Adjoint and inverse of matrices Questions in English

(A) Let $A = \left[\begin{array}{ll}2 & 1 \\ 1 & 1\end{array}\right]$.
First,we find the determinant of $A$:
$|A| = (2)(1) - (1)(1) = 2 - 1 = 1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The inverse of a $2 \times 2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is given by $\frac{1}{|A|} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Substituting the values $a=2, b=1, c=1, d=1$:
$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right] = \left[\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right]$.

(A) Let $A = \left[\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right]$.
First,we find the determinant of $A$:
$|A| = (1 \times 7) - (3 \times 2) = 7 - 6 = 1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The formula for the inverse of a $2 \times 2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is $\frac{1}{ad-bc} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Substituting the values $a=1, b=3, c=2, d=7$:
$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}7 & -3 \\ -2 & 1\end{array}\right] = \left[\begin{array}{cc}7 & -3 \\ -2 & 1\end{array}\right]$.

(A) Let $A = \left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]$.
First,we find the determinant of $A$:
$|A| = (2 \times 7) - (3 \times 5) = 14 - 15 = -1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The inverse of a $2 \times 2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is given by $\frac{1}{|A|} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Substituting the values:
$A^{-1} = \frac{1}{-1} \left[\begin{array}{cc}7 & -3 \\ -5 & 2\end{array}\right]$
$A^{-1} = \left[\begin{array}{cc}-7 & 3 \\ 5 & -2\end{array}\right]$.

(A) Let $A = \left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$.
First,we find the determinant of $A$:
$|A| = (2 \times 4) - (1 \times 7) = 8 - 7 = 1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The formula for the inverse of a $2 \times 2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is $\frac{1}{|A|} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Substituting the values:
$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right] = \left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right]$.

(A) Let $A = \left[\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\right]$.
First,we find the determinant of $A$:
$|A| = (2 \times 3) - (5 \times 1) = 6 - 5 = 1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The formula for the inverse of a $2 \times 2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is given by:
$A^{-1} = \frac{1}{|A|} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Substituting the values $a=2, b=5, c=1, d=3$ and $|A|=1$:
$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}3 & -5 \\ -1 & 2\end{array}\right] = \left[\begin{array}{cc}3 & -5 \\ -1 & 2\end{array}\right]$.

(A) Let $A = \left[\begin{array}{ll}3 & 1 \\ 5 & 2\end{array}\right]$.
First,we find the determinant of $A$:
$|A| = (3 \times 2) - (1 \times 5) = 6 - 5 = 1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The formula for the inverse of a $2 \times 2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is $\frac{1}{|A|} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Applying this to matrix $A$:
$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}2 & -1 \\ -5 & 3\end{array}\right] = \left[\begin{array}{cc}2 & -1 \\ -5 & 3\end{array}\right]$.

(A) Let $A = \left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]$.
First,calculate the determinant of $A$:
$|A| = (4 \times 4) - (5 \times 3) = 16 - 15 = 1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The inverse of a $2 \times 2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is given by $\frac{1}{|A|} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Substituting the values:
$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right] = \left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right]$.

(A) Let $A = \left[\begin{array}{cc}3 & 10 \\ 2 & 7\end{array}\right]$.
First,we find the determinant of $A$:
$|A| = (3 \times 7) - (10 \times 2) = 21 - 20 = 1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The formula for the inverse of a $2 \times 2$ matrix $\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$ is $\frac{1}{|A|} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Substituting the values:
$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}7 & -10 \\ -2 & 3\end{array}\right] = \left[\begin{array}{cc}7 & -10 \\ -2 & 3\end{array}\right]$.

(A) Let $A = \left[\begin{array}{cc}3 & -1 \\ -4 & 2\end{array}\right]$.
First,calculate the determinant of $A$:
$|A| = (3)(2) - (-1)(-4) = 6 - 4 = 2$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The formula for the inverse of a $2 \times 2$ matrix $\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$ is $\frac{1}{ad-bc} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Applying this to matrix $A$:
$A^{-1} = \frac{1}{2} \left[\begin{array}{cc}2 & 1 \\ 4 & 3\end{array}\right] = \left[\begin{array}{cc}1 & 1/2 \\ 2 & 3/2\end{array}\right]$.

(A) Let $A = \left[\begin{array}{ll}2 & -6 \\ 1 & -2\end{array}\right]$.
First,we find the determinant of $A$:
$|A| = (2)(-2) - (-6)(1) = -4 + 6 = 2$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The formula for the inverse of a $2 \times 2$ matrix $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is $\frac{1}{|A|} \left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right]$.
Here,$a=2, b=-6, c=1, d=-2$.
$A^{-1} = \frac{1}{2} \left[\begin{array}{rr}-2 & 6 \\ -1 & 2\end{array}\right] = \left[\begin{array}{rr}-1 & 3 \\ -\frac{1}{2} & 1\end{array}\right]$.

(C) Let $A = \left[\begin{array}{cc}6 & -3 \\ -2 & 1\end{array}\right]$.
First,calculate the determinant of $A$,denoted as $|A|$.
$|A| = (6 \times 1) - (-3 \times -2) = 6 - 6 = 0$.
Since the determinant of the matrix is $0$,the matrix $A$ is singular.
Therefore,the inverse of the matrix $A$ does not exist.

(A) Given matrix $A = \left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$.
First,calculate the determinant $|A| = (2)(2) - (-3)(-1) = 4 - 3 = 1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
The formula for the inverse of a $2 \times 2$ matrix $\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$ is $\frac{1}{ad-bc} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Applying this to matrix $A$:
$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}2 & 3 \\ 1 & 2\end{array}\right] = \left[\begin{array}{cc}2 & 3 \\ 1 & 2\end{array}\right]$.

(C) Let $A = \left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]$.
First,we calculate the determinant of $A$,denoted by $|A|$.
$|A| = (2 \times 2) - (1 \times 4) = 4 - 4 = 0$.
Since the determinant of the matrix is $0$,the matrix $A$ is a singular matrix.
Therefore,the inverse of the matrix $A$ does not exist.

