Adjoint and inverse of matrices Questions in English

(C) Given $|A| = 2$ and the order of matrix $A$ is $n = 4$.
We know that $|Adj(Adj(M))| = |M|^{(n-1)^2}$ for any square matrix $M$ of order $n$.
Here,$M = 2A$. Since $A$ is of order $4$,$|2A| = 2^n |A| = 2^4 \cdot 2 = 2^5 = 32$.
Now,$|Adj(Adj(2A))| = |2A|^{(4-1)^2} = |2A|^9$.
Substituting $|2A| = 2^5$,we get:
$|Adj(Adj(2A))| = (2^5)^9 = 2^{45}$.

(D) We are given that $A$ and $C$ are involutory matrices,which means $A^2 = I$ and $C^2 = I$. This implies $A^{-1} = A$ and $C^{-1} = C$.
Using the property of matrix inversion $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$,we have:
$(AB^{-1}C)^{-1} = C^{-1}(B^{-1})^{-1}A^{-1}$.
Since $(B^{-1})^{-1} = B$,we get:
$(AB^{-1}C)^{-1} = C^{-1}BA^{-1}$.
Substituting $C^{-1} = C$ and $A^{-1} = A$,we obtain:
$(AB^{-1}C)^{-1} = CBA$.

(A) Given $A = f(x) = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
First,calculate the determinant $|A|$:
$|A| = \cos x(\cos x - 0) - \sin x(-\sin x - 0) + 0 = \cos^2 x + \sin^2 x = 1$.
Since $|A| = 1$,$A^{-1} = \text{adj}(A)$.
The cofactor matrix $C$ is calculated by finding the cofactor of each element.
For a matrix of this form (orthogonal matrix),the inverse is equal to its transpose:
$A^{-1} = A^T = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Now,evaluate $f(-x)$:
$f(-x) = \begin{bmatrix} \cos(-x) & \sin(-x) & 0 \\ -\sin(-x) & \cos(-x) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Thus,$A^{-1} = f(-x)$.

(B) Given $P(\theta) = \begin{bmatrix} 1 & \cot \theta \\ -\cot \theta & 1 \end{bmatrix}$.
First,we find the determinant of $P(\theta)$:
$|P(\theta)| = (1)(1) - (\cot \theta)(-\cot \theta) = 1 + \cot^2 \theta = \csc^2 \theta$.
Since $PQ = I$,$Q$ is the inverse of $P$,i.e.,$Q = P^{-1} = \frac{1}{|P(\theta)|} \text{adj}(P(\theta))$.
The adjoint of $P(\theta)$ is $\begin{bmatrix} 1 & -\cot \theta \\ \cot \theta & 1 \end{bmatrix}$.
Thus,$Q = \frac{1}{\csc^2 \theta} \begin{bmatrix} 1 & -\cot \theta \\ \cot \theta & 1 \end{bmatrix}$.
We need to find $(\csc^2 \theta)Q$:
$(\csc^2 \theta)Q = \csc^2 \theta \cdot \frac{1}{\csc^2 \theta} \begin{bmatrix} 1 & -\cot \theta \\ \cot \theta & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\cot \theta \\ \cot \theta & 1 \end{bmatrix}$.
Now,observe $P(-\theta) = \begin{bmatrix} 1 & \cot(-\theta) \\ -\cot(-\theta) & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\cot \theta \\ \cot \theta & 1 \end{bmatrix}$.
Therefore,$(\csc^2 \theta)Q = P(-\theta)$.

(D) Given $A$ is a square matrix of order $n = 4$.
We know that $|\text{Adj}(A)| = |A|^{n-1} = |A|^{4-1} = |A|^3$.
Given $|B| = 27$,so $|A|^3 = 27$,which implies $|A| = 3$.
We need to find $|A^{-1} \text{Adj}(3AB)|$.
Using the property $|XY| = |X||Y|$,we have $|A^{-1} \text{Adj}(3AB)| = |A^{-1}| |\text{Adj}(3AB)| = \frac{1}{|A|} |\text{Adj}(3AB)|$.
Since $3AB$ is a matrix of order $4$,$|\text{Adj}(3AB)| = |3AB|^{4-1} = |3AB|^3$.
$|3AB| = 3^4 |A| |B| = 81 \times 3 \times 27 = 3^4 \times 3^1 \times 3^3 = 3^8$.
Therefore,$|\text{Adj}(3AB)| = (3^8)^3 = 3^{24}$.
Finally,$|A^{-1} \text{Adj}(3AB)| = \frac{1}{3} \times 3^{24} = 3^{23}$.

(D) Given the equation $A^T + 2A = I$.
Taking the transpose of both sides,we get $(A^T)^T + 2A^T = I^T$,which simplifies to $A + 2A^T = I$.
Now we have a system of two equations:
$1) A^T + 2A = I$
$2) 2A^T + A = I$
Multiply equation $(1)$ by $2$: $2A^T + 4A = 2I$.
Subtract equation $(2)$ from this: $(2A^T + 4A) - (2A^T + A) = 2I - I$,which gives $3A = I$,so $A = \frac{1}{3}I$.
Then $A^{-1} = (\frac{1}{3}I)^{-1} = 3I$.
Since $A$ is a $3 \times 3$ matrix,$A^{-1} = \text{diag}(3, 3, 3)$.
Therefore,$\det(A^{-1}) = 3 \times 3 \times 3 = 27$.

(D) Given that $P$ is a $4 \times 4$ matrix,so $n = 4$.
We know that for any $n \times n$ matrix $A$,$|adj(A)| = |A|^{n-1}$.
Here,$A = 3P$,so $|adj(3P)| = |3P|^{4-1} = |3P|^3$.
Since $P$ is a $4 \times 4$ matrix,$|3P| = 3^4 |P|$.
Substituting the value of $|P| = -2$,we get $|3P| = 3^4 \times (-2) = -2 \cdot 3^4$.
Now,$|adj(3P)| = (-2 \cdot 3^4)^3 = (-2)^3 \cdot (3^4)^3 = -8 \cdot 3^{12} = -3^{12} \cdot 2^3$.
Comparing this with the given options,none of the options match the result $-3^{12} \cdot 2^3$.

