8]$

$(AB)C = [x^2 - 48]$

The equation is $(AB)C = 0$. The result of $(AB)C$ is a scalar, so we equate the scalar $x^2 - 48$ to the scalar 0.

Solve the quadratic equation for $x$:

$x^2 = 48$

$x = \pm \sqrt{48}$

Simplify the square root:

$\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}$

$x = \pm 4\sqrt{3}$

The values of $x$ are $4\sqrt{3}$ and $-4\sqrt{3}$.

Given:

Annual sales matrix A = $\begin{bmatrix}10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}$ (order $2 \times 3$). The rows represent the markets (Row 1: Market I, Row 2: Market II) and the columns represent the products (Column 1: product x, Column 2: product y, Column 3: product z).

To Find:

(a) Total revenue in each market using matrix algebra, given unit sale prices.

(b) Gross profit, given unit costs.

Solution:

(a) Total revenue in each market:

The unit sale prices of products x, y, and z are $\textsf{₹}2.50$, $\textsf{₹}1.50$, and $\textsf{₹}1.00$, respectively.

We can represent these prices as a column matrix (or a row matrix). Let the prices be represented by the column matrix $P = \begin{bmatrix}2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$ (order $3 \times 1$). The entries correspond to the prices of products x, y, and z.

To find the total revenue in each market using matrix multiplication, we need a matrix where each entry represents the total revenue for a market. The annual sales matrix A has markets as rows and products sold as columns (order $2 \times 3$). The price matrix P has products as rows and price as a column (order $3 \times 1$).

The product AP will give a matrix with markets as rows and total revenue as a column (order $2 \times 1$).

Total Revenue Matrix = AP

$AP = \begin{bmatrix}10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix}2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$

• Element (1,1) (Total revenue in Market I): $(10000)(2.50) + (2000)(1.50) + (18000)(1.00) = 25000 + 3000 + 18000 = 46000$
• Element (2,1) (Total revenue in Market II): $(6000)(2.50) + (20000)(1.50) + (8000)(1.00) = 15000 + 30000 + 8000 = 53000$

The Total Revenue Matrix is $\begin{bmatrix}46000 \\ 53000 \end{bmatrix}$.

Total revenue in Market I = $\textsf{₹}46000$.

Total revenue in Market II = $\textsf{₹}53000$.

(b) Gross profit:

Gross profit = Total Revenue - Total Cost.

We need to calculate the total cost in each market first. The unit costs of products x, y, and z are $\textsf{₹}2.00$, $\textsf{₹}1.00$, and 50 paise ($\textsf{₹}0.50$), respectively.

We can represent these costs as a column matrix (or a row matrix). Let the costs be represented by the column matrix $C = \begin{bmatrix}2.00 \\ 1.00 \\ 0.50 \end{bmatrix}$ (order $3 \times 1$). The entries correspond to the costs of products x, y, and z.

The product AC will give a matrix with markets as rows and total cost as a column (order $2 \times 1$).

Total Cost Matrix = AC

$AC = \begin{bmatrix}10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix}2.00 \\ 1.00 \\ 0.50 \end{bmatrix}$

• Element (1,1) (Total cost in Market I): $(10000)(2.00) + (2000)(1.00) + (18000)(0.50) = 20000 + 2000 + 9000 = 31000$
• Element (2,1) (Total cost in Market II): $(6000)(2.00) + (20000)(1.00) + (8000)(0.50) = 12000 + 20000 + 4000 = 36000$

The Total Cost Matrix is $\begin{bmatrix}31000 \\ 36000 \end{bmatrix}$.

Total cost in Market I = $\textsf{₹}31000$.

Total cost in Market II = $\textsf{₹}36000$.

Now, calculate the gross profit in each market.

Gross Profit in Market I = Total Revenue in Market I - Total Cost in Market I = $\textsf{₹}46000 - \textsf{₹}31000 = \textsf{₹}15000$.

Gross Profit in Market II = Total Revenue in Market II - Total Cost in Market II = $\textsf{₹}53000 - \textsf{₹}36000 = \textsf{₹}17000$.

Alternatively, we can find the profit per unit for each product: Product x: $\textsf{₹}2.50 - \textsf{₹}2.00 = \textsf{₹}0.50$. Product y: $\textsf{₹}1.50 - \textsf{₹}1.00 = \textsf{₹}0.50$. Product z: $\textsf{₹}1.00 - \textsf{₹}0.50 = \textsf{₹}0.50$. Let the unit profits be represented by the column matrix $U = \begin{bmatrix}0.50 \\ 0.50 \\ 0.50 \end{bmatrix}$. Total Profit Matrix = AU $AU = \begin{bmatrix}10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix}0.50 \\ 0.50 \\ 0.50 \end{bmatrix} = \begin{bmatrix}(10000)(0.5)+(2000)(0.5)+(18000)(0.5) \\ (6000)(0.5)+(20000)(0.5)+(8000)(0.5) \end{bmatrix} = \begin{bmatrix}5000+1000+9000 \\ 3000+10000+4000 \end{bmatrix} = \begin{bmatrix}15000 \\ 17000 \end{bmatrix}$. This gives the gross profit in each market directly.

