Adjoint and inverse of matrices Questions in English

(A) We know that $(AB)^{-1} = B^{-1} A^{-1}$.
First,we find the product $AB$:
$AB = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} (2)(1) + (3)(3) & (2)(0) + (3)(1) \\ (1)(1) + (2)(3) & (1)(0) + (2)(1) \end{bmatrix} = \begin{bmatrix} 11 & 3 \\ 7 & 2 \end{bmatrix}$.
Now,we find $(AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB)$.
$|AB| = (11)(2) - (3)(7) = 22 - 21 = 1$.
$\text{adj}(AB) = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix}$.
Therefore,$B^{-1} A^{-1} = (AB)^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$.
We know that $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
First,calculate the determinant $|A| = \omega \cdot \omega - 0 \cdot 0 = \omega^{2}$.
Next,find the adjoint of $A$ for a diagonal matrix $\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$ is $\begin{bmatrix} b & 0 \\ 0 & a \end{bmatrix}$.
Thus,$\text{adj}(A) = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{\omega^{2}} \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} = \begin{bmatrix} \frac{\omega}{\omega^{2}} & 0 \\ 0 & \frac{\omega}{\omega^{2}} \end{bmatrix} = \begin{bmatrix} \frac{1}{\omega} & 0 \\ 0 & \frac{1}{\omega} \end{bmatrix}$.
Since $\omega^{3} = 1$,we have $\frac{1}{\omega} = \omega^{2}$.
So,$A^{-1} = \begin{bmatrix} \omega^{2} & 0 \\ 0 & \omega^{2} \end{bmatrix}$.
Now,check the options:
$A^{2} = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} = \begin{bmatrix} \omega^{2} & 0 \\ 0 & \omega^{2} \end{bmatrix}$.
Thus,$A^{-1} = A^{2}$.

(C) Given the characteristic equation $A^{2}-5A-6I=0$.
Multiplying by $A^{-1}$ on both sides,we get:
$A^{-1}(A^{2}-5A-6I) = A^{-1}(0)$
$A - 5I - 6A^{-1} = 0$
$6A^{-1} = A - 5I$
Substituting the matrices:
$6A^{-1} = \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix} - 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$6A^{-1} = \begin{bmatrix} 4-5 & 5-0 \\ 2-0 & 1-5 \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 2 & -4 \end{bmatrix}$
$A^{-1} = \frac{1}{6}\begin{bmatrix} -1 & 5 \\ 2 & -4 \end{bmatrix}$
Thus,the correct option is $C$.

(C) Let $A = \left[\begin{array}{ccc}1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]$.
First,we calculate the determinant $|A|$:
$|A| = 1(0 - 1) - (-3)(0 - (-2)) + 2(3 - 6)$
$|A| = 1(-1) + 3(2) + 2(-3) = -1 + 6 - 6 = -1$.
The element in the third row and first column of $A^{-1}$ is given by $\frac{C_{13}}{|A|}$,where $C_{13}$ is the cofactor of the element in the first row and third column of $A$.
$C_{13} = (-1)^{1+3} \left|\begin{array}{cc}-3 & 3 \\ 2 & -1\end{array}\right| = 1(3 - 6) = -3$.
Therefore,the element in the third row and first column of $A^{-1}$ is $\frac{-3}{-1} = 3$.

(B) First,find the determinant of $A$: $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.
Next,find the adjoint of $A$: $\text{adj } A = \left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]$.
Therefore,$A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{11} \left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]$.
Given the equation $A^{-1} = xA + yI$,we substitute the matrices:
$\frac{1}{11} \left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right] = x \left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right] + y \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$.
$\left[\begin{array}{cc}1/11 & -2/11 \\ 5/11 & 1/11\end{array}\right] = \left[\begin{array}{cc}x+y & 2x \\ -5x & x+y\end{array}\right]$.
Comparing the elements,we get $2x = -2/11$,which implies $x = -1/11$.
Also,$x+y = 1/11$. Substituting $x = -1/11$,we get $-1/11 + y = 1/11$,so $y = 2/11$.
Thus,the values are $x = -1/11$ and $y = 2/11$.