Let $A = \left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$.
First,we find the determinant of $A$ $(|A|)$:
$|A| = 2(4 - (-6)) - (-3)(4 - 9) + 3(-4 - 6)$
$|A| = 2(10) + 3(-5) + 3(-10)$
$|A| = 20 - 15 - 30 = -25$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
Next,we find the cofactor matrix $C_{ij}$:
$C_{11} = +(4+6) = 10, C_{12} = -(4-9) = 5, C_{13} = +(-4-6) = -10$
$C_{21} = -(-6+6) = 0, C_{22} = +(4-9) = -5, C_{23} = -(-4+9) = -5$
$C_{31} = +(-9-6) = -15, C_{32} = -(6-6) = 0, C_{33} = +(4+6) = 10$
The adjoint matrix $adj(A)$ is the transpose of the cofactor matrix:
$adj(A) = \left[\begin{array}{rrr}10 & 0 & -15 \\ 5 & -5 & 0 \\ -10 & -5 & 10\end{array}\right]$
Finally,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-25} \left[\begin{array}{rrr}10 & 0 & -15 \\ 5 & -5 & 0 \\ -10 & -5 & 10\end{array}\right] = \left[\begin{array}{rrr}-2/5 & 0 & 3/5 \\ -1/5 & 1/5 & 0 \\ 2/5 & 1/5 & -2/5\end{array}\right]$.

(A) Let $A=\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]$.
First,calculate the determinant $|A|$:
$|A| = 1(0 - (-25)) - 3(0 - (-10)) - 2(-15 - 0)$
$|A| = 1(25) - 3(10) - 2(-15) = 25 - 30 + 30 = 25$.
Since $|A| \neq 0$,the inverse exists.
Using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$:
Find the matrix of cofactors $C_{ij}$:
$C_{11} = +(0 - (-25)) = 25, C_{12} = -(0 - (-10)) = -10, C_{13} = +(-15 - 0) = -15$
$C_{21} = -(0 - (-10)) = -10, C_{22} = +(0 - (-4)) = 4, C_{23} = -(5 - 6) = 1$
$C_{31} = +(-15 - 0) = -15, C_{32} = -(-5 - 6) = 11, C_{33} = +(0 - (-9)) = 9$
$\text{adj}(A) = \left[\begin{array}{ccc}25 & -10 & -15 \\ -10 & 4 & 11 \\ -15 & 1 & 9\end{array}\right]^T = \left[\begin{array}{ccc}25 & -10 & -15 \\ -10 & 4 & 1 \\ -15 & 11 & 9\end{array}\right]$
Wait,recalculating cofactors:
$C_{23} = -(5 - 6) = 1$. Correct.
$C_{32} = -(-5 - 6) = 11$. Correct.
$A^{-1} = \frac{1}{25} \left[\begin{array}{ccc}25 & -10 & -15 \\ -10 & 4 & 11 \\ -15 & 1 & 9\end{array}\right] = \left[\begin{array}{ccc}1 & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}\right]$.

(A) Let $A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$.
We know that $A=IA$.
$\therefore \left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$.
Applying $R_{1} \rightarrow \frac{1}{2} R_{1}$,we have:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$.
Applying $R_{2} \rightarrow R_{2}-5 R_{1}$,we have:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$.
Applying $R_{3} \rightarrow R_{3}-R_{2}$,we have:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1\end{array}\right] A$.
Applying $R_{3} \rightarrow 2 R_{3}$,we have:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 5 & -2 & 2\end{array}\right] A$.
Applying $R_{1} \rightarrow R_{1}+\frac{1}{2} R_{3}$ and $R_{2} \rightarrow R_{2}-\frac{5}{2} R_{3}$,we have:
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] A$.
$\therefore A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$.

(A) By definition,if $A$ and $B$ are square matrices of the same order $n$ such that their product is the identity matrix $I$,then $B$ is the inverse of $A$ and $A$ is the inverse of $B$.
This relationship is expressed as $A B = B A = I$.
Therefore,matrices $A$ and $B$ are inverses of each other if and only if $A B = B A = I$.

(A) Let $A = \left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]$.
We write $A = IA$,where $I = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$.
$\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] A$.
Apply $R_2 \to R_2 - 2R_1$:
$\left[\begin{array}{cc}1 & 2 \\ 0 & -5\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right] A$.
Apply $R_2 \to -\frac{1}{5} R_2$:
$\left[\begin{array}{cc}1 & 2 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ \frac{2}{5} & -\frac{1}{5}\end{array}\right] A$.
Apply $R_1 \to R_1 - 2R_2$:
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 - 2(\frac{2}{5}) & 0 - 2(-\frac{1}{5}) \\ \frac{2}{5} & -\frac{1}{5}\end{array}\right] A$.
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5}\end{array}\right] A$.
Thus,$A^{-1} = \left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5}\end{array}\right]$.

(A) Let $A = \left[\begin{array}{cc}7 & 4 \\ 1 & -2\end{array}\right]$.
To find the inverse using elementary row transformations,we write $A = IA$,where $I = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$.
$\left[\begin{array}{cc}7 & 4 \\ 1 & -2\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] A$.
Apply $R_1 \leftrightarrow R_2$:
$\left[\begin{array}{cc}1 & -2 \\ 7 & 4\end{array}\right] = \left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right] A$.
Apply $R_2 \to R_2 - 7R_1$:
$\left[\begin{array}{cc}1 & -2 \\ 0 & 18\end{array}\right] = \left[\begin{array}{cc}0 & 1 \\ 1 & -7\end{array}\right] A$.
Apply $R_2 \to \frac{1}{18}R_2$:
$\left[\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}0 & 1 \\ \frac{1}{18} & -\frac{7}{18}\end{array}\right] A$.
Apply $R_1 \to R_1 + 2R_2$:
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}\frac{2}{18} & \frac{4}{18} \\ \frac{1}{18} & -\frac{7}{18}\end{array}\right] A$.
Thus,$A^{-1} = \frac{1}{18}\left[\begin{array}{cc}2 & 4 \\ 1 & -7\end{array}\right]$.

