Chapter 4 Determinants

We are asked to evaluate the determinant: $\begin{vmatrix} 2& 4\\-1& 2 \end{vmatrix}$

The determinant of a 2x2 matrix $\begin{vmatrix} a& b\\c& d \end{vmatrix}$ is given by the formula $ad - bc$.

Applying this formula to the given determinant, we have $a=2$, $b=4$, $c=-1$, and $d=2$.

So, the evaluation is: $(2)(2) - (4)(-1)$

Calculating the expression:

$4 - (-4)$

$4 + 4$

$8$

Thus, the value of the determinant is 8.

$\begin{vmatrix} 2& 4\\-1& 2 \end{vmatrix} = 8$

We are asked to evaluate the determinant:

$\begin{vmatrix} x& x+1\\x-1& x \end{vmatrix}$

The determinant of a $2 \times 2$ matrix $\begin{vmatrix} a& b\\c& d \end{vmatrix}$ is calculated using the formula: $ad - bc$

Applying this formula to the given determinant, we identify $a=x$, $b=x+1$, $c=x-1$, and $d=x$.

So, the evaluation becomes:

$(x)(x) - (x+1)(x-1)$

Now, we simplify the expression:

$= x^2 - (x^2 - 1^2)$

$= x^2 - (x^2 - 1)$

$= x^2 - x^2 + 1$

$= 1$

Thus, the value of the determinant is 1.

We are asked to evaluate the determinant:

$\Delta = \begin{vmatrix} 1& 2& 4\\-1& 3& 0\\4& 1& 0 \end{vmatrix}$

We can evaluate this determinant by expanding along any row or column. Expanding along the third column (C3) is convenient because it contains two zero entries.

The formula for expansion along the third column is: $\Delta = a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33}$

where $a_{ij}$ are the elements and $C_{ij}$ are the cofactors.

The elements in the third column are $a_{13}=4$, $a_{23}=0$, and $a_{33}=0$.

The cofactors are given by $C_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor determinant.

Since $a_{23}=0$ and $a_{33}=0$, we only need to calculate $C_{13}$.

$C_{13} = (-1)^{1+3}M_{13} = (-1)^4 M_{13} = 1 \cdot M_{13}$

$M_{13}$ is the determinant of the $2 \times 2$ matrix obtained by deleting the 1st row and 3rd column of the original matrix:

$M_{13} = \begin{vmatrix} -1& 3\\4& 1 \end{vmatrix}$

Evaluating the $2 \times 2$ determinant: $M_{13} = (-1)(1) - (3)(4)$

$M_{13} = -1 - 12$

$M_{13} = -13$

Now, we calculate $C_{13}$:

$C_{13} = 1 \cdot M_{13} = 1 \cdot (-13) = -13$

Finally, we calculate the determinant $\Delta$ using the expansion formula:

$\Delta = a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33}$

$\Delta = (4)(-13) + (0)(C_{23}) + (0)(C_{33})$

$\Delta = -52 + 0 + 0$

$\Delta = -52$

The value of the determinant is -52.

We are asked to evaluate the determinant:

$\Delta = \begin{vmatrix} 0& \sin \alpha & -\cos \alpha \\ -\sin \alpha&0&\sin \beta \\ \cos\alpha&-\sin\beta&0 \end{vmatrix}$

We can evaluate this determinant by expanding along any row or column. Let's expand along the first row ($R_1$).

The expansion of a $3 \times 3$ determinant $\begin{vmatrix} a_{11}& a_{12}& a_{13}\\a_{21}& a_{22}& a_{23}\\a_{31}& a_{32}& a_{33} \end{vmatrix}$ along the first row is given by: $\Delta = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$

where $a_{ij}$ are the elements and $C_{ij}$ are the cofactors.

Here, $a_{11}=0$, $a_{12}=\sin \alpha$, and $a_{13}=-\cos \alpha$.

The cofactor $C_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor determinant obtained by deleting the $i$-th row and $j$-th column.

Let's calculate the terms:

The first term is $a_{11}C_{11}$: $a_{11} = 0$ So, $a_{11}C_{11} = 0 \times C_{11} = 0$.

The second term is $a_{12}C_{12}$: $a_{12} = \sin \alpha$ $C_{12} = (-1)^{1+2} M_{12} = -M_{12}$

$M_{12}$ is the determinant of the matrix obtained by removing the 1st row and 2nd column:

$M_{12} = \begin{vmatrix} -\sin \alpha & \sin \beta \\ \cos \alpha & 0 \end{vmatrix}$

Evaluating $M_{12}$: $M_{12} = (-\sin \alpha)(0) - (\sin \beta)(\cos \alpha) = 0 - \sin \beta \cos \alpha = -\sin \beta \cos \alpha$

So, $C_{12} = -(-\sin \beta \cos \alpha) = \sin \beta \cos \alpha$.

The second term is $a_{12}C_{12} = (\sin \alpha)(\sin \beta \cos \alpha) = \sin \alpha \sin \beta \cos \alpha$.

The third term is $a_{13}C_{13}$: $a_{13} = -\cos \alpha$ $C_{13} = (-1)^{1+3} M_{13} = M_{13}$

$M_{13}$ is the determinant of the matrix obtained by removing the 1st row and 3rd column:

$M_{13} = \begin{vmatrix} -\sin \alpha & 0 \\ \cos \alpha & -\sin \beta \end{vmatrix}$

Evaluating $M_{13}$: $M_{13} = (-\sin \alpha)(-\sin \beta) - (0)(\cos \alpha) = \sin \alpha \sin \beta - 0 = \sin \alpha \sin \beta$

So, $C_{13} = \sin \alpha \sin \beta$.

The third term is $a_{13}C_{13} = (-\cos \alpha)(\sin \alpha \sin \beta) = -\cos \alpha \sin \alpha \sin \beta$.

Now, we sum the terms to find the determinant $\Delta$:

$\Delta = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$

$\Delta = 0 + (\sin \alpha \sin \beta \cos \alpha) + (-\cos \alpha \sin \alpha \sin \beta)$

$\Delta = \sin \alpha \sin \beta \cos \alpha - \sin \alpha \sin \beta \cos \alpha$

$\Delta = 0$

Thus, the value of the determinant is 0.

We are given the equation involving two determinants:

$\begin{vmatrix} 3& x\\x& 1 \end{vmatrix}$ = $\begin{vmatrix} 3& 2\\4& 1 \end{vmatrix}$

First, we evaluate the determinant on the left side of the equation. For a $2 \times 2$ matrix $\begin{vmatrix} a& b\\c& d \end{vmatrix}$, the determinant is $ad - bc$.

Left side determinant: $\begin{vmatrix} 3& x\\x& 1 \end{vmatrix} = (3)(1) - (x)(x) = 3 - x^2$

Next, we evaluate the determinant on the right side of the equation.

Right side determinant: $\begin{vmatrix} 3& 2\\4& 1 \end{vmatrix} = (3)(1) - (2)(4) = 3 - 8 = -5$

Now, we set the two evaluated determinants equal to each other, as given in the original equation:

$3 - x^2 = -5$

We solve this equation for $x$:

Subtract 3 from both sides:

$3 - x^2 - 3 = -5 - 3$

$-x^2 = -8$

Multiply both sides by -1:

$(-1)(-x^2) = (-1)(-8)$

$x^2 = 8$

Take the square root of both sides. Remember that the square root of a positive number has both a positive and a negative solution.

$x = \pm \sqrt{8}$

Simplify the square root:

$x = \pm \sqrt{4 \times 2}$

$x = \pm \sqrt{4} \times \sqrt{2}$

$x = \pm 2\sqrt{2}$

Thus, the values of $x$ for which the given equation holds are $\pm 2\sqrt{2}$.

The values of $x$ are $\mathbf{2\sqrt{2}}$ and $\mathbf{-2\sqrt{2}}$.

We need to evaluate the determinant:

$\begin{vmatrix} 2& 4\\-5& -1 \end{vmatrix}$

The determinant of a $2 \times 2$ matrix $\begin{vmatrix} a& b\\c& d \end{vmatrix}$ is given by the formula $ad - bc$.

Using this formula for the given determinant, we have $a=2$, $b=4$, $c=-5$, and $d=-1$.

The evaluation is:

$(2)(-1) - (4)(-5)$

Calculating the value:

$= -2 - (-20)$

$= -2 + 20$

$= 18$

Thus, the value of the determinant is 18.

$\begin{vmatrix} 2& 4\\-5& -1 \end{vmatrix} = 18$

We need to evaluate the determinants given in parts (i) and (ii).

Part (i)

Evaluate $\begin{vmatrix} \cos\theta& -\sin\theta\\\sin\theta& \cos\theta \end{vmatrix}$.

The determinant of a $2 \times 2$ matrix $\begin{vmatrix} a& b\\c& d \end{vmatrix}$ is calculated as $ad - bc$.

Applying the formula:

$(\cos\theta)(\cos\theta) - (-\sin\theta)(\sin\theta)$

$= \cos^2\theta - (-\sin^2\theta)$

$= \cos^2\theta + \sin^2\theta$

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

$= 1$

The value of the determinant is 1.

Part (ii)

Evaluate $\begin{vmatrix} x^2−x+1& x−1\\x+1& x+1 \end{vmatrix}$.

Using the determinant formula for a $2 \times 2$ matrix, $ad - bc$:

$a = x^2-x+1$, $b = x-1$, $c = x+1$, $d = x+1$

The determinant is:

$(x^2-x+1)(x+1) - (x-1)(x+1)$

Expanding the terms:

The first term is $(x^2-x+1)(x+1)$. This is a known algebraic identity for the sum of cubes, $a^3+b^3 = (a+b)(a^2-ab+b^2)$. Here, $a=x$ and $b=1$.

So, $(x^2-x+1)(x+1) = x^3 + 1^3 = x^3+1$.

The second term is $(x-1)(x+1)$. This is a known algebraic identity for the difference of squares, $(a-b)(a+b) = a^2-b^2$. Here, $a=x$ and $b=1$.

So, $(x-1)(x+1) = x^2 - 1^2 = x^2 - 1$.

Substitute these expanded forms back into the determinant expression:

$(x^3+1) - (x^2-1)$

Simplify the expression:

$= x^3 + 1 - x^2 + 1$

$= x^3 - x^2 + 2$

The value of the determinant is $\mathbf{x^3 - x^2 + 2}$.

We are given the matrix $A = \begin{bmatrix} 1& 2\\4& 2 \end{bmatrix}$.

We need to show that $|2A| = 4|A|$.

Step 1: Calculate $|A|$

The determinant of a $2 \times 2$ matrix $\begin{bmatrix} a& b\\c& d \end{bmatrix}$ is $ad - bc$.

For matrix $A$, we have $a=1$, $b=2$, $c=4$, and $d=2$.

$|A| = (1)(2) - (2)(4)$

$|A| = 2 - 8$

$|A| = -6$

Step 2: Calculate $2A$

To find $2A$, we multiply each element of matrix $A$ by 2.

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

$2A = \begin{bmatrix} 2& 4\\8& 4 \end{bmatrix}$

Step 3: Calculate $|2A|$

Now, we find the determinant of the matrix $2A$. For this matrix, we have $a=2$, $b=4$, $c=8$, and $d=4$.

$|2A| = (2)(4) - (4)(8)$

$|2A| = 8 - 32$

$|2A| = -24$

Step 4: Compare $|2A|$ and $4|A|$

We found $|2A| = -24$.

We need to calculate $4|A|$.

$4|A| = 4 \times (-6)$

$4|A| = -24$

Comparing the values:

$|2A| = -24$

$4|A| = -24$

Since $-24 = -24$, we have $|2A| = 4|A|$.

Hence, it is shown that $|2A| = 4|A|$ for the given matrix $A$.

We are given the matrix $A = \begin{bmatrix} 1& 0& 1\\0& 1& 2\\0& 0& 4 \end{bmatrix}$.

We need to show that $|3A| = 27|A|$.

Step 1: Calculate $|A|$

We can evaluate the determinant of matrix $A$ by expanding along the first column ($C_1$), as it contains two zero entries.

$|A| = a_{11}C_{11} + a_{21}C_{21} + a_{31}C_{31}$

The elements in the first column are $a_{11}=1$, $a_{21}=0$, and $a_{31}=0$.

So, $|A| = 1 \cdot C_{11} + 0 \cdot C_{21} + 0 \cdot C_{31} = C_{11}$.

The cofactor $C_{11}$ is $(-1)^{1+1}M_{11} = M_{11}$.

$M_{11}$ is the determinant of the matrix obtained by deleting the 1st row and 1st column of $A$:

$M_{11} = \begin{vmatrix} 1& 2\\0& 4 \end{vmatrix}$

Evaluating the $2 \times 2$ determinant: $M_{11} = (1)(4) - (2)(0) = 4 - 0 = 4$

So, $|A| = M_{11} = 4$.

Step 2: Calculate $3A$

To find $3A$, we multiply each element of matrix $A$ by 3.

$3A = 3 \times \begin{bmatrix} 1& 0& 1\\0& 1& 2\\0& 0& 4 \end{bmatrix} = \begin{bmatrix} 3 \times 1& 3 \times 0& 3 \times 1\\3 \times 0& 3 \times 1& 3 \times 2\\3 \times 0& 3 \times 0& 3 \times 4 \end{bmatrix}$

$3A = \begin{bmatrix} 3& 0& 3\\0& 3& 6\\0& 0& 12 \end{bmatrix}$

Step 3: Calculate $|3A|$

We find the determinant of the matrix $3A$. We can again expand along the first column ($C_1$).

$|3A| = a_{11}C_{11} + a_{21}C_{21} + a_{31}C_{31}$ (elements of $3A$)

The elements in the first column of $3A$ are $a_{11}=3$, $a_{21}=0$, and $a_{31}=0$.

So, $|3A| = 3 \cdot C_{11} + 0 \cdot C_{21} + 0 \cdot C_{31} = 3 \cdot C_{11}$.

The cofactor $C_{11}$ for the matrix $3A$ is $(-1)^{1+1}M_{11} = M_{11}$.

$M_{11}$ is the determinant of the matrix obtained by deleting the 1st row and 1st column of $3A$:

$M_{11} = \begin{vmatrix} 3& 6\\0& 12 \end{vmatrix}$

Evaluating the $2 \times 2$ determinant: $M_{11} = (3)(12) - (6)(0) = 36 - 0 = 36$

So, $|3A| = 3 \cdot M_{11} = 3 \cdot 36 = 108$.

Step 4: Calculate $27|A|$

We found $|A| = 4$. Now we calculate $27|A|$.

$27|A| = 27 \times 4$

$27 \times 4 = 108$

Step 5: Compare $|3A|$ and $27|A|$

We found $|3A| = 108$.

We found $27|A| = 108$.

Since $108 = 108$, we have $|3A| = 27|A|$.

Hence, it is shown that $|3A| = 27|A|$ for the given matrix $A$.

Alternate Solution: Using properties of Determinants

For any square matrix $A$ of order $n$ and any scalar $k$, the property of determinants states that $|kA| = k^n |A|$.

In this question, the matrix $A$ is of order $3 \times 3$, so $n=3$. The scalar is $k=3$.

According to the property, $|3A| = 3^3 |A|$.

$3^3 = 3 \times 3 \times 3 = 27$.

Therefore, $|3A| = 27|A|$.

To verify this property for the given matrix, we calculated $|A|=4$ and $|3A|=108$.

And $27|A| = 27 \times 4 = 108$.

Since $108 = 108$, the relationship $|3A| = 27|A|$ is confirmed for the given matrix.

We need to evaluate the given determinants.

Part (i)

Evaluate $\begin{vmatrix} 3& -1& -2\\0& 0& -1\\3& -5& 0 \end{vmatrix}$.

We can evaluate the determinant by expanding along the second row ($R_2$), as it contains two zero entries. The elements of the second row are $a_{21}=0$, $a_{22}=0$, and $a_{23}=-1$.

The formula for expansion along $R_2$ is: $\Delta = a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23}$

Since $a_{21}=0$ and $a_{22}=0$, the expression simplifies to:

$\Delta = (0)C_{21} + (0)C_{22} + (-1)C_{23} = -C_{23}$

Now we calculate the cofactor $C_{23}$. The cofactor $C_{ij} = (-1)^{i+j}M_{ij}$.

$C_{23} = (-1)^{2+3}M_{23} = (-1)^5 M_{23} = -M_{23}$

$M_{23}$ is the minor determinant obtained by deleting the 2nd row and 3rd column of the original matrix:

$M_{23} = \begin{vmatrix} 3& -1\\3& -5 \end{vmatrix}$

Evaluating the $2 \times 2$ determinant:

$M_{23} = (3)(-5) - (-1)(3) = -15 - (-3) = -15 + 3 = -12$

Now we find $C_{23}$:

$C_{23} = -M_{23} = -(-12) = 12$

Finally, we calculate the determinant $\Delta$:

$\Delta = -C_{23} = -(12) = -12$

The value of the determinant is $\mathbf{-12}$.

Part (ii)

Evaluate $\begin{vmatrix} 3& -4& 5\\1& 1& -2\\2& 3& 1 \end{vmatrix}$.

We can expand the determinant along the first row ($R_1$). The elements are $a_{11}=3$, $a_{12}=-4$, and $a_{13}=5$.

The formula for expansion along $R_1$ is: $\Delta = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$

Calculate the cofactors $C_{11}$, $C_{12}$, and $C_{13}$.

$C_{11} = (-1)^{1+1}M_{11} = M_{11}$

$M_{11} = \begin{vmatrix} 1& -2\\3& 1 \end{vmatrix} = (1)(1) - (-2)(3) = 1 - (-6) = 1 + 6 = 7$

$C_{11} = 7$

$C_{12} = (-1)^{1+2}M_{12} = -M_{12}$

$M_{12} = \begin{vmatrix} 1& -2\\2& 1 \end{vmatrix} = (1)(1) - (-2)(2) = 1 - (-4) = 1 + 4 = 5$

$C_{12} = -(5) = -5$

$C_{13} = (-1)^{1+3}M_{13} = M_{13}$

$M_{13} = \begin{vmatrix} 1& 1\\2& 3 \end{vmatrix} = (1)(3) - (1)(2) = 3 - 2 = 1$

$C_{13} = 1$

Now substitute the values into the expansion formula:

$\Delta = (3)(C_{11}) + (-4)(C_{12}) + (5)(C_{13})$

$\Delta = (3)(7) + (-4)(-5) + (5)(1)$

$\Delta = 21 + 20 + 5$

$\Delta = 46$

The value of the determinant is $\mathbf{46}$.

Part (iii)

Evaluate $\begin{vmatrix} 0& 1& 2\\-1& 0& -3\\-2& 3& 0 \end{vmatrix}$.

Let's expand the determinant along the first row ($R_1$), which contains a zero. The elements are $a_{11}=0$, $a_{12}=1$, and $a_{13}=2$.

The expansion formula along $R_1$ is: $\Delta = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$

Since $a_{11}=0$, the expression simplifies to:

$\Delta = (0)C_{11} + (1)C_{12} + (2)C_{13} = C_{12} + 2C_{13}$

Calculate the cofactors $C_{12}$ and $C_{13}$.

$C_{12} = (-1)^{1+2}M_{12} = -M_{12}$

$M_{12}$ is the minor determinant obtained by deleting the 1st row and 2nd column:

$M_{12} = \begin{vmatrix} -1& -3\\-2& 0 \end{vmatrix} = (-1)(0) - (-3)(-2) = 0 - 6 = -6$

$C_{12} = -(-6) = 6$

$C_{13} = (-1)^{1+3}M_{13} = M_{13}$

$M_{13}$ is the minor determinant obtained by deleting the 1st row and 3rd column:

$M_{13} = \begin{vmatrix} -1& 0\\-2& 3 \end{vmatrix} = (-1)(3) - (0)(-2) = -3 - 0 = -3$

$C_{13} = -3$

Now substitute the values into the simplified expansion formula:

$\Delta = C_{12} + 2C_{13}$

$\Delta = (6) + 2(-3)$

$\Delta = 6 - 6$

$\Delta = 0$

The value of the determinant is $\mathbf{0}$.

Part (iv)

Evaluate $\begin{vmatrix} 2& -1& -2\\0& 2& -1\\3& -5& 0 \end{vmatrix}$.

We can expand the determinant along the second row ($R_2$), as it contains a zero. The elements are $a_{21}=0$, $a_{22}=2$, and $a_{23}=-1$.

The expansion formula along $R_2$ is: $\Delta = a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23}$

Since $a_{21}=0$, the expression simplifies to:

$\Delta = (0)C_{21} + (2)C_{22} + (-1)C_{23} = 2C_{22} - C_{23}$

Calculate the cofactors $C_{22}$ and $C_{23}$.

$C_{22} = (-1)^{2+2}M_{22} = M_{22}$

$M_{22}$ is the minor determinant obtained by deleting the 2nd row and 2nd column:

$M_{22} = \begin{vmatrix} 2& -2\\3& 0 \end{vmatrix} = (2)(0) - (-2)(3) = 0 - (-6) = 6$

$C_{22} = 6$

$C_{23} = (-1)^{2+3}M_{23} = -M_{23}$

$M_{23}$ is the minor determinant obtained by deleting the 2nd row and 3rd column:

$M_{23} = \begin{vmatrix} 2& -1\\3& -5 \end{vmatrix} = (2)(-5) - (-1)(3) = -10 - (-3) = -10 + 3 = -7$

$C_{23} = -(-7) = 7$

Now substitute the values into the simplified expansion formula:

$\Delta = 2C_{22} - C_{23}$

$\Delta = 2(6) - (7)$

$\Delta = 12 - 7$

$\Delta = 5$

The value of the determinant is $\mathbf{5}$.

We are given the matrix $A = \begin{bmatrix} 1& 1& -2\\2& 1& -3\\5& 4& -9 \end{bmatrix}$.

We need to find the determinant $|A|$.

We will evaluate the determinant by expanding along the first row ($R_1$). The elements of the first row are $a_{11}=1$, $a_{12}=1$, and $a_{13}=-2$.

The formula for expansion along $R_1$ is: $|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$

where $C_{ij} = (-1)^{i+j}M_{ij}$ is the cofactor and $M_{ij}$ is the minor determinant.

Calculate the minors and cofactors:

For $a_{11}=1$:

$M_{11} = \begin{vmatrix} 1& -3\\4& -9 \end{vmatrix} = (1)(-9) - (-3)(4) = -9 - (-12) = -9 + 12 = 3$

$C_{11} = (-1)^{1+1}M_{11} = (1)(3) = 3$

For $a_{12}=1$:

$M_{12} = \begin{vmatrix} 2& -3\\5& -9 \end{vmatrix} = (2)(-9) - (-3)(5) = -18 - (-15) = -18 + 15 = -3$

$C_{12} = (-1)^{1+2}M_{12} = (-1)(-3) = 3$

For $a_{13}=-2$:

$M_{13} = \begin{vmatrix} 2& 1\\5& 4 \end{vmatrix} = (2)(4) - (1)(5) = 8 - 5 = 3$

$C_{13} = (-1)^{1+3}M_{13} = (1)(3) = 3$

Substitute the values of the elements and cofactors into the determinant formula:

$|A| = (1)(C_{11}) + (1)(C_{12}) + (-2)(C_{13})$

$|A| = (1)(3) + (1)(3) + (-2)(3)$

$|A| = 3 + 3 - 6$

$|A| = 6 - 6$

$|A| = 0$

The value of the determinant $|A|$ is $\mathbf{0}$.

We need to find the value(s) of $x$ for the given equations involving determinants.

Part (i)

Given equation: $\begin{vmatrix} 2& 4\\5& 1 \end{vmatrix}$ = $\begin{vmatrix} 2x& 4\\6& x \end{vmatrix}$

Evaluate the determinant on the left side:

$\begin{vmatrix} 2& 4\\5& 1 \end{vmatrix} = (2)(1) - (4)(5)$

$= 2 - 20$

$= -18$

Evaluate the determinant on the right side:

$\begin{vmatrix} 2x& 4\\6& x \end{vmatrix} = (2x)(x) - (4)(6)$

$= 2x^2 - 24$

Equate the two determinants:

$-18 = 2x^2 - 24$

Solve for $x$:

Add 24 to both sides:

$-18 + 24 = 2x^2$

$6 = 2x^2$

Divide both sides by 2:

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

$3 = x^2$

Take the square root of both sides:

$x = \pm\sqrt{3}$

The values of $x$ are $\mathbf{\sqrt{3}}$ and $\mathbf{-\sqrt{3}}$.

Part (ii)

Given equation: $\begin{vmatrix} 2& 3\\4& 5 \end{vmatrix}$ = $\begin{vmatrix} x& 3\\2x& 5 \end{vmatrix}$

Evaluate the determinant on the left side:

$\begin{vmatrix} 2& 3\\4& 5 \end{vmatrix} = (2)(5) - (3)(4)$

$= 10 - 12$

$= -2$

Evaluate the determinant on the right side:

$\begin{vmatrix} x& 3\\2x& 5 \end{vmatrix} = (x)(5) - (3)(2x)$

$= 5x - 6x$

$= -x$

Equate the two determinants:

$-2 = -x$

Multiply both sides by -1:

$(-1)(-2) = (-1)(-x)$

$2 = x$

The value of $x$ is $\mathbf{2}$.

We are given the equation:

$\begin{vmatrix} x& 2\\18& x \end{vmatrix}$ = $\begin{vmatrix} 6& 2\\18& 6 \end{vmatrix}$

We need to evaluate the determinant on the left side of the equation. The determinant of a $2 \times 2$ matrix $\begin{vmatrix} a& b\\c& d \end{vmatrix}$ is $ad - bc$.