(a) The total revenue in Market I is $\textsf{₹}46000$ and in Market II is $\textsf{₹}53000$.

(b) The gross profit in Market I is $\textsf{₹}15000$ and in Market II is $\textsf{₹}17000$. The total gross profit across both markets is $\textsf{₹}15000 + \textsf{₹}17000 = \textsf{₹}32000$. The question asks for "the gross profit", which could mean the total gross profit.

Assuming the question asks for the total gross profit:

Total Gross Profit = Gross Profit in Market I + Gross Profit in Market II = $\textsf{₹}15000 + \textsf{₹}17000 = \textsf{₹}32000$.

Given:

The matrix equation $X A = B$, where $A = \begin{bmatrix}1&2&3\\4&5&6 \end{bmatrix}$ and $B = \begin{bmatrix}−7&−8&−9\\2&4&6 \end{bmatrix}$.

To Find:

The matrix X.

Solution:

Let the dimensions of matrix X be $m \times n$.

Matrix A has dimensions $2 \times 3$. Matrix B has dimensions $2 \times 3$.

For the matrix multiplication XA to be possible, the number of columns in X must equal the number of rows in A. So, $n=2$.

The resulting matrix XA will have dimensions $m \times 3$. For XA to be equal to B, their dimensions must match. So, $m \times 3 = 2 \times 3$. This implies $m=2$.

Thus, matrix X must be a $2 \times 2$ matrix.

Let $X = \begin{bmatrix}a&b\\c&d \end{bmatrix}$.

Substitute this into the given equation $X A = B$:

$\begin{bmatrix}a&b\\c&d \end{bmatrix} \begin{bmatrix}1&2&3\\4&5&6 \end{bmatrix} = \begin{bmatrix}−7&−8&−9\\2&4&6 \end{bmatrix}$

Perform the matrix multiplication on the left side:

$\begin{bmatrix}a(1)+b(4)&a(2)+b(5)&a(3)+b(6)\\c(1)+d(4)&c(2)+d(5)&c(3)+d(6) \end{bmatrix} = \begin{bmatrix}−7&−8&−9\\2&4&6 \end{bmatrix}$

$\begin{bmatrix}a+4b&2a+5b&3a+6b\\c+4d&2c+5d&3c+6d \end{bmatrix} = \begin{bmatrix}−7&−8&−9\\2&4&6 \end{bmatrix}$

For the two matrices to be equal, their corresponding elements must be equal. This gives us a system of linear equations:

Solve the system for a and b using the first two equations:

Multiply the first equation by 2:

Subtract the second equation ($2a+5b = -8$) from this new equation:

Substitute the value of b into the first equation ($a+4b = -7$):

Solve the system for c and d using the equations involving c and d. Notice that the third equation $3c+6d = 6$ can be divided by 3 to get $c+2d = 2$. This is inconsistent with $c+4d=2$ and $2c+5d=4$. Let's check the initial setup.

The equations are correct as derived from matrix multiplication. Let's re-solve the second system.

Multiply the first equation ($c+4d=2$) by 2:

Subtract the second equation ($2c+5d = 4$) from this new equation:

Substitute the value of d into the first equation ($c+4d = 2$):

Verify these values with the third equation ($3c+6d = 6$):

The values $c=2$ and $d=0$ are consistent with all three equations.

The values are $a=1$, $b=-2$, $c=2$, and $d=0$.

Substituting these values into the matrix X:

$X = \begin{bmatrix}1&−2\\2&0 \end{bmatrix}$

The matrix X is $\begin{bmatrix}1&−2\\2&0 \end{bmatrix}$.

Choose the correct answer in the following questions:

Given:

Matrix A = $\begin{bmatrix}α& β\\γ& −α \end{bmatrix}$ and the equation $A^2 = I$, where I is the identity matrix of order $2 \times 2$, $I = \begin{bmatrix}1& 0\\0& 1 \end{bmatrix}$.

To Find:

The relationship between $\alpha, \beta,$ and $\gamma$ from the given options.

Solution:

First, calculate $A^2 = A \times A$. The order of A is $2 \times 2$, so the order of $A^2$ will also be $2 \times 2$.