(A) To determine if $A$ is invertible,we check if the determinant $|A| \neq 0$. We consider the parity of the elements in the matrix by taking the matrix modulo $2$.
Modulo $2$,the matrix $A$ becomes:
$A \equiv \begin{bmatrix} 0+1 & 0+0 & 0+0 \\ 0+0 & 0+1 & 0+0 \\ 0+0 & 0+0 & 0+1 \end{bmatrix} \pmod{2}$
$A \equiv \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \pmod{2}$
The determinant $|A| \pmod{2}$ is the determinant of the identity matrix,which is $1$.
Since $|A| \equiv 1 \pmod{2}$,the determinant $|A|$ must be an odd number.
An odd number cannot be $0$.
Therefore,$|A| \neq 0$ for all $n \in N$.
Thus,the matrix $A$ is invertible for all $n \in N$.

(D) Given $A = \begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix} = aI$,where $I$ is the identity matrix of order $3 \times 3$.
The determinant $|A| = a^3$.
We know that $|adj A| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here $n = 3$,so $|adj A| = |A|^{3-1} = |A|^2$.
Therefore,$|A| |adj A| = |A| \cdot |A|^2 = |A|^3$.
Substituting $|A| = a^3$,we get $|A|^3 = (a^3)^3 = a^9$.

(C) We are given the expression $(B^{-1}AB)^5$.
This can be written as the product of $5$ terms:
$(B^{-1}AB)^5 = (B^{-1}AB)(B^{-1}AB)(B^{-1}AB)(B^{-1}AB)(B^{-1}AB)$.
Since matrix multiplication is associative,we can group the terms as:
$= B^{-1}A(BB^{-1})A(BB^{-1})A(BB^{-1})A(BB^{-1})AB$.
Since $BB^{-1} = I$ (the identity matrix),the expression simplifies to:
$= B^{-1}A(I)A(I)A(I)A(I)AB$.
$= B^{-1}AAAAAB$.
$= B^{-1}A^5B$.

(D) For a $3 \times 3$ matrix $A$,we have the following properties:
$1$. $adj(adj(A)) = |A|^{n-2} A$. Since $n=3$,$adj(adj(A)) = |A|^{3-2} A = |A| A$. Thus,option $B$ is true.
$2$. We know that $adj(A) = |A| A^{-1}$. Replacing $A$ with $adj(A)$,we get $adj(adj(A)) = |adj(A)| (adj(A))^{-1}$.
$3$. Since $|adj(A)| = |A|^{n-1} = |A|^{3-1} = |A|^2$,we have $adj(adj(A)) = |A|^2 (adj(A))^{-1}$. Thus,option $C$ is true.
$4$. Comparing the results,$adj(adj(A)) = |A| A$ and $adj(adj(A)) = |A|^2 (adj(A))^{-1}$.
$5$. Option $D$ states $adj(adj(A)) = |A| (adj(A))^{-1}$,which is generally false.
$6$. Option $A$ states $adj(A) = |A| (adj(A))^{-1}$. Since $adj(A) = |A| A^{-1}$,this implies $A^{-1} = (adj(A))^{-1}$,which is true. Therefore,option $D$ is the one that is not always true.

(A) Given that $A$ is a $3 \times 3$ matrix,so $n = 3$.
We know the property $|k \cdot M| = k^n |M|$,where $M$ is an $n \times n$ matrix.
Applying this to the given equation: $|5 \cdot \text{adj } A| = 5^3 |\text{adj } A| = 125 |\text{adj } A|$.
We also know the property $|\text{adj } A| = |A|^{n-1}$.
Substituting $n = 3$,we get $|\text{adj } A| = |A|^{3-1} = |A|^2$.
Therefore,the equation becomes $125 |A|^2 = 5$.
$|A|^2 = \frac{5}{125} = \frac{1}{25}$.
Taking the square root on both sides,$|A| = \pm \sqrt{\frac{1}{25}} = \pm \frac{1}{5}$.

(A) Given $A \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$.
Let $B = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$. So,$AB = C$.
We know that $A^{-1} = B C^{-1}$.
First,find $C^{-1}$. Since $C$ is a permutation matrix,$C^{-1} = C^T = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$.
Now,$A^{-1} = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$.
Performing matrix multiplication:
$A^{-1} = \begin{bmatrix} (1)(0)+(2)(0)+(3)(1) & (1)(1)+(2)(0)+(3)(0) & (1)(0)+(2)(1)+(3)(0) \\ (0)(0)+(2)(0)+(3)(1) & (0)(1)+(2)(0)+(3)(0) & (0)(0)+(2)(1)+(3)(0) \\ (0)(0)+(1)(0)+(1)(1) & (0)(1)+(1)(0)+(1)(0) & (0)(0)+(1)(1)+(1)(0) \end{bmatrix} = \begin{bmatrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{bmatrix}$.

(C) Given $A = \begin{bmatrix} 0 & 0 & a \\ 0 & b & c \\ d & e & f \end{bmatrix}$. The determinant $|A| = 0(bf - ce) - 0(0 - cd) + a(0 - bd) = -abd$.
Since $abd \neq 0$,the matrix is invertible.
The cofactor matrix is calculated as:
$C_{11} = bf - ce, C_{12} = -(-cd) = cd, C_{13} = -bd$
$C_{21} = -(-ae) = ae, C_{22} = -ad, C_{23} = 0$
$C_{31} = -ab, C_{32} = 0, C_{33} = 0$
Thus,$\text{adj}(A) = \begin{bmatrix} bf-ce & ae & -ab \\ cd & -ad & 0 \\ -bd & 0 & 0 \end{bmatrix}$.
$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{-abd} \begin{bmatrix} bf-ce & ae & -ab \\ cd & -ad & 0 \\ -bd & 0 & 0 \end{bmatrix}$.
Given $A^{-1} = A^T$,where $A^T = \begin{bmatrix} 0 & 0 & d \\ 0 & b & e \\ a & c & f \end{bmatrix}$.
Equating $A^{-1} = A^T$ leads to the condition $A A^T = I$ (since $A^{-1} = A^T \implies A A^T = I$).
Calculating $A A^T = \begin{bmatrix} 0 & 0 & a \\ 0 & b & c \\ d & e & f \end{bmatrix} \begin{bmatrix} 0 & 0 & d \\ 0 & b & e \\ a & c & f \end{bmatrix} = \begin{bmatrix} a^2 & ac & af \\ ac & b^2+c^2 & be+cf \\ ad & ae+cf & d^2+e^2+f^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
From $a^2 = 1$,$a = \pm 1$.
From $ac = 0$,$c = 0$.
From $af = 0$,$f = 0$.
From $b^2+c^2 = 1$,$b^2 = 1 \implies b = \pm 1$.
From $be+cf = 0$,$be = 0 \implies e = 0$.
From $d^2+e^2+f^2 = 1$,$d^2 = 1 \implies d = \pm 1$.
Thus,$a, b, d \in \{1, -1\}$ and $c=e=f=0$.
The number of such matrices is $2 \times 2 \times 2 = 2^3 = 8$.