Left side determinant: $\begin{vmatrix} x& 2\\18& x \end{vmatrix} = (x)(x) - (2)(18)$

$= x^2 - 36$

Next, we evaluate the determinant on the right side of the equation.

Right side determinant: $\begin{vmatrix} 6& 2\\18& 6 \end{vmatrix} = (6)(6) - (2)(18)$

$= 36 - 36$

$= 0$

Now, we set the two evaluated determinants equal to each other, as given in the original equation:

$x^2 - 36 = 0$

We solve this equation for $x$:

Add 36 to both sides:

$x^2 = 36$

Take the square root of both sides. Remember that the square root of a positive number has both a positive and a negative solution.

$x = \pm\sqrt{36}$

$x = \pm 6$

Thus, the values of $x$ are 6 and -6.

Looking at the given options:

(A) 6

(B) $\pm 6$

(C) – 6

(D) 0

The solution $x = \pm 6$ matches option (B).

The correct option is $\mathbf{(B) \pm 6}$.

Let the vertices of the triangle be A(3, 8), B(–4, 2), and C(5, 1).

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

Area $= \frac{1}{2} |\begin{vmatrix} x_1& y_1& 1\\x_2& y_2& 1\\x_3& y_3& 1 \end{vmatrix}|$

Substitute the coordinates of the vertices into the determinant:

Area $= \frac{1}{2} |\begin{vmatrix} 3& 8& 1\\-4& 2& 1\\5& 1& 1 \end{vmatrix}|$

Evaluate the determinant by expanding along the first row ($R_1$). The elements are $a_{11}=3$, $a_{12}=8$, and $a_{13}=1$. The expansion formula along $R_1$ is:

$\Delta = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$

where $C_{ij} = (-1)^{i+j}M_{ij}$ is the cofactor and $M_{ij}$ is the minor determinant.

Calculate the minors and cofactors:

For $a_{11}=3$:

$M_{11} = \begin{vmatrix} 2& 1\\1& 1 \end{vmatrix} = (2)(1) - (1)(1) = 2 - 1 = 1$

$C_{11} = (-1)^{1+1}M_{11} = (1)(1) = 1$

For $a_{12}=8$:

$M_{12} = \begin{vmatrix} -4& 1\\5& 1 \end{vmatrix} = (-4)(1) - (1)(5) = -4 - 5 = -9$

$C_{12} = (-1)^{1+2}M_{12} = (-1)(-9) = 9$

For $a_{13}=1$:

$M_{13} = \begin{vmatrix} -4& 2\\5& 1 \end{vmatrix} = (-4)(1) - (2)(5) = -4 - 10 = -14$

$C_{13} = (-1)^{1+3}M_{13} = (1)(-14) = -14$

Substitute the values into the determinant expansion:

$\Delta = (3)(C_{11}) + (8)(C_{12}) + (1)(C_{13})$

$\Delta = (3)(1) + (8)(9) + (1)(-14)$

$\Delta = 3 + 72 - 14$

$\Delta = 75 - 14$

$\Delta = 61$

Now calculate the area using the formula:

Area $= \frac{1}{2} |\Delta| = \frac{1}{2} |61| = \frac{61}{2}$

The area of the triangle is $\mathbf{\frac{61}{2}}$ square units.

This problem consists of two parts.

Part 1: Find the equation of the line joining A(1, 3) and B(0, 0) using determinants.

We know that three points are collinear if the area of the triangle formed by these points is zero.

Let the equation of the line joining A(1, 3) and B(0, 0) be represented by including a general point P(x, y) on the line.

Since the points A(1, 3), B(0, 0), and P(x, y) are collinear, the area of the triangle formed by these points is 0.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

Area $= \frac{1}{2} |\begin{vmatrix} x_1& y_1& 1\\x_2& y_2& 1\\x_3& y_3& 1 \end{vmatrix}|$

Substitute the coordinates A(1, 3), B(0, 0), and P(x, y) into the determinant formula and set the area to 0:

$\frac{1}{2} |\begin{vmatrix} 1& 3& 1\\0& 0& 1\\x& y& 1 \end{vmatrix}| = 0$

This implies that the determinant must be zero:

$\begin{vmatrix} 1& 3& 1\\0& 0& 1\\x& y& 1 \end{vmatrix} = 0$

Evaluate the determinant by expanding along the second row ($R_2$) for simplicity (due to zeros):

The expansion along $R_2$ is $a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23}$.

Here $a_{21}=0$, $a_{22}=0$, $a_{23}=1$.

So, $0 \cdot C_{21} + 0 \cdot C_{22} + 1 \cdot C_{23} = 0$, which means $C_{23} = 0$.

Calculate the cofactor $C_{23}$: $C_{23} = (-1)^{2+3}M_{23} = -M_{23}$

$M_{23}$ is the minor obtained by deleting the 2nd row and 3rd column:

$M_{23} = \begin{vmatrix} 1& 3\\x& y \end{vmatrix} = (1)(y) - (3)(x) = y - 3x$

So, $C_{23} = -(y - 3x) = 3x - y$.

Setting $C_{23} = 0$ gives the equation of the line:

$3x - y = 0$

or

$y = 3x$

The equation of the line joining A(1, 3) and B(0, 0) is $\mathbf{y = 3x}$ or $\mathbf{3x - y = 0}$.

Part 2: Find k if D(k, 0) is a point such that area of triangle ABD is 3 sq units.

The vertices of triangle ABD are A(1, 3), B(0, 0), and D(k, 0).

The area of triangle ABD is given as 3 square units.

Using the area formula with vertices A(1, 3), B(0, 0), and D(k, 0):

Area $= \frac{1}{2} |\begin{vmatrix} 1& 3& 1\\0& 0& 1\\k& 0& 1 \end{vmatrix}|$

Given Area = 3, so:

$\frac{1}{2} |\begin{vmatrix} 1& 3& 1\\0& 0& 1\\k& 0& 1 \end{vmatrix}| = 3$

$|\begin{vmatrix} 1& 3& 1\\0& 0& 1\\k& 0& 1 \end{vmatrix}| = 6$

Evaluate the determinant. Expanding along the second row ($R_2$) is again simplest:

Expansion along $R_2$: $a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23}$.

Here $a_{21}=0$, $a_{22}=0$, $a_{23}=1$.

So, the determinant value is $0 \cdot C_{21} + 0 \cdot C_{22} + 1 \cdot C_{23} = C_{23}$.

Calculate the cofactor $C_{23}$: $C_{23} = (-1)^{2+3}M_{23} = -M_{23}$

$M_{23}$ is the minor obtained by deleting the 2nd row and 3rd column of $\begin{vmatrix} 1& 3& 1\\0& 0& 1\\k& 0& 1 \end{vmatrix}$:

$M_{23} = \begin{vmatrix} 1& 3\\k& 0 \end{vmatrix} = (1)(0) - (3)(k) = 0 - 3k = -3k$

So, the determinant value is $C_{23} = -(-3k) = 3k$.

We have the equation from the area calculation:

$|3k| = 6$

This absolute value equation gives two possibilities:

Case 1: $3k = 6$

$k = \frac{6}{3}$

$k = 2$

Case 2: $3k = -6$

$k = \frac{-6}{3}$

$k = -2$

The possible values of k are $\mathbf{2}$ and $\mathbf{-2}$.

We need to find the area of the triangle with the given vertices in each part.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

Area $= \frac{1}{2} |\begin{vmatrix} x_1& y_1& 1\\x_2& y_2& 1\\x_3& y_3& 1 \end{vmatrix}|$

Part (i)

Vertices: (1, 0), (6, 0), (4, 3)

Let $(x_1, y_1) = (1, 0)$, $(x_2, y_2) = (6, 0)$, and $(x_3, y_3) = (4, 3)$.

The determinant is:

$\Delta = \begin{vmatrix} 1& 0& 1\\6& 0& 1\\4& 3& 1 \end{vmatrix}$

Expanding along the second column ($C_2$) (as it has two zeros):

$\Delta = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32}$

$\Delta = (0)C_{12} + (0)C_{22} + (3)C_{32}$

$\Delta = 3C_{32}$

Calculate $C_{32} = (-1)^{3+2}M_{32} = -M_{32}$.

$M_{32} = \begin{vmatrix} 1& 1\\6& 1 \end{vmatrix} = (1)(1) - (1)(6) = 1 - 6 = -5$

$C_{32} = -(-5) = 5$

So, $\Delta = 3 \times 5 = 15$.

The area is: Area $= \frac{1}{2} |\Delta| = \frac{1}{2} |15| = \frac{15}{2}$

The area of the triangle is $\mathbf{\frac{15}{2}}$ square units.

Part (ii)

Vertices: (2, 7), (1, 1), (10, 8)

Let $(x_1, y_1) = (2, 7)$, $(x_2, y_2) = (1, 1)$, and $(x_3, y_3) = (10, 8)$.

The determinant is:

$\Delta = \begin{vmatrix} 2& 7& 1\\1& 1& 1\\10& 8& 1 \end{vmatrix}$

Expanding along the first row ($R_1$):

$\Delta = 2C_{11} + 7C_{12} + 1C_{13}$

Calculate the cofactors:

$C_{11} = (-1)^{1+1}M_{11} = \begin{vmatrix} 1& 1\\8& 1 \end{vmatrix} = (1)(1) - (1)(8) = 1 - 8 = -7$

$C_{12} = (-1)^{1+2}M_{12} = -\begin{vmatrix} 1& 1\\10& 1 \end{vmatrix} = -((1)(1) - (1)(10)) = -(1 - 10) = -(-9) = 9$

$C_{13} = (-1)^{1+3}M_{13} = \begin{vmatrix} 1& 1\\10& 8 \end{vmatrix} = (1)(8) - (1)(10) = 8 - 10 = -2$

Substitute the cofactor values:

$\Delta = 2(-7) + 7(9) + 1(-2)$

$\Delta = -14 + 63 - 2$

$\Delta = 49 - 2 = 47$

The area is: Area $= \frac{1}{2} |\Delta| = \frac{1}{2} |47| = \frac{47}{2}$

The area of the triangle is $\mathbf{\frac{47}{2}}$ square units.

Part (iii)

Vertices: (–2, –3), (3, 2), (–1, –8)

Let $(x_1, y_1) = (-2, -3)$, $(x_2, y_2) = (3, 2)$, and $(x_3, y_3) = (-1, -8)$.

The determinant is:

$\Delta = \begin{vmatrix} -2& -3& 1\\3& 2& 1\\-1& -8& 1 \end{vmatrix}$

Expanding along the first row ($R_1$):

$\Delta = (-2)C_{11} + (-3)C_{12} + (1)C_{13}$

Calculate the cofactors:

$C_{11} = (-1)^{1+1}M_{11} = \begin{vmatrix} 2& 1\\-8& 1 \end{vmatrix} = (2)(1) - (1)(-8) = 2 - (-8) = 2 + 8 = 10$

$C_{12} = (-1)^{1+2}M_{12} = -\begin{vmatrix} 3& 1\\-1& 1 \end{vmatrix} = -((3)(1) - (1)(-1)) = -(3 - (-1)) = -(3 + 1) = -4$

$C_{13} = (-1)^{1+3}M_{13} = \begin{vmatrix} 3& 2\\-1& -8 \end{vmatrix} = (3)(-8) - (2)(-1) = -24 - (-2) = -24 + 2 = -22$

Substitute the cofactor values:

$\Delta = (-2)(10) + (-3)(-4) + (1)(-22)$

$\Delta = -20 + 12 - 22$

$\Delta = -8 - 22 = -30$

The area is: Area $= \frac{1}{2} |\Delta| = \frac{1}{2} |-30| = \frac{1}{2} \times 30 = 15$

The area of the triangle is $\mathbf{15}$ square units.

Given:

The three points are A (a, b + c), B (b, c + a), and C (c, a + b).

To Show:

Points A, B, and C are collinear.

Solution:

Points are collinear if and only if the area of the triangle formed by these points is zero.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

Area $= \frac{1}{2} |\begin{vmatrix} x_1& y_1& 1\\x_2& y_2& 1\\x_3& y_3& 1 \end{vmatrix}|$

Substitute the coordinates of points A(a, b+c), B(b, c+a), and C(c, a+b) into the determinant:

Area $= \frac{1}{2} |\begin{vmatrix} a& b+c& 1\\b& c+a& 1\\c& a+b& 1 \end{vmatrix}|$

To show that the points are collinear, we need to show that the determinant value is zero.

Let $\Delta = \begin{vmatrix} a& b+c& 1\\b& c+a& 1\\c& a+b& 1 \end{vmatrix}$.

We can use properties of determinants to evaluate this. Apply the column operation $C_1 \to C_1 + C_2$. This operation does not change the value of the determinant.

$\Delta = \begin{vmatrix} a + (b+c)& b+c& 1\\b + (c+a)& c+a& 1\\c + (a+b)& a+b& 1 \end{vmatrix}$

$\Delta = \begin{vmatrix} a+b+c& b+c& 1\\a+b+c& c+a& 1\\a+b+c& a+b& 1 \end{vmatrix}$

Now, take the common factor $(a+b+c)$ out from the first column ($C_1$).

$\Delta = (a+b+c) \begin{vmatrix} 1& b+c& 1\\1& c+a& 1\\1& a+b& 1 \end{vmatrix}$

Observe the resulting determinant matrix $\begin{vmatrix} 1& b+c& 1\\1& c+a& 1\\1& a+b& 1 \end{vmatrix}$. The first column ($C_1$) and the third column ($C_3$) are identical.

A property of determinants states that if any two columns (or rows) are identical, the value of the determinant is zero.

Thus, $\begin{vmatrix} 1& b+c& 1\\1& c+a& 1\\1& a+b& 1 \end{vmatrix} = 0$.

Substitute this value back into the expression for $\Delta$:

$\Delta = (a+b+c) \times 0 = 0$

Since the determinant $\Delta = 0$, the area of the triangle formed by points A, B, and C is:

Area $= \frac{1}{2} |\Delta| = \frac{1}{2} |0| = 0$

As the area of the triangle formed by points A, B, and C is 0, the points A, B, and C are collinear.

Hence, the points A (a, b + c), B (b, c + a), and C (c, a + b) are collinear.

Solution:

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Given that the area is 4 sq. units, we have:

$\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 4$

This implies:

$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = \pm 8$

(i) Vertices are (k, 0), (4, 0), (0, 2)

Let $(x_1, y_1) = (k, 0)$, $(x_2, y_2) = (4, 0)$, and $(x_3, y_3) = (0, 2)$.

Substituting these values into the expression $x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$:

$k(0 - 2) + 4(2 - 0) + 0(0 - 0)$

$= k(-2) + 4(2) + 0$

$= -2k + 8$

According to the area condition, we have:

$-2k + 8 = \pm 8$

We consider two cases:

Case 1: $-2k + 8 = 8$

$-2k = 8 - 8$

$-2k = 0$

$k = 0$

Case 2: $-2k + 8 = -8$

$-2k = -8 - 8$

$-2k = -16$

$k = \frac{-16}{-2}$

$k = 8$

Thus, the possible values of k for part (i) are 0 and 8.

(ii) Vertices are (–2, 0), (0, 4), (0, k)

Let $(x_1, y_1) = (-2, 0)$, $(x_2, y_2) = (0, 4)$, and $(x_3, y_3) = (0, k)$.

Substituting these values into the expression $x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$:

$-2(4 - k) + 0(k - 0) + 0(0 - 4)$

$= -2(4 - k) + 0 + 0$

$= -8 + 2k$

According to the area condition, we have:

$-8 + 2k = \pm 8$

We consider two cases:

Case 1: $-8 + 2k = 8$

$2k = 8 + 8$

$2k = 16$

$k = \frac{16}{2}$

$k = 8$

Case 2: $-8 + 2k = -8$

$2k = -8 + 8$

$2k = 0$

$k = 0$

Thus, the possible values of k for part (ii) are 8 and 0.

Solution:

The equation of the line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ can be found using the determinant formula. If a third point $(x, y)$ lies on the same line, then the three points are collinear, and the area of the triangle formed by these three points is zero.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

Area $= \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$

For the points $(x_1, y_1)$, $(x_2, y_2)$, and $(x, y)$ to be collinear, the area must be zero:

$\frac{1}{2} \begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0$

This implies:

$\begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0$

(i) Find equation of line joining (1, 2) and (3, 6) using determinants.

Let $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 6)$. Let $(x, y)$ be a general point on the line.

Using the determinant formula for collinear points:

$\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0$

Expanding the determinant along the first row:

$x \begin{vmatrix} 2 & 1 \\ 6 & 1 \end{vmatrix} - y \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 3 & 6 \end{vmatrix} = 0$

$x((2)(1) - (6)(1)) - y((1)(1) - (3)(1)) + 1((1)(6) - (3)(2)) = 0$

$x(2 - 6) - y(1 - 3) + 1(6 - 6) = 0$

$x(-4) - y(-2) + 1(0) = 0$

$-4x + 2y = 0$

Dividing by 2:

$-2x + y = 0$

Or:

$y = 2x$

The equation of the line joining (1, 2) and (3, 6) is $2x - y = 0$.

(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Let $(x_1, y_1) = (3, 1)$ and $(x_2, y_2) = (9, 3)$. Let $(x, y)$ be a general point on the line.

Using the determinant formula for collinear points:

$\begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0$

Expanding the determinant along the first row:

$x \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} - y \begin{vmatrix} 3 & 1 \\ 9 & 1 \end{vmatrix} + 1 \begin{vmatrix} 3 & 1 \\ 9 & 3 \end{vmatrix} = 0$

$x((1)(1) - (3)(1)) - y((3)(1) - (9)(1)) + 1((3)(3) - (9)(1)) = 0$

$x(1 - 3) - y(3 - 9) + 1(9 - 9) = 0$

$x(-2) - y(-6) + 1(0) = 0$

$-2x + 6y = 0$

Dividing by 2:

$-x + 3y = 0$

Or:

$x - 3y = 0$

The equation of the line joining (3, 1) and (9, 3) is $x - 3y = 0$.

Solution:

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

Area $= \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

Given vertices are $(x_1, y_1) = (2, -6)$, $(x_2, y_2) = (5, 4)$, and $(x_3, y_3) = (k, 4)$. The area is given as 35 sq units.

So, $35 = \frac{1}{2} \left| \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} \right|$

This means, the determinant value can be $\pm 2 \times 35 = \pm 70$.

$\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 70$

Expanding the determinant:

$2(4 \times 1 - 4 \times 1) - (-6)(5 \times 1 - k \times 1) + 1(5 \times 4 - k \times 4)$

$2(4 - 4) + 6(5 - k) + 1(20 - 4k)$

$2(0) + 30 - 6k + 20 - 4k$

$0 + 50 - 10k$

$50 - 10k$

So, we have the equation:

$50 - 10k = \pm 70$

We consider two cases:

Case 1: $50 - 10k = 70$

$-10k = 70 - 50$

$-10k = 20$

$k = \frac{20}{-10}$

$k = -2$

Case 2: $50 - 10k = -70$

$-10k = -70 - 50$

$-10k = -120$

$k = \frac{-120}{-10}$

$k = 12$

The possible values of k are 12 and -2.

Comparing with the given options, the correct option is (D).

The final answer is $\boxed{12, -2}$.

Solution:

The given determinant is $\Delta = \begin{vmatrix} 1& 2& 3\\4& 5& 6\\7& 8& 9 \end{vmatrix}$.

We need to find the minor of the element 6.

The element 6 is located in the second row and the third column of the determinant. Let's denote this element as $a_{23}$.

The minor of an element $a_{ij}$, denoted by $M_{ij}$, is the determinant of the square submatrix obtained by deleting the $i$-th row and $j$-th column of the original determinant.

To find the minor of element 6 ($a_{23}$), we delete the 2nd row and the 3rd column from the determinant $\Delta$.

Deleting the 2nd row and 3rd column leaves us with the submatrix: $\begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix}$.

The minor of element 6, $M_{23}$, is the determinant of this submatrix:

$M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix}$

Now, we calculate the value of this $2 \times 2$ determinant:

$M_{23} = (1 \times 8) - (2 \times 7)$

$M_{23} = 8 - 14$

$M_{23} = -6$

Therefore, the minor of element 6 in the given determinant is -6.

Solution:

The given determinant is $\Delta = \begin{vmatrix} 1& -2\\4& 3 \end{vmatrix}$.

The elements of the determinant are denoted by $a_{ij}$, where $i$ is the row number and $j$ is the column number.

$a_{11} = 1$, $a_{12} = -2$

$a_{21} = 4$, $a_{22} = 3$

The minor of an element $a_{ij}$, denoted by $M_{ij}$, is the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column.

The cofactor of an element $a_{ij}$, denoted by $C_{ij}$, is given by $C_{ij} = (-1)^{i+j} M_{ij}$.

Minor of element $a_{11} = 1$:

Deleting the 1st row and 1st column, the remaining element is 3.

$M_{11} = \det([3]) = 3$

Cofactor of element $a_{11} = 1$:

$C_{11} = (-1)^{1+1} M_{11} = (-1)^2 \times 3 = 1 \times 3 = 3$

Minor of element $a_{12} = -2$:

Deleting the 1st row and 2nd column, the remaining element is 4.

$M_{12} = \det([4]) = 4$

Cofactor of element $a_{12} = -2$:

$C_{12} = (-1)^{1+2} M_{12} = (-1)^3 \times 4 = -1 \times 4 = -4$

Minor of element $a_{21} = 4$:

Deleting the 2nd row and 1st column, the remaining element is -2.

$M_{21} = \det([-2]) = -2$

Cofactor of element $a_{21} = 4$:

$C_{21} = (-1)^{2+1} M_{21} = (-1)^3 \times (-2) = -1 \times (-2) = 2$

Minor of element $a_{22} = 3$:

Deleting the 2nd row and 2nd column, the remaining element is 1.

$M_{22} = \det([1]) = 1$

Cofactor of element $a_{22} = 3$:

$C_{22} = (-1)^{2+2} M_{22} = (-1)^4 \times 1 = 1 \times 1 = 1$

Summary of Minors and Cofactors:

Minor of 1 ($M_{11}$) = 3, Cofactor of 1 ($C_{11}$) = 3

Minor of -2 ($M_{12}$) = 4, Cofactor of -2 ($C_{12}$) = -4

Minor of 4 ($M_{21}$) = -2, Cofactor of 4 ($C_{21}$) = 2

Minor of 3 ($M_{22}$) = 1, Cofactor of 3 ($C_{22}$) = 1

Solution:

The given determinant is $\Delta = \begin{vmatrix} a_{11}& a_{12}& a_{13}\\a_{21}& a_{22}& a_{23}\\a_{31}& a_{32}& a_{33} \end{vmatrix}$.

We need to find the minors and cofactors of the elements $a_{11}$ and $a_{21}$.

The minor of an element $a_{ij}$, denoted by $M_{ij}$, is the determinant of the square submatrix obtained by deleting the $i$-th row and $j$-th column.

The cofactor of an element $a_{ij}$, denoted by $C_{ij}$, is given by $C_{ij} = (-1)^{i+j} M_{ij}$.

For element $a_{11}$:

Element $a_{11}$ is in the 1st row and 1st column.

To find the minor $M_{11}$, we delete the 1st row and 1st column:

The remaining submatrix is $\begin{vmatrix} a_{22}& a_{23} \\ a_{32}& a_{33} \end{vmatrix}$.

The minor $M_{11}$ is the determinant of this submatrix:

$M_{11} = \det \begin{vmatrix} a_{22}& a_{23} \\ a_{32}& a_{33} \end{vmatrix} = a_{22}a_{33} - a_{23}a_{32}$

The cofactor $C_{11}$ is calculated using the formula $C_{ij} = (-1)^{i+j} M_{ij}$:

$C_{11} = (-1)^{1+1} M_{11} = (-1)^2 (a_{22}a_{33} - a_{23}a_{32})$

$C_{11} = 1 \times (a_{22}a_{33} - a_{23}a_{32})$

$C_{11} = a_{22}a_{33} - a_{23}a_{32}$

For element $a_{21}$:

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

To find the minor $M_{21}$, we delete the 2nd row and 1st column:

The remaining submatrix is $\begin{vmatrix} a_{12}& a_{13} \\ a_{32}& a_{33} \end{vmatrix}$.

The minor $M_{21}$ is the determinant of this submatrix:

$M_{21} = \det \begin{vmatrix} a_{12}& a_{13} \\ a_{32}& a_{33} \end{vmatrix} = a_{12}a_{33} - a_{13}a_{32}$

The cofactor $C_{21}$ is calculated using the formula $C_{ij} = (-1)^{i+j} M_{ij}$:

$C_{21} = (-1)^{2+1} M_{21} = (-1)^3 (a_{12}a_{33} - a_{13}a_{32})$

$C_{21} = -1 \times (a_{12}a_{33} - a_{13}a_{32})$

$C_{21} = -(a_{12}a_{33} - a_{13}a_{32})$

$C_{21} = a_{13}a_{32} - a_{12}a_{33}$

Therefore:

Minor of $a_{11}$, $M_{11} = a_{22}a_{33} - a_{23}a_{32}$

Cofactor of $a_{11}$, $C_{11} = a_{22}a_{33} - a_{23}a_{32}$

Minor of $a_{21}$, $M_{21} = a_{12}a_{33} - a_{13}a_{32}$

Cofactor of $a_{21}$, $C_{21} = a_{13}a_{32} - a_{12}a_{33}$

Solution:

The given determinant is $\Delta = \begin{vmatrix} 2& -3& 5\\6& 0& 4\\1& 5& -7 \end{vmatrix}$.

The elements are $a_{11}=2, a_{12}=-3, a_{13}=5$, $a_{21}=6, a_{22}=0, a_{23}=4$, and $a_{31}=1, a_{32}=5, a_{33}=-7$.

The minor of an element $a_{ij}$, denoted by $M_{ij}$, is the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column.