(A) Let $A = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix}$. We write $A = IA$:
$\begin{bmatrix} 1 & 2 & 1 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A$
Applying $R_2 \to R_2 - 3R_1$ and $R_3 \to R_3 - R_1$:
$\begin{bmatrix} 1 & 2 & 1 \\ 0 & -4 & 0 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -3 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} A$
Applying $R_2 \to R_2 / (-4)$:
$\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3/4 & -1/4 & 0 \\ -1 & 0 & 1 \end{bmatrix} A$
Applying $R_1 \to R_1 - 2R_2$ and $R_3 \to R_3 + R_2$:
$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -1/2 & 1/2 & 0 \\ 3/4 & -1/4 & 0 \\ -1/4 & -1/4 & 1 \end{bmatrix} A$
Applying $R_1 \to R_1 - R_3$:
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -1/4 & 3/4 & -1 \\ 3/4 & -1/4 & 0 \\ -1/4 & -1/4 & 1 \end{bmatrix} A$
Thus,$A^{-1} = \begin{bmatrix} -1/4 & 3/4 & -1 \\ 3/4 & -1/4 & 0 \\ -1/4 & -1/4 & 1 \end{bmatrix}$.

(A) Let $A = \begin{bmatrix} -1 & 2 \\ -3 & 5 \end{bmatrix}$. We write $A = IA$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
$\begin{bmatrix} -1 & 2 \\ -3 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A$
Apply $R_1 \to -R_1$:
$\begin{bmatrix} 1 & -2 \\ -3 & 5 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} A$
Apply $R_2 \to R_2 + 3R_1$:
$\begin{bmatrix} 1 & -2 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ -3 & 1 \end{bmatrix} A$
Apply $R_2 \to -R_2$:
$\begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 3 & -1 \end{bmatrix} A$
Apply $R_1 \to R_1 + 2R_2$:
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -2 \\ 3 & -1 \end{bmatrix} A$
Thus,$A^{-1} = \begin{bmatrix} 5 & -2 \\ 3 & -1 \end{bmatrix}$.

(N/A) To find the inverse using elementary row operations,we write $A = IA$,where $I$ is the identity matrix: $\left[\begin{array}{ccc} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$.
Applying $R_1 \leftrightarrow R_2$: $\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{array}\right] = \left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] A$.
Applying $R_3 \to R_3 - 3R_1$: $\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{array}\right] = \left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{array}\right] A$.
Applying $R_3 \to R_3 + 5R_2$: $\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{array}\right] = \left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{array}\right] A$.
Applying $R_3 \to \frac{1}{2}R_3$: $\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right] A$.
Applying $R_2 \to R_2 - 2R_3$ and $R_1 \to R_1 - 3R_3$: $\left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{ccc} -\frac{15}{2} & \frac{11}{2} & -\frac{3}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right] A$.
Applying $R_1 \to R_1 - 2R_2$: $\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right] A$.
Thus,$A^{-1} = \left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right]$.

(D) Suppose $A = \left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right]$.
To find the inverse using elementary row operations,we use the relation $A = IA$.
$\left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] A$
Apply $R_2 \rightarrow R_2 + 5R_1$:
$\left[\begin{array}{cc}1 & 3 \\ 0 & 22\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 5 & 1\end{array}\right] A$
Apply $R_2 \rightarrow \frac{1}{22} R_2$:
$\left[\begin{array}{cc}1 & 3 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
Apply $R_1 \rightarrow R_1 - 3R_2$:
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 - 3(\frac{5}{22}) & 0 - 3(\frac{1}{22}) \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}\frac{7}{22} & -\frac{3}{22} \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
Thus,$A^{-1} = \frac{1}{22} \left[\begin{array}{cc}7 & -3 \\ 5 & 1\end{array}\right]$.

(C) Let $A = \left[\begin{array}{cc}1 & -3 \\ -2 & 6\end{array}\right]$.
To find the inverse using elementary row operations,we write $A = IA$,where $I$ is the identity matrix.
$\left[\begin{array}{cc}1 & -3 \\ -2 & 6\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] A$
Applying the row operation $R_2 \rightarrow R_2 + 2R_1$:
$\left[\begin{array}{cc}1 & -3 \\ -2 + 2(1) & 6 + 2(-3)\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 0 + 2(1) & 1 + 2(0)\end{array}\right] A$
$\left[\begin{array}{cc}1 & -3 \\ 0 & 0\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 2 & 1\end{array}\right] A$
Since all elements in the second row of the matrix on the left-hand side are zero,the matrix $A$ is singular (non-invertible).
Therefore,$A^{-1}$ does not exist.

(A) Let $A = \left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]$. To find $A^{-1}$ using elementary row transformations,we write $A = IA$.
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_2 \rightarrow R_2 + 2R_1$ and $R_3 \rightarrow R_3 + R_1$:
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -1 & 1 & 7 \\ -1 & 1 & 6\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1\end{array}\right] A$
Applying $R_3 \rightarrow R_3 - R_2$:
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -1 & 1 & 7 \\ 0 & 0 & -1\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 2 & 1 & 0 \\ -1 & -1 & 1\end{array}\right] A$
Applying $R_1 \rightarrow R_1 + R_2$:
$\left[\begin{array}{ccc}1 & 0 & 10 \\ -1 & 1 & 7 \\ 0 & 0 & -1\end{array}\right] = \left[\begin{array}{ccc}3 & 1 & 0 \\ 2 & 1 & 0 \\ -1 & -1 & 1\end{array}\right] A$
Applying $R_2 \rightarrow R_2 + R_1$:
$\left[\begin{array}{ccc}1 & 0 & 10 \\ 0 & 1 & 17 \\ 0 & 0 & -1\end{array}\right] = \left[\begin{array}{ccc}3 & 1 & 0 \\ 5 & 2 & 0 \\ -1 & -1 & 1\end{array}\right] A$
Applying $R_1 \rightarrow R_1 + 10R_3$,$R_2 \rightarrow R_2 + 17R_3$,and $R_3 \rightarrow -R_3$:
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}-7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1\end{array}\right] A$
Thus,$A^{-1} = \left[\begin{array}{ccc}-7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1\end{array}\right]$.

(A) To find the inverse of the matrix $A = \left[\begin{array}{ccc}2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & 1\end{array}\right]$ using elementary row transformations,we write $A = IA$:
$\left[\begin{array}{ccc}2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & 1\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_1 \rightarrow R_1 + 2R_2$:
$\left[\begin{array}{ccc}0 & -1 & 1 \\ -1 & -2 & 2 \\ 1 & 1 & 1\end{array}\right] = \left[\begin{array}{ccc}1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_1 \rightarrow R_1 + R_2$ and $R_3 \rightarrow R_3 + R_2$:
$\left[\begin{array}{ccc}-1 & -3 & 3 \\ -1 & -2 & 2 \\ 0 & -1 & 3\end{array}\right] = \left[\begin{array}{ccc}1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{array}\right] A$
Alternatively,we can calculate the determinant $|A|$:
$|A| = 2(-2 - 2) - 3(-1 - 2) - 3(-1 + 2) = 2(-4) - 3(-3) - 3(1) = -8 + 9 - 3 = -2$.
Since $|A| \neq 0$,the inverse exists. However,the question asks for row transformations. Performing row operations to reach the identity matrix on the left,we find that the matrix is invertible. The inverse is $A^{-1} = \left[\begin{array}{ccc}4 & 6 & 0 \\ -3 & -5 & -1 \\ -1 & -1 & 1\end{array}\right]$.