(C) Given $A = \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix}$.
First,we find the determinant $|A| = (2)(-7) - (-3)(5) = -14 + 15 = 1$.
Since $|A| \neq 0$,$A^{-1}$ exists.
The inverse is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1} \begin{bmatrix} -7 & 3 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 3 \\ -5 & 2 \end{bmatrix}$.
Now,calculate $2A = 2 \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix} = \begin{bmatrix} 4 & -6 \\ 10 & -14 \end{bmatrix}$.
Next,calculate $3A^{-1} = 3 \begin{bmatrix} -7 & 3 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} -21 & 9 \\ -15 & 6 \end{bmatrix}$.
Finally,$2A - 3A^{-1} = \begin{bmatrix} 4 & -6 \\ 10 & -14 \end{bmatrix} - \begin{bmatrix} -21 & 9 \\ -15 & 6 \end{bmatrix} = \begin{bmatrix} 4 - (-21) & -6 - 9 \\ 10 - (-15) & -14 - 6 \end{bmatrix} = \begin{bmatrix} 25 & -15 \\ 25 & -20 \end{bmatrix}$.

(D) Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $AX = I$.
Since $AX = I$,it follows that $X = A^{-1}$.
First,find the determinant of $A$: $|A| = (1)(4) - (2)(3) = 4 - 6 = -2$.
The inverse of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$.
$X = \begin{bmatrix} \frac{4}{-2} & \frac{-2}{-2} \\ \frac{-3}{-2} & \frac{1}{-2} \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$.

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$.
The inverse of a diagonal matrix $A = \text{diag}(a, b, c)$ is given by $A^{-1} = \text{diag}(a^{-1}, b^{-1}, c^{-1})$.
Here,$A = \begin{bmatrix} \omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Therefore,$A^{-1} = \begin{bmatrix} \omega^{-1} & 0 & 0 \\ 0 & (\omega^2)^{-1} & 0 \\ 0 & 0 & 1^{-1} \end{bmatrix} = \begin{bmatrix} \frac{1}{\omega} & 0 & 0 \\ 0 & \frac{1}{\omega^2} & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Since $\omega^3 = 1$,we have $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$.
Thus,$A^{-1} = \begin{bmatrix} \omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

(A) Given that $A$ is a non-singular matrix,so $|A| \neq 0$.
Given equation: $(A+I)(A-I) = 0$.
Expanding the product: $A^2 - AI + IA - I^2 = 0$.
Since $AI = IA = A$ and $I^2 = I$,we get $A^2 - I = 0$.
Therefore,$A^2 = I$.
Multiplying both sides by $A^{-1}$,we get $A^{-1}A^2 = A^{-1}I$.
This simplifies to $A = A^{-1}$.
Now,substituting $A^{-1} = A$ into the expression $A + A^{-1}$,we get $A + A = 2A$.

(A) Given $A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$.
We know that $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
First,calculate the determinant $|A| = (x \times 0) - (1 \times 1) = -1$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} 0 & -1 \\ -1 & x \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{-1} \begin{bmatrix} 0 & -1 \\ -1 & x \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -x \end{bmatrix}$.
Given $A = A^{-1}$,we equate the matrices:
$\begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -x \end{bmatrix}$.
Comparing the corresponding elements,we get $x = 0$ and $0 = -x$,which implies $x = 0$.

(D) Given that $A$ is a non-singular matrix,so $|A| \neq 0$.
Given the equation $(A-2I)(A-4I) = 0$.
Expanding the product,we get $A^2 - 4A - 2A + 8I = 0$,which simplifies to $A^2 - 6A + 8I = 0$.
Since $A$ is non-singular,$A^{-1}$ exists. Multiplying the entire equation by $A^{-1}$ on both sides:
$A^{-1}(A^2 - 6A + 8I) = A^{-1}(0)$
$A - 6I + 8A^{-1} = 0$
Rearranging the terms to solve for $A + 8A^{-1}$,we get:
$A + 8A^{-1} = 6I$.