The cofactor of an element $a_{ij}$, denoted by $A_{ij}$ (or $C_{ij}$), is given by $A_{ij} = (-1)^{i+j} M_{ij}$.

Calculating Minors ($M_{ij}$):

$M_{11} = \begin{vmatrix} 0 & 4 \\ 5 & -7 \end{vmatrix} = (0)(-7) - (4)(5) = 0 - 20 = -20$

$M_{12} = \begin{vmatrix} 6 & 4 \\ 1 & -7 \end{vmatrix} = (6)(-7) - (4)(1) = -42 - 4 = -46$

$M_{13} = \begin{vmatrix} 6 & 0 \\ 1 & 5 \end{vmatrix} = (6)(5) - (0)(1) = 30 - 0 = 30$

$M_{21} = \begin{vmatrix} -3 & 5 \\ 5 & -7 \end{vmatrix} = (-3)(-7) - (5)(5) = 21 - 25 = -4$

$M_{22} = \begin{vmatrix} 2 & 5 \\ 1 & -7 \end{vmatrix} = (2)(-7) - (5)(1) = -14 - 5 = -19$

$M_{23} = \begin{vmatrix} 2 & -3 \\ 1 & 5 \end{vmatrix} = (2)(5) - (-3)(1) = 10 - (-3) = 13$

$M_{31} = \begin{vmatrix} -3 & 5 \\ 0 & 4 \end{vmatrix} = (-3)(4) - (5)(0) = -12 - 0 = -12$

$M_{32} = \begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = (2)(4) - (5)(6) = 8 - 30 = -22$

$M_{33} = \begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix} = (2)(0) - (-3)(6) = 0 - (-18) = 18$

Calculating Cofactors ($A_{ij}$):

$A_{11} = (-1)^{1+1} M_{11} = (1)(-20) = -20$

$A_{12} = (-1)^{1+2} M_{12} = (-1)(-46) = 46$

$A_{13} = (-1)^{1+3} M_{13} = (1)(30) = 30$

$A_{21} = (-1)^{2+1} M_{21} = (-1)(-4) = 4$

$A_{22} = (-1)^{2+2} M_{22} = (1)(-19) = -19$

$A_{23} = (-1)^{2+3} M_{23} = (-1)(13) = -13$

$A_{31} = (-1)^{3+1} M_{31} = (1)(-12) = -12$

$A_{32} = (-1)^{3+2} M_{32} = (-1)(-22) = 22$

$A_{33} = (-1)^{3+3} M_{33} = (1)(18) = 18$

Verification:

We need to verify that $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33} = 0$.

The elements from the first row are $a_{11}=2$, $a_{12}=-3$, $a_{13}=5$.

The cofactors from the third row are $A_{31}=-12$, $A_{32}=22$, $A_{33}=18$.

Let's calculate the expression $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$:

$a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33} = (2)(-12) + (-3)(22) + (5)(18)$

$= -24 - 66 + 90$

$= -90 + 90$

$= 0$

Since the sum is 0, the property $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33} = 0$ is verified.

This demonstrates a general property: the sum of the products of elements of any row (or column) with the cofactors of the corresponding elements of *another* row (or column) is zero.

Solution:

For a determinant $\begin{vmatrix} a_{11}& a_{12} \\ a_{21}& a_{22} \end{vmatrix}$, the minors and cofactors of the elements are defined as follows:

Minor of $a_{ij}$ ($M_{ij}$) is the determinant obtained by deleting the $i$-th row and $j$-th column.

Cofactor of $a_{ij}$ ($A_{ij}$ or $C_{ij}$) is given by $A_{ij} = (-1)^{i+j} M_{ij}$.

(i) Determinant $\begin{vmatrix} 2& -4\\0& 3 \end{vmatrix}$

The elements are $a_{11}=2$, $a_{12}=-4$, $a_{21}=0$, and $a_{22}=3$.

Minor of $a_{11} = 2$:

Delete 1st row and 1st column. The remaining element is 3.

$M_{11} = 3$

Cofactor of $a_{11} = 2$:

$A_{11} = (-1)^{1+1} M_{11} = (-1)^2 (3) = 1 \times 3 = 3$

Minor of $a_{12} = -4$:

Delete 1st row and 2nd column. The remaining element is 0.

$M_{12} = 0$

Cofactor of $a_{12} = -4$:

$A_{12} = (-1)^{1+2} M_{12} = (-1)^3 (0) = -1 \times 0 = 0$

Minor of $a_{21} = 0$:

Delete 2nd row and 1st column. The remaining element is -4.

$M_{21} = -4$

Cofactor of $a_{21} = 0$:

$A_{21} = (-1)^{2+1} M_{21} = (-1)^3 (-4) = -1 \times (-4) = 4$

Minor of $a_{22} = 3$:

Delete 2nd row and 2nd column. The remaining element is 2.

$M_{22} = 2$

Cofactor of $a_{22} = 3$:

$A_{22} = (-1)^{2+2} M_{22} = (-1)^4 (2) = 1 \times 2 = 2$

(ii) Determinant $\begin{vmatrix} a& c\\b& d \end{vmatrix}$

The elements are $a_{11}=a$, $a_{12}=c$, $a_{21}=b$, and $a_{22}=d$.

Minor of $a_{11} = a$:

Delete 1st row and 1st column. The remaining element is d.

$M_{11} = d$

Cofactor of $a_{11} = a$:

$A_{11} = (-1)^{1+1} M_{11} = (-1)^2 (d) = 1 \times d = d$

Minor of $a_{12} = c$:

Delete 1st row and 2nd column. The remaining element is b.

$M_{12} = b$

Cofactor of $a_{12} = c$:

$A_{12} = (-1)^{1+2} M_{12} = (-1)^3 (b) = -1 \times b = -b$

Minor of $a_{21} = b$:

Delete 2nd row and 1st column. The remaining element is c.

$M_{21} = c$

Cofactor of $a_{21} = b$:

$A_{21} = (-1)^{2+1} M_{21} = (-1)^3 (c) = -1 \times c = -c$

Minor of $a_{22} = d$:

Delete 2nd row and 2nd column. The remaining element is a.

$M_{22} = a$

Cofactor of $a_{22} = d$:

$A_{22} = (-1)^{2+2} M_{22} = (-1)^4 (a) = 1 \times a = a$

Solution:

For a determinant $\begin{vmatrix} a_{11}& a_{12}& a_{13}\\a_{21}& a_{22}& a_{23}\\a_{31}& a_{32}& a_{33} \end{vmatrix}$, the minor of an element $a_{ij}$, denoted by $M_{ij}$, is the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column. The cofactor of an element $a_{ij}$, denoted by $A_{ij}$, is given by $A_{ij} = (-1)^{i+j} M_{ij}$.

(i) Determinant $\begin{vmatrix} 1& 0& 0\\0& 1& 0\\0& 0& 1 \end{vmatrix}$

The elements are $a_{11}=1, a_{12}=0, a_{13}=0, a_{21}=0, a_{22}=1, a_{23}=0, a_{31}=0, a_{32}=0, a_{33}=1$.

Minors:

$M_{11} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1$

$M_{12} = \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = (0)(1) - (0)(0) = 0$

$M_{13} = \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = (0)(0) - (1)(0) = 0$

$M_{21} = \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = (0)(1) - (0)(0) = 0$

$M_{22} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1$

$M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = (1)(0) - (0)(0) = 0$

$M_{31} = \begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} = (0)(0) - (0)(1) = 0$

$M_{32} = \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = (1)(0) - (0)(0) = 0$

$M_{33} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1$

Cofactors:

$A_{11} = (-1)^{1+1} M_{11} = (1)(1) = 1$

$A_{12} = (-1)^{1+2} M_{12} = (-1)(0) = 0$

$A_{13} = (-1)^{1+3} M_{13} = (1)(0) = 0$

$A_{21} = (-1)^{2+1} M_{21} = (-1)(0) = 0$

$A_{22} = (-1)^{2+2} M_{22} = (1)(1) = 1$

$A_{23} = (-1)^{2+3} M_{23} = (-1)(0) = 0$

$A_{31} = (-1)^{3+1} M_{31} = (1)(0) = 0$

$A_{32} = (-1)^{3+2} M_{32} = (-1)(0) = 0$

$A_{33} = (-1)^{3+3} M_{33} = (1)(1) = 1$

(ii) Determinant $\begin{vmatrix} 1& 0& 4\\3& 5& -1\\0& 1& 2 \end{vmatrix}$

The elements are $a_{11}=1, a_{12}=0, a_{13}=4, a_{21}=3, a_{22}=5, a_{23}=-1, a_{31}=0, a_{32}=1, a_{33}=2$.

Minors:

$M_{11} = \begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix} = (5)(2) - (-1)(1) = 10 - (-1) = 11$

$M_{12} = \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} = (3)(2) - (-1)(0) = 6 - 0 = 6$

$M_{13} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = (3)(1) - (5)(0) = 3 - 0 = 3$

$M_{21} = \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} = (0)(2) - (4)(1) = 0 - 4 = -4$

$M_{22} = \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} = (1)(2) - (4)(0) = 2 - 0 = 2$

$M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1 - 0 = 1$

$M_{31} = \begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix} = (0)(-1) - (4)(5) = 0 - 20 = -20$

$M_{32} = \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (4)(3) = -1 - 12 = -13$

$M_{33} = \begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix} = (1)(5) - (0)(3) = 5 - 0 = 5$

Cofactors:

$A_{11} = (-1)^{1+1} M_{11} = (1)(11) = 11$

$A_{12} = (-1)^{1+2} M_{12} = (-1)(6) = -6$

$A_{13} = (-1)^{1+3} M_{13} = (1)(3) = 3$

$A_{21} = (-1)^{2+1} M_{21} = (-1)(-4) = 4$

$A_{22} = (-1)^{2+2} M_{22} = (1)(2) = 2$

$A_{23} = (-1)^{2+3} M_{23} = (-1)(1) = -1$

$A_{31} = (-1)^{3+1} M_{31} = (1)(-20) = -20$

$A_{32} = (-1)^{3+2} M_{32} = (-1)(-13) = 13$

$A_{33} = (-1)^{3+3} M_{33} = (1)(5) = 5$

Solution:

The given determinant is $\Delta = \begin{vmatrix} 5& 3& 8\\2& 0& 1\\1& 2& 3 \end{vmatrix}$.

We need to evaluate the determinant using the cofactors of the elements of the second row.

The elements of the second row are $a_{21}=2$, $a_{22}=0$, and $a_{23}=1$.

The determinant $\Delta$ can be evaluated as the sum of the product of the elements of the second row with their corresponding cofactors ($A_{ij}$):

$\Delta = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$

First, we need to find the cofactors $A_{21}$, $A_{22}$, and $A_{23}$.

The cofactor $A_{ij}$ is given by $A_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$.

Calculating Cofactors of the second row:

Cofactor of $a_{21} = 2$:

$M_{21}$ is the determinant obtained by deleting the 2nd row and 1st column:

$M_{21} = \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = (3)(3) - (8)(2) = 9 - 16 = -7$

$A_{21} = (-1)^{2+1} M_{21} = (-1)^3 (-7) = -1 \times (-7) = 7$

Cofactor of $a_{22} = 0$:

$M_{22}$ is the determinant obtained by deleting the 2nd row and 2nd column:

$M_{22} = \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = (5)(3) - (8)(1) = 15 - 8 = 7$

$A_{22} = (-1)^{2+2} M_{22} = (-1)^4 (7) = 1 \times 7 = 7$

Cofactor of $a_{23} = 1$:

$M_{23}$ is the determinant obtained by deleting the 2nd row and 3rd column:

$M_{23} = \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = (5)(2) - (3)(1) = 10 - 3 = 7$

$A_{23} = (-1)^{2+3} M_{23} = (-1)^5 (7) = -1 \times 7 = -7$

Evaluating the determinant using the second row cofactors:

$\Delta = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$

Substitute the values of the elements and their cofactors:

$\Delta = (2)(7) + (0)(7) + (1)(-7)$

$\Delta = 14 + 0 - 7$

$\Delta = 7$

The value of the determinant is 7.

Solution:

The given determinant is $\Delta = \begin{vmatrix} 1& x& yz\\1& y& zx\\1& z& xy \end{vmatrix}$.

We need to evaluate the determinant using the cofactors of the elements of the third column.

The elements of the third column are $a_{13}=yz$, $a_{23}=zx$, and $a_{33}=xy$.

The value of the determinant $\Delta$ using the third column cofactors is given by the formula:

$\Delta = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33}$

where $A_{ij}$ is the cofactor of element $a_{ij}$, and $A_{ij} = (-1)^{i+j} M_{ij}$.

Calculating Cofactors of the third column:

Cofactor of $a_{13} = yz$:

$M_{13}$ is the determinant obtained by deleting the 1st row and 3rd column:

$M_{13} = \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = (1)(z) - (y)(1) = z - y$

$A_{13} = (-1)^{1+3} M_{13} = (-1)^4 (z - y) = 1 \times (z - y) = z - y$

Cofactor of $a_{23} = zx$:

$M_{23}$ is the determinant obtained by deleting the 2nd row and 3rd column:

$M_{23} = \begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = (1)(z) - (x)(1) = z - x$

$A_{23} = (-1)^{2+3} M_{23} = (-1)^5 (z - x) = -1 \times (z - x) = x - z$

Cofactor of $a_{33} = xy$:

$M_{33}$ is the determinant obtained by deleting the 3rd row and 3rd column:

$M_{33} = \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = (1)(y) - (x)(1) = y - x$

$A_{33} = (-1)^{3+3} M_{33} = (-1)^6 (y - x) = 1 \times (y - x) = y - x$

Evaluating the determinant using the third column cofactors:

$\Delta = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33}$

Substitute the values of the elements and their cofactors:

$\Delta = (yz)(z - y) + (zx)(x - z) + (xy)(y - x)$

Expand the terms:

$\Delta = yz^2 - y^2z + zx^2 - z^2x + xy^2 - x^2y$

Rearranging the terms to facilitate factorization (grouping terms with similar variables or powers):

$\Delta = x^2z - x^2y + xy^2 - y^2z + yz^2 - z^2x$

Group terms and factor out common factors:

$\Delta = x^2(z - y) + x(y^2 - z^2) + yz(z - y)$

Recognize the difference of squares: $y^2 - z^2 = (y - z)(y + z) = -(z - y)(y + z)$

$\Delta = x^2(z - y) + x[-(z - y)(y + z)] + yz(z - y)$

$\Delta = x^2(z - y) - x(z - y)(y + z) + yz(z - y)$

Factor out the common term $(z - y)$:

$\Delta = (z - y) [x^2 - x(y + z) + yz]$

Expand the term inside the square bracket:

$\Delta = (z - y) [x^2 - xy - xz + yz]$

Factor the quadratic expression inside the bracket by grouping:

$x^2 - xy - xz + yz = x(x - y) - z(x - y)$

$= (x - y)(x - z)$

Substitute this factorization back into the expression for $\Delta$:

$\Delta = (z - y) (x - y) (x - z)$

Rearranging the factors to a standard cyclic order $(x-y)(y-z)(z-x)$ requires changing the sign of two factors:

$\Delta = (x - y) (- (y - z)) (- (z - x))$

$\Delta = (x - y) (y - z) (z - x)$

The value of the determinant is $(x - y)(y - z)(z - x)$.

Solution:

The value of a determinant can be calculated by summing the products of the elements of any row or column with their corresponding cofactors.

Let $\Delta$ be the determinant and $a_{ij}$ be the element in the $i$-th row and $j$-th column. Let $A_{ij}$ be the cofactor of the element $a_{ij}$.

The value of the determinant $\Delta$ can be expressed in several ways:

Expanding along the 1st row: $\Delta = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}$

Expanding along the 2nd row: $\Delta = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$

Expanding along the 3rd row: $\Delta = a_{31} A_{31} + a_{32} A_{32} + a_{33} A_{33}$

Expanding along the 1st column: $\Delta = a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$

Expanding along the 2nd column: $\Delta = a_{12} A_{12} + a_{22} A_{22} + a_{32} A_{32}$

Expanding along the 3rd column: $\Delta = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33}$

We also know that if we multiply the elements of a row (or column) by the cofactors of a different row (or column), the sum is zero.

For example, $a_{11} A_{21} + a_{12} A_{22} + a_{13} A_{23} = 0$ (elements of row 1 with cofactors of row 2).

Let's examine the given options:

(A) $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$: This involves elements of row 1 ($a_{11}, a_{12}, a_{13}$) and cofactors of row 3 ($A_{31}, A_{32}, A_{33}$). This sum is 0.

(B) $a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31}$: This mixes elements from row 1 with cofactors from column 1. This is not a standard valid expansion and does not equal $\Delta$.

(C) $a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13}$: This involves elements of row 2 ($a_{21}, a_{22}, a_{23}$) and cofactors of row 1 ($A_{11}, A_{12}, A_{13}$). This sum is 0.

(D) $a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$: This involves elements of column 1 ($a_{11}, a_{21}, a_{31}$) and their corresponding cofactors ($A_{11}, A_{21}, A_{31}$). This is a valid expansion of the determinant along the first column.

Therefore, the value of $\Delta$ is correctly given by option (D).

The final answer is $\boxed{(D)}$.

Given:

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

To Find:

The adjoint of matrix A (adj A).

Solution:

For a $2 \times 2$ matrix A = $\begin{bmatrix}a&b\\c&d \end{bmatrix}$, the adjoint of A is given by swapping the elements on the main diagonal and changing the sign of the elements on the off-diagonal.

So, adj A = $\begin{bmatrix}d&-b\\-c&a \end{bmatrix}$.

In the given matrix A = $\begin{bmatrix}2&3\\1&4 \end{bmatrix}$, we have:

$a = 2$

$b = 3$

$c = 1$

$d = 4$

Now, we apply the formula for adj A:

adj A = $\begin{bmatrix}4&-(3)\\-(1)&2 \end{bmatrix}$

adj A = $\begin{bmatrix}4&-3\\-1&2 \end{bmatrix}$.

Final Answer:

The adjoint of matrix A is $\begin{bmatrix}4&-3\\-1&2 \end{bmatrix}$.

Given the matrix:

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

To Find:

1. Verify that A adj A = | A| I

2. Find A^–1

Solution:

1. Calculate the determinant of A, |A|:

$|A| = \begin{vmatrix}1&3&3\\1&4&3\\1&3&4 \end{vmatrix}$

Expanding along the first row:

$|A| = 1(4 \times 4 - 3 \times 3) - 3(1 \times 4 - 3 \times 1) + 3(1 \times 3 - 4 \times 1)$

$|A| = 1(16 - 9) - 3(4 - 3) + 3(3 - 4)$

$|A| = 1(7) - 3(1) + 3(-1)$

$|A| = 7 - 3 - 3$

$|A| = 1$

2. Calculate the cofactors of the elements of A:

$C_{11} = + \begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} = (4)(4) - (3)(3) = 16 - 9 = 7$

$C_{12} = - \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = -((1)(4) - (3)(1)) = -(4 - 3) = -1$

$C_{13} = + \begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix} = (1)(3) - (4)(1) = 3 - 4 = -1$

$C_{21} = - \begin{vmatrix} 3 & 3 \\ 3 & 4 \end{vmatrix} = -((3)(4) - (3)(3)) = -(12 - 9) = -3$

$C_{22} = + \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = (1)(4) - (3)(1) = 4 - 3 = 1$

$C_{23} = - \begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix} = -((1)(3) - (3)(1)) = -(3 - 3) = 0$

$C_{31} = + \begin{vmatrix} 3 & 3 \\ 4 & 3 \end{vmatrix} = (3)(3) - (3)(4) = 9 - 12 = -3$

$C_{32} = - \begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix} = -((1)(3) - (3)(1)) = -(3 - 3) = 0$

$C_{33} = + \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = (1)(4) - (3)(1) = 4 - 3 = 1$

3. Form the matrix of cofactors and find the adjoint of A (adj A):

The matrix of cofactors is:

$\begin{bmatrix} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix}$

The adjoint of A is the transpose of the cofactor matrix:

$\text{adj A} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$

4. Verify A adj A = | A| I:

Calculate A adj A:

$A \text{ adj A} = \begin{bmatrix}1&3&3\\1&4&3\\1&3&4 \end{bmatrix} \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$

$A \text{ adj A} = \begin{bmatrix} (1)(7)+(3)(-1)+(3)(-1) & (1)(-3)+(3)(1)+(3)(0) & (1)(-3)+(3)(0)+(3)(1) \\ (1)(7)+(4)(-1)+(3)(-1) & (1)(-3)+(4)(1)+(3)(0) & (1)(-3)+(4)(0)+(3)(1) \\ (1)(7)+(3)(-1)+(4)(-1) & (1)(-3)+(3)(1)+(4)(0) & (1)(-3)+(3)(0)+(4)(1) \end{bmatrix}$

$A \text{ adj A} = \begin{bmatrix} 7 - 3 - 3 & -3 + 3 + 0 & -3 + 0 + 3 \\ 7 - 4 - 3 & -3 + 4 + 0 & -3 + 0 + 3 \\ 7 - 3 - 4 & -3 + 3 + 0 & -3 + 0 + 4 \end{bmatrix}$

$A \text{ adj A} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Calculate |A| I:

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

Since $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, we have A adj A = |A| I.

Thus, the property A adj A = |A| I is verified.

5. Find A^–1:

The inverse of a matrix A is given by the formula $A^{-1} = \frac{1}{|A|} \text{ adj A}$, provided $|A| \neq 0$.

Since $|A| = 1 \neq 0$, the inverse A^-1 exists.

$A^{-1} = \frac{1}{1} \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$

Final Answers:

The property A adj A = |A| I is verified as both sides equal $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

The inverse of matrix A is $\mathbf{A^{-1} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}}$.

Given matrices:

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

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

To Verify:

$(AB)^{-1} = B^{-1}A^{-1}$

Solution:

Step 1: Calculate AB

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

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

$AB = \begin{bmatrix} 2-3 & -4+9 \\ 1+4 & -2-12 \end{bmatrix}$

$AB = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix}$

Step 2: Calculate (AB)$^{-1}$

Let $C = AB = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix}$.

Determinant of C:

$|C| = (-1)(-14) - (5)(5) = 14 - 25 = -11$

Since $|C| = -11 \neq 0$, the inverse exists.

Adjoint of C:

$\text{adj } C = \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix}$

Inverse of C:

$(AB)^{-1} = C^{-1} = \frac{1}{|C|} \text{ adj } C = \frac{1}{-11} \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix}$

$(AB)^{-1} = \begin{bmatrix} \frac{-14}{-11} & \frac{-5}{-11} \\ \frac{-5}{-11} & \frac{-1}{-11} \end{bmatrix} = \begin{bmatrix} 14/11 & 5/11 \\ 5/11 & 1/11 \end{bmatrix}$

Step 3: Calculate A$^{-1}$

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

Determinant of A:

$|A| = (2)(-4) - (3)(1) = -8 - 3 = -11$

Since $|A| = -11 \neq 0$, the inverse exists.

Adjoint of A:

$\text{adj } A = \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix}$

Inverse of A:

$A^{-1} = \frac{1}{|A|} \text{ adj } A = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} \frac{-4}{-11} & \frac{-3}{-11} \\ \frac{-1}{-11} & \frac{2}{-11} \end{bmatrix} = \begin{bmatrix} 4/11 & 3/11 \\ 1/11 & -2/11 \end{bmatrix}$

Step 4: Calculate B$^{-1}$

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

Determinant of B:

$|B| = (1)(3) - (-2)(-1) = 3 - 2 = 1$

Since $|B| = 1 \neq 0$, the inverse exists.

Adjoint of B:

$\text{adj } B = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$

Inverse of B:

$B^{-1} = \frac{1}{|B|} \text{ adj } B = \frac{1}{1} \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$

$B^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$

Step 5: Calculate B$^{-1}$A$^{-1}$

$B^{-1}A^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 4/11 & 3/11 \\ 1/11 & -2/11 \end{bmatrix}$

$B^{-1}A^{-1} = \begin{bmatrix} (3)(4/11)+(2)(1/11) & (3)(3/11)+(2)(-2/11) \\ (1)(4/11)+(1)(1/11) & (1)(3/11)+(1)(-2/11) \end{bmatrix}$

$B^{-1}A^{-1} = \begin{bmatrix} 12/11+2/11 & 9/11-4/11 \\ 4/11+1/11 & 3/11-2/11 \end{bmatrix}$

$B^{-1}A^{-1} = \begin{bmatrix} 14/11 & 5/11 \\ 5/11 & 1/11 \end{bmatrix}$

Step 6: Verification

From Step 2, we have $(AB)^{-1} = \begin{bmatrix} 14/11 & 5/11 \\ 5/11 & 1/11 \end{bmatrix}$.

From Step 5, we have $B^{-1}A^{-1} = \begin{bmatrix} 14/11 & 5/11 \\ 5/11 & 1/11 \end{bmatrix}$.

Since $(AB)^{-1}$ and $B^{-1}A^{-1}$ are equal, the property $(AB)^{-1} = B^{-1}A^{-1}$ is verified.

Given matrix:

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

Given equation to verify:

$A^2 – 4A + I = O$

where $I = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$ is the 2 × 2 identity matrix and $O = \begin{bmatrix}0&0\\0&0 \end{bmatrix}$ is the 2 × 2 zero matrix.