(N/A) Let $A = \left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$. To find $A^{-1}$ using elementary row transformations,we use $A = IA$.
$\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_1 \rightarrow \frac{1}{2}R_1$:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_2 \rightarrow R_2 - 5R_1$:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_3 \rightarrow R_3 - R_2$:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2}\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1\end{array}\right] A$
Applying $R_3 \rightarrow 2R_3$:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 5 & -2 & 2\end{array}\right] A$
Applying $R_1 \rightarrow R_1 + \frac{1}{2}R_3$ and $R_2 \rightarrow R_2 - \frac{5}{2}R_3$:
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] A$
Thus,$A^{-1} = \left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$.

(C) Given $C = \operatorname{adj} A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1 \end{bmatrix}$.
Calculating the determinant $|C| = |\operatorname{adj} A| = 2(0 - (-4)) - (-1)(1 - 2) + 1(2 - 0) = 2(4) + 1(-1) + 1(2) = 8 - 1 + 2 = 9$.
We know that $|\operatorname{adj} A| = |A|^{n-1}$,where $n=3$. So,$|A|^2 = 9$,which implies $|A| = \pm 3$. Thus,$\lambda = \pm 3$ and $|\lambda| = 3$.
Given $B = \operatorname{adj} C = \operatorname{adj}(\operatorname{adj} A)$.
Using the property $|\operatorname{adj} M| = |M|^{n-1}$,we have $|B| = |\operatorname{adj} C| = |C|^{n-1} = |C|^2 = 9^2 = 81$.
We need to find $\mu = |(B^{-1})^T|$. Since $|(B^{-1})^T| = |B^{-1}| = \frac{1}{|B|}$,we get $\mu = \frac{1}{81}$.
Therefore,the ordered pair $(|\lambda|, \mu) = (3, 1/81)$.

(A) Given $PQ = kI_3$. Taking the determinant on both sides,$|P||Q| = |kI_3| = k^3|I_3| = k^3$.
Given $|Q| = \frac{k^2}{2}$,so $|P| \cdot \frac{k^2}{2} = k^3$. Since $k \neq 0$,$|P| = 2k$.
Calculating $|P|$: $|P| = 3(0 - (-5\alpha)) - (-1)(0 - 3\alpha) + (-2)(-10 - 0) = 3(5\alpha) + (-3\alpha) + 20 = 15\alpha - 3\alpha + 20 = 12\alpha + 20$.
Thus,$12\alpha + 20 = 2k$,or $k = 6\alpha + 10$.
Since $PQ = kI_3$,$Q = kP^{-1} = k \cdot \frac{\text{adj}(P)}{|P|} = k \cdot \frac{\text{adj}(P)}{2k} = \frac{1}{2} \text{adj}(P)$.
The element $q_{23}$ is the $(2,3)$ entry of $\frac{1}{2} \text{adj}(P)$.
The $(2,3)$ entry of $\text{adj}(P)$ is the cofactor $C_{32} = -\begin{vmatrix} 3 & -2 \\ 2 & \alpha \end{vmatrix} = -(3\alpha - (-4)) = -(3\alpha + 4)$.
So,$q_{23} = \frac{-(3\alpha + 4)}{2} = -\frac{k}{8}$.
This implies $4(3\alpha + 4) = k$.
Substituting $k = 6\alpha + 10$: $12\alpha + 16 = 6\alpha + 10 \Rightarrow 6\alpha = -6 \Rightarrow \alpha = -1$.
Then $k = 6(-1) + 10 = 4$.
Finally,$\alpha^2 + k^2 = (-1)^2 + 4^2 = 1 + 16 = 17$.

(B) First,observe that $A$ is an orthogonal matrix because $AA^{T} = I$.
$AA^{T} = \begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{-2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Given $Q = A^{T}BA$,we find $Q^{n} = (A^{T}BA)(A^{T}BA)...(A^{T}BA) = A^{T}B^{n}A$.
Thus,$Q^{2021} = A^{T}B^{2021}A$.
Now,calculate powers of $B = \begin{bmatrix} 1 & 0 \\ i & 1 \end{bmatrix}$.
$B^{2} = \begin{bmatrix} 1 & 0 \\ i & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ i & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2i & 1 \end{bmatrix}$.
By induction,$B^{n} = \begin{bmatrix} 1 & 0 \\ ni & 1 \end{bmatrix}$,so $B^{2021} = \begin{bmatrix} 1 & 0 \\ 2021i & 1 \end{bmatrix}$.
Now,$AQ^{2021}A^{T} = A(A^{T}B^{2021}A)A^{T} = (AA^{T})B^{2021}(AA^{T}) = I B^{2021} I = B^{2021}$.
Therefore,$AQ^{2021}A^{T} = \begin{bmatrix} 1 & 0 \\ 2021i & 1 \end{bmatrix}$.
The inverse of a matrix $\begin{bmatrix} 1 & 0 \\ k & 1 \end{bmatrix}$ is $\begin{bmatrix} 1 & 0 \\ -k & 1 \end{bmatrix}$.
Thus,$(AQ^{2021}A^{T})^{-1} = \begin{bmatrix} 1 & 0 \\ -2021i & 1 \end{bmatrix}$.