(B) Given the expression $(A^2 - 5A)A^{-1}$.
Distributing $A^{-1}$ inside the parentheses,we get:
$(A^2 \cdot A^{-1}) - (5A \cdot A^{-1})$
Since $A^2 \cdot A^{-1} = A$ and $A \cdot A^{-1} = I$ (where $I$ is the identity matrix),the expression simplifies to:
$A - 5I$
Now,substitute the matrix $A$ and the identity matrix $I$:
$\begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$= \begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$= \begin{bmatrix} 1-5 & 2-0 & 3-0 \\ -1-0 & 1-5 & 2-0 \\ 1-0 & 2-0 & 4-5 \end{bmatrix}$
$= \begin{bmatrix} -4 & 2 & 3 \\ -1 & -4 & 2 \\ 1 & 2 & -1 \end{bmatrix}$

(B) Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$.
First,calculate the determinant $|A| = 1(3(-1) - 0(2)) - 0 + 0 = -3$.
Next,find the matrix of cofactors $C_{ij}$:
$C_{11} = +((-1)(3) - 0) = -3$,$C_{12} = -(3(-1) - 0) = 3$,$C_{13} = +(3(2) - 3(5)) = -9$.
$C_{21} = -(0(-1) - 0(2)) = 0$,$C_{22} = +(1(-1) - 0(5)) = -1$,$C_{23} = -(1(2) - 0(5)) = -2$.
$C_{31} = +(0(0) - 3(0)) = 0$,$C_{32} = -(1(0) - 0(3)) = 0$,$C_{33} = +(1(3) - 0(3)) = 3$.
Thus,$\text{Adj}(A) = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$.
Finally,$A^{-1} = \frac{1}{|A|} \text{Adj}(A) = \frac{1}{-3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$.

(A) We use the property of matrix inversion: $(XY)^{-1} = Y^{-1} X^{-1}$.
Applying this to the given expression:
$(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1}$
Since $(M^{-1})^{-1} = M$,we have:
$(B^{-1} A^{-1})^{-1} = A B$
Now,calculate the product $AB$:
$AB = \begin{bmatrix} 2 & 3 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$
$AB = \begin{bmatrix} (2)(0) + (3)(1) & (2)(-1) + (3)(0) \\ (-3)(0) + (2)(1) & (-3)(-1) + (2)(0) \end{bmatrix}$
$AB = \begin{bmatrix} 0 + 3 & -2 + 0 \\ 0 + 2 & 3 + 0 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ 2 & 3 \end{bmatrix}$

(C) Given $AX = I$,then $X = A^{-1}$.
First,calculate the determinant of $A$:
$|A| = (1 \times 3) - (2 \times 4) = 3 - 8 = -5$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj } A = \begin{bmatrix} 3 & -2 \\ -4 & 1 \end{bmatrix}$.
Now,calculate $X = A^{-1} = \frac{1}{|A|} \text{adj } A$:
$X = \frac{1}{-5} \begin{bmatrix} 3 & -2 \\ -4 & 1 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} -3 & 2 \\ 4 & -1 \end{bmatrix}$.

(B) To find the multiplicative inverse $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
First,calculate the determinant $|A|$:
$|A| = (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta) = \cos^2 \theta + \sin^2 \theta = 1$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{1} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{bmatrix}$.
First,we find the determinant of $A$:
$|A| = (\cos \theta)(-\cos \theta) - (-\sin \theta)(-\sin \theta) = -\cos^2 \theta - \sin^2 \theta = -(\cos^2 \theta + \sin^2 \theta) = -1$.
Next,we find the adjoint of $A$,$\text{adj}(A)$,by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} -\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
The inverse of $A$ is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$:
$A^{-1} = \frac{1}{-1} \begin{bmatrix} -\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{bmatrix}$.
Thus,$A^{-1} = A$.