Solution:

Part 1: Show that A satisfies the equation A² – 4A + I = O

First, calculate $A^2 = A \times A$:

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

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

$A^2 = \begin{bmatrix} 4+3 & 6+6 \\ 2+2 & 3+4 \end{bmatrix}$

$A^2 = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}$

Next, calculate $4A$:

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

$4A = \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix}$

Now, substitute $A^2$, $4A$, and $I$ into the expression $A^2 – 4A + I$:

$A^2 – 4A + I = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$A^2 – 4A + I = \begin{bmatrix} 7-8 & 12-12 \\ 4-4 & 7-8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$A^2 – 4A + I = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$A^2 – 4A + I = \begin{bmatrix} -1+1 & 0+0 \\ 0+0 & -1+1 \end{bmatrix}$

$A^2 – 4A + I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Since the result is the zero matrix $O$, the matrix A satisfies the equation $A^2 – 4A + I = O$.

Part 2: Using the equation to find A^–1

We have the equation:

We want to find A^–1. Since $|A| = (2)(2) - (3)(1) = 4 - 3 = 1 \neq 0$, A is invertible and A^–1 exists.

Multiply equation (i) by A^–1 on the right:

$(A^2 – 4A + I)A^{-1} = OA^{-1}$

$A^2 A^{-1} – 4A A^{-1} + I A^{-1} = O$

Using the properties of matrix multiplication ($A^2 A^{-1} = A$, $A A^{-1} = I$, $I A^{-1} = A^{-1}$, $OA^{-1} = O$):

$A – 4I + A^{-1} = O$

Now, isolate A^–1:

$A^{-1} = 4I - A$

Substitute the matrices for $I$ and $A$:

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

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

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

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

Final Answer:

The matrix A satisfies the equation $A^2 – 4A + I = O$.

Using this equation, the inverse of matrix A is $\mathbf{A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}}$.

Given matrix:

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

To Find:

Adjoint of matrix A (adj A)

Solution:

For a 2 × 2 matrix $A = \begin{bmatrix}a&b\\c&d \end{bmatrix}$, the adjoint is given by $\text{adj } A = \begin{bmatrix}d&-b\\-c&a \end{bmatrix}$.

In the given matrix $A = \begin{bmatrix}1&2\\3&4 \end{bmatrix}$, we have $a=1$, $b=2$, $c=3$, and $d=4$.

Using the formula for the adjoint of a 2 × 2 matrix:

$\text{adj } A = \begin{bmatrix}4&-2\\-3&1 \end{bmatrix}$

Final Answer:

The adjoint of the given matrix is $\mathbf{\begin{bmatrix}4&-2\\-3&1 \end{bmatrix}}$.

Given matrix:

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

To Find:

Adjoint of matrix A (adj A)

Solution:

The adjoint of a matrix A is the transpose of the cofactor matrix of A.

Let $C_{ij}$ be the cofactor of the element $a_{ij}$ in matrix A.

The cofactors are calculated as $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of $a_{ij}$.

Calculate the cofactors:

$C_{11} = (-1)^{1+1} \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = +(3 \times 1 - 5 \times 0) = 3 - 0 = 3$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix} = -(2 \times 1 - 5 \times (-2)) = -(2 + 10) = -12$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} = +(2 \times 0 - 3 \times (-2)) = 0 + 6 = 6$

$C_{21} = (-1)^{2+1} \begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = -((-1) \times 1 - 2 \times 0) = -(-1 - 0) = 1$

$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} = +(1 \times 1 - 2 \times (-2)) = 1 + 4 = 5$

$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = -(1 \times 0 - (-1) \times (-2)) = -(0 - 2) = 2$

$C_{31} = (-1)^{3+1} \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = +((-1) \times 5 - 2 \times 3) = -5 - 6 = -11$

$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = -(1 \times 5 - 2 \times 2) = -(5 - 4) = -1$

$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = +(1 \times 3 - (-1) \times 2) = 3 + 2 = 5$

The matrix of cofactors is:

$\text{Cofactor matrix} = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} = \begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}$

The adjoint of A is the transpose of the cofactor matrix:

$\text{adj A} = (\text{Cofactor matrix})^T = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$

Final Answer:

The adjoint of the given matrix is $\mathbf{\begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}}$.

Given matrix:

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

Identity matrix of order 2:

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

Zero matrix of order 2:

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

To Verify:

$A (\text{adj } A) = (\text{adj } A) A = | A | I$

Solution:

Step 1: Calculate the determinant of A, |A|

$|A| = (2)(-6) - (3)(-4)$

$|A| = -12 - (-12)$

$|A| = -12 + 12$

$|A| = 0$

Step 2: Calculate the adjoint of A, adj A

For a 2 × 2 matrix $A = \begin{bmatrix}a&b\\c&d \end{bmatrix}$, the adjoint is $\text{adj } A = \begin{bmatrix}d&-b\\-c&a \end{bmatrix}$.

Here, $a=2$, $b=3$, $c=-4$, $d=-6$.

$\text{adj } A = \begin{bmatrix}-6&-3\\-(-4)&2 \end{bmatrix} = \begin{bmatrix}-6&-3\\4&2 \end{bmatrix}$

Step 3: Calculate A (adj A)

$A (\text{adj } A) = \begin{bmatrix}2&3\\−4&−6 \end{bmatrix} \begin{bmatrix}-6&-3\\4&2 \end{bmatrix}$

$A (\text{adj } A) = \begin{bmatrix} (2)(-6)+(3)(4) & (2)(-3)+(3)(2) \\ (-4)(-6)+(-6)(4) & (-4)(-3)+(-6)(2) \end{bmatrix}$

$A (\text{adj } A) = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix}$

$A (\text{adj } A) = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Step 4: Calculate (adj A) A

$(\text{adj } A) A = \begin{bmatrix}-6&-3\\4&2 \end{bmatrix} \begin{bmatrix}2&3\\−4&−6 \end{bmatrix}$

$(\text{adj } A) A = \begin{bmatrix} (-6)(2)+(-3)(-4) & (-6)(3)+(-3)(-6) \\ (4)(2)+(2)(-4) & (4)(3)+(2)(-6) \end{bmatrix}$

$(\text{adj } A) A = \begin{bmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{bmatrix}$

$(\text{adj } A) A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Step 5: Calculate |A| I

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

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

Step 6: Verification

From Step 3, $A (\text{adj } A) = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.

From Step 4, $(\text{adj } A) A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.

From Step 5, $|A| I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.

Since $A (\text{adj } A) = (\text{adj } A) A = | A | I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, the property is verified.

Given matrix:

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

Identity matrix of order 3:

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

To Verify:

$A (\text{adj } A) = (\text{adj } A) A = | A | I$

Solution:

Step 1: Calculate the determinant of A, |A|

Expanding along the first row:

$|A| = 1 \begin{vmatrix} 0 & -2 \\ 0 & 3 \end{vmatrix} - (-1) \begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} + 2 \begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix}$

$|A| = 1( (0)(3) - (-2)(0) ) + 1( (3)(3) - (-2)(1) ) + 2( (3)(0) - (0)(1) )$

$|A| = 1(0) + 1(9 + 2) + 2(0)$

$|A| = 0 + 1(11) + 0$

$|A| = 11$

Step 2: Calculate the cofactors of A

$C_{11} = (-1)^{1+1} \begin{vmatrix} 0 & -2 \\ 0 & 3 \end{vmatrix} = +( (0)(3) - (-2)(0) ) = 0 - 0 = 0$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} = -( (3)(3) - (-2)(1) ) = -(9 + 2) = -11$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix} = +( (3)(0) - (0)(1) ) = 0 - 0 = 0$

$C_{21} = (-1)^{2+1} \begin{vmatrix} -1 & 2 \\ 0 & 3 \end{vmatrix} = -( (-1)(3) - (2)(0) ) = -(-3 - 0) = 3$

$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = +( (1)(3) - (2)(1) ) = 3 - 2 = 1$

$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} = -( (1)(0) - (-1)(1) ) = -(0 + 1) = -1$

$C_{31} = (-1)^{3+1} \begin{vmatrix} -1 & 2 \\ 0 & -2 \end{vmatrix} = +( (-1)(-2) - (2)(0) ) = 2 - 0 = 2$

$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 2 \\ 3 & -2 \end{vmatrix} = -( (1)(-2) - (2)(3) ) = -(-2 - 6) = -(-8) = 8$

$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -1 \\ 3 & 0 \end{vmatrix} = +( (1)(0) - (-1)(3) ) = 0 + 3 = 3$

The matrix of cofactors is:

$\begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} = \begin{bmatrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{bmatrix}$

Step 3: Calculate the adjoint of A, adj A

The adjoint of A is the transpose of the cofactor matrix:

$\text{adj } A = (\text{Cofactor matrix})^T = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$

Step 4: Calculate A (adj A)

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

$A (\text{adj } A) = \begin{bmatrix} (1)(0)+(-1)(-11)+(2)(0) & (1)(3)+(-1)(1)+(2)(-1) & (1)(2)+(-1)(8)+(2)(3) \\ (3)(0)+(0)(-11)+(-2)(0) & (3)(3)+(0)(1)+(-2)(-1) & (3)(2)+(0)(8)+(-2)(3) \\ (1)(0)+(0)(-11)+(3)(0) & (1)(3)+(0)(1)+(3)(-1) & (1)(2)+(0)(8)+(3)(3) \end{bmatrix}$

$A (\text{adj } A) = \begin{bmatrix} 0+11+0 & 3-1-2 & 2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9 \end{bmatrix}$

$A (\text{adj } A) = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$

Step 5: Calculate (adj A) A

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

$(\text{adj } A) A = \begin{bmatrix} (0)(1)+(3)(3)+(2)(1) & (0)(-1)+(3)(0)+(2)(0) & (0)(2)+(3)(-2)+(2)(3) \\ (-11)(1)+(1)(3)+(8)(1) & (-11)(-1)+(1)(0)+(8)(0) & (-11)(2)+(1)(-2)+(8)(3) \\ (0)(1)+(-1)(3)+(3)(1) & (0)(-1)+(-1)(0)+(3)(0) & (0)(2)+(-1)(-2)+(3)(3) \end{bmatrix}$

$(\text{adj } A) A = \begin{bmatrix} 0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0+0+0 & 0+2+9 \end{bmatrix}$

$(\text{adj } A) A = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$

Step 6: Calculate |A| I

We have $|A| = 11$ and $I = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$.

$|A| I = 11 \times \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = \begin{bmatrix}11 \times 1 & 11 \times 0 & 11 \times 0\\11 \times 0 & 11 \times 1 & 11 \times 0\\11 \times 0 & 11 \times 0 & 11 \times 1 \end{bmatrix} = \begin{bmatrix}11&0&0\\0&11&0\\0&0&11 \end{bmatrix}$

Step 7: Verification

From Step 4, $A (\text{adj } A) = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$.

From Step 5, $(\text{adj } A) A = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$.

From Step 6, $|A| I = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$.

Since $A (\text{adj } A) = (\text{adj } A) A = | A | I$, the property is verified.

Given matrix:

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

To Find:

The inverse of matrix A (A^–1), if it exists.

Solution:

For a matrix to have an inverse, its determinant must be non-zero.

Calculate the determinant of A, |A|:

$|A| = (2)(3) - (-2)(4)$

$|A| = 6 - (-8)$

$|A| = 6 + 8$

$|A| = 14$

Since $|A| = 14 \neq 0$, the inverse of matrix A exists.

The formula for the inverse of a 2 × 2 matrix $A = \begin{bmatrix}a&b\\c&d \end{bmatrix}$ is $A^{-1} = \frac{1}{|A|} \text{ adj } A$, where $\text{adj } A = \begin{bmatrix}d&-b\\-c&a \end{bmatrix}$.

First, calculate the adjoint of A:

$\text{adj } A = \begin{bmatrix}3&-(-2)\\-4&2 \end{bmatrix} = \begin{bmatrix}3&2\\-4&2 \end{bmatrix}$

Now, use the inverse formula:

$A^{-1} = \frac{1}{14} \begin{bmatrix}3&2\\-4&2 \end{bmatrix}$

Distribute the scalar $\frac{1}{14}$ to each element of the adjoint matrix:

$A^{-1} = \begin{bmatrix}\frac{3}{14}&\frac{2}{14}\\\frac{-4}{14}&\frac{2}{14} \end{bmatrix}$

Simplify the fractions:

$A^{-1} = \begin{bmatrix}3/14&1/7\\-2/7&1/7 \end{bmatrix}$

Final Answer:

The inverse of the given matrix is $\mathbf{A^{-1} = \begin{bmatrix}3/14&1/7\\-2/7&1/7 \end{bmatrix}}$.

Given matrix:

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

To Find:

The inverse of matrix A (A^–1), if it exists.

Solution:

For a matrix to have an inverse, its determinant must be non-zero.

Calculate the determinant of A, |A|:

$|A| = (-1)(2) - (5)(-3)$

$|A| = -2 - (-15)$

$|A| = -2 + 15$

$|A| = 13$

Since $|A| = 13 \neq 0$, the inverse of matrix A exists.

The formula for the inverse of a 2 × 2 matrix $A = \begin{bmatrix}a&b\\c&d \end{bmatrix}$ is $A^{-1} = \frac{1}{|A|} \text{ adj } A$, where $\text{adj } A = \begin{bmatrix}d&-b\\-c&a \end{bmatrix}$.

First, calculate the adjoint of A:

$\text{adj } A = \begin{bmatrix}2&-(5)\\-(-3)&-1 \end{bmatrix} = \begin{bmatrix}2&-5\\3&-1 \end{bmatrix}$

Now, use the inverse formula:

$A^{-1} = \frac{1}{13} \begin{bmatrix}2&-5\\3&-1 \end{bmatrix}$

Distribute the scalar $\frac{1}{13}$ to each element of the adjoint matrix:

$A^{-1} = \begin{bmatrix}\frac{2}{13}&\frac{-5}{13}\\\frac{3}{13}&\frac{-1}{13} \end{bmatrix}$

Final Answer:

The inverse of the given matrix is $\mathbf{A^{-1} = \begin{bmatrix}2/13&-5/13\\3/13&-1/13 \end{bmatrix}}$.

Given matrix:

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

To Find:

The inverse of matrix A (A^–1), if it exists.

Solution:

For a matrix to have an inverse, its determinant must be non-zero.

Calculate the determinant of A, |A|:

Since A is an upper triangular matrix, its determinant is the product of the diagonal elements.

$|A| = (1)(2)(5)$

$|A| = 10$

Since $|A| = 10 \neq 0$, the inverse of matrix A exists.

To find the inverse $A^{-1}$, we use the formula $A^{-1} = \frac{1}{|A|} \text{ adj } A$.

First, calculate the cofactors of the elements of A:

$C_{11} = (-1)^{1+1} \begin{vmatrix} 2 & 4 \\ 0 & 5 \end{vmatrix} = (2)(5) - (4)(0) = 10$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 0 & 4 \\ 0 & 5 \end{vmatrix} = -((0)(5) - (4)(0)) = 0$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 0 & 2 \\ 0 & 0 \end{vmatrix} = (0)(0) - (2)(0) = 0$

$C_{21} = (-1)^{2+1} \begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} = -((2)(5) - (3)(0)) = -10$

$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 3 \\ 0 & 5 \end{vmatrix} = (1)(5) - (3)(0) = 5$

$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 0 & 0 \end{vmatrix} = -((1)(0) - (2)(0)) = 0$

$C_{31} = (-1)^{3+1} \begin{vmatrix} 2 & 3 \\ 2 & 4 \end{vmatrix} = (2)(4) - (3)(2) = 8 - 6 = 2$

$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = -((1)(4) - (3)(0)) = -4$

$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 2 \\ 0 & 2 \end{vmatrix} = (1)(2) - (2)(0) = 2$

The matrix of cofactors is:

$\text{Cofactor matrix} = \begin{bmatrix} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{bmatrix}$

The adjoint of A is the transpose of the cofactor matrix:

$\text{adj A} = (\text{Cofactor matrix})^T = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$

Now, use the formula for the inverse:

$A^{-1} = \frac{1}{|A|} \text{ adj A} = \frac{1}{10} \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$

Distribute the scalar $\frac{1}{10}$ to each element:

$A^{-1} = \begin{bmatrix} \frac{10}{10} & \frac{-10}{10} & \frac{2}{10} \\ \frac{0}{10} & \frac{5}{10} & \frac{-4}{10} \\ \frac{0}{10} & \frac{0}{10} & \frac{2}{10} \end{bmatrix}$

Simplify the fractions:

$A^{-1} = \begin{bmatrix} 1 & -1 & 1/5 \\ 0 & 1/2 & -2/5 \\ 0 & 0 & 1/5 \end{bmatrix}$

Final Answer:

The inverse of the given matrix is $\mathbf{A^{-1} = \begin{bmatrix} 1 & -1 & 1/5 \\ 0 & 1/2 & -2/5 \\ 0 & 0 & 1/5 \end{bmatrix}}$.

Given matrix:

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

To Find:

The inverse of matrix A (A^–1), if it exists.

Solution:

For a matrix to have an inverse, its determinant must be non-zero.

Calculate the determinant of A, |A|:

Since A is a lower triangular matrix, the determinant is the product of the diagonal elements.

$|A| = (1)(3)(-1)$

$|A| = -3$

Since $|A| = -3 \neq 0$, the inverse of matrix A exists.

To find the inverse $A^{-1}$, we use the formula $A^{-1} = \frac{1}{|A|} \text{ adj } A$.

First, calculate the cofactors of the elements of A:

$C_{11} = (-1)^{1+1} \begin{vmatrix} 3 & 0 \\ 2 & -1 \end{vmatrix} = (3)(-1) - (0)(2) = -3 - 0 = -3$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 3 & 0 \\ 5 & -1 \end{vmatrix} = -((3)(-1) - (0)(5)) = -(-3 - 0) = 3$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 3 & 3 \\ 5 & 2 \end{vmatrix} = (3)(2) - (3)(5) = 6 - 15 = -9$

$C_{21} = (-1)^{2+1} \begin{vmatrix} 0 & 0 \\ 2 & -1 \end{vmatrix} = -((0)(-1) - (0)(2)) = -(0 - 0) = 0$

$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 0 \\ 5 & -1 \end{vmatrix} = (1)(-1) - (0)(5) = -1 - 0 = -1$

$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 0 \\ 5 & 2 \end{vmatrix} = -((1)(2) - (0)(5)) = -(2 - 0) = -2$

$C_{31} = (-1)^{3+1} \begin{vmatrix} 0 & 0 \\ 3 & 0 \end{vmatrix} = (0)(0) - (0)(3) = 0 - 0 = 0$

$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 0 \\ 3 & 0 \end{vmatrix} = -((1)(0) - (0)(3)) = -(0 - 0) = 0$

$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 0 \\ 3 & 3 \end{vmatrix} = (1)(3) - (0)(3) = 3 - 0 = 3$

The matrix of cofactors is:

$\text{Cofactor matrix} = \begin{bmatrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{bmatrix}$

The adjoint of A is the transpose of the cofactor matrix:

$\text{adj A} = (\text{Cofactor matrix})^T = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$

Now, use the formula for the inverse:

$A^{-1} = \frac{1}{|A|} \text{ adj A} = \frac{1}{-3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$

Distribute the scalar $\frac{1}{-3}$ to each element:

$A^{-1} = \begin{bmatrix} \frac{-3}{-3} & \frac{0}{-3} & \frac{0}{-3} \\ \frac{3}{-3} & \frac{-1}{-3} & \frac{0}{-3} \\ \frac{-9}{-3} & \frac{-2}{-3} & \frac{3}{-3} \end{bmatrix}$

Simplify the fractions:

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

Final Answer:

The inverse of the given matrix is $\mathbf{A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1/3 & 0 \\ 3 & 2/3 & -1 \end{bmatrix}}$.

Given:

The matrix A = $\begin{bmatrix}2&1&3\\4&−1&0\\−7&2&1 \end{bmatrix}$.

To Find:

The inverse of the matrix A, denoted by $A^{-1}$, if it exists.

Solution:

To find the inverse of a matrix A, we first need to calculate its determinant. If the determinant is non-zero, the inverse exists and is given by the formula:

$A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A)$

where det(A) is the determinant of A and adj(A) is the adjoint of A.

Step 1: Calculate the Determinant of A (det(A)).

For a $3 \times 3$ matrix $\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{bmatrix}$, the determinant can be calculated as:

det(A) = $a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$

For the given matrix A = $\begin{bmatrix}2&1&3\\4&−1&0\\−7&2&1 \end{bmatrix}$:

det(A) = $2((-1)(1) - (0)(2)) - 1((4)(1) - (0)(-7)) + 3((4)(2) - (-1)(-7))$

det(A) = $2(-1 - 0) - 1(4 - 0) + 3(8 - 7)$

det(A) = $2(-1) - 1(4) + 3(1)$

det(A) = $-2 - 4 + 3$

det(A) = $-6 + 3$

det(A) = $-3$

Since det(A) = $-3 \neq 0$, the inverse of matrix A exists.

Step 2: Calculate the Matrix of Cofactors.

The cofactor $C_{ij}$ of an element $a_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of $a_{ij}$. The minor $M_{ij}$ is the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column.

$C_{11} = (-1)^{1+1} \begin{vmatrix}-1&0\\2&1 \end{vmatrix} = (1)((-1)(1) - (0)(2)) = 1(-1 - 0) = -1$

$C_{12} = (-1)^{1+2} \begin{vmatrix}4&0\\-7&1 \end{vmatrix} = (-1)((4)(1) - (0)(-7)) = -1(4 - 0) = -4$

$C_{13} = (-1)^{1+3} \begin{vmatrix}4&-1\\-7&2 \end{vmatrix} = (1)((4)(2) - (-1)(-7)) = 1(8 - 7) = 1$

$C_{21} = (-1)^{2+1} \begin{vmatrix}1&3\\2&1 \end{vmatrix} = (-1)((1)(1) - (3)(2)) = -1(1 - 6) = -1(-5) = 5$

$C_{22} = (-1)^{2+2} \begin{vmatrix}2&3\\-7&1 \end{vmatrix} = (1)((2)(1) - (3)(-7)) = 1(2 + 21) = 23$

$C_{23} = (-1)^{2+3} \begin{vmatrix}2&1\\-7&2 \end{vmatrix} = (-1)((2)(2) - (1)(-7)) = -1(4 + 7) = -1(11) = -11$

$C_{31} = (-1)^{3+1} \begin{vmatrix}1&3\\-1&0 \end{vmatrix} = (1)((1)(0) - (3)(-1)) = 1(0 + 3) = 3$

$C_{32} = (-1)^{3+2} \begin{vmatrix}2&3\\4&0 \end{vmatrix} = (-1)((2)(0) - (3)(4)) = -1(0 - 12) = -1(-12) = 12$

$C_{33} = (-1)^{3+3} \begin{vmatrix}2&1\\4&-1 \end{vmatrix} = (1)((2)(-1) - (1)(4)) = 1(-2 - 4) = -6$

The matrix of cofactors, C, is:

$C = \begin{bmatrix}-1&-4&1\\5&23&-11\\3&12&-6 \end{bmatrix}$

Step 3: Calculate the Adjoint of A (adj(A)).

The adjoint of A is the transpose of the matrix of cofactors.

adj(A) = $C^T = \begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6 \end{bmatrix}$

Step 4: Calculate the Inverse of A ($A^{-1}$).

Using the formula $A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A)$:

$A^{-1} = \frac{1}{-3} \begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6 \end{bmatrix}$

Multiply each element of the adjoint matrix by $-\frac{1}{3}$:

$A^{-1} = \begin{bmatrix}(-\frac{1}{3})(-1)&(-\frac{1}{3})(5)&(-\frac{1}{3})(3)\\(-\frac{1}{3})(-4)&(-\frac{1}{3})(23)&(-\frac{1}{3})(12)\\(-\frac{1}{3})(1)&(-\frac{1}{3})(-11)&(-\frac{1}{3})(-6) \end{bmatrix}$

$A^{-1} = \begin{bmatrix}\frac{1}{3}&-\frac{5}{3}&-1\\\frac{4}{3}&-\frac{23}{3}&-4\\-\frac{1}{3}&\frac{11}{3}&2 \end{bmatrix}$

Final Answer:

The inverse of the given matrix A is:

$A^{-1} = \begin{bmatrix}\frac{1}{3}&-\frac{5}{3}&-1\\\frac{4}{3}&-\frac{23}{3}&-4\\-\frac{1}{3}&\frac{11}{3}&2 \end{bmatrix}$

Given matrix:

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

To Find:

The inverse of matrix A (A^–1), if it exists.

Solution:

For a matrix to have an inverse, its determinant must be non-zero.

Calculate the determinant of A, |A|:

Expanding along the first column:

$|A| = 1 \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} - 0 \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} + 3 \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix}$

$|A| = 1( (2)(4) - (-3)(-2) ) - 0 + 3( (-1)(-3) - (2)(2) )$

$|A| = (8 - 6) + 3(3 - 4)$

$|A| = 2 + 3(-1)$

$|A| = 2 - 3$

$|A| = -1$

Since $|A| = -1 \neq 0$, the inverse of matrix A exists.

To find the inverse $A^{-1}$, we use the formula $A^{-1} = \frac{1}{|A|} \text{ adj } A$.