(C) Given $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
By the property of rotation matrices,$A^n = \begin{bmatrix} \cos(n\theta) & -\sin(n\theta) \\ \sin(n\theta) & \cos(n\theta) \end{bmatrix}$ for any integer $n$.
Therefore,$A^{-50} = \begin{bmatrix} \cos(-50\theta) & -\sin(-50\theta) \\ \sin(-50\theta) & \cos(-50\theta) \end{bmatrix} = \begin{bmatrix} \cos(50\theta) & \sin(50\theta) \\ -\sin(50\theta) & \cos(50\theta) \end{bmatrix}$.
Given $\theta = \frac{\pi}{12}$,we have $50\theta = 50 \times \frac{\pi}{12} = \frac{25\pi}{6} = 4\pi + \frac{\pi}{6}$.
Since $\cos(4\pi + \frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(4\pi + \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$,
$A^{-50} = \begin{bmatrix} \cos(\frac{\pi}{6}) & \sin(\frac{\pi}{6}) \\ -\sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$.

(C) To check if $A$ is invertible,we calculate its determinant $|A|$.
$|A| = e^t \cdot e^{-t} \cdot e^{-t} \begin{vmatrix} 1 & \cos t & \sin t \\ 1 & -(\cos t + \sin t) & \cos t - \sin t \\ 1 & 2 \sin t & -2 \cos t \end{vmatrix} = e^{-t} \begin{vmatrix} 1 & \cos t & \sin t \\ 1 & -\cos t - \sin t & \cos t - \sin t \\ 1 & 2 \sin t & -2 \cos t \end{vmatrix}$.
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$|A| = e^{-t} \begin{vmatrix} 1 & \cos t & \sin t \\ 0 & -2 \cos t - \sin t & \cos t - 2 \sin t \\ 0 & 2 \sin t - \cos t & -2 \cos t - \sin t \end{vmatrix}$.
Expanding along the first column:
$|A| = e^{-t} [(-2 \cos t - \sin t)^2 - (\cos t - 2 \sin t)(2 \sin t - \cos t)]$.
$|A| = e^{-t} [4 \cos^2 t + \sin^2 t + 4 \sin t \cos t - (2 \sin t \cos t - \cos^2 t - 4 \sin^2 t + 2 \sin t \cos t)]$.
$|A| = e^{-t} [4 \cos^2 t + \sin^2 t + 4 \sin t \cos t + \cos^2 t + 4 \sin^2 t - 4 \sin t \cos t] = e^{-t} [5 \cos^2 t + 5 \sin^2 t] = 5e^{-t}$.
Since $5e^{-t} \neq 0$ for all $t \in \mathbb{R}$,the matrix $A$ is invertible for all $t \in \mathbb{R}$.

(C) Given that $AA^T = I_3$,$A$ is an orthogonal matrix.
Thus,the product $AA^T$ is:
$\begin{bmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{bmatrix} \begin{bmatrix} 0 & p & p \\ 2q & q & -q \\ r & -r & r \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Performing matrix multiplication:
Row $1$ $\cdot$ Col $1$: $0^2 + (2q)^2 + r^2 = 4q^2 + r^2 = 1$ (Eq. $1$)
Row $2$ $\cdot$ Col $2$: $p^2 + q^2 + (-r)^2 = p^2 + q^2 + r^2 = 1$ (Eq. $2$)
Row $2$ $\cdot$ Col $3$: $p^2 - q^2 - r^2 = 0$ (Eq. $3$)
Adding (Eq. $2$) and (Eq. $3$):
$2p^2 = 1 \implies p^2 = \frac{1}{2} \implies |p| = \frac{1}{\sqrt{2}}$.

(D) We know that $\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a+b \\ 0 & 1 \end{bmatrix}$.
Applying this property to the given product:
$\begin{bmatrix} 1 & 1+2+3+\dots+(n-1) \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix}$.
The sum of the first $(n-1)$ natural numbers is $\frac{(n-1)n}{2}$.
So,$\frac{n(n-1)}{2} = 78 \Rightarrow n^2 - n - 156 = 0$.
Solving the quadratic equation: $(n-13)(n+12) = 0$. Since $n$ must be positive,$n = 13$.
We need to find the inverse of $A = \begin{bmatrix} 1 & 13 \\ 0 & 1 \end{bmatrix}$.
For a matrix $A = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$,the inverse is $A^{-1} = \begin{bmatrix} 1 & -k \\ 0 & 1 \end{bmatrix}$.
Thus,the inverse is $\begin{bmatrix} 1 & -13 \\ 0 & 1 \end{bmatrix}$.

(C) Given that $B = A^{-1}$,we know that $\det(B) = \frac{1}{\det(A)}$.
First,calculate the determinant of matrix $B$:
$\det(B) = 5(2(-1) - 3(1)) - 2\alpha(0(-1) - \alpha(1)) + 1(0(3) - \alpha(2))$
$\det(B) = 5(-2 - 3) - 2\alpha(-\alpha) + 1(-2\alpha)$
$\det(B) = 5(-5) + 2\alpha^2 - 2\alpha = 2\alpha^2 - 2\alpha - 25$.
Given the condition $\det(A) + 1 = 0$,we have $\det(A) = -1$.
Since $\det(B) = \frac{1}{\det(A)}$,we have $\det(B) = \frac{1}{-1} = -1$.
Equating the two expressions for $\det(B)$:
$2\alpha^2 - 2\alpha - 25 = -1$
$2\alpha^2 - 2\alpha - 24 = 0$
Dividing by $2$:
$\alpha^2 - \alpha - 12 = 0$
$(\alpha - 4)(\alpha + 3) = 0$.
The values of $\alpha$ are $4$ and $-3$.
The sum of these values is $4 + (-3) = 1$.