(B) Given $A$ is a $3 \times 3$ matrix,so $|A| = \Delta$.
Using the property $\operatorname{adj}(kA) = k^{n-1} \operatorname{adj}(A)$,where $n=3$,we have $\operatorname{adj}(2A) = 2^{2} \operatorname{adj}(A) = 4 \operatorname{adj}(A)$.
Next,$\operatorname{adj}(\operatorname{adj}(2A)) = \operatorname{adj}(4 \operatorname{adj}(A)) = 4^{3-1} \operatorname{adj}(\operatorname{adj}(A)) = 16 \operatorname{adj}(\operatorname{adj}(A))$.
Using $\operatorname{adj}(\operatorname{adj}(A)) = |A|^{n-2} A = |A| A$,we get $16 |A| A$.
Now,the expression is $\det(2 \operatorname{adj}(2 \operatorname{adj}(\operatorname{adj}(2A)))) = \det(2 \operatorname{adj}(2(16 |A| A))) = \det(2 \operatorname{adj}(32 |A| A))$.
Using $\operatorname{adj}(kA) = k^{n-1} \operatorname{adj}(A)$,we have $\operatorname{adj}(32 |A| A) = (32 |A|)^{2} \operatorname{adj}(A)$.
So,$\det(2 \cdot (32 |A|)^{2} \operatorname{adj}(A)) = \det(2^{1} \cdot 2^{10} |A|^{2} \operatorname{adj}(A)) = \det(2^{11} |A|^{2} \operatorname{adj}(A))$.
Since $\det(kM) = k^{n} \det(M)$,we have $(2^{11} |A|^{2})^{3} \det(\operatorname{adj}(A)) = 2^{33} |A|^{6} |A|^{2} = 2^{33} |A|^{8}$.
Given $2^{33} |A|^{8} = 2^{41}$,so $|A|^{8} = 2^{8}$,which implies $|A| = \pm 2$.
Thus,$\det(A^{2}) = |A|^{2} = (\pm 2)^{2} = 4$.

(B) Given $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$.
First,we find the determinant of $A$: $|A| = (1)(4) - (2)(-1) = 4 + 2 = 6$.
Next,we find the adjoint of $A$: $\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix}$.
Thus,$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{6} \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix}$.
We are given $A^{-1} = \alpha I + \beta A$,so:
$\begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix} = \alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix}$.
Comparing the elements,we get:
$2\beta = -\frac{1}{3} \implies \beta = -\frac{1}{6}$.
$\alpha + \beta = \frac{2}{3} \implies \alpha - \frac{1}{6} = \frac{2}{3} \implies \alpha = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}$.
Finally,calculate $4(\alpha - \beta) = 4(\frac{5}{6} - (-\frac{1}{6})) = 4(\frac{5}{6} + \frac{1}{6}) = 4(1) = 4$.

(A) Given $X = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$,we find $X^{2} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and $X^{3} = O$.
Then $Y = \alpha I + \beta X + \gamma X^{2} = \begin{bmatrix} \alpha & \beta & \gamma \\ 0 & \alpha & \beta \\ 0 & 0 & \alpha \end{bmatrix}$.
Since $Y \cdot Y^{-1} = I$,we have:
$\begin{bmatrix} \alpha & \beta & \gamma \\ 0 & \alpha & \beta \\ 0 & 0 & \alpha \end{bmatrix} \begin{bmatrix} \frac{1}{5} & \frac{-2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{-2}{5} \\ 0 & 0 & \frac{1}{5} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Comparing the elements:
$1$. $\frac{\alpha}{5} = 1 \Rightarrow \alpha = 5$.
$2$. $-\frac{2\alpha}{5} + \frac{\beta}{5} = 0 \Rightarrow -2(5) + \beta = 0 \Rightarrow \beta = 10$.
$3$. $\frac{\alpha}{5} - \frac{2\beta}{5} + \frac{\gamma}{5} = 0 \Rightarrow 5 - 2(10) + \gamma = 0 \Rightarrow 5 - 20 + \gamma = 0 \Rightarrow \gamma = 15$.
Finally,calculate $(\alpha - \beta + \gamma)^{2} = (5 - 10 + 15)^{2} = (10)^{2} = 100$.

(B) We know that for a square matrix $A$ of order $n$,$\operatorname{Adj}(\operatorname{Adj}(A)) = |A|^{n-2} A$.
Taking the determinant on both sides,$|\operatorname{Adj}(\operatorname{Adj}(A))| = (|A|^{n-2})^n = |A|^{(n-1)^2}$.
Here,the matrix is of order $n=3$,so $|\operatorname{Adj}(\operatorname{Adj}(A))| = |A|^{(3-1)^2} = |A|^4$.
Now,calculate the determinant of the given matrix:
$|\operatorname{Adj}(\operatorname{Adj}(A))| = \begin{vmatrix} 14 & 28 & -14 \\ -14 & 14 & 28 \\ 28 & -14 & 14 \end{vmatrix} = 14^3 \begin{vmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{vmatrix}$.
$= 14^3 [1(1 - (-2)) - 2(-1 - 4) - 1(1 - 2)] = 14^3 [1(3) - 2(-5) - 1(-1)] = 14^3 [3 + 10 + 1] = 14^3 \times 14 = 14^4$.
Thus,$|A|^4 = 14^4$.
Since we need the positive value,$|A| = 14$.

(A) Given $A$ is a $3 \times 3$ matrix,so $n = 3$. We have $\det(A) = 2$.
We need to find $\det(\det(A) \cdot \operatorname{adj}(5 \operatorname{adj}(A^3)))$.
Since $\det(A) = 2$,the expression becomes $\det(2 \cdot \operatorname{adj}(5 \operatorname{adj}(A^3)))$.
Using the property $\det(kA) = k^n \det(A)$,we get $2^3 \det(\operatorname{adj}(5 \operatorname{adj}(A^3)))$.
Using $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$,where $n=3$,we have $8 \cdot (\det(5 \operatorname{adj}(A^3)))^2$.
Using $\det(kM) = k^n \det(M)$,we have $8 \cdot (5^3 \det(\operatorname{adj}(A^3)))^2 = 8 \cdot 5^6 \cdot (\det(\operatorname{adj}(A^3)))^2$.
Using $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$,we have $8 \cdot 5^6 \cdot ((\det(A^3))^2)^2 = 8 \cdot 5^6 \cdot (\det(A)^3)^4$.
Substituting $\det(A) = 2$,we get $2^3 \cdot 5^6 \cdot (2^3)^4 = 2^3 \cdot 5^6 \cdot 2^{12} = 2^{15} \cdot 5^6$.
$2^{15} \cdot 5^6 = 2^9 \cdot 2^6 \cdot 5^6 = 512 \cdot 10^6$.