(A) Let $A = \left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$.
First,we calculate the determinant $|A|$:
$|A| = 0(2-3) - 1(1-9) + 2(1-6) = 0 - 1(-8) + 2(-5) = 8 - 10 = -2$.
Since $|A| \neq 0$,the inverse exists.
Next,we find the matrix of cofactors $C_{ij}$:
$C_{11} = +(2-3) = -1, C_{12} = -(1-9) = 8, C_{13} = +(1-6) = -5$.
$C_{21} = -(1-2) = 1, C_{22} = +(0-6) = -6, C_{23} = -(0-3) = 3$.
$C_{31} = +(3-4) = -1, C_{32} = -(0-2) = 2, C_{33} = +(0-1) = -1$.
Thus,$\operatorname{adj}(A) = \left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]^T = \left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]$.
Finally,$A^{-1} = \frac{1}{|A|} \operatorname{adj}(A) = \frac{1}{-2} \left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right] = \left[\begin{array}{rrr}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}\right]$.

(C) Given the diagonal matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix}$.
Since $A$ is a diagonal matrix,its inverse $A^{-1}$ is also a diagonal matrix where the diagonal elements are the reciprocals of the diagonal elements of $A$.
Thus,$A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{1} & 0 \\ 0 & 0 & \frac{1}{3} \end{bmatrix}$.
The sum of all elements of $A^{-1}$ is $\frac{1}{2} + 1 + \frac{1}{3}$.
To add these,find the common denominator,which is $6$.
Sum $= \frac{3}{6} + \frac{6}{6} + \frac{2}{6} = \frac{11}{6}$.

(C) Given $AB = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}$ and $11B^{-1} = \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix}$.
We know that $A = (AB)B^{-1}$.
From the second equation,$B^{-1} = \frac{1}{11} \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix}$.
Substituting these into the expression for $A$:
$A = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix} \times \frac{1}{11} \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix}$
$A = \frac{1}{11} \begin{bmatrix} (-6)(5) + (26)(2) & (-6)(-3) + (26)(1) \\ (-1)(5) + (19)(2) & (-1)(-3) + (19)(1) \end{bmatrix}$
$A = \frac{1}{11} \begin{bmatrix} -30 + 52 & 18 + 26 \\ -5 + 38 & 3 + 19 \end{bmatrix}$
$A = \frac{1}{11} \begin{bmatrix} 22 & 44 \\ 33 & 22 \end{bmatrix}$
$A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$.
Thus,the correct option is $C$.

(B) First,calculate the sum of matrices $A$ and $B$:
$A+B = \begin{bmatrix} 4+1 & 0 & 0 \\ 0 & 3+2 & 0 \\ 0 & 0 & 2+3 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} = 5I_3$.
Now,find the inverse of the resulting matrix:
$(A+B)^{-1} = (5I_3)^{-1}$.
Using the property $(kA)^{-1} = \frac{1}{k} A^{-1}$,we get:
$(A+B)^{-1} = \frac{1}{5} I_3^{-1} = \frac{1}{5} I_3$.

(D) By definition,if two matrices $X$ and $Y$ are inverses of each other,then their product must result in the identity matrix $I$.
Therefore,$XY = I$ and $YX = I$.
Thus,$XY = YX = I$.

(C) Given the equation $AB = 4I$,where $I$ is the identity matrix.
To find $A^{-1}$,we multiply both sides of the equation by $A^{-1}$ from the left:
$A^{-1}(AB) = A^{-1}(4I)$
$(A^{-1}A)B = 4A^{-1}$
Since $A^{-1}A = I$,we have:
$IB = 4A^{-1}$
$B = 4A^{-1}$
Dividing both sides by $4$,we get:
$A^{-1} = \frac{1}{4}B$
Therefore,the correct option is $C$.

(D) To find the inverse of a matrix $A$,we first calculate its determinant,$|A|$.
Given $A = \begin{bmatrix} 2 & -4 \\ -3 & 6 \end{bmatrix}$.
$|A| = (2 \times 6) - (-4 \times -3) = 12 - 12 = 0$.
Since the determinant of matrix $A$ is $0$,the matrix is singular.
$A$ singular matrix does not have an inverse.
Therefore,$A^{-1}$ does not exist.