(B) Given $A = \begin{bmatrix} 2 & 2 \\ 9 & 4 \end{bmatrix}$.
First,calculate the determinant $|A| = (2 \times 4) - (2 \times 9) = 8 - 18 = -10$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements: $\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ -9 & 2 \end{bmatrix}$.
The inverse is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{-10} \begin{bmatrix} 4 & -2 \\ -9 & 2 \end{bmatrix}$.
Therefore,$10 A^{-1} = 10 \times \left( \frac{1}{-10} \begin{bmatrix} 4 & -2 \\ -9 & 2 \end{bmatrix} \right) = -1 \begin{bmatrix} 4 & -2 \\ -9 & 2 \end{bmatrix} = \begin{bmatrix} -4 & 2 \\ 9 & -2 \end{bmatrix}$.
Now,check the options:
$A - 6I = \begin{bmatrix} 2 & 2 \\ 9 & 4 \end{bmatrix} - \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 2-6 & 2-0 \\ 9-0 & 4-6 \end{bmatrix} = \begin{bmatrix} -4 & 2 \\ 9 & -2 \end{bmatrix}$.
Thus,$10 A^{-1} = A - 6I$.

(C) Given $A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3 \end{bmatrix}$.
First,calculate the determinant of $A$:
$|A| = 1(3 \times 3 - 4 \times -1) - 1(1 \times 3 - 4 \times 1) + 2(1 \times -1 - 3 \times 1)$
$|A| = 1(9 + 4) - 1(3 - 4) + 2(-1 - 3)$
$|A| = 13 + 1 - 8 = 6$.
We are given $B = \operatorname{adj} A$ and $C = 3A$. We need to find $\frac{|\operatorname{adj} B|}{|C|}$.
Since $B = \operatorname{adj} A$,then $\operatorname{adj} B = \operatorname{adj}(\operatorname{adj} A)$.
Using the property $|\operatorname{adj} M| = |M|^{n-1}$ where $n$ is the order of the matrix:
$|\operatorname{adj} B| = |\operatorname{adj}(\operatorname{adj} A)| = |A|^{(n-1)^2} = |A|^{(3-1)^2} = |A|^4$.
For the denominator,$|C| = |3A| = 3^n |A| = 3^3 |A| = 27 |A|$.
Thus,$\frac{|\operatorname{adj} B|}{|C|} = \frac{|A|^4}{27 |A|} = \frac{|A|^3}{27}$.
Substituting $|A| = 6$:
$\frac{6^3}{27} = \frac{216}{27} = 8$.

(C) The adjoint of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $adj$ $A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Given $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$,we have $a=2, b=3, c=1, d=4$.
Substituting these values into the formula:
$adj$ $A = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$.

(D) First,calculate the determinant $|A|$:
$|A| = 1(16 - 9) - 3(4 - 3) + 3(3 - 4) = 1(7) - 3(1) + 3(-1) = 7 - 3 - 3 = 1 \neq 0$.
Next,find the cofactor matrix and its transpose to get $\text{adj } A$:
$A_{11} = +(16-9) = 7, A_{12} = -(4-3) = -1, A_{13} = +(3-4) = -1$
$A_{21} = -(12-9) = -3, A_{22} = +(4-3) = 1, A_{23} = -(3-3) = 0$
$A_{31} = +(9-12) = -3, A_{32} = -(3-3) = 0, A_{33} = +(4-3) = 1$
Thus,$\text{adj } A = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$.
Verify $A(\text{adj } A) = |A|I$:
$A(\text{adj } A) = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 7-3-3 & -3+3+0 & -3+0+3 \\ 7-4-3 & -3+4+0 & -3+0+3 \\ 7-3-4 & -3+3+0 & -3+0+4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 1 \cdot I = |A|I$.
Finally,$A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{1} \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$.

(A) First,we calculate the product $AB$:
$AB = \left[\begin{array}{cc}2 & 3 \\ 1 & -4\end{array}\right] \left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right] = \left[\begin{array}{cc}2(1)+3(-1) & 2(-2)+3(3) \\ 1(1)+(-4)(-1) & 1(-2)+(-4)(3)\end{array}\right] = \left[\begin{array}{cc}-1 & 5 \\ 5 & -14\end{array}\right]$.
Next,we find the determinant $|AB| = (-1)(-14) - (5)(5) = 14 - 25 = -11$.
Since $|AB| \neq 0$,$(AB)^{-1}$ exists.
$(AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB) = \frac{1}{-11} \left[\begin{array}{cc}-14 & -5 \\ -5 & -1\end{array}\right] = \frac{1}{11} \left[\begin{array}{cc}14 & 5 \\ 5 & 1\end{array}\right]$.
Now,we calculate $A^{-1}$ and $B^{-1}$:
$|A| = 2(-4) - 3(1) = -8 - 3 = -11$.
$A^{-1} = \frac{1}{-11} \left[\begin{array}{cc}-4 & -3 \\ -1 & 2\end{array}\right] = \frac{1}{11} \left[\begin{array}{cc}4 & 3 \\ 1 & -2\end{array}\right]$.
$|B| = 1(3) - (-2)(-1) = 3 - 2 = 1$.
$B^{-1} = \frac{1}{1} \left[\begin{array}{cc}3 & 2 \\ 1 & 1\end{array}\right] = \left[\begin{array}{cc}3 & 2 \\ 1 & 1\end{array}\right]$.
Finally,calculate $B^{-1} A^{-1}$:
$B^{-1} A^{-1} = \left[\begin{array}{cc}3 & 2 \\ 1 & 1\end{array}\right] \left( \frac{1}{11} \left[\begin{array}{cc}4 & 3 \\ 1 & -2\end{array}\right] \right) = \frac{1}{11} \left[\begin{array}{cc}3(4)+2(1) & 3(3)+2(-2) \\ 1(4)+1(1) & 1(3)+1(-2)\end{array}\right] = \frac{1}{11} \left[\begin{array}{cc}14 & 5 \\ 5 & 1\end{array}\right]$.
Since $(AB)^{-1} = B^{-1} A^{-1}$,the identity is verified.