(C) Given $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$.
We observe $A^2 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ and $A^3 = I$.
Since $49 = 3 \times 16 + 1$,$A^{49} = A^1 = A$.
Since $98 = 3 \times 32 + 2$,$A^{98} = A^2$.
Thus,$B_{0} = A + 2A^2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 2 \\ 2 & 0 & 0 \\ 0 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{bmatrix}$.
Calculating $\det(B_{0}) = 0(0-2) - 1(0-1) + 2(4-0) = 0 + 1 + 8 = 9$.
We know $\det(\text{Adj}(M)) = (\det(M))^{k-1}$ where $k$ is the order of the matrix. Here $k=3$,so $\det(\text{Adj}(M)) = (\det(M))^2$.
For $B_{n} = \text{Adj}(B_{n-1})$,$\det(B_{n}) = (\det(B_{n-1}))^2$.
Thus,$\det(B_{4}) = (\det(B_{0}))^{2^4} = (\det(B_{0}))^{16}$.
$\det(B_{4}) = 9^{16} = (3^2)^{16} = 3^{32}$.

(C) The matrix $A$ is given by $A = \begin{bmatrix} a & b \\ 1 & 1 \end{bmatrix}$.
For $A^{-1}$ to exist,the determinant $|A| = a - b$ must be non-zero,i.e.,$a \neq b$.
The inverse of $A$ is given by $A^{-1} = \frac{1}{a-b} \begin{bmatrix} 1 & -b \\ -1 & a \end{bmatrix} = \begin{bmatrix} \frac{1}{a-b} & -\frac{b}{a-b} \\ -\frac{1}{a-b} & \frac{a}{a-b} \end{bmatrix}$.
For $A^{-1}$ to contain only integer entries,each entry must be an integer.
This requires $(a-b)$ to be a divisor of $1$,$-b$,$-1$,and $a$.
Specifically,$(a-b)$ must divide $1$,which implies $a-b = 1$ or $a-b = -1$.
Case $1$: $a-b = 1 \implies a = b+1$.
Since $-50 \leq b \leq 50$,there are $101$ possible values for $b$,and for each $b$,$a$ is uniquely determined.
Case $2$: $a-b = -1 \implies a = b-1$.
Similarly,for $-50 \leq b \leq 50$,there are $101$ possible values for $b$,and for each $b$,$a$ is uniquely determined.
Thus,the total number of such matrices is $101 + 101 = 202$.

(B) Given that $A$ is a $3 \times 3$ matrix such that the sum of elements in each row is $1$. This can be written as $A \cdot \mathbf{u} = \mathbf{u}$,where $\mathbf{u} = [1, 1, 1]^T$ is a column vector of ones.
Since $A$ is invertible,we can multiply both sides by $A^{-1}$:
$A^{-1} (A \cdot \mathbf{u}) = A^{-1} \cdot \mathbf{u}$
$(A^{-1} A) \cdot \mathbf{u} = A^{-1} \cdot \mathbf{u}$
$I \cdot \mathbf{u} = A^{-1} \cdot \mathbf{u}$
$\mathbf{u} = A^{-1} \cdot \mathbf{u}$
This equation implies that the sum of the elements in each row of $A^{-1}$ is equal to the corresponding element in the vector $\mathbf{u}$,which is $1$.
Therefore,the sum of each row of $A^{-1}$ is $1$.

(A) Given $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))| = 12^4$.
We know that for a matrix $A$ of order $n$,$|\operatorname{adj}(\operatorname{adj}(\dots \operatorname{adj} A))|$ ($k$ times) is $|A|^{(n-1)^k}$.
Here $n = 3$ and $k = 3$,so $|A|^{(3-1)^3} = 12^4$.
$|A|^{2^3} = 12^4 \Rightarrow |A|^8 = 12^4$.
Taking the square root on both sides,$|A|^4 = 12^2 = 144$.
$|A|^2 = 12 \Rightarrow |A| = \sqrt{12} = 2\sqrt{3}$.
We need to find $|A^{-1} \operatorname{adj} A|$.
Using the property $|XY| = |X||Y|$,we have $|A^{-1}| |\operatorname{adj} A|$.
Since $|A^{-1}| = \frac{1}{|A|}$ and $|\operatorname{adj} A| = |A|^{n-1} = |A|^{3-1} = |A|^2$.
Thus,$|A^{-1} \operatorname{adj} A| = \frac{1}{|A|} \cdot |A|^2 = |A|$.
Therefore,$|A^{-1} \operatorname{adj} A| = 2\sqrt{3}$.

(D) First,observe that $A$ is an orthogonal matrix,meaning $AA^{T} = A^{T}A = I$,where $I$ is the identity matrix.
Next,calculate the powers of $B$:
$B^2 = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2i \\ 0 & 1 \end{bmatrix}$
$B^3 = \begin{bmatrix} 1 & -2i \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -3i \\ 0 & 1 \end{bmatrix}$
By induction,$B^{n} = \begin{bmatrix} 1 & -ni \\ 0 & 1 \end{bmatrix}$,so $B^{2023} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix}$.
Given $M = A^{T}BA$,we find $M^{n} = (A^{T}BA)(A^{T}BA)...(A^{T}BA) = A^{T}B^{n}A$.
Thus,$M^{2023} = A^{T}B^{2023}A$.
Now,calculate $AM^{2023}A^{T}$:
$AM^{2023}A^{T} = A(A^{T}B^{2023}A)A^{T} = (AA^{T})B^{2023}(AA^{T}) = I \cdot B^{2023} \cdot I = B^{2023} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix}$.
The inverse of $\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$ is $\begin{bmatrix} 1 & -k \\ 0 & 1 \end{bmatrix}$.
Therefore,the inverse of $B^{2023}$ is $\begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix}$.

(D) Given $P^{T} = aP + (a - 1)I$.
Taking transpose on both sides,we get $(P^{T})^{T} = (aP + (a - 1)I)^{T}$.
$P = aP^{T} + (a - 1)I$.
Substitute $P^{T} = aP + (a - 1)I$ into the equation:
$P = a(aP + (a - 1)I) + (a - 1)I$.
$P = a^{2}P + a(a - 1)I + (a - 1)I$.
$P = a^{2}P + (a^{2} - a + a - 1)I$.
$P = a^{2}P + (a^{2} - 1)I$.
$(1 - a^{2})P = (a^{2} - 1)I$.
Since $a > 1$,$a^{2} - 1 \neq 0$,so $-(a^{2} - 1)P = (a^{2} - 1)I$.
$P = -I$.
Now,$|P| = |-I| = (-1)^{3} |I| = -1$.
We know that $|\operatorname{Adj} P| = |P|^{n-1}$,where $n = 3$.
$|\operatorname{Adj} P| = (-1)^{3-1} = (-1)^{2} = 1$.