First, calculate the cofactors of the elements of A:

$C_{11} = (-1)^{1+1} \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} = +( (2)(4) - (-3)(-2) ) = 8 - 6 = 2$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 0 & -3 \\ 3 & 4 \end{vmatrix} = -( (0)(4) - (-3)(3) ) = -(0 + 9) = -9$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 0 & 2 \\ 3 & -2 \end{vmatrix} = +( (0)(-2) - (2)(3) ) = 0 - 6 = -6$

$C_{21} = (-1)^{2+1} \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} = -( (-1)(4) - (2)(-2) ) = -(-4 + 4) = 0$

$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = +( (1)(4) - (2)(3) ) = 4 - 6 = -2$

$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix} = -( (1)(-2) - (-1)(3) ) = -(-2 + 3) = -(1) = -1$

$C_{31} = (-1)^{3+1} \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix} = +( (-1)(-3) - (2)(2) ) = 3 - 4 = -1$

$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = -( (1)(-3) - (2)(0) ) = -(-3 - 0) = 3$

$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} = +( (1)(2) - (-1)(0) ) = 2 - 0 = 2$

The matrix of cofactors is:

$\text{Cofactor matrix} = \begin{bmatrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{bmatrix}$

The adjoint of A is the transpose of the cofactor matrix:

$\text{adj A} = (\text{Cofactor matrix})^T = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$

Now, use the formula for the inverse:

$A^{-1} = \frac{1}{|A|} \text{ adj A} = \frac{1}{-1} \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$

$A^{-1} = -1 \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$

Multiply each element by -1:

$A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$

Final Answer:

The inverse of the given matrix is $\mathbf{A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}}$.

Given matrix:

$A = \begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&−\cos\alpha \end{bmatrix}$

To Find:

The inverse of matrix A (A^–1), if it exists.

Solution:

For a matrix to have an inverse, its determinant must be non-zero.

Calculate the determinant of A, |A|:

Expanding along the first row:

$|A| = 1 \begin{vmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix} - 0 \begin{vmatrix} 0 & \sin\alpha \\ 0 & -\cos\alpha \end{vmatrix} + 0 \begin{vmatrix} 0 & \cos\alpha \\ 0 & \sin\alpha \end{vmatrix}$

$|A| = 1( (\cos\alpha)(-\cos\alpha) - (\sin\alpha)(\sin\alpha) ) - 0 + 0$

$|A| = (-\cos^2\alpha - \sin^2\alpha)$

$|A| = -(\cos^2\alpha + \sin^2\alpha)$

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

$|A| = -(1)$

$|A| = -1$

Since $|A| = -1 \neq 0$, the inverse of matrix A exists.

To find the inverse $A^{-1}$, we use the formula $A^{-1} = \frac{1}{|A|} \text{ adj } A$.

First, calculate the cofactors of the elements of A:

$C_{11} = (-1)^{1+1} \begin{vmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix} = (\cos\alpha)(-\cos\alpha) - (\sin\alpha)(\sin\alpha) = -\cos^2\alpha - \sin^2\alpha = -1$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 0 & \sin\alpha \\ 0 & -\cos\alpha \end{vmatrix} = -((0)(-\cos\alpha) - (\sin\alpha)(0)) = 0$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 0 & \cos\alpha \\ 0 & \sin\alpha \end{vmatrix} = (0)(\sin\alpha) - (\cos\alpha)(0) = 0$

$C_{21} = (-1)^{2+1} \begin{vmatrix} 0 & 0 \\ \sin\alpha & -\cos\alpha \end{vmatrix} = -((0)(-\cos\alpha) - (0)(\sin\alpha)) = 0$

$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 0 \\ 0 & -\cos\alpha \end{vmatrix} = (1)(-\cos\alpha) - (0)(0) = -\cos\alpha$

$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 0 \\ 0 & \sin\alpha \end{vmatrix} = -((1)(\sin\alpha) - (0)(0)) = -\sin\alpha$

$C_{31} = (-1)^{3+1} \begin{vmatrix} 0 & 0 \\ \cos\alpha & \sin\alpha \end{vmatrix} = (0)(\sin\alpha) - (0)(\cos\alpha) = 0$

$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 0 \\ 0 & \sin\alpha \end{vmatrix} = -((1)(\sin\alpha) - (0)(0)) = -\sin\alpha$

$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 0 \\ 0 & \cos\alpha \end{vmatrix} = (1)(\cos\alpha) - (0)(0) = \cos\alpha$

The matrix of cofactors is:

$\text{Cofactor matrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$

The adjoint of A is the transpose of the cofactor matrix:

$\text{adj A} = (\text{Cofactor matrix})^T = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$

Now, use the formula for the inverse:

$A^{-1} = \frac{1}{|A|} \text{ adj A} = \frac{1}{-1} \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$

$A^{-1} = -1 \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$

Multiply each element by -1:

$A^{-1} = \begin{bmatrix} (-1)(-1) & (-1)(0) & (-1)(0) \\ (-1)(0) & (-1)(-\cos\alpha) & (-1)(-\sin\alpha) \\ (-1)(0) & (-1)(-\sin\alpha) & (-1)(\cos\alpha) \end{bmatrix}$

$A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$

Final Answer:

The inverse of the given matrix is $\mathbf{A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}}$.

Given:

Matrices A = $\begin{bmatrix}3&7\\2&5 \end{bmatrix}$ and B = $\begin{bmatrix}6&8\\7&9 \end{bmatrix}$.

To Verify:

$(AB)^{-1} = B^{-1}A^{-1}$.

Solution:

Step 1: Calculate AB

$AB = \begin{bmatrix}3&7\\2&5 \end{bmatrix} \begin{bmatrix}6&8\\7&9 \end{bmatrix}$

$AB = \begin{bmatrix} (3)(6)+(7)(7) & (3)(8)+(7)(9) \\ (2)(6)+(5)(7) & (2)(8)+(5)(9) \end{bmatrix}$

$AB = \begin{bmatrix} 18+49 & 24+63 \\ 12+35 & 16+45 \end{bmatrix}$

$AB = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$

Step 2: Calculate (AB)$^{-1}$

Let $C = AB = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$.

Determinant of C:

$|C| = |AB| = (67)(61) - (87)(47)$

$|C| = 4087 - 4089$

$|C| = -2$

Since $|C| = -2 \neq 0$, the inverse exists.

Adjoint of C:

$\text{adj } C = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$

Inverse of C:

$(AB)^{-1} = C^{-1} = \frac{1}{|C|} \text{ adj } C = \frac{1}{-2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$

$(AB)^{-1} = \begin{bmatrix} \frac{61}{-2} & \frac{-87}{-2} \\ \frac{-47}{-2} & \frac{67}{-2} \end{bmatrix} = \begin{bmatrix} \frac{-61}{2} & \frac{87}{2} \\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix}$

Step 3: Calculate A$^{-1}$

$A = \begin{bmatrix}3&7\\2&5 \end{bmatrix}$

Determinant of A:

$|A| = (3)(5) - (7)(2) = 15 - 14 = 1$

Since $|A| = 1 \neq 0$, the inverse exists.

Adjoint of A:

$\text{adj } A = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$

Inverse of A:

$A^{-1} = \frac{1}{|A|} \text{ adj } A = \frac{1}{1} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$

Step 4: Calculate B$^{-1}$

$B = \begin{bmatrix}6&8\\7&9 \end{bmatrix}$

Determinant of B:

$|B| = (6)(9) - (8)(7) = 54 - 56 = -2$

Since $|B| = -2 \neq 0$, the inverse exists.

Adjoint of B:

$\text{adj } B = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$

Inverse of B:

$B^{-1} = \frac{1}{|B|} \text{ adj } B = \frac{1}{-2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$

$B^{-1} = \begin{bmatrix} \frac{9}{-2} & \frac{-8}{-2} \\ \frac{-7}{-2} & \frac{6}{-2} \end{bmatrix} = \begin{bmatrix} \frac{-9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix}$

Step 5: Calculate B$^{-1}$A$^{-1}$

$B^{-1}A^{-1} = \begin{bmatrix} \frac{-9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$

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

$B^{-1}A^{-1} = \begin{bmatrix} \frac{-45}{2} - 8 & \frac{63}{2} + 12 \\ \frac{35}{2} + 6 & \frac{-49}{2} - 9 \end{bmatrix}$

$B^{-1}A^{-1} = \begin{bmatrix} \frac{-45}{2} - \frac{16}{2} & \frac{63}{2} + \frac{24}{2} \\ \frac{35}{2} + \frac{12}{2} & \frac{-49}{2} - \frac{18}{2} \end{bmatrix}$

$B^{-1}A^{-1} = \begin{bmatrix} \frac{-61}{2} & \frac{87}{2} \\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix}$

Step 6: Verification

From Step 2, we have $(AB)^{-1} = \begin{bmatrix} \frac{-61}{2} & \frac{87}{2} \\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix}$.

From Step 5, we have $B^{-1}A^{-1} = \begin{bmatrix} \frac{-61}{2} & \frac{87}{2} \\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix}$.

Since $(AB)^{-1}$ and $B^{-1}A^{-1}$ are equal, the property $(AB)^{-1} = B^{-1}A^{-1}$ is verified.

Given matrix:

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

Identity matrix of order 2:

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

Zero matrix of order 2:

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

To Show:

$A^2 – 5A + 7I = O$

To Find:

A^–1 using the given equation.

Solution:

Part 1: Show that A satisfies the equation A² – 5A + 7I = O

First, calculate $A^2 = A \times A$:

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

$A^2 = \begin{bmatrix} (3)(3)+(1)(-1) & (3)(1)+(1)(2) \\ (-1)(3)+(2)(-1) & (-1)(1)+(2)(2) \end{bmatrix}$

$A^2 = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix}$

$A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

Next, calculate $5A$:

$5A = 5 \begin{bmatrix}3&1\\−1&2 \end{bmatrix} = \begin{bmatrix}5 \times 3 & 5 \times 1 \\ 5 \times -1 & 5 \times 2 \end{bmatrix}$

$5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$

Calculate $7I$:

$7I = 7 \begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}7 \times 1 & 7 \times 0 \\ 7 \times 0 & 7 \times 1 \end{bmatrix}$

$7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

Now, substitute $A^2$, $5A$, and $7I$ into the expression $A^2 – 5A + 7I$:

$A^2 – 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$A^2 – 5A + 7I = \begin{bmatrix} 8-15 & 5-5 \\ -5-(-5) & 3-10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$A^2 – 5A + 7I = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$A^2 – 5A + 7I = \begin{bmatrix} -7+7 & 0+0 \\ 0+0 & -7+7 \end{bmatrix}$

$A^2 – 5A + 7I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Since the result is the zero matrix $O$, the matrix A satisfies the equation $A^2 – 5A + 7I = O$.

Part 2: Find A^–1 using the equation

We have the equation:

We want to find A^–1. First, check if A is invertible by calculating its determinant:

$|A| = (3)(2) - (1)(-1) = 6 - (-1) = 6 + 1 = 7$

Since $|A| = 7 \neq 0$, A is invertible and A^–1 exists.

Multiply equation (i) by A^–1 on the right:

Using the properties of matrix multiplication ($A^2 A^{-1} = A$, $A A^{-1} = I$, $I A^{-1} = A^{-1}$, $OA^{-1} = O$):

Now, isolate $7A^{-1}$:

Divide by 7:

Substitute the matrices for $I$ and $A$ into equation (vi):

$A^{-1} = \frac{1}{7} \left( 5 \begin{bmatrix}1&0\\0&1 \end{bmatrix} - \begin{bmatrix}3&1\\−1&2 \end{bmatrix} \right)$

$A^{-1} = \frac{1}{7} \left( \begin{bmatrix}5&0\\0&5 \end{bmatrix} - \begin{bmatrix}3&1\\−1&2 \end{bmatrix} \right)$

$A^{-1} = \frac{1}{7} \begin{bmatrix} 5-3 & 0-1 \\ 0-(-1) & 5-2 \end{bmatrix}$

$A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

Distribute the scalar $\frac{1}{7}$:

$A^{-1} = \begin{bmatrix} 2/7 & -1/7 \\ 1/7 & 3/7 \end{bmatrix}$

Final Answer:

The matrix A satisfies the equation $A^2 – 5A + 7I = O$.

Using this equation, the inverse of matrix A is $\mathbf{A^{-1} = \begin{bmatrix} 2/7 & -1/7 \\ 1/7 & 3/7 \end{bmatrix}}$.

Given matrix:

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

Given equation:

$A^2 + aA + bI = O$

where $I = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$ is the 2 × 2 identity matrix and $O = \begin{bmatrix}0&0\\0&0 \end{bmatrix}$ is the 2 × 2 zero matrix.

To Find:

The values of numbers a and b.

Solution:

First, calculate $A^2 = A \times A$:

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

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

$A^2 = \begin{bmatrix} 9+2 & 6+2 \\ 3+1 & 2+1 \end{bmatrix}$

Next, calculate $aA$:

$aA = a \begin{bmatrix}3&2\\1&1 \end{bmatrix}$

Calculate $bI$:

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

Substitute $A^2$, $aA$, $bI$, and $O$ into the given equation $A^2 + aA + bI = O$:

Perform the matrix addition on the left side of equation (iv):

Equating the corresponding elements of the matrices in equation (vi), we get a system of linear equations:

From equation (viii), we have:

$2a = -8$

$a = \frac{-8}{2}$

$a = -4$

From equation (ix), we have:

$a = -4$

Both equations (viii) and (ix) give the same value for $a$.

Now substitute the value of $a = -4$ into equation (vii):

$11 + 3(-4) + b = 0$

$11 - 12 + b = 0$

$-1 + b = 0$

$b = 1$

Now substitute the value of $a = -4$ into equation (x) to verify the value of $b$:

$3 + (-4) + b = 0$

$3 - 4 + b = 0$

$-1 + b = 0$

$b = 1$

Both equations (vii) and (x) give the same value for $b$.

Thus, the values of a and b that satisfy the equation are $a=-4$ and $b=1$.

Final Answer:

The numbers a and b such that A² + aA + bI = O are $\mathbf{a = -4}$ and $\mathbf{b = 1}$.

Solution:

The given matrix is:

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

Verification of $A^3 – 6A^2 + 5A + 11 I = O$

First, we calculate $A^2$:

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

$A^2 = \begin{bmatrix} (1)(1)+(1)(1)+(1)(2) & (1)(1)+(1)(2)+(1)(-1) & (1)(1)+(1)(-3)+(1)(3) \\ (1)(1)+(2)(1)+(-3)(2) & (1)(1)+(2)(2)+(-3)(-1) & (1)(1)+(2)(-3)+(-3)(3) \\ (2)(1)+(-1)(1)+(3)(2) & (2)(1)+(-1)(2)+(3)(-1) & (2)(1)+(-1)(-3)+(3)(3) \end{bmatrix}$

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

$A^2 = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}$

Next, we calculate $A^3$:

$A^3 = A^2 \times A = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} \begin{bmatrix}1&1&1\\1&2&−3\\2&−1&3 \end{bmatrix}$

$A^3 = \begin{bmatrix} (4)(1)+(2)(1)+(1)(2) & (4)(1)+(2)(2)+(1)(-1) & (4)(1)+(2)(-3)+(1)(3) \\ (-3)(1)+(8)(1)+(-14)(2) & (-3)(1)+(8)(2)+(-14)(-1) & (-3)(1)+(8)(-3)+(-14)(3) \\ (7)(1)+(-3)(1)+(14)(2) & (7)(1)+(-3)(2)+(14)(-1) & (7)(1)+(-3)(-3)+(14)(3) \end{bmatrix}$

$A^3 = \begin{bmatrix} 4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+16+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42 \end{bmatrix}$

$A^3 = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}$

Now, we calculate $6A^2$, $5A$, and $11I$:

$6A^2 = 6 \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} = \begin{bmatrix} 24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84 \end{bmatrix}$

$5A = 5 \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15 \end{bmatrix}$

$11I = 11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$

Now, we compute $A^3 – 6A^2 + 5A + 11 I$:

$A^3 – 6A^2 + 5A + 11 I = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix} - \begin{bmatrix} 24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84 \end{bmatrix} + \begin{bmatrix} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15 \end{bmatrix} + \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$

$= \begin{bmatrix} 8 - 24 + 5 + 11 & 7 - 12 + 5 + 0 & 1 - 6 + 5 + 0 \\ -23 - (-18) + 5 + 0 & 27 - 48 + 10 + 11 & -69 - (-84) - 15 + 0 \\ 32 - 42 + 10 + 0 & -13 - (-18) - 5 + 0 & 58 - 84 + 15 + 11 \end{bmatrix}$

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

Thus, $A^3 – 6A^2 + 5A + 11 I = O$ is verified.

Finding $A^{-1}$ using the equation $A^3 – 6A^2 + 5A + 11 I = O$

We have the equation:

$A^3 – 6A^2 + 5A + 11 I = O$

Multiply by $A^{-1}$ on both sides (since $\det(A) = -11 \neq 0$, $A^{-1}$ exists):

$A^{-1}(A^3 – 6A^2 + 5A + 11 I) = A^{-1}O$

$A^{-1}A^3 – 6A^{-1}A^2 + 5A^{-1}A + 11A^{-1}I = O$

Using the properties $A^{-1}A = I$ and $IA = A$, $I^2=I$, $AI = A$, $A^{-1}I = A^{-1}$:

$(A^{-1}A)A^2 – 6(A^{-1}A)A + 5(A^{-1}A) + 11A^{-1} = O$

$I A^2 – 6 I A + 5 I + 11A^{-1} = O$

$A^2 – 6A + 5I + 11A^{-1} = O$

Now, isolate $11A^{-1}$:

$11A^{-1} = -A^2 + 6A - 5I$

$A^{-1} = \frac{1}{11}(-A^2 + 6A - 5I)$

We already have $A^2$, $A$, and $I$. We calculate $-A^2$, $6A$, and $-5I$:

$-A^2 = - \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} = \begin{bmatrix} -4 & -2 & -1 \\ 3 & -8 & 14 \\ -7 & 3 & -14 \end{bmatrix}$

$6A = \begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix}$

$-5I = \begin{bmatrix} -5 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{bmatrix}$

Now, we compute $-A^2 + 6A - 5I$:

$-A^2 + 6A - 5I = \begin{bmatrix} -4 & -2 & -1 \\ 3 & -8 & 14 \\ -7 & 3 & -14 \end{bmatrix} + \begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix} + \begin{bmatrix} -5 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{bmatrix}$

$= \begin{bmatrix} -4 + 6 - 5 & -2 + 6 + 0 & -1 + 6 + 0 \\ 3 + 6 + 0 & -8 + 12 - 5 & 14 - 18 + 0 \\ -7 + 12 + 0 & 3 - 6 + 0 & -14 + 18 - 5 \end{bmatrix}$

$= \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}$

Therefore, $A^{-1}$ is:

$A^{-1} = \frac{1}{11}(-A^2 + 6A - 5I) = \frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}$

Solution:

The given matrix is:

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

Verification of $A^3 – 6A^2 + 9A – 4I = O$

First, we calculate $A^2$:

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

$A^2 = \begin{bmatrix} (2)(2)+(-1)(-1)+(1)(1) & (2)(-1)+(-1)(2)+(1)(-1) & (2)(1)+(-1)(-1)+(1)(2) \\ (-1)(2)+(2)(-1)+(-1)(1) & (-1)(-1)+(2)(2)+(-1)(-1) & (-1)(1)+(2)(-1)+(-1)(2) \\ (1)(2)+(-1)(-1)+(2)(1) & (1)(-1)+(-1)(2)+(2)(-1) & (1)(1)+(-1)(-1)+(2)(2) \end{bmatrix}$

$A^2 = \begin{bmatrix} 4+1+1 & -2-2-1 & 2+1+2 \\ -2-2-1 & 1+4+1 & -1-2-2 \\ 2+1+2 & -1-2-2 & 1+1+4 \end{bmatrix}$

$A^2 = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}$

Next, we calculate $A^3$:

$A^3 = A^2 \times A = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} \begin{bmatrix}2&−1&1\\−1&2&−1\\1&−1&2 \end{bmatrix}$

$A^3 = \begin{bmatrix} (6)(2)+(-5)(-1)+(5)(1) & (6)(-1)+(-5)(2)+(5)(-1) & (6)(1)+(-5)(-1)+(5)(2) \\ (-5)(2)+(6)(-1)+(-5)(1) & (-5)(-1)+(6)(2)+(-5)(-1) & (-5)(1)+(6)(-1)+(-5)(2) \\ (5)(2)+(-5)(-1)+(6)(1) & (5)(-1)+(-5)(2)+(6)(-1) & (5)(1)+(-5)(-1)+(6)(2) \end{bmatrix}$

$A^3 = \begin{bmatrix} 12+5+5 & -6-10-5 & 6+5+10 \\ -10-6-5 & 5+12+5 & -5-6-10 \\ 10+5+6 & -5-10-6 & 5+5+12 \end{bmatrix}$

$A^3 = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix}$

Now, we calculate $6A^2$, $9A$, and $4I$:

$6A^2 = 6 \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} = \begin{bmatrix} 36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36 \end{bmatrix}$

$9A = 9 \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} = \begin{bmatrix} 18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18 \end{bmatrix}$

$4I = 4 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$

Now, we compute $A^3 – 6A^2 + 9A – 4I$:

$A^3 – 6A^2 + 9A – 4I = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} - \begin{bmatrix} 36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36 \end{bmatrix} + \begin{bmatrix} 18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$

$= \begin{bmatrix} 22 - 36 + 18 - 4 & -21 - (-30) - 9 - 0 & 21 - 30 + 9 - 0 \\ -21 - (-30) - 9 - 0 & 22 - 36 + 18 - 4 & -21 - (-30) - 9 - 0 \\ 21 - 30 + 9 - 0 & -21 - (-30) - 9 - 0 & 22 - 36 + 18 - 4 \end{bmatrix}$

$= \begin{bmatrix} 40 - 40 & -21 + 30 - 9 & 30 - 30 \\ -21 + 30 - 9 & 40 - 40 & -21 + 30 - 9 \\ 30 - 30 & -21 + 30 - 9 & 40 - 40 \end{bmatrix}$

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

Thus, $A^3 – 6A^2 + 9A – 4I = O$ is verified.

Finding $A^{-1}$ using the equation $A^3 – 6A^2 + 9A – 4I = O$

We have the equation:

$A^3 – 6A^2 + 9A – 4I = O$

Multiply by $A^{-1}$ on both sides (we can do this because $\det(A) = 4 \neq 0$, so $A^{-1}$ exists):

$A^{-1}(A^3 – 6A^2 + 9A – 4I) = A^{-1}O$

$A^{-1}A^3 – 6A^{-1}A^2 + 9A^{-1}A – 4A^{-1}I = O$

Using the properties $A^{-1}A = I$ and $IA = A$, $I^2=I$, $AI = A$, $A^{-1}I = A^{-1}$:

$(A^{-1}A)A^2 – 6(A^{-1}A)A + 9(A^{-1}A) – 4A^{-1} = O$

$I A^2 – 6 I A + 9 I – 4A^{-1} = O$

$A^2 – 6A + 9I – 4A^{-1} = O$

Now, isolate $4A^{-1}$:

$4A^{-1} = A^2 – 6A + 9I$

$A^{-1} = \frac{1}{4}(A^2 – 6A + 9I)$

We already have $A^2$, $A$, and $I$. We calculate $6A$ and $9I$:

$A^2 = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}$

$6A = 6 \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} = \begin{bmatrix} 12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12 \end{bmatrix}$

$9I = 9 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$

Now, we compute $A^2 – 6A + 9I$:

$A^2 – 6A + 9I = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} - \begin{bmatrix} 12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$

$= \begin{bmatrix} 6 - 12 + 9 & -5 - (-6) + 0 & 5 - 6 + 0 \\ -5 - (-6) + 0 & 6 - 12 + 9 & -5 - (-6) + 0 \\ 5 - 6 + 0 & -5 - (-6) + 0 & 6 - 12 + 9 \end{bmatrix}$

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

Therefore, $A^{-1}$ is:

$A^{-1} = \frac{1}{4}(A^2 – 6A + 9I) = \frac{1}{4} \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$

Solution:

Let A be a nonsingular square matrix of order $n \times n$. The given matrix A is of order $3 \times 3$, so $n=3$.

We know the property relating a matrix, its adjoint, and its determinant:

$A (\text{adj} A) = (\text{adj} A) A = |A| I$

Taking the determinant of both sides of $A (\text{adj} A) = |A| I$:

$|A (\text{adj} A)| = ||A| I|$

Using the property $|AB| = |A||B|$ on the left side and the property $|kI| = k^n$ on the right side, where $k = |A|$ and $n$ is the order of the matrix (here $n=3$):

$|A| |\text{adj} A| = |A|^3$

Since A is a nonsingular matrix, $|A| \neq 0$. We can divide both sides by $|A|$:

$|\text{adj} A| = \frac{|A|^3}{|A|}$

$|\text{adj} A| = |A|^{3-1}$

$|\text{adj} A| = |A|^2$

Thus, for a nonsingular square matrix A of order $3 \times 3$, $|adj A|$ is equal to $|A|^2$.

The correct option is (B) $| A |^{2}$.

Solution:

Let A be an invertible matrix of order 2. This means A is a $2 \times 2$ matrix and its inverse $A^{-1}$ exists.

By the definition of an inverse matrix, we have:

$A A^{-1} = I$

where I is the identity matrix of the same order as A, which is a $2 \times 2$ identity matrix in this case:

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

Taking the determinant of both sides of the equation $A A^{-1} = I$:

$|A A^{-1}| = |I|$

Using the property that the determinant of a product of matrices is the product of their determinants, i.e., $|AB| = |A||B|$, we get:

$|A| |A^{-1}| = |I|$

The determinant of the identity matrix $I$ is always 1, regardless of its order. So, $|I|=1$.

$|A| |A^{-1}| = 1$

Since A is an invertible matrix, it is non-singular, which means its determinant $|A| \neq 0$. Therefore, we can divide both sides of the equation by $|A|$:

$|A^{-1}| = \frac{1}{|A|}$

Thus, the determinant of the inverse of an invertible matrix A is equal to the reciprocal of the determinant of A.

The correct option is (B) $\frac{1}{det (A)}$.

Solution:

The given system of linear equations is:

$2x + 5y = 1$

$3x + 2y = 7$

This system can be written in matrix form as $AX = B$, where:

$A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}$

To solve for the variables x and y, we need to find the inverse of matrix A, if it exists. The solution is given by $X = A^{-1}B$.