(B) First,we calculate $A^{2} = A \cdot A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4+3 & 6+6 \\ 2+2 & 3+4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}$.
Now,substitute into the equation $A^{2} - 4A + I$:
$\begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} - 4 \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7-8+1 & 12-12+0 \\ 4-4+0 & 7-8+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
To find $A^{-1}$,start with $A^{2} - 4A + I = O$.
Subtract $I$ from both sides: $A^{2} - 4A = -I$.
Multiply by $A^{-1}$ on both sides: $A^{-1}(A^{2} - 4A) = A^{-1}(-I)$.
$(A^{-1}A)A - 4(A^{-1}A) = -A^{-1}$.
$IA - 4I = -A^{-1}$.
$A - 4I = -A^{-1}$,which implies $A^{-1} = 4I - A$.
$A^{-1} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$.

(C) Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$.
The adjoint of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Here,$a = 1, b = 2, c = 3, d = 4$.
Therefore,$\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$.

(D) Let $A = \left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$.
The adjoint of a matrix is the transpose of the cofactor matrix. Let $C_{ij}$ be the cofactor of the element $a_{ij}$.
$C_{11} = +\left|\begin{array}{cc}3 & 5 \\ 0 & 1\end{array}\right| = 3 - 0 = 3$
$C_{12} = -\left|\begin{array}{cc}2 & 5 \\ -2 & 1\end{array}\right| = -(2 + 10) = -12$
$C_{13} = +\left|\begin{array}{cc}2 & 3 \\ -2 & 0\end{array}\right| = 0 - (-6) = 6$
$C_{21} = -\left|\begin{array}{cc}-1 & 2 \\ 0 & 1\end{array}\right| = -(-1 - 0) = 1$
$C_{22} = +\left|\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right| = 1 - (-4) = 5$
$C_{23} = -\left|\begin{array}{cc}1 & -1 \\ -2 & 0\end{array}\right| = -(0 - 2) = 2$
$C_{31} = +\left|\begin{array}{cc}-1 & 2 \\ 3 & 5\end{array}\right| = -5 - 6 = -11$
$C_{32} = -\left|\begin{array}{cc}1 & 2 \\ 2 & 5\end{array}\right| = -(5 - 4) = -1$
$C_{33} = +\left|\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right| = 3 - (-2) = 5$
The cofactor matrix is $\left[\begin{array}{ccc}3 & -12 & 6 \\ 1 & 5 & -1 \\ -11 & -1 & 5\end{array}\right]$.
Taking the transpose,we get $\text{adj}(A) = \left[\begin{array}{ccc}3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5\end{array}\right]$.

(A) Given $A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$.
First,calculate the determinant $|A|$:
$|A| = (2)(-6) - (3)(-4) = -12 + 12 = 0$.
Thus,$|A| I = 0 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Next,find the adjoint of $A$ $(\text{adj } A)$:
The cofactors are $C_{11} = -6, C_{12} = 4, C_{21} = -3, C_{22} = 2$.
Therefore,$\text{adj } A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}$.
Now,calculate $A(\text{adj } A)$:
$A(\text{adj } A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Finally,calculate $(\text{adj } A) A$:
$(\text{adj } A) A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} = \begin{bmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Since $A(\text{adj } A) = (\text{adj } A) A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = |A| I$,the identity is verified.

(A) $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$
$|A| = 1(0 - 0) - (-1)(9 - (-2)) + 2(0 - 0) = 1(0) + 1(11) + 2(0) = 11$
$\therefore |A| I = 11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$
Now,calculating the cofactors $A_{ij}$:
$A_{11} = (0 - 0) = 0, A_{12} = -(9 - (-2)) = -11, A_{13} = (0 - 0) = 0$
$A_{21} = -(-3 - 0) = 3, A_{22} = (3 - 2) = 1, A_{23} = -(0 - (-1)) = -1$
$A_{31} = (2 - 0) = 2, A_{32} = -(-2 - 6) = 8, A_{33} = (0 - (-3)) = 3$
$\therefore \text{adj } A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$
Now,$A(\text{adj } A) = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$
Also,$(\text{adj } A) A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$
Hence,$A(\text{adj } A) = (\text{adj } A) A = |A| I$ is verified.

Let $A = \left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$.
We know that the inverse of a matrix $A$ is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$,which exists if $|A| \neq 0$.
First,calculate the determinant $|A|$:
$|A| = \left|\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right| = (2 \times 3) - (4 \times -2) = 6 + 8 = 14$.
Since $|A| = 14 \neq 0$,the inverse $A^{-1}$ exists.
Next,calculate the adjoint of $A$ (adj $A$):
For a $2 \times 2$ matrix $\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$,the adjoint is $\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Thus,$\text{adj}(A) = \left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right]$.
Finally,calculate $A^{-1}$:
$A^{-1} = \frac{1}{14} \left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right] = \left[\begin{array}{cc}\frac{3}{14} & \frac{2}{14} \\ -\frac{4}{14} & \frac{2}{14}\end{array}\right] = \left[\begin{array}{cc}\frac{3}{14} & \frac{1}{7} \\ -\frac{2}{7} & \frac{1}{7}\end{array}\right]$.

(D) Let $A = \left[\begin{array}{cc}-1 & 5 \\ -3 & 2\end{array}\right]$.
First,calculate the determinant of $A$:
$|A| = (-1)(2) - (5)(-3) = -2 + 15 = 13$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The adjoint of a $2 \times 2$ matrix $\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$ is given by $\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Applying this to matrix $A$:
$adj(A) = \left[\begin{array}{cc}2 & -5 \\ 3 & -1\end{array}\right]$.
The inverse is given by $A^{-1} = \frac{1}{|A|} adj(A)$:
$A^{-1} = \frac{1}{13} \left[\begin{array}{cc}2 & -5 \\ 3 & -1\end{array}\right]$.

