Adjoint and inverse of matrices Questions in English

(D) Given $A = \begin{bmatrix} 1 & \cot \frac{\theta}{2} \\ -\cot \frac{\theta}{2} & 1 \end{bmatrix}$.
First,calculate the determinant $|A| = (1)(1) - (\cot \frac{\theta}{2})(-\cot \frac{\theta}{2}) = 1 + \cot^2 \frac{\theta}{2} = \csc^2 \frac{\theta}{2}$.
The adjoint of $A$ is $\text{adj}(A) = \begin{bmatrix} 1 & -\cot \frac{\theta}{2} \\ \cot \frac{\theta}{2} & 1 \end{bmatrix} = A^T$.
Thus,$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{\csc^2 \frac{\theta}{2}} A^T = \sin^2 \frac{\theta}{2} A^T$.
Since $\sin^2 \frac{\theta}{2} = \frac{1-\cos \theta}{2}$,the correct option is $D$.

(C) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 \times 3$.
Given $A = \frac{1}{11} \begin{bmatrix} -1 & 7 & -24 \\ 2 & a & 4 \\ 2 & -3 & 15 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & 3 & 4 \\ 2 & -3 & 4 \\ b & -1 & c \end{bmatrix}$.
So,$A \cdot A^{-1} = \frac{1}{11} \begin{bmatrix} -1 & 7 & -24 \\ 2 & a & 4 \\ 2 & -3 & 15 \end{bmatrix} \begin{bmatrix} 3 & 3 & 4 \\ 2 & -3 & 4 \\ b & -1 & c \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Multiplying the matrices:
Row $1 \times$ Column $1$: $\frac{1}{11} [(-1)(3) + (7)(2) + (-24)(b)] = 1 \implies -3 + 14 - 24b = 11 \implies 11 - 24b = 11 \implies b = 0$.
Row $2 \times$ Column $2$: $\frac{1}{11} [(2)(3) + (a)(-3) + (4)(-1)] = 1 \implies 6 - 3a - 4 = 11 \implies 2 - 3a = 11 \implies -3a = 9 \implies a = -3$.
Row $3 \times$ Column $3$: $\frac{1}{11} [(2)(4) + (-3)(4) + (15)(c)] = 1 \implies 8 - 12 + 15c = 11 \implies -4 + 15c = 11 \implies 15c = 15 \implies c = 1$.
Thus,the values are $a = -3, b = 0, c = 1$.

(C) We know that $(AB)^{-1} = B^{-1} A^{-1}$.
Given $(AB)^{-1} = \frac{1}{6} \begin{bmatrix} -7 & -3 \\ 2 & 3 \end{bmatrix}$ and $A^{-1} = \frac{1}{3} \begin{bmatrix} 4 & 3 \\ -1 & 0 \end{bmatrix}$.
Substituting these into the formula,we get:
$B^{-1} A^{-1} = \frac{1}{6} \begin{bmatrix} -7 & -3 \\ 2 & 3 \end{bmatrix}$
$B^{-1} \left( \frac{1}{3} \begin{bmatrix} 4 & 3 \\ -1 & 0 \end{bmatrix} \right) = \frac{1}{6} \begin{bmatrix} -7 & -3 \\ 2 & 3 \end{bmatrix}$
$B^{-1} \begin{bmatrix} 4 & 3 \\ -1 & 0 \end{bmatrix} = \frac{3}{6} \begin{bmatrix} -7 & -3 \\ 2 & 3 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -7 & -3 \\ 2 & 3 \end{bmatrix}$
Let $M = \begin{bmatrix} 4 & 3 \\ -1 & 0 \end{bmatrix}$. Then $B^{-1} M = \frac{1}{2} \begin{bmatrix} -7 & -3 \\ 2 & 3 \end{bmatrix}$.
$B^{-1} = \frac{1}{2} \begin{bmatrix} -7 & -3 \\ 2 & 3 \end{bmatrix} M^{-1}$.
First,find $M^{-1} = \frac{1}{|M|} \text{adj}(M)$.
$|M| = (4)(0) - (3)(-1) = 0 + 3 = 3$.
$\text{adj}(M) = \begin{bmatrix} 0 & -3 \\ 1 & 4 \end{bmatrix}$.
$M^{-1} = \frac{1}{3} \begin{bmatrix} 0 & -3 \\ 1 & 4 \end{bmatrix}$.
Now,$B^{-1} = \frac{1}{2} \begin{bmatrix} -7 & -3 \\ 2 & 3 \end{bmatrix} \left( \frac{1}{3} \begin{bmatrix} 0 & -3 \\ 1 & 4 \end{bmatrix} \right) = \frac{1}{6} \begin{bmatrix} (-7)(0) + (-3)(1) & (-7)(-3) + (-3)(4) \\ (2)(0) + (3)(1) & (2)(-3) + (3)(4) \end{bmatrix}$
$B^{-1} = \frac{1}{6} \begin{bmatrix} -3 & 21 - 12 \\ 3 & -6 + 12 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} -3 & 9 \\ 3 & 6 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -1 & 3 \\ 1 & 2 \end{bmatrix}$.