First, we calculate the determinant of matrix A:

$\det(A) = (2)(2) - (5)(3) = 4 - 15 = -11$

Since $\det(A) = -11 \neq 0$, matrix A is nonsingular, and its inverse $A^{-1}$ exists. Therefore, the system has a unique solution.

Next, we find the adjoint of matrix A. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

$\text{adj} A = \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$

Now, we calculate the inverse of matrix A using the formula $A^{-1} = \frac{1}{\det(A)} \text{adj} A$:

$A^{-1} = \frac{1}{-11} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{2}{11} & \frac{5}{11} \\ \frac{3}{11} & -\frac{2}{11} \end{bmatrix}$

Finally, we find the matrix X by multiplying $A^{-1}$ by B:

$X = A^{-1}B = \begin{bmatrix} -\frac{2}{11} & \frac{5}{11} \\ \frac{3}{11} & -\frac{2}{11} \end{bmatrix} \begin{bmatrix} 1 \\ 7 \end{bmatrix}$

$X = \begin{bmatrix} (-\frac{2}{11})(1) + (\frac{5}{11})(7) \\ (\frac{3}{11})(1) + (-\frac{2}{11})(7) \end{bmatrix}$

$X = \begin{bmatrix} -\frac{2}{11} + \frac{35}{11} \\ \frac{3}{11} - \frac{14}{11} \end{bmatrix}$

$X = \begin{bmatrix} \frac{-2+35}{11} \\ \frac{3-14}{11} \end{bmatrix}$

$X = \begin{bmatrix} \frac{33}{11} \\ \frac{-11}{11} \end{bmatrix}$

$X = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$

Since $X = \begin{bmatrix} x \\ y \end{bmatrix}$, by comparing the elements, we get:

$x = 3$

$y = -1$

Thus, the solution to the given system of equations is $x=3$ and $y=-1$.

Solution:

The given system of linear equations is:

$3x – 2y + 3z = 8$

$2x + y – z = 1$

$4x – 3y + 2z = 4$

This system can be written in matrix form as $AX = B$, where:

$A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$

To solve for the variables x, y, and z, we need to find the inverse of matrix A, if it exists. The solution is given by $X = A^{-1}B$.

First, we calculate the determinant of matrix A:

$\det(A) = 3((1)(2) - (-1)(-3)) - (-2)((2)(2) - (-1)(4)) + 3((2)(-3) - (1)(4))$

$\det(A) = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4)$

$\det(A) = 3(-1) + 2(8) + 3(-10)$

$\det(A) = -3 + 16 - 30$

$\det(A) = 13 - 30$

$\det(A) = -17$

Since $\det(A) = -17 \neq 0$, matrix A is nonsingular, and its inverse $A^{-1}$ exists. Therefore, the system has a unique solution.

Next, we find the adjoint of matrix A. This requires finding the matrix of cofactors.

The cofactor $C_{ij}$ is given by $(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the determinant of the minor matrix obtained by removing the $i$-th row and $j$-th column.

$C_{11} = (-1)^{1+1} \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} = (1)((1)(2) - (-1)(-3)) = 2 - 3 = -1$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} = (-1)((2)(2) - (-1)(4)) = -(4 + 4) = -8$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} = (1)((2)(-3) - (1)(4)) = -6 - 4 = -10$

$C_{21} = (-1)^{2+1} \begin{vmatrix} -2 & 3 \\ -3 & 2 \end{vmatrix} = (-1)((-2)(2) - (3)(-3)) = -(-4 + 9) = -5$

$C_{22} = (-1)^{2+2} \begin{vmatrix} 3 & 3 \\ 4 & 2 \end{vmatrix} = (1)((3)(2) - (3)(4)) = 6 - 12 = -6$

$C_{23} = (-1)^{2+3} \begin{vmatrix} 3 & -2 \\ 4 & -3 \end{vmatrix} = (-1)((3)(-3) - (-2)(4)) = -(-9 + 8) = -(-1) = 1$

$C_{31} = (-1)^{3+1} \begin{vmatrix} -2 & 3 \\ 1 & -1 \end{vmatrix} = (1)((-2)(-1) - (3)(1)) = 2 - 3 = -1$

$C_{32} = (-1)^{3+2} \begin{vmatrix} 3 & 3 \\ 2 & -1 \end{vmatrix} = (-1)((3)(-1) - (3)(2)) = -(-3 - 6) = -(-9) = 9$

$C_{33} = (-1)^{3+3} \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} = (1)((3)(1) - (-2)(2)) = 3 - (-4) = 3 + 4 = 7$

The matrix of cofactors C is:

$C = \begin{bmatrix} -1 & -8 & -10 \\ -5 & -6 & 1 \\ -1 & 9 & 7 \end{bmatrix}$

The adjoint of A is the transpose of the cofactor matrix:

$\text{adj} A = C^T = \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}$

Now, we calculate the inverse of matrix A using the formula $A^{-1} = \frac{1}{\det(A)} \text{adj} A$:

$A^{-1} = \frac{1}{-17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} = \begin{bmatrix} \frac{1}{17} & \frac{5}{17} & \frac{1}{17} \\ \frac{8}{17} & \frac{6}{17} & -\frac{9}{17} \\ \frac{10}{17} & -\frac{1}{17} & -\frac{7}{17} \end{bmatrix}$

Finally, we find the matrix X by multiplying $A^{-1}$ by B:

$X = A^{-1}B = \frac{1}{-17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$

$X = \frac{1}{-17} \begin{bmatrix} (-1)(8) + (-5)(1) + (-1)(4) \\ (-8)(8) + (-6)(1) + (9)(4) \\ (-10)(8) + (1)(1) + (7)(4) \end{bmatrix}$

$X = \frac{1}{-17} \begin{bmatrix} -8 - 5 - 4 \\ -64 - 6 + 36 \\ -80 + 1 + 28 \end{bmatrix}$

$X = \frac{1}{-17} \begin{bmatrix} -17 \\ -70 + 36 \\ -79 + 28 \end{bmatrix}$

$X = \frac{1}{-17} \begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix}$

$X = \begin{bmatrix} \frac{-17}{-17} \\ \frac{-34}{-17} \\ \frac{-51}{-17} \end{bmatrix}$

$X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

Since $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, by comparing the elements, we get:

$x = 1$

$y = 2$

$z = 3$

Thus, the solution to the given system of equations is $x=1$, $y=2$, and $z=3$.

Solution:

Let the three numbers be $x$, $y$, and $z$.

We represent the given information algebraically as a system of linear equations:

1. The sum of three numbers is 6:

$x + y + z = 6$

2. If we multiply third number by 3 and add second number to it, we get 11:

$y + 3z = 11$

This can be written as $0x + y + 3z = 11$.

3. By adding first and third numbers, we get double of the second number:

$x + z = 2y$

This can be rearranged as $x - 2y + z = 0$.

The system of equations is:

$x + y + z = 6$

$0x + y + 3z = 11$

$x - 2y + z = 0$

This system can be written in matrix form as $AX = B$, where:

$A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}$

To solve for the variables x, y, and z, we need to find the inverse of matrix A, if it exists. The solution is given by $X = A^{-1}B$.

First, we calculate the determinant of matrix A:

$\det(A) = 1 \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix}$

$\det(A) = 1((1)(1) - (3)(-2)) - 1((0)(1) - (3)(1)) + 1((0)(-2) - (1)(1))$

$\det(A) = 1(1 + 6) - 1(0 - 3) + 1(0 - 1)$

$\det(A) = 1(7) - 1(-3) + 1(-1)$

$\det(A) = 7 + 3 - 1$

$\det(A) = 9$

Since $\det(A) = 9 \neq 0$, matrix A is nonsingular, and its inverse $A^{-1}$ exists. Therefore, the system has a unique solution.

Next, we find the adjoint of matrix A. This requires finding the matrix of cofactors.

The cofactors $C_{ij}$ are:

$C_{11} = (-1)^{1+1} \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} = 1(1 - (-6)) = 7$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} = -1(0 - 3) = 3$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix} = 1(0 - 1) = -1$

$C_{21} = (-1)^{2+1} \begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} = -1(1 - (-2)) = -3$

$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 1(1 - 1) = 0$

$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = -1(-2 - 1) = 3$

$C_{31} = (-1)^{3+1} \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} = 1(3 - 1) = 2$

$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} = -1(3 - 0) = -3$

$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1(1 - 0) = 1$

The matrix of cofactors C is:

$C = \begin{bmatrix} 7 & 3 & -1 \\ -3 & 0 & 3 \\ 2 & -3 & 1 \end{bmatrix}$

The adjoint of A is the transpose of the cofactor matrix:

$\text{adj} A = C^T = \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}$

Now, we calculate the inverse of matrix A using the formula $A^{-1} = \frac{1}{\det(A)} \text{adj} A$:

$A^{-1} = \frac{1}{9} \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}$

Finally, we find the matrix X by multiplying $A^{-1}$ by B:

$X = A^{-1}B = \frac{1}{9} \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix} \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}$

$X = \frac{1}{9} \begin{bmatrix} (7)(6) + (-3)(11) + (2)(0) \\ (3)(6) + (0)(11) + (-3)(0) \\ (-1)(6) + (3)(11) + (1)(0) \end{bmatrix}$

$X = \frac{1}{9} \begin{bmatrix} 42 - 33 + 0 \\ 18 + 0 + 0 \\ -6 + 33 + 0 \end{bmatrix}$

$X = \frac{1}{9} \begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix}$

$X = \begin{bmatrix} \frac{9}{9} \\ \frac{18}{9} \\ \frac{27}{9} \end{bmatrix}$

$X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

Since $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, by comparing the elements, we get:

$x = 1$

$y = 2$

$z = 3$

Thus, the three numbers are 1, 2, and 3.

Solution:

The given system of equations is:

$x + 2y = 2$

$2x + 3y = 3$

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$

To examine the consistency of the system, we first calculate the determinant of the coefficient matrix A.

$\det(A) = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix}$

$\det(A) = (1)(3) - (2)(2)$

$\det(A) = 3 - 4$

$\det(A) = -1$

Since $\det(A) = -1 \neq 0$, the matrix A is nonsingular. A system of linear equations $AX=B$ is consistent if and only if $\det(A) \neq 0$ or if $\det(A) = 0$ and $(\text{adj} A)B = O$.

In this case, $\det(A) \neq 0$. Therefore, the system of equations is consistent and has a unique solution.

Solution:

The given system of equations is:

$2x – y = 5$

$x + y = 4$

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$

To examine the consistency of the system, we first calculate the determinant of the coefficient matrix A.

$\det(A) = \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}$

$\det(A) = (2)(1) - (-1)(1)$

$\det(A) = 2 - (-1)$

$\det(A) = 2 + 1$

$\det(A) = 3$

Since $\det(A) = 3 \neq 0$, the matrix A is nonsingular. A system of linear equations $AX=B$ is consistent if and only if $\det(A) \neq 0$ or if $\det(A) = 0$ and $(\text{adj} A)B = O$.

In this case, $\det(A) \neq 0$. Therefore, the system of equations is consistent and has a unique solution.

Solution:

The given system of equations is:

$x + 3y = 5$

$2x + 6y = 8$

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$

To examine the consistency of the system, we first calculate the determinant of the coefficient matrix A.

$\det(A) = \begin{vmatrix} 1 & 3 \\ 2 & 6 \end{vmatrix}$

$\det(A) = (1)(6) - (3)(2)$

$\det(A) = 6 - 6$

$\det(A) = 0$

Since $\det(A) = 0$, the matrix A is singular. In this case, we need to calculate $(\text{adj} A)B$ to determine the consistency.

First, find the adjoint of A. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\text{adj} A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

$\text{adj} A = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$

Next, calculate $(\text{adj} A)B$:

$(\text{adj} A)B = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \end{bmatrix}$

$(\text{adj} A)B = \begin{bmatrix} (6)(5) + (-3)(8) \\ (-2)(5) + (1)(8) \end{bmatrix}$

$(\text{adj} A)B = \begin{bmatrix} 30 - 24 \\ -10 + 8 \end{bmatrix}$

$(\text{adj} A)B = \begin{bmatrix} 6 \\ -2 \end{bmatrix}$

Since $\det(A) = 0$ and $(\text{adj} A)B = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ (the zero matrix O), the system of equations is inconsistent and has no solution.

Solution:

The given system of equations is:

$x + y + z = 1$

$2x + 3y + 2z = 2$

$ax + ay + 2az = 4$

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$

To examine the consistency of the system, we first calculate the determinant of the coefficient matrix A.

$\det(A) = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a \end{vmatrix}$

Using the property that if two rows or columns are identical, the determinant is 0. If we take 'a' common from the third row:

$\det(A) = a \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 1 & 2 \end{vmatrix}$

Expanding along the first row:

$\det(A) = a \left[ 1 \begin{vmatrix} 3 & 2 \\ 1 & 2 \end{vmatrix} - 1 \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} + 1 \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} \right]$

$\det(A) = a \left[ (3)(2) - (2)(1) - ((2)(2) - (2)(1)) + ((2)(1) - (3)(1)) \right]$

$\det(A) = a \left[ (6 - 2) - (4 - 2) + (2 - 3) \right]$

$\det(A) = a \left[ 4 - 2 - 1 \right]$

$\det(A) = a \left[ 1 \right]$

$\det(A) = a$

We now examine the consistency based on the value of $\det(A)$.

Case 1: $\det(A) \neq 0$

This occurs when $a \neq 0$.

If $\det(A) \neq 0$, then matrix A is nonsingular. The system of equations $AX=B$ has a unique solution given by $X = A^{-1}B$.

Therefore, if $a \neq 0$, the system of equations is consistent.

Case 2: $\det(A) = 0$

This occurs when $a = 0$.

If $\det(A) = 0$, then matrix A is singular. In this case, we need to calculate $(\text{adj} A)B$ to determine consistency.

When $a=0$, the matrix A becomes $A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 0 & 0 & 0 \end{bmatrix}$.

The adjoint of A is the transpose of the matrix of cofactors. The cofactors are calculated as follows:

$C_{11} = \begin{vmatrix} 3 & 2 \\ 0 & 0 \end{vmatrix} = 0$, $C_{12} = - \begin{vmatrix} 2 & 2 \\ 0 & 0 \end{vmatrix} = 0$, $C_{13} = \begin{vmatrix} 2 & 3 \\ 0 & 0 \end{vmatrix} = 0$

$C_{21} = - \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} = 0$, $C_{22} = \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} = 0$, $C_{23} = - \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} = 0$

$C_{31} = \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} = 2-3 = -1$, $C_{32} = - \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} = -(2-2) = 0$, $C_{33} = \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = 3-2 = 1$

The matrix of cofactors is $C = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$.

The adjoint of A is $\text{adj} A = C^T = \begin{bmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

Now, we calculate $(\text{adj} A)B$:

$(\text{adj} A)B = \begin{bmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$

$(\text{adj} A)B = \begin{bmatrix} (0)(1) + (0)(2) + (-1)(4) \\ (0)(1) + (0)(2) + (0)(4) \\ (0)(1) + (0)(2) + (1)(4) \end{bmatrix} = \begin{bmatrix} -4 \\ 0 \\ 4 \end{bmatrix}$

Since $(\text{adj} A)B = \begin{bmatrix} -4 \\ 0 \\ 4 \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ (the zero matrix O), the system of equations has no solution when $\det(A) = 0$.

Therefore, if $a = 0$, the system of equations is inconsistent.

Conclusion:

The system of equations is consistent if $a \neq 0$, and inconsistent if $a = 0$.

Solution:

We are given the following system of linear equations:

We can use the elimination method to solve this system.

Subtract Equation (3) from Equation (1) to eliminate the variable $x$:

Now we have a system of two linear equations with two variables, $y$ and $z$, using Equation (2) and Equation (4):

Multiply Equation (2) by 2:

Now compare Equation (4) and Equation (5):

Subtract Equation (5) from Equation (4):

We have arrived at a contradiction ($0 = 1$). This indicates that the system of equations is inconsistent.

Therefore, there is no solution that satisfies all three equations simultaneously.

Given:

The system of linear equations:

$5x - y + 4z = 5$

$2x + 3y + 5z = 2$

$5x - 2y + 6z = -1$

To Examine:

The consistency of the given system of equations.

Solution:

We can write the given system of equations in the matrix form $AX = B$, where:

A is the coefficient matrix,

X is the variable matrix, and

B is the constant matrix.

$A = \begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6 \end{bmatrix}$, $X = \begin{bmatrix}x\\y\\z \end{bmatrix}$, $B = \begin{bmatrix}5\\2\\-1 \end{bmatrix}$

To determine the consistency of the system, we first calculate the determinant of the coefficient matrix A.

det(A) = $\begin{vmatrix}5&-1&4\\2&3&5\\5&-2&6 \end{vmatrix}$

Expanding along the first row:

det(A) = $5 \begin{vmatrix}3&5\\-2&6 \end{vmatrix} - (-1) \begin{vmatrix}2&5\\5&6 \end{vmatrix} + 4 \begin{vmatrix}2&3\\5&-2 \end{vmatrix}$

det(A) = $5((3)(6) - (5)(-2)) + 1((2)(6) - (5)(5)) + 4((2)(-2) - (3)(5))$

det(A) = $5(18 + 10) + 1(12 - 25) + 4(-4 - 15)$

det(A) = $5(28) + 1(-13) + 4(-19)$

det(A) = $140 - 13 - 76$

det(A) = $140 - 89$

det(A) = $51$

Since det(A) = $51 \neq 0$, the matrix A is non-singular.

For a system of linear equations $AX = B$, if det(A) $\neq 0$, the system is consistent and has a unique solution.

Conclusion:

Since the determinant of the coefficient matrix is non-zero (det(A) = 51), the given system of equations is consistent and has a unique solution.

Solution:

The given system of linear equations is:

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$.

To solve for $X$, we need to find the inverse of matrix $A$, denoted by $A^{-1}$, such that $X = A^{-1}B$.

First, calculate the determinant of matrix $A$:

Since $|A| = 1 \neq 0$, the matrix $A$ is non-singular, and its inverse $A^{-1}$ exists.

Next, find the adjoint of matrix $A$, denoted by $\text{adj}(A)$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

So, $\text{adj}(A) = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$.

Now, calculate the inverse of matrix $A$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Finally, find the matrix $X$ by multiplying $A^{-1}$ and $B$ ($X = A^{-1}B$).

By comparing the corresponding elements of the matrices, we get the values of $x$ and $y$.

The solution to the system of equations is $x = 2$ and $y = -3$.

Solution:

The given system of linear equations is:

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} -2 \\ 3 \end{bmatrix}$.

To solve for $X$, we need to find the inverse of matrix $A$, denoted by $A^{-1}$, such that $X = A^{-1}B$.

First, calculate the determinant of matrix $A$:

Since $|A| = 11 \neq 0$, the matrix $A$ is non-singular, and its inverse $A^{-1}$ exists.

Next, find the adjoint of matrix $A$, denoted by $\text{adj}(A)$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

So, $\text{adj}(A) = \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}$.

Now, calculate the inverse of matrix $A$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Finally, find the matrix $X$ by multiplying $A^{-1}$ and $B$ ($X = A^{-1}B$).

By comparing the corresponding elements of the matrices, we get the values of $x$ and $y$.

The solution to the system of equations is $x = -\frac{5}{11}$ and $y = \frac{12}{11}$.

Solution:

The given system of linear equations is:

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$.

To solve for $X$, we need to find the inverse of matrix $A$, denoted by $A^{-1}$, such that $X = A^{-1}B$.

First, calculate the determinant of matrix $A$:

Since $|A| = -11 \neq 0$, the matrix $A$ is non-singular, and its inverse $A^{-1}$ exists.

Next, find the adjoint of matrix $A$, denoted by $\text{adj}(A)$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

So, $\text{adj}(A) = \begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}$.

Now, calculate the inverse of matrix $A$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Finally, find the matrix $X$ by multiplying $A^{-1}$ and $B$ ($X = A^{-1}B$).

By comparing the corresponding elements of the matrices, we get the values of $x$ and $y$.

The solution to the system of equations is $x = -\frac{6}{11}$ and $y = -\frac{19}{11}$.

Solution:

The given system of linear equations is:

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 3 \\ 5 \end{bmatrix}$.

To solve for $X$, we need to find the inverse of matrix $A$, denoted by $A^{-1}$, such that $X = A^{-1}B$.

First, calculate the determinant of matrix $A$:

Since $|A| = 4 \neq 0$, the matrix $A$ is non-singular, and its inverse $A^{-1}$ exists.

Next, find the adjoint of matrix $A$, denoted by $\text{adj}(A)$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

So, $\text{adj}(A) = \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}$.

Now, calculate the inverse of matrix $A$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Finally, find the matrix $X$ by multiplying $A^{-1}$ and $B$ ($X = A^{-1}B$).

By comparing the corresponding elements of the matrices, we get the values of $x$ and $y$.

The solution to the system of equations is $x = -1$ and $y = 4$.

Solution:

The given system of linear equations is:

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 1 \\ 3/2 \\ 9 \end{bmatrix}$.

To solve for $X$, we need to find the inverse of matrix $A$, denoted by $A^{-1}$, such that $X = A^{-1}B$.

First, calculate the determinant of matrix $A$:

Since $|A| = 34 \neq 0$, the matrix $A$ is non-singular, and its inverse $A^{-1}$ exists.

Next, find the adjoint of matrix $A$, denoted by $\text{adj}(A)$. This requires calculating the cofactor matrix $C$ and then transposing it ($C^T$).

The cofactor matrix $C$ is given by $C_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor determinant.

The cofactor matrix is $C = \begin{bmatrix} 13 & 5 & 3 \\ 8 & -10 & -6 \\ 1 & 3 & -5 \end{bmatrix}$.

The adjoint matrix is $\text{adj}(A) = C^T = \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}$.

Now, calculate the inverse of matrix $A$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Finally, find the matrix $X$ by multiplying $A^{-1}$ and $B$ ($X = A^{-1}B$).

By comparing the corresponding elements of the matrices, we get the values of $x$, $y$, and $z$.

The solution to the system of equations is $x = 1, y = \frac{1}{2}, z = -\frac{3}{2}$.

Solution:

The given system of linear equations is:

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$.

To solve for $X$, we need to find the inverse of matrix $A$, denoted by $A^{-1}$, such that $X = A^{-1}B$.

First, calculate the determinant of matrix $A$:

Since $|A| = 10 \neq 0$, the matrix $A$ is non-singular, and its inverse $A^{-1}$ exists.

Next, find the adjoint of matrix $A$, denoted by $\text{adj}(A)$. This requires calculating the cofactor matrix $C$ and then transposing it ($C^T$).

The cofactor matrix $C$ is given by $C_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor determinant.

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

The adjoint matrix is $\text{adj}(A) = C^T = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$.

Now, calculate the inverse of matrix $A$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Finally, find the matrix $X$ by multiplying $A^{-1}$ and $B$ ($X = A^{-1}B$).

By comparing the corresponding elements of the matrices, we get the values of $x$, $y$, and $z$.

The solution to the system of equations is $x = 2, y = -1, z = 1$.

Solution:

The given system of linear equations is:

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}$.

To solve for $X$, we need to find the inverse of matrix $A$, denoted by $A^{-1}$, such that $X = A^{-1}B$.

First, calculate the determinant of matrix $A$:

Since $|A| = 40 \neq 0$, the matrix $A$ is non-singular, and its inverse $A^{-1}$ exists.

Next, find the adjoint of matrix $A$, denoted by $\text{adj}(A)$. This requires calculating the cofactor matrix $C$ and then transposing it ($C^T$).

The cofactor matrix $C$ is given by $C_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor determinant.

The cofactor matrix is $C = \begin{bmatrix} 5 & 5 & 5 \\ 3 & -13 & 11 \\ 9 & 1 & -7 \end{bmatrix}$.

The adjoint matrix is $\text{adj}(A) = C^T = \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}$.

Now, calculate the inverse of matrix $A$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Finally, find the matrix $X$ by multiplying $A^{-1}$ and $B$ ($X = A^{-1}B$).

By comparing the corresponding elements of the matrices, we get the values of $x$, $y$, and $z$.

The solution to the system of equations is $x = 1, y = 2, z = -1$.

Solution:

The given system of linear equations is:

This system can be written in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$.

To solve for $X$, we need to find the inverse of matrix $A$, denoted by $A^{-1}$, such that $X = A^{-1}B$.

First, calculate the determinant of matrix $A$:

Since $|A| = 4 \neq 0$, the matrix $A$ is non-singular, and its inverse $A^{-1}$ exists.

Next, find the adjoint of matrix $A$, denoted by $\text{adj}(A)$. This requires calculating the cofactor matrix $C$ and then transposing it ($C^T$).

The cofactor matrix $C$ is given by $C_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor determinant.

The cofactor matrix is $C = \begin{bmatrix} 7 & -19 & -11 \\ 1 & -1 & -1 \\ -3 & 11 & 7 \end{bmatrix}$.

The adjoint matrix is $\text{adj}(A) = C^T = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$.

Now, calculate the inverse of matrix $A$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Finally, find the matrix $X$ by multiplying $A^{-1}$ and $B$ ($X = A^{-1}B$).

By comparing the corresponding elements of the matrices, we get the values of $x$, $y$, and $z$.

The solution to the system of equations is $x = 2, y = 1, z = 3$.

Given:

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

System of equations:

$2x - 3y + 5z = 11$

$3x + 2y - 4z = -5$

$x + y - 2z = -3$

To Find:

A^–1 and the solution (x, y, z) for the given system of equations using A^–1.

Solution:

First, we find the inverse of matrix A (A^–1).