$A^2 = \begin{bmatrix}α& β\\γ& −α \end{bmatrix} \begin{bmatrix}α& β\\γ& −α \end{bmatrix}$

• Element (1,1): $(α)(α) + (β)(γ) = α^2 + βγ$
• Element (1,2): $(α)(β) + (β)(−α) = αβ - αβ = 0$
• Element (2,1): $(γ)(α) + (−α)(γ) = γα - αγ = 0$
• Element (2,2): $(γ)(β) + (−α)(−α) = γβ + α^2 = α^2 + βγ$

$A^2 = \begin{bmatrix}α^2 + βγ& 0\\0& α^2 + βγ \end{bmatrix}$

The given equation is $A^2 = I$. Substitute the computed $A^2$ and the identity matrix:

$\begin{bmatrix}α^2 + βγ& 0\\0& α^2 + βγ \end{bmatrix} = \begin{bmatrix}1& 0\\0& 1 \end{bmatrix}$

For these two matrices to be equal, their corresponding elements must be equal.

Equating the elements in the first row, first column (or second row, second column):

Equating the off-diagonal elements ($0=0$) gives no additional information.

From equation (1), we can rearrange the terms to match the options:

$α^2 + βγ - 1 = 0$

Or, $1 - α^2 - βγ = 0$.

Let's examine the given options:

(A) $1 + α^2 + βγ = 0$

(B) $1 – α^2 + βγ = 0$

(C) $1 – α^2 – βγ = 0$: This matches our derived equation.

(D) $1 + α^2 – βγ = 0$

The relationship is $α^2 + βγ = 1$, which can be written as $1 - α^2 - βγ = 0$.

The correct option is (C).

Given:

Matrix A is both symmetric and skew-symmetric.

To Determine:

The type of matrix A.

Solution:

Since matrix A is symmetric, by definition, its transpose is equal to the matrix itself:

Since matrix A is skew-symmetric, by definition, its transpose is equal to the negative of the matrix itself:

From equations (1) and (2), we have:

$A = -A$

Add A to both sides of the equation:

$A + A = -A + A$

$2A = O$

where O is the zero matrix of the same order as A. Multiply by $\frac{1}{2}$:

$\frac{1}{2}(2A) = \frac{1}{2}O$

$A = O$

This means that every element of matrix A must be 0.

Therefore, the matrix A must be a zero matrix.

Let's examine the given options:

(A) A is a diagonal matrix: A zero matrix is a special case of a diagonal matrix where all diagonal elements are zero. However, not all diagonal matrices are zero matrices.

(B) A is a zero matrix: This matches our finding.

(C) A is a square matrix: Symmetric and skew-symmetric matrices must be square matrices. So, A is indeed a square matrix. However, this condition alone does not restrict the matrix to being both symmetric and skew-symmetric.

(D) None of these: Incorrect, as option (B) is correct.

The condition that A is both symmetric and skew-symmetric uniquely determines that A must be the zero matrix.

The correct option is (B).

Given:

A is a square matrix such that $A^2 = A$. Such a matrix is called an idempotent matrix.

To Compute:

The expression $(I + A)^3 – 7 A$, where I is the identity matrix of the same order as A.

Solution:

We need to expand the term $(I + A)^3$. We can use the binomial expansion formula, or simply multiply it out:

$(I + A)^3 = (I + A)(I + A)(I + A)$

First, calculate $(I + A)^2$:

$(I + A)^2 = (I + A)(I + A) = I \cdot I + I \cdot A + A \cdot I + A \cdot A$

Using the properties of the identity matrix ($I \cdot I = I$, $I \cdot A = A$, $A \cdot I = A$) and the given condition ($A \cdot A = A^2 = A$):

$(I + A)^2 = I + A + A + A = I + 3A$

Now, calculate $(I + A)^3 = (I + A)^2 (I + A) = (I + 3A)(I + A)$:

$(I + A)^3 = I \cdot I + I \cdot A + 3A \cdot I + 3A \cdot A$

Using the properties of the identity matrix and the given condition $A^2 = A$:

$(I + A)^3 = I + A + 3A + 3A^2 = I + 4A + 3A$

$(I + A)^3 = I + 7A$

Now, substitute this result into the given expression $(I + A)^3 – 7 A$:

$(I + A)^3 – 7 A = (I + 7A) - 7A$

$(I + A)^3 – 7 A = I + 7A - 7A$

$(I + A)^3 – 7 A = I + (7A - 7A)$

$(I + A)^3 – 7 A = I + O$

$(I + A)^3 – 7 A = I$

The expression simplifies to the identity matrix I.

Let's examine the given options:

(A) A

(B) I – A

(C) I: This matches our result.

(D) 3A

The correct option is (C).