(C) To find $A^{-1}$,we first calculate the determinant $|A|$.
$|A| = 3((-3)(1) - (4)(-1)) - (-3)((2)(1) - (4)(0)) + 4((2)(-1) - (-3)(0))$
$|A| = 3(-3 + 4) + 3(2) + 4(-2) = 3(1) + 6 - 8 = 3 + 6 - 8 = 1$.
Since $|A| = 1 \neq 0$,$A^{-1}$ exists.
Next,we find the matrix of cofactors $C_{ij}$.
$C_{11} = +((-3)(1) - (4)(-1)) = 1$,$C_{12} = -((2)(1) - (4)(0)) = -2$,$C_{13} = +((2)(-1) - (-3)(0)) = -2$.
$C_{21} = -((-3)(1) - (4)(-1)) = -1$,$C_{22} = +((3)(1) - (4)(0)) = 3$,$C_{23} = -((3)(-1) - (-3)(0)) = 3$.
$C_{31} = +((-3)(4) - (-3)(4)) = 0$,$C_{32} = -((3)(4) - (2)(4)) = -4$,$C_{33} = +((3)(-3) - (2)(-3)) = -3$.
Thus,$adj(A) = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.
Calculating $A^2 = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}$.
Calculating $A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.
Since $A^{-1} = \frac{1}{|A|} adj(A) = 1 \cdot adj(A) = adj(A)$,and we found $A^3 = adj(A)$,therefore $A^{-1} = A^3$.

(C) Given the equation $(A-3I)(A-5I) = 0$.
Expanding the expression,we get $A^2 - 5A - 3A + 15I = 0$,which simplifies to $A^2 - 8A + 15I = 0$.
Since $A$ is a non-singular matrix,we can multiply the entire equation by $A^{-1}$ on both sides:
$A^{-1}(A^2 - 8A + 15I) = A^{-1}(0)$.
This gives $A - 8I + 15A^{-1} = 0$.
Rearranging the terms to isolate $15A^{-1}$,we get $15A^{-1} = 8I - A$.
Dividing both sides by $8$,we obtain $\frac{15}{8} A^{-1} = \frac{8I - A}{8} = I - \frac{1}{8} A$.

(C) Given the characteristic equation $A^2 - 4A + 3I = 0$.
We need to find $(A + 3I)^{-1}$.
Let $f(A) = A^2 - 4A + 3I = 0$.
We can write $A^2 - 4A = -3I$.
Since $A^2 - 4A + 3I = 0$,we have $A(A - 4I) = -3I$,which implies $A(4I - A) = 3I$.
However,we need the inverse of $(A + 3I)$.
Let $P(x) = x^2 - 4x + 3 = (x - 1)(x - 3) = 0$.
By Cayley-Hamilton theorem,$A$ satisfies its characteristic equation.
We want to express $(A + 3I)^{-1}$ in terms of $A$ and $I$.
Let $(A + 3I)^{-1} = xA + yI$.
Then $(A + 3I)(xA + yI) = I$.
$xA^2 + yA + 3xA + 3yI = I$.
$xA^2 + (y + 3x)A + 3yI = I$.
Substitute $A^2 = 4A - 3I$:
$x(4A - 3I) + (y + 3x)A + 3yI = I$.
$(4x + y + 3x)A + (-3x + 3y)I = I$.
$(7x + y)A + (3y - 3x)I = I$.
Comparing coefficients:
$7x + y = 0 \implies y = -7x$.
$3y - 3x = 1$.
Substitute $y = -7x$ into the second equation:
$3(-7x) - 3x = 1 \implies -21x - 3x = 1 \implies -24x = 1 \implies x = -\frac{1}{24}$.
Then $y = -7(-\frac{1}{24}) = \frac{7}{24}$.
Thus,$(A + 3I)^{-1} = -\frac{1}{24}A + \frac{7}{24}I = \frac{7I}{24} - \frac{A}{24}$.
Therefore,the correct option is $C$.

(D) Given $A = \begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}$.
For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,the inverse is given by $A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$,where $|A| = ad - bc$.
First,calculate the determinant $|A| = (3)(2) - (-1)(-4) = 6 - 4 = 2$.
Since $|A| \neq 0$,the inverse exists.
Now,find the adjoint matrix: $\text{adj}(A) = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{2} \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix} = \begin{bmatrix} \frac{2}{2} & \frac{1}{2} \\ \frac{4}{2} & \frac{3}{2} \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{2} \\ 2 & \frac{3}{2} \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$.
First,we find the determinant of $A$:
$|A| = (2)(3) - (-2)(4) = 6 + 8 = 14$.
Since $|A| \neq 0$,$A^{-1}$ exists.
The adjoint of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{Adj } A = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$.
Using the formula $A^{-1} = \frac{1}{|A|} \text{Adj } A$:
$A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$.
First,calculate the determinant $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.
Since $|A| \neq 0$,$A^{-1}$ exists.
The inverse is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{11} \begin{bmatrix} 1 & -2 \\ 5 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{bmatrix}$.
We are given $A^{-1} = xA + yI_2$.
Substituting the matrices:
$\begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{bmatrix} = x \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix} + y \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} x+y & 2x \\ -5x & x+y \end{bmatrix}$.
Comparing the corresponding elements:
$2x = \frac{-2}{11} \implies x = \frac{-1}{11}$.
$x + y = \frac{1}{11} \implies \frac{-1}{11} + y = \frac{1}{11} \implies y = \frac{2}{11}$.
Thus,$x = \frac{-1}{11}$ and $y = \frac{2}{11}$.