(C) Given $B = A^{-1}$,we have $AB = I$,where $I$ is the identity matrix.
Given $10B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$,so $B = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$.
Since $AB = I$,we have $A(10B) = 10I$.
$\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}$.
We need to find $\alpha$. Let us multiply the second row of $A$ with the third column of $10B$:
$(2)(2) + (1)(\alpha) + (-3)(3) = 0$.
$4 + \alpha - 9 = 0$.
$\alpha - 5 = 0$.
$\alpha = 5$.

(D) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 \times 3$.
Given $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -x & 3 & -1 \end{bmatrix}$.
Let $B = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -x & 3 & -1 \end{bmatrix}$. Then $A^{-1} = -\frac{1}{2} B$.
Thus,$A \cdot (-\frac{1}{2} B) = I$,which implies $A \cdot B = -2I = \begin{bmatrix} -2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix}$.
Multiplying the third row of $A$ with the first column of $B$ to find the element at position $(3, 1)$ of the product matrix:
$(3 \times -1) + (1 \times 8) + (1 \times -x) = -2$.
$-3 + 8 - x = -2$.
$5 - x = -2$.
$x = 5 + 2 = 7$.
Wait,let us re-calculate the determinant $|A|$:
$|A| = 0(2-3) - 1(1-9) + 2(1-6) = 0 - 1(-8) + 2(-5) = 8 - 10 = -2$.
Since $A^{-1} = \frac{1}{|A|} \text{adj}(A)$,the matrix $B$ must be $\text{adj}(A)$.
The element at $(3, 1)$ of $\text{adj}(A)$ is the cofactor $C_{13}$ of matrix $A$.
$C_{13} = (-1)^{1+3} \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = 1(1 - 6) = -5$.
Comparing this with the matrix $B$,the element at $(3, 1)$ is $-x$.
So,$-x = -5$,which means $x = 5$.
Therefore,the correct option is $D$.

(C) We know that $A \times A^{-1} = I$,where $I$ is the identity matrix of order $3 \times 3$.
Given $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{5} \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & \alpha \\ 2 & 2 & -3 \end{bmatrix}$.
Multiplying the second row of $A$ with the third column of $A^{-1}$ must result in the element at position $(2, 3)$ of the identity matrix,which is $0$.
So,$\frac{1}{5} [ (2 \times 2) + (1 \times \alpha) + (2 \times -3) ] = 0$.
$4 + \alpha - 6 = 0$.
$\alpha - 2 = 0$.
$\alpha = 2$.

(C) Given the diagonal matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$.
Since $A$ is a diagonal matrix,its inverse $A^{-1}$ is given by the reciprocal of its diagonal elements:
$A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix}$.
Now,we need to find $(A^{-1})^2 = A^{-1} \times A^{-1}$:
$(A^{-1})^2 = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix} \times \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & 0 & 0 \\ 0 & \frac{1}{9} & 0 \\ 0 & 0 & \frac{1}{16} \end{bmatrix}$.
Thus,the correct option is $C$.

(B) We know that for any non-singular matrix $A$,the product of $A$ and its inverse $A^{-1}$ is the identity matrix $I$.
$A \cdot A^{-1} = I$
Taking the determinant on both sides:
$\det(A \cdot A^{-1}) = \det(I)$
Using the property $\det(AB) = \det(A) \cdot \det(B)$,we get:
$\det(A) \cdot \det(A^{-1}) = \det(I)$
Since $\det(I) = 1$ for an identity matrix,we have:
$\det(A) \cdot \det(A^{-1}) = 1$
Therefore,$\det(A^{-1}) = \frac{1}{\det(A)}$.

(C) We know that for any square matrix $A$ of order $n$,the property $A(\operatorname{adj} A) = |A| I$ holds,where $I$ is the identity matrix of order $n$.
Given $A = \begin{bmatrix} 5 & -2 \\ 4 & 3 \end{bmatrix}$.
The determinant of $A$ is $|A| = (5 \times 3) - (-2 \times 4) = 15 - (-8) = 15 + 8 = 23$.
Since the order of matrix $A$ is $2 \times 2$,we have $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Therefore,$A(\operatorname{adj} A) = |A| I = 23 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 23 I$.