(D) Given $|A|=2$ and $|\operatorname{Adj}(2 \cdot \operatorname{Adj}(2A^{-1}))| = 2^{84}$.
Using the property $|\operatorname{Adj}(M)| = |M|^{n-1}$,we have:
$|\operatorname{Adj}(2 \cdot \operatorname{Adj}(2A^{-1}))| = |2 \cdot \operatorname{Adj}(2A^{-1})|^{n-1} = (2^n |\operatorname{Adj}(2A^{-1})|)^{n-1}$.
Since $|\operatorname{Adj}(2A^{-1})| = |2A^{-1}|^{n-1} = (2^n |A|^{-1})^{n-1} = (2^n \cdot 2^{-1})^{n-1} = (2^{n-1})^{n-1} = 2^{(n-1)^2}$.
Substituting this back:
$|\operatorname{Adj}(2 \cdot \operatorname{Adj}(2A^{-1}))| = (2^n \cdot 2^{(n-1)^2})^{n-1} = (2^{n + n^2 - 2n + 1})^{n-1} = (2^{n^2 - n + 1})^{n-1} = 2^{(n-1)(n^2 - n + 1)}$.
Given $2^{(n-1)(n^2 - n + 1)} = 2^{84}$,we have $(n-1)(n^2 - n + 1) = 84$.
For $n=5$,$(5-1)(25-5+1) = 4 \times 21 = 84$.
Thus,$n=5$.

(A) Given $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$.
First,calculate the determinant $|A|$:
$|A| = 2(4 - 1) - 1(2 - 0) + 0 = 2(3) - 2 = 4$.
Since $A$ is a $3 \times 3$ matrix,for any matrix $M$,$|\operatorname{adj}(M)| = |M|^{3-1} = |M|^2$.
Therefore,$|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(2A)))| = |2A|^{(3-1)^3} = |2A|^{2^3} = |2A|^8$.
Since $|kA| = k^n|A|$ for an $n \times n$ matrix,$|2A| = 2^3|A| = 8 \times 4 = 32 = 2^5$.
Substituting this into the expression:
$|2A|^8 = (2^5)^8 = 2^{40}$.
We are given that this equals $(16)^n = (2^4)^n = 2^{4n}$.
Equating the exponents: $4n = 40 \Rightarrow n = 10$.

(D) By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation $|A - xI| = 0$.
$|A - xI| = \begin{vmatrix} 1 - x & 5 \\ \lambda & 10 - x \end{vmatrix} = (1 - x)(10 - x) - 5\lambda = x^2 - 11x + 10 - 5\lambda = 0$.
Thus,$A^2 - 11A + (10 - 5\lambda)I = 0$.
Multiplying by $A^{-1}$,we get $A - 11I + (10 - 5\lambda)A^{-1} = 0$.
$(10 - 5\lambda)A^{-1} = -A + 11I$.
$A^{-1} = \frac{-1}{10 - 5\lambda}A + \frac{11}{10 - 5\lambda}I$.
Comparing this with $A^{-1} = \alpha A + \beta I$,we have $\alpha = \frac{-1}{10 - 5\lambda}$ and $\beta = \frac{11}{10 - 5\lambda}$.
Given $\alpha + \beta = -2$,so $\frac{-1 + 11}{10 - 5\lambda} = -2 \Rightarrow \frac{10}{10 - 5\lambda} = -2$.
$10 = -20 + 10\lambda \Rightarrow 10\lambda = 30 \Rightarrow \lambda = 3$.
Then $\alpha = \frac{-1}{10 - 15} = \frac{-1}{-5} = \frac{1}{5}$ and $\beta = \frac{11}{10 - 15} = \frac{11}{-5} = -\frac{11}{5}$.
Finally,$4\alpha^2 + \beta^2 + \lambda^2 = 4(\frac{1}{25}) + (\frac{121}{25}) + 3^2 = \frac{4 + 121}{25} + 9 = \frac{125}{25} + 9 = 5 + 9 = 14$.

(A) Given $A$ is a $3 \times 3$ matrix,so $n=3$ and $|A|=2$.
First,calculate $|3A|$. Since $A$ is $3 \times 3$,$|3A| = 3^3 |A| = 27 \times 2 = 54$.
Now,we need to find $|3 \operatorname{adj}(|3A|A^2)| = |3 \operatorname{adj}(54A^2)|$.
Using the property $|kM| = k^n |M|$ for an $n \times n$ matrix $M$,we have $|3 \operatorname{adj}(54A^2)| = 3^3 |\operatorname{adj}(54A^2)| = 27 |\operatorname{adj}(54A^2)|$.
Using the property $|\operatorname{adj}(M)| = |M|^{n-1}$,we get $27 |54A^2|^{3-1} = 27 |54A^2|^2$.
Since $|54A^2| = 54^3 |A^2| = 54^3 |A|^2 = 54^3 \times 2^2 = 54^3 \times 4$.
Substituting this back: $27 \times (54^3 \times 4)^2 = 27 \times 54^6 \times 16 = (3^3) \times (2 \times 3^3)^6 \times 2^4 = 3^3 \times 2^6 \times 3^{18} \times 2^4 = 3^{21} \times 2^{10} = 3^{11} \times 3^{10} \times 2^{10} = 3^{11} \times (3 \times 2)^{10} = 3^{11} \times 6^{10}$.