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

(B) Let $A = \left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$.
First,we find the determinant $|A|$:
$|A| = 1((-3)(1) - (0)(2)) - 0 + 0 = -3$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
Next,we find the cofactor matrix $C_{ij}$:
$C_{11} = +((-3) - 0) = -3$
$C_{12} = -((-3) - 0) = 3$
$C_{13} = +((6) - (15)) = -9$
$C_{21} = -(0 - 0) = 0$
$C_{22} = +((-1) - 0) = -1$
$C_{23} = -(2 - 0) = -2$
$C_{31} = +(0 - 0) = 0$
$C_{32} = -(0 - 0) = 0$
$C_{33} = +(3 - 0) = 3$
The adjoint matrix $adj(A)$ is the transpose of the cofactor matrix:
$adj(A) = \left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$.
Finally,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-3} \left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right] = -\frac{1}{3} \left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$.

(C) Let $A = \left[\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]$.
First,we calculate the determinant $|A|$:
$|A| = 2((-1)(1) - (0)(2)) - 1((4)(1) - (0)(-7)) + 3((4)(2) - (-1)(-7))$
$|A| = 2(-1 - 0) - 1(4 - 0) + 3(8 - 7)$
$|A| = 2(-1) - 1(4) + 3(1) = -2 - 4 + 3 = -3$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
Next,we find the cofactor matrix $C_{ij}$:
$C_{11} = +((-1)(1) - (0)(2)) = -1$
$C_{12} = -((4)(1) - (0)(-7)) = -4$
$C_{13} = +((4)(2) - (-1)(-7)) = 8 - 7 = 1$
$C_{21} = -((1)(1) - (3)(2)) = -(1 - 6) = 5$
$C_{22} = +((2)(1) - (3)(-7)) = 2 + 21 = 23$
$C_{23} = -((2)(2) - (1)(-7)) = -(4 + 7) = -11$
$C_{31} = +((1)(0) - (3)(-1)) = 0 + 3 = 3$
$C_{32} = -((2)(0) - (3)(4)) = -(0 - 12) = 12$
$C_{33} = +((2)(-1) - (1)(4)) = -2 - 4 = -6$
The adjoint matrix $adj(A)$ is the transpose of the cofactor matrix:
$adj(A) = \left[\begin{array}{ccc}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{array}\right]$.
Finally,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-3} \left[\begin{array}{ccc}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{array}\right] = -\frac{1}{3} \left[\begin{array}{ccc}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{array}\right]$.

(D) Let $A = \left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$.
First,calculate the determinant $|A|$ by expanding along the first column:
$|A| = 1(8 - 6) - 0( -4 + 4) + 3(3 - 4) = 1(2) - 0 + 3(-1) = 2 - 3 = -1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
Next,find the cofactor matrix $C_{ij}$:
$C_{11} = +(8-6) = 2, C_{12} = -(0+9) = -9, C_{13} = +(0-6) = -6$
$C_{21} = -(-4+4) = 0, C_{22} = +(4-6) = -2, C_{23} = -(-2+3) = -1$
$C_{31} = +(3-4) = -1, C_{32} = -(-3-0) = 3, C_{33} = +(2-0) = 2$
The adjoint matrix $\text{adj } A$ is the transpose of the cofactor matrix:
$\text{adj } A = \left[\begin{array}{ccc}2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2\end{array}\right]$.
Finally,$A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{-1} \left[\begin{array}{ccc}2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2\end{array}\right] = \left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$.

(A) Let $A = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos a & \sin a \\ 0 & \sin a & -\cos a\end{array}\right]$.
First,we calculate the determinant $|A|$:
$|A| = 1(\cos a(-\cos a) - \sin a(\sin a)) = -\cos^2 a - \sin^2 a = -(\cos^2 a + \sin^2 a) = -1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
Next,we find the cofactor matrix $C$ where $C_{ij}$ is the cofactor of element $a_{ij}$:
$C_{11} = +(-\cos^2 a - \sin^2 a) = -1$,$C_{12} = -(0) = 0$,$C_{13} = +(0) = 0$.
$C_{21} = -(0) = 0$,$C_{22} = +(-\cos a) = -\cos a$,$C_{23} = -(\sin a) = -\sin a$.
$C_{31} = +(0) = 0$,$C_{32} = -(\sin a) = -\sin a$,$C_{33} = +(\cos a) = \cos a$.
The adjoint matrix $\operatorname{adj} A$ is the transpose of the cofactor matrix:
$\operatorname{adj} A = \left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos a & -\sin a \\ 0 & -\sin a & \cos a\end{array}\right]$.
Finally,$A^{-1} = \frac{1}{|A|} \operatorname{adj} A = \frac{1}{-1} \left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos a & -\sin a \\ 0 & -\sin a & \cos a\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos a & \sin a \\ 0 & \sin a & -\cos a\end{array}\right]$.

(A) Given $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$.
$|A| = (3 \times 5) - (7 \times 2) = 15 - 14 = 1$.
$adj(A) = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.
$A^{-1} = \frac{1}{|A|} adj(A) = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.
Given $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$.
$|B| = (6 \times 9) - (8 \times 7) = 54 - 56 = -2$.
$adj(B) = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$.
$B^{-1} = \frac{1}{|B|} adj(B) = -\frac{1}{2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix}$.
Now,$B^{-1} A^{-1} = \begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} -\frac{45}{2} - 8 & \frac{63}{2} + 12 \\ \frac{35}{2} + 6 & -\frac{49}{2} - 9 \end{bmatrix} = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \dots (1)$.
Now,$AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 18+49 & 24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$.
$|AB| = (67 \times 61) - (87 \times 47) = 4087 - 4089 = -2$.
$adj(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$.
$(AB)^{-1} = \frac{1}{|AB|} adj(AB) = -\frac{1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \dots (2)$.
From $(1)$ and $(2)$,$(AB)^{-1} = B^{-1} A^{-1}$ is verified.