(B) Given the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 3 \\ 1 & 0 & 1 \end{bmatrix}$.
First,we calculate the determinant of $A$,denoted as $|A|$.
Expanding along the first row: $|A| = 1(2 \times 1 - 3 \times 0) - 0 + 0 = 1(2) = 2$.
The order of the matrix $A$ is $n = 3$.
We use the property of the adjoint of a matrix: $|\operatorname{adj} A| = |A|^{n-1}$.
Substituting the values: $|\operatorname{adj} A| = 2^{3-1} = 2^2 = 4$.
Therefore,the correct option is $B$.

(C) We know that for any square matrix $A$ of order $n \times n$,the property of the adjoint of a matrix is given by:
$|\operatorname{adj} A| = |A|^{n-1}$
Given that the order of the matrix $A$ is $3 \times 3$,so $n = 3$.
Substituting $n = 3$ into the formula:
$|\operatorname{adj} A| = |A|^{3-1} = |A|^2$
Therefore,the correct option is $C$.

(D) To find the inverse of a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,we first calculate its determinant $|A| = ad - bc$.
Given $A = \begin{bmatrix} 8 & -2 \\ -4 & 1 \end{bmatrix}$,the determinant is $|A| = (8)(1) - (-2)(-4) = 8 - 8 = 0$.
Since the determinant of the matrix is $0$,the matrix $A$ is a singular matrix.
$A$ singular matrix does not have an inverse.
Therefore,$A^{-1}$ does not exist.

(D) The inverse of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$,where $|A| = ad - bc$ and $\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Given $A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$.
First,calculate the determinant $|A| = (2)(3) - (-2)(4) = 6 - (-8) = 6 + 8 = 14$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$.
This matches option $D$.

(B) First,we calculate the determinant of matrix $A$:
$|A| = 1(1 \times 1 - 2 \times 1) - 3(2 \times 1 - 2 \times 5) + 4(2 \times 1 - 1 \times 5)$
$|A| = 1(1 - 2) - 3(2 - 10) + 4(2 - 5)$
$|A| = 1(-1) - 3(-8) + 4(-3)$
$|A| = -1 + 24 - 12 = 11$
We know the property $|\text{adj } A| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $|\text{adj } A| = |A|^{3-1} = |A|^2$.
$|\text{adj } A| = (11)^2 = 121$.

(B) To find the inverse of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,we use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$,where $|A| = ad - bc$ and $\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Given $A = \begin{bmatrix} 2 & -3 \\ 5 & 4 \end{bmatrix}$.
First,calculate the determinant $|A| = (2)(4) - (-3)(5) = 8 + 15 = 23$.
Next,find the adjoint of $A$: $\text{adj}(A) = \begin{bmatrix} 4 & 3 \\ -5 & 2 \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{23} \begin{bmatrix} 4 & 3 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} \frac{4}{23} & \frac{3}{23} \\ -\frac{5}{23} & \frac{2}{23} \end{bmatrix}$.
Thus,the correct option is $B$.

(D) We know that for any invertible matrix $A$,the product $A \cdot A^{-1} = I$,where $I$ is the identity matrix of the same order.
Given $A = \begin{bmatrix} 2x & 0 \\ x & x \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$.
Therefore,$A \cdot A^{-1} = \begin{bmatrix} 2x & 0 \\ x & x \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Performing matrix multiplication:
$\begin{bmatrix} (2x)(1) + (0)(-1) & (2x)(0) + (0)(2) \\ (x)(1) + (x)(-1) & (x)(0) + (x)(2) \end{bmatrix} = \begin{bmatrix} 2x & 0 \\ 0 & 2x \end{bmatrix}$.
Equating this to the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we get:
$2x = 1$.
Solving for $x$,we get $x = \frac{1}{2}$.
Thus,the correct option is $D$.