To find A^–1, we need to calculate the determinant of A, the matrix of cofactors, and the adjoint of A.

Calculate the determinant of A, $|A|$:

$|A| = \begin{vmatrix}2&−3&5\\3&2&−4\\1&1&−2 \end{vmatrix}$

$|A| = 2((2)(-2) - (-4)(1)) - (-3)((3)(-2) - (-4)(1)) + 5((3)(1) - (2)(1))$

$|A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2)$

$|A| = 2(0) + 3(-2) + 5(1)$

$|A| = 0 - 6 + 5$

$|A| = -1$

Since $|A| = -1 \neq 0$, the inverse of A exists.

Calculate the matrix of cofactors, C:

$C_{11} = + \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (-4)(1) = -4 + 4 = 0$

$C_{12} = - \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} = -((3)(-2) - (-4)(1)) = -(-6 + 4) = -(-2) = 2$

$C_{13} = + \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} = (3)(1) - (2)(1) = 3 - 2 = 1$

$C_{21} = - \begin{vmatrix} -3 & 5 \\ 1 & -2 \end{vmatrix} = -((-3)(-2) - (5)(1)) = -(6 - 5) = -1$

$C_{22} = + \begin{vmatrix} 2 & 5 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (5)(1) = -4 - 5 = -9$

$C_{23} = - \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = -((2)(1) - (-3)(1)) = -(2 + 3) = -5$

$C_{31} = + \begin{vmatrix} -3 & 5 \\ 2 & -4 \end{vmatrix} = (-3)(-4) - (5)(2) = 12 - 10 = 2$

$C_{32} = - \begin{vmatrix} 2 & 5 \\ 3 & -4 \end{vmatrix} = -((2)(-4) - (5)(3)) = -(-8 - 15) = -(-23) = 23$

$C_{33} = + \begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix} = (2)(2) - (-3)(3) = 4 + 9 = 13$

The matrix of cofactors is:

$C = \begin{bmatrix}0&2&1\\−1&−9&−5\\2&23&13 \end{bmatrix}$

Calculate the adjoint of A, adj(A), which is the transpose of the cofactor matrix:

$\text{adj(A)} = C^T = \begin{bmatrix}0&−1&2\\2&−9&23\\1&−5&13 \end{bmatrix}$

Calculate A^–1 using the formula $A^{-1} = \frac{1}{|A|} \text{adj(A)}$:

$A^{-1} = \frac{1}{-1} \begin{bmatrix}0&−1&2\\2&−9&23\\1&−5&13 \end{bmatrix}$

$A^{-1} = -1 \begin{bmatrix}0&−1&2\\2&−9&23\\1&−5&13 \end{bmatrix}$

$A^{-1} = \begin{bmatrix}0&1&−2\\−2&9&−23\\−1&5&−13 \end{bmatrix}$

Now, we use A^–1 to solve the system of equations.

The given system of equations can be written in matrix form as $AX = B$, where:

$A = \begin{bmatrix}2&−3&5\\3&2&−4\\1&1&−2 \end{bmatrix}$, $X = \begin{bmatrix}x\\y\\z \end{bmatrix}$, and $B = \begin{bmatrix}11\\−5\\−3 \end{bmatrix}$

To solve for X, we use the relation $X = A^{-1}B$:

$X = \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}0&1&−2\\−2&9&−23\\−1&5&−13 \end{bmatrix} \begin{bmatrix}11\\−5\\−3 \end{bmatrix}$

Perform the matrix multiplication:

$x = (0)(11) + (1)(-5) + (-2)(-3) = 0 - 5 + 6 = 1$

$y = (-2)(11) + (9)(-5) + (-23)(-3) = -22 - 45 + 69 = -67 + 69 = 2$

$z = (-1)(11) + (5)(-5) + (-13)(-3) = -11 - 25 + 39 = -36 + 39 = 3$

So, we have:

$\begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}1\\2\\3 \end{bmatrix}$

Therefore, the solution to the system of equations is $x=1$, $y=2$, and $z=3$.

Given:

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is $\textsf{₹}$ 60.

The cost of 2 kg onion, 4 kg wheat and 6 kg rice is $\textsf{₹}$ 90.

The cost of 6 kg onion, 2 kg wheat and 3 kg rice is $\textsf{₹}$ 70.

To Find:

The cost of each item per kg using the matrix method.

Solution:

Let $x$, $y$, and $z$ be the cost per kg of onion, wheat, and rice, respectively.

Based on the given information, we can write the following system of linear equations:

$4x + 3y + 2z = 60$

$2x + 4y + 6z = 90$

$6x + 2y + 3z = 70$

This system of equations can be written in matrix form as $AX = B$, where:

$A = \begin{bmatrix}4&3&2\\2&4&6\\6&2&3 \end{bmatrix}$, $X = \begin{bmatrix}x\\y\\z \end{bmatrix}$, and $B = \begin{bmatrix}60\\90\\70 \end{bmatrix}$

To solve for X using the matrix method, we need to find the inverse of matrix A (A^–1) and then calculate $X = A^{-1}B$.

First, calculate the determinant of A, $|A|$:

$|A| = \begin{vmatrix}4&3&2\\2&4&6\\6&2&3 \end{vmatrix}$

$|A| = 4((4)(3) - (6)(2)) - 3((2)(3) - (6)(6)) + 2((2)(2) - (4)(6))$

$|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24)$

$|A| = 4(0) - 3(-30) + 2(-20)$

$|A| = 0 + 90 - 40$

$|A| = 50$

Since $|A| = 50 \neq 0$, the inverse of A exists.

Next, calculate the matrix of cofactors, C:

$C_{11} = + \begin{vmatrix} 4 & 6 \\ 2 & 3 \end{vmatrix} = (4)(3) - (6)(2) = 12 - 12 = 0$

$C_{12} = - \begin{vmatrix} 2 & 6 \\ 6 & 3 \end{vmatrix} = -((2)(3) - (6)(6)) = -(6 - 36) = 30$

$C_{13} = + \begin{vmatrix} 2 & 4 \\ 6 & 2 \end{vmatrix} = (2)(2) - (4)(6) = 4 - 24 = -20$

$C_{21} = - \begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix} = -((3)(3) - (2)(2)) = -(9 - 4) = -5$

$C_{22} = + \begin{vmatrix} 4 & 2 \\ 6 & 3 \end{vmatrix} = (4)(3) - (2)(6) = 12 - 12 = 0$

$C_{23} = - \begin{vmatrix} 4 & 3 \\ 6 & 2 \end{vmatrix} = -((4)(2) - (3)(6)) = -(8 - 18) = 10$

$C_{31} = + \begin{vmatrix} 3 & 2 \\ 4 & 6 \end{vmatrix} = (3)(6) - (2)(4) = 18 - 8 = 10$

$C_{32} = - \begin{vmatrix} 4 & 2 \\ 2 & 6 \end{vmatrix} = -((4)(6) - (2)(2)) = -(24 - 4) = -20$

$C_{33} = + \begin{vmatrix} 4 & 3 \\ 2 & 4 \end{vmatrix} = (4)(4) - (3)(2) = 16 - 6 = 10$

The matrix of cofactors is:

$C = \begin{bmatrix}0&30&−20\\−5&0&10\\10&−20&10 \end{bmatrix}$

The adjoint of A, adj(A), is the transpose of the cofactor matrix:

$\text{adj(A)} = C^T = \begin{bmatrix}0&−5&10\\30&0&−20\\−20&10&10 \end{bmatrix}$

Calculate A^–1 using the formula $A^{-1} = \frac{1}{|A|} \text{adj(A)}$:

$A^{-1} = \frac{1}{50} \begin{bmatrix}0&−5&10\\30&0&−20\\−20&10&10 \end{bmatrix}$

Now, solve for X using $X = A^{-1}B$:

$X = \begin{bmatrix}x\\y\\z \end{bmatrix} = \frac{1}{50} \begin{bmatrix}0&−5&10\\30&0&−20\\−20&10&10 \end{bmatrix} \begin{bmatrix}60\\90\\70 \end{bmatrix}$

Perform the matrix multiplication $\begin{bmatrix}0&−5&10\\30&0&−20\\−20&10&10 \end{bmatrix} \begin{bmatrix}60\\90\\70 \end{bmatrix}$:

Row 1: $(0)(60) + (-5)(90) + (10)(70) = 0 - 450 + 700 = 250$

Row 2: $(30)(60) + (0)(90) + (-20)(70) = 1800 + 0 - 1400 = 400$

Row 3: $(-20)(60) + (10)(90) + (10)(70) = -1200 + 900 + 700 = -1200 + 1600 = 400$

So, the result of the multiplication is $\begin{bmatrix}250\\400\\400 \end{bmatrix}$.

Now, multiply by $\frac{1}{50}$:

$X = \begin{bmatrix}x\\y\\z \end{bmatrix} = \frac{1}{50} \begin{bmatrix}250\\400\\400 \end{bmatrix} = \begin{bmatrix}250/50\\400/50\\400/50 \end{bmatrix} = \begin{bmatrix}5\\8\\8 \end{bmatrix}$

Thus, $x = 5$, $y = 8$, and $z = 8$.

Answer:

The cost of onion per kg is $\textsf{₹}5$.

The cost of wheat per kg is $\textsf{₹}8$.

The cost of rice per kg is $\textsf{₹}8$.

Given:

System of equations:

$x - y + 2z = 1$

$2y - 3z = 1$

$3x - 2y + 4z = 2$

Product of matrices: $\begin{bmatrix}1&1&2\\0&2&3\\3&2&4 \end{bmatrix} \begin{bmatrix}2&0&1\\9&2&3\\6&1&2 \end{bmatrix}$

To Find:

The solution (x, y, z) for the given system of equations using the matrix method and the provided product.

Solution:

The given system of equations can be written in matrix form as $CX = D$, where:

$C = \begin{bmatrix}1&−1&2\\0&2&−3\\3&−2&4 \end{bmatrix}$, $X = \begin{bmatrix}x\\y\\z \end{bmatrix}$, and $D = \begin{bmatrix}1\\1\\2 \end{bmatrix}$

To solve the system $CX = D$ using the matrix method, we need to find the inverse of the coefficient matrix C, denoted as $C^{-1}$. The solution is then given by $X = C^{-1}D$. The instruction to "use the product" suggests that the provided matrix product is related to finding $C^{-1}$.

Let's calculate the determinant of C, $|C|$:

$|C| = \begin{vmatrix}1&−1&2\\0&2&−3\\3&−2&4 \end{vmatrix}$

$|C| = 1((2)(4) - (-3)(-2)) - (-1)((0)(4) - (-3)(3)) + 2((0)(-2) - (2)(3))$

$|C| = 1(8 - 6) + 1(0 + 9) + 2(0 - 6)$

$|C| = 1(2) + 1(9) + 2(-6)$

$|C| = 2 + 9 - 12 = -1$

Since $|C| = -1 \neq 0$, the inverse of C exists.

Calculate the matrix of cofactors of C, denoted by Adj(C):

Adj(C)$_{11} = + \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} = 8 - 6 = 2$

Adj(C)$_{12} = - \begin{vmatrix} 0 & -3 \\ 3 & 4 \end{vmatrix} = -(0 - (-9)) = -9$

Adj(C)$_{13} = + \begin{vmatrix} 0 & 2 \\ 3 & -2 \end{vmatrix} = 0 - 6 = -6$

Adj(C)$_{21} = - \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} = -(-4 - (-4)) = 0$

Adj(C)$_{22} = + \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 4 - 6 = -2$

Adj(C)$_{23} = - \begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix} = -(-2 - (-3)) = -1$

Adj(C)$_{31} = + \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix} = 3 - 4 = -1$

Adj(C)$_{32} = - \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = -(-3 - 0) = 3$

Adj(C)$_{33} = + \begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} = 2 - 0 = 2$

The adjoint matrix of C is the transpose of the cofactor matrix:

$\text{adj(C)} = \begin{bmatrix}2&−9&−6\\0&−2&−1\\−1&3&2 \end{bmatrix}^T = \begin{bmatrix}2&0&−1\\−9&−2&3\\−6&−1&2 \end{bmatrix}$

Calculate $C^{-1}$ using the formula $C^{-1} = \frac{1}{|C|} \text{adj(C)}$:

$C^{-1} = \frac{1}{-1} \begin{bmatrix}2&0&−1\\−9&−2&3\\−6&−1&2 \end{bmatrix} = -1 \begin{bmatrix}2&0&−1\\−9&−2&3\\−6&−1&2 \end{bmatrix} = \begin{bmatrix}−2&0&1\\9&2&−3\\6&1&−2 \end{bmatrix}$

Now, let's consider the provided product $\begin{bmatrix}1&1&2\\0&2&3\\3&2&4 \end{bmatrix} \begin{bmatrix}2&0&1\\9&2&3\\6&1&2 \end{bmatrix}$. A common technique in such problems is that one of the matrices in the product is the coefficient matrix (or its transpose) and the other is its inverse (or a multiple), such that their product is the identity matrix or a scalar multiple of it. While the provided matrices do not directly fit this pattern with the coefficient matrix C or its inverse $C^{-1}$ calculated above, the instruction implies using a product resulting in an identity matrix to find the inverse.

Let's demonstrate the product of the coefficient matrix $C$ and its inverse $C^{-1}$:

$C^{-1}C = \begin{bmatrix}−2&0&1\\9&2&−3\\6&1&−2 \end{bmatrix} \begin{bmatrix}1&−1&2\\0&2&−3\\3&−2&4 \end{bmatrix}$

$\begin{bmatrix} (−2)(1)+0(0)+1(3) & (−2)(−1)+0(2)+1(−2) & (−2)(2)+0(−3)+1(4) \\ 9(1)+2(0)+(−3)(3) & 9(−1)+2(2)+(−3)(−2) & 9(2)+2(−3)+(−3)(4) \\ 6(1)+1(0)+(−2)(3) & 6(−1)+1(2)+(−2)(−2) & 6(2)+1(−3)+(−2)(4) \end{bmatrix}$

$= \begin{bmatrix} −2+0+3 & 2+0−2 & −4+0+4 \\ 9+0−9 & −9+4+6 & 18−6−12 \\ 6+0−6 & −6+2+4 & 12−3−8 \end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = I$

This shows that $\begin{bmatrix}−2&0&1\\9&2&−3\\6&1&−2 \end{bmatrix}$ is indeed the inverse of the coefficient matrix C.

Now, we solve $CX = D$ using $X = C^{-1}D$:

$X = \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}−2&0&1\\9&2&−3\\6&1&−2 \end{bmatrix} \begin{bmatrix}1\\1\\2 \end{bmatrix}$

Perform the matrix multiplication:

$x = (−2)(1) + (0)(1) + (1)(2) = −2 + 0 + 2 = 0$

$y = (9)(1) + (2)(1) + (−3)(2) = 9 + 2 - 6 = 5$

$z = (6)(1) + (1)(1) + (−2)(2) = 6 + 1 - 4 = 3$

So, we have:

$\begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}0\\5\\3 \end{bmatrix}$

Therefore, the solution to the system of equations is $x=0$, $y=5$, and $z=3$. The provided product matrices in the question appear to be incorrect or irrelevant to the standard method of using a product to find the inverse of the coefficient matrix derived from the given system.

Given:

The determinant $\begin{vmatrix} x&\sinθ&\cosθ\\−\sinθ&−x&1\\\cosθ&1&x \end{vmatrix}$.

To Prove:

The determinant is independent of $\theta$.

Proof:

Let $\Delta$ be the given determinant.

$\Delta = \begin{vmatrix} x&\sinθ&\cosθ\\−\sinθ&−x&1\\\cosθ&1&x \end{vmatrix}$

We will expand the determinant along the first row (R1).

$\Delta = x \begin{vmatrix} −x & 1 \\ 1 & x \end{vmatrix} - \sin\theta \begin{vmatrix} −\sin\theta & 1 \\ \cos\theta & x \end{vmatrix} + \cos\theta \begin{vmatrix} −\sin\theta & −x \\ \cos\theta & 1 \end{vmatrix}$

$\Delta = x((-x)(x) - (1)(1)) - \sin\theta((-sin\theta)(x) - (1)(cos\theta)) + \cos\theta((-sin\theta)(1) - (-x)(cos\theta))$

$\Delta = x(-x^2 - 1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)$

$\Delta = -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^2\theta$

Combine the terms with $\sin\theta\cos\theta$:

$\Delta = -x^3 - x + x\sin^2\theta + x\cos^2\theta$

Factor out $x$ from the last two terms:

$\Delta = -x^3 - x + x(\sin^2\theta + \cos^2\theta)$

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

$\Delta = -x^3 - x + x(1)$

$\Delta = -x^3 - x + x$

$\Delta = -x^3$

The determinant simplifies to $-x^3$, which is an expression that depends only on $x$ and does not contain the variable $\theta$.

Therefore, the determinant is independent of $\theta$.

Hence, proved.

Given:

The determinant $\begin{vmatrix} \cosα \cosβ &\cosα \sinβ &−\sinα\\−\sinβ &\cosβ &0\\\sinα \cosβ &\sinα \sinβ &\cosα \end{vmatrix}$.

To Evaluate:

The value of the given determinant.

Solution:

Let $\Delta$ be the given determinant.

$\Delta = \begin{vmatrix} \cosα \cosβ &\cosα \sinβ &−\sinα\\−\sinβ &\cosβ &0\\\sinα \cosβ &\sinα \sinβ &\cosα \end{vmatrix}$

We can evaluate this determinant by expanding along any row or column. Let's expand along the third row (R3) since it contains a zero element, simplifying the calculation.

$\Delta = (\sinα \cosβ) \times \text{(Cofactor of element in R3, C1)} + (\sinα \sinβ) \times \text{(Cofactor of element in R3, C2)} + (\cosα) \times \text{(Cofactor of element in R3, C3)}$

The cofactor of an element in row $i$ and column $j$ is $(-1)^{i+j}$ times the determinant of the minor matrix (the matrix obtained by deleting row $i$ and column $j$).

Cofactor of $(\sinα \cosβ)$ (element in R3, C1, so $i=3, j=1$):

$(-1)^{3+1} \begin{vmatrix} \cosα \sinβ & −\sinα \\ \cosβ & 0 \end{vmatrix} = +1 ((\cosα \sinβ)(0) - (-\sinα)(\cosβ))$

$= 0 - (-\sinα \cosβ) = \sinα \cosβ$

Term 1: $(\sinα \cosβ) \times (\sinα \cosβ) = \sin^2α \cos^2β$

Cofactor of $(\sinα \sinβ)$ (element in R3, C2, so $i=3, j=2$):

$(-1)^{3+2} \begin{vmatrix} \cosα \cosβ & −\sinα \\ −\sinβ & 0 \end{vmatrix} = -1 ((\cosα \cosβ)(0) - (-\sinα)(−\sinβ))$

$= -1 (0 - (\sinα \sinβ)) = -1 (-\sinα \sinβ) = \sinα \sinβ$

Term 2: $(\sinα \sinβ) \times (\sinα \sinβ) = \sin^2α \sin^2β$

Cofactor of $(\cosα)$ (element in R3, C3, so $i=3, j=3$):

$(-1)^{3+3} \begin{vmatrix} \cosα \cosβ & \cosα \sinβ \\ −\sinβ & \cosβ \end{vmatrix} = +1 ((\cosα \cosβ)(\cosβ) - (\cosα \sinβ)(−\sinβ))$

$= (\cosα \cos^2β) - (-\cosα \sin^2β) = \cosα \cos^2β + \cosα \sin^2β$

$= \cosα (\cos^2β + \sin^2β)$

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

$= \cosα (1) = \cosα$

Term 3: $(\cosα) \times (\cosα) = \cos^2α$

Summing the terms to get the value of the determinant:

$\Delta = \sin^2α \cos^2β + \sin^2α \sin^2β + \cos^2α$

Factor out $\sin^2α$ from the first two terms:

$\Delta = \sin^2α (\cos^2β + \sin^2β) + \cos^2α$

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

$\Delta = \sin^2α (1) + \cos^2α$

$\Delta = \sin^2α + \cos^2α$

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

$\Delta = 1$

Answer:

The value of the determinant is 1.

Given:

$A^{-1} = \begin{bmatrix} 3&−1&1\\−15&6&−5\\5&−2&2 \end{bmatrix}$

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

To Find:

$(AB)^{-1}$

Solution:

We use the property of matrix inverses which states that for two invertible matrices A and B, $(AB)^{-1} = B^{-1}A^{-1}$.

We are given $A^{-1}$. We need to find $B^{-1}$.

To find the inverse of matrix B, we first calculate its determinant, $|B|$.

$|B| = \begin{vmatrix} 1&2&−2\\−1&3&0\\0&−2&1 \end{vmatrix}$

Expand along the third column (C3):

$|B| = (−2) \times (-1)^{1+3} \begin{vmatrix} −1 & 3 \\ 0 & −2 \end{vmatrix} + (0) \times (-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 0 & −2 \end{vmatrix} + (1) \times (-1)^{3+3} \begin{vmatrix} 1 & 2 \\ −1 & 3 \end{vmatrix}$

$|B| = (−2) \times 1 \times ((-1)(-2) - (3)(0)) + 0 + (1) \times 1 \times ((1)(3) - (2)(-1))$

$|B| = −2(2 - 0) + (3 - (−2))$

$|B| = −2(2) + (3 + 2)$

$|B| = −4 + 5$

$|B| = 1$

Since $|B| = 1 \neq 0$, the inverse of B exists.

Next, we find the matrix of cofactors of B, denoted by C_B.

$C_{11} = + \begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix} = 3 - 0 = 3$

$C_{12} = - \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} = -(-1 - 0) = 1$

$C_{13} = + \begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} = 2 - 0 = 2$

$C_{21} = - \begin{vmatrix} 2 & -2 \\ -2 & 1 \end{vmatrix} = -(2 - 4) = 2$

$C_{22} = + \begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1$

$C_{23} = - \begin{vmatrix} 1 & 2 \\ 0 & -2 \end{vmatrix} = -(-2 - 0) = 2$

$C_{31} = + \begin{vmatrix} 2 & -2 \\ 3 & 0 \end{vmatrix} = 0 - (-6) = 6$

$C_{32} = - \begin{vmatrix} 1 & -2 \\ -1 & 0 \end{vmatrix} = -(0 - 2) = 2$

$C_{33} = + \begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} = 3 - (-2) = 5$

The matrix of cofactors is $C_B = \begin{bmatrix}3&1&2\\2&1&2\\6&2&5 \end{bmatrix}$.

The adjoint of B, adj(B), is the transpose of the cofactor matrix.

$\text{adj(B)} = C_B^T = \begin{bmatrix}3&2&6\\1&1&2\\2&2&5 \end{bmatrix}$

Now, calculate $B^{-1}$ using the formula $B^{-1} = \frac{1}{|B|} \text{adj(B)}$:

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

Finally, calculate $(AB)^{-1} = B^{-1}A^{-1}$:

$(AB)^{-1} = \begin{bmatrix}3&2&6\\1&1&2\\2&2&5 \end{bmatrix} \begin{bmatrix}3&−1&1\\−15&6&−5\\5&−2&2 \end{bmatrix}$

Perform the matrix multiplication:

$(AB)^{-1} = \begin{bmatrix} (3)(3)+(2)(-15)+(6)(5) & (3)(-1)+(2)(6)+(6)(-2) & (3)(1)+(2)(-5)+(6)(2) \\ (1)(3)+(1)(-15)+(2)(5) & (1)(-1)+(1)(6)+(2)(-2) & (1)(1)+(1)(-5)+(2)(2) \\ (2)(3)+(2)(-15)+(5)(5) & (2)(-1)+(2)(6)+(5)(-2) & (2)(1)+(2)(-5)+(5)(2) \end{bmatrix}$

$(AB)^{-1} = \begin{bmatrix} 9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10 \end{bmatrix}$

$(AB)^{-1} = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}$

Answer:

$(AB)^{-1} = \begin{bmatrix} 9&−3&5\\−2&1&0\\1&0&2 \end{bmatrix}$

Given:

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

To Verify:

(i) $[\text{adj A}]^{-1} = \text{adj} (A^{-1})$

(ii) $(A^{-1})^{-1} = A$

Solution:

First, we find the determinant of A, $|A|$.

$|A| = \begin{vmatrix} 1&2&1\\2&3&1\\1&1&5 \end{vmatrix}$

Expand along the first row:

$|A| = 1 \begin{vmatrix} 3 & 1 \\ 1 & 5 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ 1 & 5 \end{vmatrix} + 1 \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix}$

$|A| = 1((3)(5) - (1)(1)) - 2((2)(5) - (1)(1)) + 1((2)(1) - (3)(1))$

$|A| = 1(15 - 1) - 2(10 - 1) + 1(2 - 3)$

$|A| = 1(14) - 2(9) + 1(-1)$

$|A| = 14 - 18 - 1 = -5$

Since $|A| = -5 \neq 0$, matrix A is invertible.

Next, we find the adjoint of A, adj(A).

First, calculate the matrix of cofactors C:

$C_{11} = + \begin{vmatrix} 3 & 1 \\ 1 & 5 \end{vmatrix} = (3)(5) - (1)(1) = 15 - 1 = 14$

$C_{12} = - \begin{vmatrix} 2 & 1 \\ 1 & 5 \end{vmatrix} = -((2)(5) - (1)(1)) = -(10 - 1) = -9$

$C_{13} = + \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = (2)(1) - (3)(1) = 2 - 3 = -1$

$C_{21} = - \begin{vmatrix} 2 & 1 \\ 1 & 5 \end{vmatrix} = -((2)(5) - (1)(1)) = -(10 - 1) = -9$

$C_{22} = + \begin{vmatrix} 1 & 1 \\ 1 & 5 \end{vmatrix} = (1)(5) - (1)(1) = 5 - 1 = 4$

$C_{23} = - \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = -((1)(1) - (2)(1)) = -(1 - 2) = 1$

$C_{31} = + \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = (2)(1) - (1)(3) = 2 - 3 = -1$

$C_{32} = - \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = -((1)(1) - (1)(2)) = -(1 - 2) = 1$

$C_{33} = + \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = (1)(3) - (2)(2) = 3 - 4 = -1$

The matrix of cofactors is $C = \begin{bmatrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \end{bmatrix}$.