(A) Given $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.
First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} (3)(3) + (1)(-1) & (3)(1) + (1)(2) \\ (-1)(3) + (2)(-1) & (-1)(1) + (2)(2) \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$.
Now,evaluate $A^{2} - 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
$= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$.
To find $A^{-1}$,multiply the equation $A^{2} - 5A + 7I = 0$ by $A^{-1}$:
$A^{2}A^{-1} - 5AA^{-1} + 7IA^{-1} = 0$.
$A - 5I + 7A^{-1} = 0$.
$7A^{-1} = 5I - A$.
$7A^{-1} = 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 5-3 & 0-1 \\ 0-(-1) & 5-2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$.
First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}$.
Next,calculate $A^{3} = A^{2} \cdot A = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}$.
Now,substitute into $A^{3} - 6A^{2} + 5A + 11I$:
$= \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix} - 6 \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} + 5 \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} + 11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0$.
To find $A^{-1}$,multiply $A^{3} - 6A^{2} + 5A + 11I = 0$ by $A^{-1}$:
$A^{2} - 6A + 5I + 11A^{-1} = 0 \Rightarrow 11A^{-1} = -(A^{2} - 6A + 5I)$.
$A^{2} - 6A + 5I = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} - \begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix} + \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix}$.
Thus,$A^{-1} = -\frac{1}{11} \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}$.

(A) Given $A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$.
First,calculate $A^{2} = A \times A = \left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right] \left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right] = \left[\begin{array}{ccc}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right]$.
Next,calculate $A^{3} = A^{2} \times A = \left[\begin{array}{ccc}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right] \left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right] = \left[\begin{array}{ccc}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{array}\right]$.
Now,verify $A^{3}-6 A^{2}+9 A-4 I = \left[\begin{array}{ccc}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{array}\right] - 6\left[\begin{array}{ccc}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right] + 9\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right] - 4\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] = 0$.
To find $A^{-1}$,multiply the equation $A^{3}-6 A^{2}+9 A-4 I=0$ by $A^{-1}$:
$A^{2}-6 A+9 I-4 A^{-1}=0 \Rightarrow 4 A^{-1} = A^{2}-6 A+9 I$.
$4 A^{-1} = \left[\begin{array}{ccc}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right] - 6\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right] + 9\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]$.
Thus,$A^{-1} = \frac{1}{4}\left[\begin{array}{ccc}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]$.

(C) We know that for any square matrix $A$ of order $n$,the property of the adjoint matrix is given by:
$A(adj\, A) = (adj\, A)A = |A|I_n$
Taking the determinant on both sides:
$|A(adj\, A)| = ||A|I_n|$
$|A| \cdot |adj\, A| = |A|^n \cdot |I_n|$
Since $|I_n| = 1$,we have:
$|A| \cdot |adj\, A| = |A|^n$
$|adj\, A| = |A|^{n-1}$
Given that the order of the matrix $A$ is $3 \times 3$,so $n = 3$.
Therefore,$|adj\, A| = |A|^{3-1} = |A|^2$.
Hence,the correct answer is $C$.

(D) Since $A$ is an invertible matrix,$A^{-1}$ exists and $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
As matrix $A$ is of order $2$,let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
Then,$|A| = ad - bc$ and $\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Now,$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \begin{bmatrix} \frac{d}{|A|} & \frac{-b}{|A|} \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{bmatrix}$.
Therefore,$\det(A^{-1}) = \begin{vmatrix} \frac{d}{|A|} & \frac{-b}{|A|} \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix}$.
$= \frac{1}{|A|^2} \begin{vmatrix} d & -b \\ -c & a \end{vmatrix}$.
$= \frac{1}{|A|^2} (ad - bc)$.
$= \frac{1}{|A|^2} \cdot |A| = \frac{1}{|A|}$.
Therefore,$\det(A^{-1}) = \frac{1}{\det(A)}$.
Hence,the correct answer is $D$.

(D) We know that $(AB)^{-1} = B^{-1}A^{-1}$.
First,we calculate $B^{-1}$. We know that $B^{-1} = \frac{1}{|B|} \text{adj}(B)$,which exists if $|B| \neq 0$.
Calculating $|B|$:
$|B| = \left|\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right| = 1(3-0) - 2(-1-0) - 2(2-0) = 3 + 2 - 4 = 1$.
Since $|B| = 1 \neq 0$,$B^{-1}$ exists.
Calculating $\text{adj}(B)$:
The matrix of cofactors is obtained by calculating the cofactor $A_{ij}$ for each element.
$A_{11} = 3, A_{12} = 1, A_{13} = 2$
$A_{21} = 2, A_{22} = 1, A_{23} = 2$
$A_{31} = 6, A_{32} = 2, A_{33} = 5$
Thus,$\text{adj}(B) = \left[\begin{array}{ccc}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$.
Since $|B| = 1$,$B^{-1} = \text{adj}(B) = \left[\begin{array}{ccc}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$.
Now,$(AB)^{-1} = B^{-1}A^{-1} = \left[\begin{array}{ccc}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right] \left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$.
Performing matrix multiplication:
$(AB)^{-1} = \left[\begin{array}{ccc}9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10\end{array}\right] = \left[\begin{array}{ccc}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{array}\right]$.

(N/A) Given $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$.
First,calculate the determinant $|A| = 1(15 - 1) + 2(-10 - 1) + 1(-2 - 3) = 14 - 22 - 5 = -13$.
The matrix of cofactors is calculated as:
$C_{11} = 14, C_{12} = 11, C_{13} = -5$
$C_{21} = 11, C_{22} = 4, C_{23} = -3$
$C_{31} = -5, C_{32} = -3, C_{33} = -1$
Thus,$adj A = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}$.
Now,$[adj A]^{-1} = \frac{1}{|adj A|} adj(adj A)$.
$|adj A| = 14(-4 - 9) - 11(-11 - 15) - 5(-33 + 20) = 14(-13) - 11(-26) - 5(-13) = -182 + 286 + 65 = 169$.
The adjoint of $adj A$ is $\begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix}$.
So,$[adj A]^{-1} = \frac{1}{169} \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} = \frac{1}{13} \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{bmatrix}$.
Next,$A^{-1} = \frac{1}{|A|} adj A = -\frac{1}{13} \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix} = \frac{1}{13} \begin{bmatrix} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{bmatrix}$.
Calculating $adj(A^{-1})$ involves finding the cofactors of $A^{-1}$,which results in $\frac{1}{13} \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{bmatrix}$.
Thus,$[adj A]^{-1} = adj(A^{-1})$ is verified.