(B) Given the diagonal matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$.
For a diagonal matrix $A = \text{diag}(a, b, c)$,the inverse is given by $A^{-1} = \text{diag}(\frac{1}{a}, \frac{1}{b}, \frac{1}{c})$.
Substituting the values $a=2, b=3, c=4$,we get:
$A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix}$.
Thus,option $B$ is the correct answer.

(A) Given,$(A+B)(A-B)=A^{2}-B^{2}$
Expanding the left side: $A^{2}-AB+BA-B^{2}=A^{2}-B^{2}$
Subtracting $A^{2}-B^{2}$ from both sides: $-AB+BA=0$
Therefore,$AB=BA$
Now,we evaluate $(ABA^{-1})^{2}$:
$(ABA^{-1})^{2} = (ABA^{-1})(ABA^{-1})$
Since $AB=BA$,we can write $AB = BA$,so $ABA^{-1} = BAA^{-1} = BI = B$
Thus,$(ABA^{-1})^{2} = B^{2}$

(B) We know that for any square matrix $A$,the property relating the adjoint and the determinant is $|\operatorname{adj}(A)| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here,$n=2$ and $|A|=2$.
So,$|\operatorname{adj}(A)| = |A|^{2-1} = |A|^1 = 2$.
Given $\operatorname{adj}(A) = B = \left[\begin{array}{ll}1 & 3 \\ 1 & \alpha\end{array}\right]$,we calculate the determinant of $B$:
$|B| = (1)(\alpha) - (3)(1) = \alpha - 3$.
Since $|\operatorname{adj}(A)| = |B|$,we have $\alpha - 3 = 2$.
Therefore,$\alpha = 2 + 3 = 5$.

(C) Given that $A$ is a $3 \times 3$ matrix,so $n = 3$.
We know that for any $n \times n$ matrix $M$,$|kM| = k^n |M|$.
Applying this to $|5 \times \text{adj}(A)| = 5$,we get $5^3 |\text{adj}(A)| = 5$.
This simplifies to $|\text{adj}(A)| = \frac{5}{5^3} = \frac{1}{5^2} = \frac{1}{25}$.
We also know the property $|\text{adj}(A)| = |A|^{n-1}$.
For $n = 3$,$|\text{adj}(A)| = |A|^{3-1} = |A|^2$.
Equating the two expressions for $|\text{adj}(A)|$,we have $|A|^2 = \frac{1}{25}$.
Taking the square root on both sides,we get $|A| = \pm \frac{1}{5}$.

(B) Given the equation $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A = I$,where $I$ is the identity matrix.
Let $B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$. Then $BA = I$,which implies $A = B^{-1}$.
The inverse of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Here,$ad-bc = (2)(2) - (1)(3) = 4 - 3 = 1$.
Thus,$B^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$.
Therefore,$A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$.

(D) The characteristic equation of a $3 \times 3$ matrix $A$ is given by $|A - \lambda I| = 0$,which results in $-\lambda^3 + \text{tr}(A)\lambda^2 - (\dots)\lambda + |A| = 0$.
Given the characteristic equation $\lambda^{3}-5 \lambda^{2}-3 \lambda+2=0$,we compare this with the standard form $\lambda^3 - \text{tr}(A)\lambda^2 + (\dots)\lambda - |A| = 0$.
Comparing the constant terms,we get $-|A| = 2$,which implies $|A| = -2$.
We know that for a square matrix $A$ of order $n$,$|\text{adj}(A)| = |A|^{n-1}$.
Here,$n = 3$,so $|\text{adj}(A)| = |A|^{3-1} = |A|^2$.
Substituting the value of $|A|$,we get $|\text{adj}(A)| = (-2)^2 = 4$.

(D) Given,$A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = 2(4 - 0) - 1(0 - 1) + 0(0 - 2) = 2(4) - 1(-1) + 0 = 8 + 1 = 9$.
We know the property of the adjoint of a matrix: $|\operatorname{adj} A| = |A|^{n-1}$,where $n$ is the order of the square matrix $A$.
Here,$n = 3$.
Therefore,$|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Substituting the value of $|A| = 9$,we get:
$|\operatorname{adj} A| = 9^2 = 81$.