The adjoint of A is the transpose of the cofactor matrix:

$\text{adj(A)} = C^T = \begin{bmatrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \end{bmatrix}$

Now, we find the inverse of A, $A^{-1}$.

$A^{-1} = \frac{1}{|A|} \text{adj(A)}$

$A^{-1} = \frac{1}{-5} \begin{bmatrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \end{bmatrix} = \begin{bmatrix} \frac{14}{-5} & \frac{-9}{-5} & \frac{-1}{-5} \\ \frac{-9}{-5} & \frac{4}{-5} & \frac{1}{-5} \\ \frac{-1}{-5} & \frac{1}{-5} & \frac{-1}{-5} \end{bmatrix} = \begin{bmatrix} \frac{-14}{5} & \frac{9}{5} & \frac{1}{5} \\ \frac{9}{5} & \frac{-4}{5} & \frac{-1}{5} \\ \frac{1}{5} & \frac{-1}{5} & \frac{1}{5} \end{bmatrix}$

Verification of (ii) $(A^{-1})^{-1} = A$:

We need to find the inverse of $A^{-1}$. Let $B = A^{-1}$.

$B = \begin{bmatrix} \frac{-14}{5} & \frac{9}{5} & \frac{1}{5} \\ \frac{9}{5} & \frac{-4}{5} & \frac{-1}{5} \\ \frac{1}{5} & \frac{-1}{5} & \frac{1}{5} \end{bmatrix}$

First, find the determinant of B, $|B|$. We know that $|B| = |A^{-1}| = \frac{1}{|A|}$.

$|B| = \frac{1}{-5} = \frac{-1}{5}$.

Next, find the adjoint of B, adj(B) = adj($A^{-1}$). Calculate the matrix of cofactors of $A^{-1}$.

Cofactor of element in (1,1): $+ \begin{vmatrix} \frac{-4}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{1}{5} \end{vmatrix} = (\frac{-4}{5})(\frac{1}{5}) - (\frac{-1}{5})(\frac{-1}{5}) = \frac{-4}{25} - \frac{1}{25} = \frac{-5}{25} = \frac{-1}{5}$

Cofactor of element in (1,2): $- \begin{vmatrix} \frac{9}{5} & \frac{-1}{5} \\ \frac{1}{5} & \frac{1}{5} \end{vmatrix} = -((\frac{9}{5})(\frac{1}{5}) - (\frac{-1}{5})(\frac{1}{5})) = -(\frac{9}{25} + \frac{1}{25}) = -\frac{10}{25} = \frac{-2}{5}$

Cofactor of element in (1,3): $+ \begin{vmatrix} \frac{9}{5} & \frac{-4}{5} \\ \frac{1}{5} & \frac{-1}{5} \end{vmatrix} = (\frac{9}{5})(\frac{-1}{5}) - (\frac{-4}{5})(\frac{1}{5}) = \frac{-9}{25} + \frac{4}{25} = \frac{-5}{25} = \frac{-1}{5}$

Cofactor of element in (2,1): $- \begin{vmatrix} \frac{9}{5} & \frac{1}{5} \\ \frac{-1}{5} & \frac{1}{5} \end{vmatrix} = -((\frac{9}{5})(\frac{1}{5}) - (\frac{1}{5})(\frac{-1}{5})) = -(\frac{9}{25} + \frac{1}{25}) = -\frac{10}{25} = \frac{-2}{5}$

Cofactor of element in (2,2): $+ \begin{vmatrix} \frac{-14}{5} & \frac{1}{5} \\ \frac{1}{5} & \frac{1}{5} \end{vmatrix} = (\frac{-14}{5})(\frac{1}{5}) - (\frac{1}{5})(\frac{1}{5}) = \frac{-14}{25} - \frac{1}{25} = \frac{-15}{25} = \frac{-3}{5}$

Cofactor of element in (2,3): $- \begin{vmatrix} \frac{-14}{5} & \frac{9}{5} \\ \frac{1}{5} & \frac{-1}{5} \end{vmatrix} = -((\frac{-14}{5})(\frac{-1}{5}) - (\frac{9}{5})(\frac{1}{5})) = -(\frac{14}{25} - \frac{9}{25}) = -\frac{5}{25} = \frac{-1}{5}$

Cofactor of element in (3,1): $+ \begin{vmatrix} \frac{9}{5} & \frac{1}{5} \\ \frac{-4}{5} & \frac{-1}{5} \end{vmatrix} = (\frac{9}{5})(\frac{-1}{5}) - (\frac{1}{5})(\frac{-4}{5}) = \frac{-9}{25} + \frac{4}{25} = \frac{-5}{25} = \frac{-1}{5}$

Cofactor of element in (3,2): $- \begin{vmatrix} \frac{-14}{5} & \frac{1}{5} \\ \frac{9}{5} & \frac{-1}{5} \end{vmatrix} = -((\frac{-14}{5})(\frac{-1}{5}) - (\frac{1}{5})(\frac{9}{5})) = -(\frac{14}{25} - \frac{9}{25}) = -\frac{5}{25} = \frac{-1}{5}$

Cofactor of element in (3,3): $+ \begin{vmatrix} \frac{-14}{5} & \frac{9}{5} \\ \frac{9}{5} & \frac{-4}{5} \end{vmatrix} = (\frac{-14}{5})(\frac{-4}{5}) - (\frac{9}{5})(\frac{9}{5}) = \frac{56}{25} - \frac{81}{25} = \frac{-25}{25} = -1$

The matrix of cofactors of $A^{-1}$ is $\begin{bmatrix} \frac{-1}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-2}{5} & \frac{-3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-1}{5} & -1 \end{bmatrix}$.

So, $\text{adj}(A^{-1}) = \begin{bmatrix} \frac{-1}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-2}{5} & \frac{-3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-1}{5} & -1 \end{bmatrix}^T = \begin{bmatrix} \frac{-1}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-2}{5} & \frac{-3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-1}{5} & -1 \end{bmatrix}$.

Now calculate $(A^{-1})^{-1} = \frac{1}{|A^{-1}|} \text{adj}(A^{-1})$:

$(A^{-1})^{-1} = \frac{1}{\frac{-1}{5}} \begin{bmatrix} \frac{-1}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-2}{5} & \frac{-3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-1}{5} & -1 \end{bmatrix}$

$(A^{-1})^{-1} = -5 \begin{bmatrix} \frac{-1}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-2}{5} & \frac{-3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-1}{5} & -1 \end{bmatrix} = \begin{bmatrix} (-5)(\frac{-1}{5}) & (-5)(\frac{-2}{5}) & (-5)(\frac{-1}{5}) \\ (-5)(\frac{-2}{5}) & (-5)(\frac{-3}{5}) & (-5)(\frac{-1}{5}) \\ (-5)(\frac{-1}{5}) & (-5)(\frac{-1}{5}) & (-5)(-1) \end{bmatrix}$

$(A^{-1})^{-1} = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$

This is equal to the original matrix A.

Thus, $(A^{-1})^{-1} = A$ is verified.

Verification of (i) $[\text{adj A}]^{-1} = \text{adj} (A^{-1})$:

We have already calculated adj(A) and adj($A^{-1}$).

$\text{adj(A)} = \begin{bmatrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \end{bmatrix}$

$\text{adj}(A^{-1}) = \begin{bmatrix} \frac{-1}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-2}{5} & \frac{-3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-1}{5} & -1 \end{bmatrix}$

Now, we find the inverse of adj(A). Let $M = \text{adj A}$.

$M = \begin{bmatrix} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \end{bmatrix}$

First, find the determinant of M, $|M|$. We know that $|M| = |\text{adj A}| = |A|^{n-1}$ where n is the order of the matrix (here n=3). So, $|M| = |A|^{3-1} = |A|^2$.

$|M| = (-5)^2 = 25$.

Next, find the adjoint of M, adj(M) = adj(adj A). Calculate the matrix of cofactors of adj(A).

Cofactor of element in (1,1): $+ \begin{vmatrix} 4 & 1 \\ 1 & -1 \end{vmatrix} = (4)(-1) - (1)(1) = -4 - 1 = -5$

Cofactor of element in (1,2): $- \begin{vmatrix} -9 & 1 \\ -1 & -1 \end{vmatrix} = (9 - (-1)) = -(9 + 1) = -10$

Cofactor of element in (1,3): $+ \begin{vmatrix} -9 & 4 \\ -1 & 1 \end{vmatrix} = (-9)(1) - (4)(-1) = -9 + 4 = -5$

Cofactor of element in (2,1): $- \begin{vmatrix} -9 & -1 \\ 1 & -1 \end{vmatrix} = -(9 - (-1)) = -(9 + 1) = -10$

Cofactor of element in (2,2): $+ \begin{vmatrix} 14 & -1 \\ -1 & -1 \end{vmatrix} = (14)(-1) - (-1)(-1) = -14 - 1 = -15$

Cofactor of element in (2,3): $- \begin{vmatrix} 14 & -9 \\ -1 & 1 \end{vmatrix} = -((14)(1) - (-9)(-1)) = -(14 - 9) = -5$

Cofactor of element in (3,1): $+ \begin{vmatrix} -9 & -1 \\ 4 & 1 \end{vmatrix} = (-9)(1) - (-1)(4) = -9 + 4 = -5$

Cofactor of element in (3,2): $- \begin{vmatrix} 14 & -1 \\ -9 & 1 \end{vmatrix} = -((14)(1) - (-1)(-9)) = -(14 - 9) = -5$

Cofactor of element in (3,3): $+ \begin{vmatrix} 14 & -9 \\ -9 & 4 \end{vmatrix} = (14)(4) - (-9)(-9) = 56 - 81 = -25$

The matrix of cofactors of adj(A) is $\begin{bmatrix} -5 & -10 & -5 \\ -10 & -15 & -5 \\ -5 & -5 & -25 \end{bmatrix}$.

So, $\text{adj(adj A)} = \begin{bmatrix} -5 & -10 & -5 \\ -10 & -15 & -5 \\ -5 & -5 & -25 \end{bmatrix}^T = \begin{bmatrix} -5 & -10 & -5 \\ -10 & -15 & -5 \\ -5 & -5 & -25 \end{bmatrix}$.

Now calculate $[\text{adj A}]^{-1} = \frac{1}{|\text{adj A}|} \text{adj(adj A)}$:

$[\text{adj A}]^{-1} = \frac{1}{25} \begin{bmatrix} -5 & -10 & -5 \\ -10 & -15 & -5 \\ -5 & -5 & -25 \end{bmatrix}$

$[\text{adj A}]^{-1} = \begin{bmatrix} \frac{-5}{25} & \frac{-10}{25} & \frac{-5}{25} \\ \frac{-10}{25} & \frac{-15}{25} & \frac{-5}{25} \\ \frac{-5}{25} & \frac{-5}{25} & \frac{-25}{25} \end{bmatrix} = \begin{bmatrix} \frac{-1}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-2}{5} & \frac{-3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-1}{5} & -1 \end{bmatrix}$

Comparing $[\text{adj A}]^{-1}$ and adj($A^{-1}$):

$[\text{adj A}]^{-1} = \begin{bmatrix} \frac{-1}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-2}{5} & \frac{-3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-1}{5} & -1 \end{bmatrix}$

$\text{adj}(A^{-1}) = \begin{bmatrix} \frac{-1}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-2}{5} & \frac{-3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-1}{5} & -1 \end{bmatrix}$

The two matrices are equal.

Thus, $[\text{adj A}]^{-1} = \text{adj} (A^{-1})$ is verified.

Conclusion:

Both the properties $(A^{-1})^{-1} = A$ and $[\text{adj A}]^{-1} = \text{adj} (A^{-1})$ are verified for the given matrix A.

Given:

The determinant $\begin{vmatrix} x&y&x+y\\y&x+y&x\\x+y&x&y \end{vmatrix}$.

To Evaluate:

The value of the given determinant.

Solution:

Let $\Delta$ be the given determinant.

$\Delta = \begin{vmatrix} x&y&x+y\\y&x+y&x\\x+y&x&y \end{vmatrix}$

Apply the column operation $C_1 \to C_1 + C_2 + C_3$:

$\Delta = \begin{vmatrix} x+y+(x+y)&y&x+y\\y+(x+y)+x&x+y&x\\x+y+x+y&x&y \end{vmatrix}$

$\Delta = \begin{vmatrix} 2(x+y)&y&x+y\\2(x+y)&x+y&x\\2(x+y)&x&y \end{vmatrix}$

Take out the common factor $2(x+y)$ from the first column ($C_1$):

$\Delta = 2(x+y) \begin{vmatrix} 1&y&x+y\\1&x+y&x\\1&x&y \end{vmatrix}$

Apply the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:

$\Delta = 2(x+y) \begin{vmatrix} 1&y&x+y\\1-1&(x+y)-y&x-(x+y)\\1-1&x-y&y-(x+y) \end{vmatrix}$

$\Delta = 2(x+y) \begin{vmatrix} 1&y&x+y\\0&x&-y\\0&x-y&−x \end{vmatrix}$

Expand the determinant along the first column ($C_1$):

$\Delta = 2(x+y) \left[ 1 \begin{vmatrix} x & -y \\ x-y & -x \end{vmatrix} - 0 + 0 \right]$

$\Delta = 2(x+y) [(x)(-x) - (-y)(x-y)]$

$\Delta = 2(x+y) [-x^2 + yx - y^2]$

$\Delta = 2(x+y) (-x^2 + xy - y^2)$

Multiply the terms:

$\Delta = 2 [ x(-x^2 + xy - y^2) + y(-x^2 + xy - y^2) ]$

$\Delta = 2 [ -x^3 + x^2y - xy^2 - x^2y + xy^2 - y^3 ]$

Combine like terms:

$\Delta = 2 [ -x^3 + (x^2y - x^2y) + (-xy^2 + xy^2) - y^3 ]$

$\Delta = 2 [ -x^3 + 0 + 0 - y^3 ]$

$\Delta = 2 (-x^3 - y^3)$

$\Delta = -2x^3 - 2y^3$

Answer:

The value of the determinant is $-2x^3 - 2y^3$.

Given:

The determinant $\begin{vmatrix} 1&x&y\\1&x+y&y\\1&x&x+y \end{vmatrix}$.

To Evaluate:

The value of the given determinant.

Solution:

Let $\Delta$ be the given determinant.

$\Delta = \begin{vmatrix} 1&x&y\\1&x+y&y\\1&x&x+y \end{vmatrix}$

We can simplify the determinant by applying row operations to create zeros in the first column.

Apply the operation $R_2 \to R_2 - R_1$:

The new element in row 2, column 1 is $1 - 1 = 0$.

The new element in row 2, column 2 is $(x+y) - x = y$.

The new element in row 2, column 3 is $y - y = 0$.

Apply the operation $R_3 \to R_3 - R_1$:

The new element in row 3, column 1 is $1 - 1 = 0$.

The new element in row 3, column 2 is $x - x = 0$.

The new element in row 3, column 3 is $(x+y) - y = x$.

The determinant becomes:

$\Delta = \begin{vmatrix} 1&x&y\\0&y&0\\0&0&x \end{vmatrix}$

Now, expand the determinant along the first column ($C_1$). Since the first column has two zeros, the expansion is straightforward.

$\Delta = 1 \times \text{(Cofactor of element in R1, C1)} - 0 \times \text{(Cofactor of element in R2, C1)} + 0 \times \text{(Cofactor of element in R3, C1)}$

$\Delta = 1 \times (-1)^{1+1} \begin{vmatrix} y & 0 \\ 0 & x \end{vmatrix} - 0 + 0$

$\Delta = 1 \times 1 \times ((y)(x) - (0)(0))$

$\Delta = yx - 0$

$\Delta = xy$

Answer:

The value of the determinant is $xy$.

Given:

The system of equations:

$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$

$\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1$

$\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$

To Solve:

The given system of equations for $x$, $y$, and $z$ using the matrix method.

Solution:

We can transform this system of equations into a linear system by substituting $u = \frac{1}{x}$, $v = \frac{1}{y}$, and $w = \frac{1}{z}$.

The system becomes:

$2u + 3v + 10w = 4$

$4u - 6v + 5w = 1$

$6u + 9v - 20w = 2$

This is a linear system in terms of $u$, $v$, and $w$. We can write this system in matrix form as $AW = B$, where:

$A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}$, $W = \begin{bmatrix} u \\ v \\ w \end{bmatrix}$, and $B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$

To solve for $W$, we need to find the inverse of matrix A ($A^{-1}$) and then calculate $W = A^{-1}B$.

First, calculate the determinant of A, $|A|$:

$|A| = \begin{vmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{vmatrix}$

Expand along the first row:

$|A| = 2 \begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} - 3 \begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} + 10 \begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix}$

$|A| = 2((-6)(-20) - (5)(9)) - 3((4)(-20) - (5)(6)) + 10((4)(9) - (-6)(6))$

$|A| = 2(120 - 45) - 3(-80 - 30) + 10(36 + 36)$

$|A| = 2(75) - 3(-110) + 10(72)$

$|A| = 150 + 330 + 720$

$|A| = 1200$

Since $|A| = 1200 \neq 0$, the inverse of A exists.

Next, calculate the matrix of cofactors, C:

$C_{11} = + \begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} = 120 - 45 = 75$

$C_{12} = - \begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} = -(-80 - 30) = 110$

$C_{13} = + \begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix} = 36 - (-36) = 72$

$C_{21} = - \begin{vmatrix} 3 & 10 \\ 9 & -20 \end{vmatrix} = -(-60 - 90) = 150$

$C_{22} = + \begin{vmatrix} 2 & 10 \\ 6 & -20 \end{vmatrix} = -40 - 60 = -100$

$C_{23} = - \begin{vmatrix} 2 & 3 \\ 6 & 9 \end{vmatrix} = -(18 - 18) = 0$

$C_{31} = + \begin{vmatrix} 3 & 10 \\ -6 & 5 \end{vmatrix} = 15 - (-60) = 75$

$C_{32} = - \begin{vmatrix} 2 & 10 \\ 4 & 5 \end{vmatrix} = -(10 - 40) = 30$

$C_{33} = + \begin{vmatrix} 2 & 3 \\ 4 & -6 \end{vmatrix} = -12 - 12 = -24$

The matrix of cofactors is $C = \begin{bmatrix} 75 & 110 & 72 \\ 150 & -100 & 0 \\ 75 & 30 & -24 \end{bmatrix}$.

The adjoint of A, adj(A), is the transpose of the cofactor matrix:

$\text{adj(A)} = C^T = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}$

Calculate $A^{-1}$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj(A)}$:

$A^{-1} = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}$

Now, solve for $W$ using $W = A^{-1}B$:

$W = \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$

Perform the matrix multiplication:

Row 1: $(75)(4) + (150)(1) + (75)(2) = 300 + 150 + 150 = 600$

Row 2: $(110)(4) + (-100)(1) + (30)(2) = 440 - 100 + 60 = 400$

Row 3: $(72)(4) + (0)(1) + (-24)(2) = 288 + 0 - 48 = 240$

So, $W = \frac{1}{1200} \begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} 600/1200 \\ 400/1200 \\ 240/1200 \end{bmatrix} = \begin{bmatrix} 1/2 \\ 1/3 \\ 1/5 \end{bmatrix}$

Thus, $u = \frac{1}{2}$, $v = \frac{1}{3}$, and $w = \frac{1}{5}$.

Substitute back $u = \frac{1}{x}$, $v = \frac{1}{y}$, $w = \frac{1}{z}$:

$\frac{1}{x} = \frac{1}{2} \implies x = 2$

$\frac{1}{y} = \frac{1}{3} \implies y = 3$

$\frac{1}{z} = \frac{1}{5} \implies z = 5$

Answer:

The solution to the system of equations is $x=2$, $y=3$, and $z=5$.

Given:

Matrix A = $\begin{bmatrix}x&0&0\\0&y&0\\0&0&z \end{bmatrix}$, where $x, y, z$ are nonzero real numbers.

To Find:

The inverse of matrix A, denoted as A^–1.

Solution:

To find the inverse of matrix A, we can use the formula $A^{-1} = \frac{1}{|A|} \text{adj(A)}$.

First, calculate the determinant of A, $|A|$. For a diagonal matrix, the determinant is the product of the diagonal elements.

$|A| = \begin{vmatrix}x&0&0\\0&y&0\\0&0&z \end{vmatrix} = x(y)(z) = xyz$

Since $x, y, z$ are nonzero, $|A| = xyz \neq 0$, so the inverse of A exists.

Next, calculate the matrix of cofactors of A, denoted by C.

$C_{11} = + \begin{vmatrix} y & 0 \\ 0 & z \end{vmatrix} = yz - 0 = yz$

$C_{12} = - \begin{vmatrix} 0 & 0 \\ 0 & z \end{vmatrix} = -(0 - 0) = 0$

$C_{13} = + \begin{vmatrix} 0 & y \\ 0 & 0 \end{vmatrix} = 0 - 0 = 0$

$C_{21} = - \begin{vmatrix} 0 & 0 \\ 0 & z \end{vmatrix} = -(0 - 0) = 0$

$C_{22} = + \begin{vmatrix} x & 0 \\ 0 & z \end{vmatrix} = xz - 0 = xz$

$C_{23} = - \begin{vmatrix} x & 0 \\ 0 & 0 \end{vmatrix} = -(0 - 0) = 0$

$C_{31} = + \begin{vmatrix} 0 & 0 \\ y & 0 \end{vmatrix} = 0 - 0 = 0$

$C_{32} = - \begin{vmatrix} x & 0 \\ 0 & 0 \end{vmatrix} = -(0 - 0) = 0$

$C_{33} = + \begin{vmatrix} x & 0 \\ 0 & y \end{vmatrix} = xy - 0 = xy$

The matrix of cofactors is $C = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}$.

The adjoint of A, adj(A), is the transpose of the cofactor matrix. Since the cofactor matrix is diagonal, its transpose is itself.

$\text{adj(A)} = C^T = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}$

Now, calculate A^–1 using the formula $A^{-1} = \frac{1}{|A|} \text{adj(A)}$:

$A^{-1} = \frac{1}{xyz} \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}$

Multiply each element inside the matrix by $\frac{1}{xyz}$:

$A^{-1} = \begin{bmatrix} \frac{yz}{xyz} & \frac{0}{xyz} & \frac{0}{xyz} \\ \frac{0}{xyz} & \frac{xz}{xyz} & \frac{0}{xyz} \\ \frac{0}{xyz} & \frac{0}{xyz} & \frac{xy}{xyz} \end{bmatrix}$

$A^{-1} = \begin{bmatrix} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{bmatrix}$

$A^{-1} = \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}$

Comparing this result with the given options, we find that it matches option (A).

Answer:

The inverse of matrix A is $\begin{bmatrix}x^{−1}&0&0\\0&y^{−1}&0\\0&0&z^{−1} \end{bmatrix}$.

The correct option is (A).

Given:

Matrix A = $\begin{bmatrix}1&\sinθ&1\\−\sinθ&1&\sinθ\\−1&−\sinθ&1 \end{bmatrix}$

The range of $\theta$ is $0 \leq \theta \leq 2\pi$.

To Find:

Which interval the determinant of A, Det(A), belongs to.

Solution:

We need to evaluate the determinant of matrix A, $|A|$.

$|A| = \begin{vmatrix}1&\sinθ&1\\−\sinθ&1&\sinθ\\−1&−\sinθ&1 \end{vmatrix}$

Let's expand the determinant along the first row (R1).

$|A| = 1 \begin{vmatrix} 1 & \sin\theta \\ -\sin\theta & 1 \end{vmatrix} - \sin\theta \begin{vmatrix} -\sin\theta & \sin\theta \\ -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} -\sin\theta & 1 \\ -1 & -\sin\theta \end{vmatrix}$

$|A| = 1((1)(1) - (\sin\theta)(-\sin\theta)) - \sin\theta((-\sin\theta)(1) - (\sin\theta)(-1)) + 1((-\sin\theta)(-\sin\theta) - (1)(-1))$

$|A| = 1(1 + \sin^2\theta) - \sin\theta(-\sin\theta + \sin\theta) + 1(\sin^2\theta + 1)$

$|A| = 1 + \sin^2\theta - \sin\theta(0) + \sin^2\theta + 1$

$|A| = 1 + \sin^2\theta + \sin^2\theta + 1$

$|A| = 2 + 2\sin^2\theta$

Now, we need to determine the range of $|A|$ given that $0 \leq \theta \leq 2\pi$.

For $0 \leq \theta \leq 2\pi$, the range of $\sin\theta$ is $[-1, 1]$.

The range of $\sin^2\theta$ is $[0, 1]$ (since $(\text{number})^2 \ge 0$ and the maximum value of $\sin\theta$ is 1, so maximum of $\sin^2\theta$ is $1^2=1$).

Now, let's find the range of $2 + 2\sin^2\theta$.

The minimum value occurs when $\sin^2\theta$ is minimum (i.e., 0):

$|A|_{min} = 2 + 2(0) = 2$

The maximum value occurs when $\sin^2\theta$ is maximum (i.e., 1):

$|A|_{max} = 2 + 2(1) = 4$

So, the determinant $|A|$ lies in the interval $[2, 4]$.

This corresponds to option (D).

Answer:

The determinant of A belongs to the interval $[2, 4]$.

The correct option is (D).