(N/A) Given $A=\left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$.
First,calculate the determinant $|A|$:
$|A| = 1(15-1) - (-2)(-10-1) + 1(-2-3) = 1(14) + 2(-11) + 1(-5) = 14 - 22 - 5 = -13$.
Next,find the matrix of cofactors $C_{ij}$:
$C_{11} = +(15-1) = 14, C_{12} = -( -10-1) = 11, C_{13} = +(-2-3) = -5$.
$C_{21} = -(-10-1) = 11, C_{22} = +(5-1) = 4, C_{23} = -(1+2) = -3$.
$C_{31} = +(-2-3) = -5, C_{32} = -(1+2) = -3, C_{33} = +(3-4) = -1$.
Thus,$adj(A) = \left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]$.
$A^{-1} = \frac{1}{|A|} adj(A) = -\frac{1}{13} \left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right] = \frac{1}{13} \left[\begin{array}{ccc}-14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1\end{array}\right]$.
Now,to find $(A^{-1})^{-1}$,we find the determinant and adjoint of $A^{-1}$.
$|A^{-1}| = \frac{1}{|A|} = -\frac{1}{13}$.
The adjoint of $A^{-1}$ is given by $adj(A^{-1}) = \frac{1}{|A|^{n-1}} A = \frac{1}{(-13)^2} A = \frac{1}{169} \left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$.
Finally,$(A^{-1})^{-1} = \frac{adj(A^{-1})}{|A^{-1}|} = \frac{\frac{1}{169} A}{-\frac{1}{13}} = -\frac{13}{169} A = -\frac{1}{13} A = -\frac{1}{13} \left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right] \times (-13) = A$.
Hence,$\left(A^{-1}\right)^{-1}=A$ is verified.

(A) Given the diagonal matrix $A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$.
The determinant of $A$ is $|A| = x(yz - 0) - 0 + 0 = xyz$.
Since $x, y, z \neq 0$,$|A| = xyz \neq 0$,so $A^{-1}$ exists.
The inverse of a diagonal matrix $A = \text{diag}(x, y, z)$ is given by $A^{-1} = \text{diag}(x^{-1}, y^{-1}, z^{-1})$.
Calculating the adjugate matrix:
$adj(A) = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}$.
Thus,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{xyz} \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix} = \begin{bmatrix} \frac{yz}{xyz} & 0 & 0 \\ 0 & \frac{xz}{xyz} & 0 \\ 0 & 0 & \frac{xy}{xyz} \end{bmatrix} = \begin{bmatrix} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{bmatrix} = \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}$.

(A) To find the inverse using elementary row operations,we write $A=IA$.
$\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$
Applying $R_{2} \rightarrow R_{2}-2 R_{1}$:
$\left[\begin{array}{rr}1 & 2 \\ 0 & -5\end{array}\right]=\left[\begin{array}{rr}1 & 0 \\ -2 & 1\end{array}\right] A$
Applying $R_{2} \rightarrow -\frac{1}{5} R_{2}$:
$\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ \frac{2}{5} & \frac{-1}{5}\end{array}\right] A$
Applying $R_{1} \rightarrow R_{1}-2 R_{2}$:
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{-1}{5}\end{array}\right] A$
Thus,$A^{-1}=\left[\begin{array}{ll}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{-1}{5}\end{array}\right]$.

(A) To find the inverse of matrix $A$ using elementary row operations,we write $A = IA$:
$\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right] = \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_1 \leftrightarrow R_2$:
$\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1\end{array}\right] = \left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_3 \to R_3 - 3R_1$:
$\left[\begin{array}{rrr}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8\end{array}\right] = \left[\begin{array}{rrr}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1\end{array}\right] A$
Applying $R_1 \to R_1 - 2R_2$:
$\left[\begin{array}{rrr}1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & -5 & -8\end{array}\right] = \left[\begin{array}{rrr}-2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1\end{array}\right] A$
Applying $R_3 \to R_3 + 5R_2$:
$\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2\end{array}\right] = \left[\begin{array}{ccc}-2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1\end{array}\right] A$
Applying $R_3 \to \frac{1}{2}R_3$:
$\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}-2 & 1 & 0 \\ 1 & 0 & 0 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right] A$
Applying $R_1 \to R_1 + R_3$ and $R_2 \to R_2 - 2R_3$:
$\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right] A$
Thus,$A^{-1} = \left[\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right]$.

(C) To find the inverse of matrix $P$,we first calculate its determinant,$|P|$.
$|P| = (10 \times 1) - (-2 \times -5) = 10 - 10 = 0$.
Since the determinant of matrix $P$ is $0$,the matrix is singular.
$A$ matrix is invertible if and only if its determinant is non-zero.
Therefore,$P^{-1}$ does not exist.

(A) Let $A = \left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$.
First,we find the determinant of $A$:
$|A| = (1)(3) - (-1)(2) = 3 + 2 = 5$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
We use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
The matrix of cofactors is:
$C_{11} = 3, C_{12} = -2, C_{21} = 1, C_{22} = 1$.
Thus,$\text{adj}(A) = \left[\begin{array}{cc}3 & 1 \\ -2 & 1\end{array}\right]^T = \left[\begin{array}{cc}3 & -2 \\ 1 & 1\end{array}\right]$.
Wait,let's re-calculate the adjoint correctly:
$C_{11} = 3, C_{12} = -2, C_{21} = -(-1) = 1, C_{22} = 1$.
So,$\text{adj}(A) = \left[\begin{array}{cc}3 & 1 \\ -2 & 1\end{array}\right]$.
Therefore,$A^{-1} = \frac{1}{5} \left[\begin{array}{cc}3 & 1 \\ -2 & 1\end{array}\right] = \left[\begin{array}{cc}\frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right]$.