(D) Given $A = \frac{1}{5! 6! 7!} \begin{bmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}$.
Factoring out $5!, 6!, 7!$ from rows $1, 2, 3$ respectively:
$A = \frac{5! 6! 7!}{5! 6! 7!} \begin{bmatrix} 1 & 6 & 6 \times 7 \\ 1 & 7 & 7 \times 8 \\ 1 & 8 & 8 \times 9 \end{bmatrix} = \begin{bmatrix} 1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{bmatrix}$.
Calculating $|A|$:
$|A| = 1(7 \times 72 - 8 \times 56) - 6(1 \times 72 - 1 \times 56) + 42(1 \times 8 - 1 \times 7)$
$|A| = 1(504 - 448) - 6(16) + 42(1) = 56 - 96 + 42 = 2$.
We need to find $|\operatorname{adj}(\operatorname{adj}(2A))|$.
Using the property $|\operatorname{adj}(\operatorname{adj}(M))| = |M|^{(n-1)^2}$,where $n=3$ is the order of matrix $A$:
$|\operatorname{adj}(\operatorname{adj}(2A))| = |2A|^{(3-1)^2} = |2A|^4$.
Since $|2A| = 2^3 |A| = 8 \times 2 = 16 = 2^4$:
$|\operatorname{adj}(\operatorname{adj}(2A))| = (2^4)^4 = 2^{16}$.

(A) Given the system of equations:
$4m + n = 22$ $(1)$
$17m + 4n = 93$ $(2)$
Multiplying $(1)$ by $4$,we get $16m + 4n = 88$.
Subtracting this from $(2)$,we get $m = 5$.
Substituting $m = 5$ into $(1)$,$20 + n = 22$,so $n = 2$.
Thus,the order of matrix $A$ is $m = 5$,and $|A| = m - n = 5 - 2 = 3$.
We need to find $\operatorname{det}(n \operatorname{adj}(\operatorname{adj}(mA)))$.
Since $n = 2$ and $m = 5$,this is $\operatorname{det}(2 \operatorname{adj}(\operatorname{adj}(5A)))$.
Using the property $\operatorname{det}(kA) = k^m \operatorname{det}(A)$ for a matrix of order $m$:
$\operatorname{det}(2 \operatorname{adj}(\operatorname{adj}(5A))) = 2^5 \operatorname{det}(\operatorname{adj}(\operatorname{adj}(5A)))$.
Using $\operatorname{det}(\operatorname{adj}(B)) = |B|^{m-1}$,we have $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(5A))) = |\operatorname{adj}(5A)|^{5-1} = |\operatorname{adj}(5A)|^4$.
Since $|\operatorname{adj}(5A)| = |5A|^{5-1} = |5A|^4 = (5^5 |A|)^4 = 5^{20} |A|^4$.
So,$\operatorname{det}(\operatorname{adj}(\operatorname{adj}(5A))) = (5^{20} |A|^4)^4 = 5^{80} |A|^{16}$.
Thus,$\operatorname{det}(2 \operatorname{adj}(\operatorname{adj}(5A))) = 2^5 \cdot 5^{80} \cdot 3^{16}$.
Since $6^5 = 2^5 \cdot 3^5$,we rewrite the expression:
$2^5 \cdot 5^{80} \cdot 3^{16} = 3^{11} \cdot 5^{80} \cdot (2^5 \cdot 3^5) = 3^{11} \cdot 5^{80} \cdot 6^5$.
Comparing with $3^a 5^b 6^c$,we get $a = 11, b = 80, c = 5$.
Therefore,$a + b + c = 11 + 80 + 5 = 96$.

(A) Given that $A$ is a $3 \times 3$ matrix,so $|A| = 2$.
The formula for the determinant of the $k$-th iterated adjoint of $A$ is given by $\det(\operatorname{adj}^k(A)) = |A|^{(n-1)^k}$,where $n$ is the order of the matrix.
Here,$n=3$ and $k=2024$.
Thus,$n = |A|^{(3-1)^{2024}} = 2^{2^{2024}}$.
We need to find $2^{2^{2024}} \pmod{9}$.
By Euler's totient theorem,$\phi(9) = 9(1 - 1/3) = 6$.
We find $2^{2024} \pmod{6}$.
$2^{2024} \equiv 0 \pmod{2}$ and $2^{2024} = (2^2)^{1012} = 4^{1012} \equiv 1^{1012} \equiv 1 \pmod{3}$.
By Chinese Remainder Theorem,$2^{2024} \equiv 4 \pmod{6}$.
So,$2^{2024} = 6k + 4$ for some integer $k$.
Then $n = 2^{6k+4} = (2^6)^k \cdot 2^4 = 64^k \cdot 16$.
Since $64 \equiv 1 \pmod{9}$,we have $n \equiv 1^k \cdot 16 \equiv 16 \equiv 7 \pmod{9}$.
Therefore,the remainder is $7$.

(C) Given $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.
First,find $\operatorname{adj}(A) = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}$.
Let $M = \operatorname{adj}(A) = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}$.
Then $M^2 = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 0 & 1 \end{bmatrix}$.
By induction,$M^k = \begin{bmatrix} 1 & -2k \\ 0 & 1 \end{bmatrix}$.
$B = I + M + M^2 + \dots + M^{10} = \sum_{k=0}^{10} M^k$.
The sum of the diagonal elements is $\sum_{k=0}^{10} 1 + \sum_{k=0}^{10} 1 = 11 + 11 = 22$.
The sum of the off-diagonal elements is $\sum_{k=0}^{10} (-2k) + 0 = -2 \times \frac{10 \times 11}{2} = -110$.
Thus,$B = \begin{bmatrix} 11 & -110 \\ 0 & 11 \end{bmatrix}$.
The sum of all elements of $B$ is $11 - 110 + 0 + 11 = -88$.

(A) Let $A = \begin{bmatrix} a & b \\ b & d \end{bmatrix}$.
Given $A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$,we have $a + b = 3$ and $b + d = 7$.
Thus,$a = 3 - b$ and $d = 7 - b$.
The determinant $|A| = ad - b^2 = 1$.
Substituting the values,$(3 - b)(7 - b) - b^2 = 1$.
$21 - 3b - 7b + b^2 - b^2 = 1 \implies 21 - 10b = 1 \implies 10b = 20 \implies b = 2$.
Then $a = 3 - 2 = 1$ and $d = 7 - 2 = 5$.
So,$A = \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix}$.
The inverse $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1} \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix}$.
Given $A^{-1} = \alpha A + \beta I$,we have $\begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} = \alpha \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\alpha \\ 2\alpha & 5\alpha + \beta \end{bmatrix}$.
Comparing elements,$2\alpha = -2 \implies \alpha = -1$.
Substituting $\alpha = -1$ into $\alpha + \beta = 5$,we get $-1 + \beta = 5 \implies \beta = 6$.
Therefore,$\alpha + \beta = -1 + 6 = 5$.