(B) We know that $A(\operatorname{adj} A) = |A| I$.
Given that $A(\operatorname{adj} A) = 5 I$,comparing both sides,we get $|A| = 5$.
For a square matrix $A$ of order $n$,the property of the adjoint matrix is $|\operatorname{adj} A| = |A|^{n-1}$.
Here,the order of the matrix is $n = 3$.
Therefore,$|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Substituting the value of $|A|$,we get $|\operatorname{adj} A| = (5)^2 = 25$.

(C) We know that for any square matrix $A$,$A \cdot \operatorname{adj}(A) = |A| I$,where $|A|$ is the determinant of $A$ and $I$ is the identity matrix of the same order.
First,calculate the determinant of $A$:
$|A| = 1(2 \times 4 - (-3) \times (-2)) - (-2)(0 \times 4 - (-3) \times 3) + 2(0 \times (-2) - 2 \times 3)$
$|A| = 1(8 - 6) + 2(0 + 9) + 2(0 - 6)$
$|A| = 1(2) + 2(9) + 2(-6)$
$|A| = 2 + 18 - 12 = 8$
Since $A$ is a $3 \times 3$ matrix,$I = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$.
Therefore,$A \cdot \operatorname{adj}(A) = |A| I = 8 \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8\end{array}\right]$.

(A) Given the equation: $A^2 - 5A + 7I = 0$.
Since $A^2 - 5A + 7I = 0$,we have $7I = 5A - A^2$.
Multiplying both sides by $A^{-1}$ (assuming $|A| \neq 0$):
$A^{-1}(A^2 - 5A + 7I) = A^{-1}(0)$.
$A^{-1}A^2 - 5A^{-1}A + 7A^{-1}I = 0$.
$A - 5I + 7A^{-1} = 0$.
$7A^{-1} = 5I - A$.
$A^{-1} = \frac{1}{7}(5I - A)$.

(D) Given,$A = \begin{bmatrix} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{bmatrix}$ and $AB = I$.
Since $AB = I$,we have $B = A^{-1}$.
The determinant of $A$ is $|A| = (1)(1) - (\tan \frac{\alpha}{2})(-\tan \frac{\alpha}{2}) = 1 + \tan^2 \frac{\alpha}{2} = \sec^2 \frac{\alpha}{2}$.
The adjoint of $A$ is $\text{adj}(A) = \begin{bmatrix} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{bmatrix}$.
Thus,$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{\sec^2 \frac{\alpha}{2}} \begin{bmatrix} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{bmatrix}$.
Since $\frac{1}{\sec^2 \frac{\alpha}{2}} = \cos^2 \frac{\alpha}{2}$,we get $B = \cos^2 \frac{\alpha}{2} \begin{bmatrix} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{bmatrix}$.
Note that $A^T = \begin{bmatrix} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{bmatrix}$.
Therefore,$B = \cos^2 \frac{\alpha}{2} \cdot A^T$.

(B) We know that for any invertible matrix $A$,the inverse of a power of a matrix is given by $(A^n)^{-1} = (A^{-1})^n$.
Applying this property to the given expression:
$(A^2)^{-1} = (A^{-1})^2$.
Thus,the correct option is $(A^{-1})^2$.

(A) Given $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = (2)(-2) - (-1)(3) = -4 + 3 = -1$.
Since $|A| \neq 0$,$A$ is invertible.
Now,calculate $A^2$:
$A^2 = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix} = \begin{bmatrix} 4-3 & -2+2 \\ 6-6 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Since $A^2 = I$,we can find $A^3$:
$A^3 = A^2 \cdot A = I \cdot A = A$.
We need to find the inverse of $A^3$,which is $(A^3)^{-1}$.
Since $A^3 = A$,then $(A^3)^{-1} = A^{-1}$.
From $A^2 = I$,we know that $A \cdot A = I$,which implies $A^{-1} = A$.
Therefore,$(A^3)^{-1} = A$.