(B) Given $B = A^{-1}$,therefore $BA = I$.
$\begin{bmatrix} -2 & 0 & b \\ 7 & -1 & -2 \\ c & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & 3 \\ 3 & 2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Multiplying the matrices on the left side:
Row $1$,Column $1$: $(-2)(1) + (0)(1) + (b)(3) = -2 + 3b = 1 \Rightarrow 3b = 3 \Rightarrow b = 1$.
Row $2$,Column $2$: $(7)(1) + (-1)(a) + (-2)(2) = 7 - a - 4 = 3 - a = 1 \Rightarrow a = 2$.
Row $3$,Column $3$: $(c)(1) + (1)(3) + (1)(2) = c + 3 + 2 = c + 5 = 1 \Rightarrow c = -4$.
Now,calculate the value of $4a + 2b - c$:
$4(2) + 2(1) - (-4) = 8 + 2 + 4 = 14$.

(A) Let $A = \left[\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]$.
The determinant of $A$ is given by $|A| = (0.8)(0.8) - (-0.6)(0.6) = 0.64 + 0.36 = 1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
For a $2 \times 2$ matrix $A = \left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$,the inverse is given by $A^{-1} = \frac{1}{|A|} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Substituting the values,we get $A^{-1} = \frac{1}{1} \left[\begin{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right] = \left[\begin{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]$.

(B) Given $A = \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix} = \begin{bmatrix} 4+7 & 2+4 \\ 14+28 & 7+16 \end{bmatrix} = \begin{bmatrix} 11 & 6 \\ 42 & 23 \end{bmatrix}$.
Next,calculate $5A = 5 \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 35 & 20 \end{bmatrix}$.
Then,$A^2 - 5A = \begin{bmatrix} 11 & 6 \\ 42 & 23 \end{bmatrix} - \begin{bmatrix} 10 & 5 \\ 35 & 20 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 7 & 3 \end{bmatrix}$.
Let $B = A^2 - 5A = \begin{bmatrix} 1 & 1 \\ 7 & 3 \end{bmatrix}$.
The determinant $|B| = (1)(3) - (1)(7) = 3 - 7 = -4$.
The inverse $B^{-1} = \frac{1}{|B|} \text{adj}(B) = \frac{1}{-4} \begin{bmatrix} 3 & -1 \\ -7 & 1 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -3 & 1 \\ 7 & -1 \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$. The determinant $|A| = (1)(4) - (2)(-1) = 4 + 2 = 6 \neq 0$.
Since $A^{-1} = \frac{1}{|A|} \text{adj}(A)$,we have $A^{-1} = \frac{1}{6} \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix}$.
Given $A^{-1} = \alpha I + \beta A$,we have:
$\begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix} = \alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix}$.
Comparing the elements,we get:
$-\beta = \frac{1}{6} \implies \beta = -\frac{1}{6}$.
$\alpha + \beta = \frac{2}{3} \implies \alpha - \frac{1}{6} = \frac{2}{3} \implies \alpha = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}$.
Therefore,$4(\alpha - \beta) = 4(\frac{5}{6} - (-\frac{1}{6})) = 4(\frac{6}{6}) = 4(1) = 4$.

(A) Given $A = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 4+1 & -2-3 \\ -2-3 & 1+9 \end{bmatrix} = \begin{bmatrix} 5 & -5 \\ -5 & 10 \end{bmatrix}$.
Now,calculate $2A^2 + 5A = 2 \begin{bmatrix} 5 & -5 \\ -5 & 10 \end{bmatrix} + 5 \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 10 & -10 \\ -10 & 20 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 15 \end{bmatrix} = \begin{bmatrix} 20 & -15 \\ -15 & 35 \end{bmatrix}$.
Let $M = 2A^2 + 5A = \begin{bmatrix} 20 & -15 \\ -15 & 35 \end{bmatrix}$.
The determinant $|M| = (20)(35) - (-15)(-15) = 700 - 225 = 475$.
The inverse $M^{-1} = \frac{1}{|M|} \text{adj}(M) = \frac{1}{475} \begin{bmatrix} 35 & 15 \\ 15 & 20 \end{bmatrix}$.
Dividing by $5$,we get $M^{-1} = \frac{1}{95} \begin{bmatrix} 7 & 3 \\ 3 & 4 \end{bmatrix}$.

(D) Given $A = \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix}$.
Calculate $A^2 = \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} i^2+1 & i \\ i & 1 \end{bmatrix} = \begin{bmatrix} 0 & i \\ i & 1 \end{bmatrix}$.
Calculate $A^3 = A^2 \times A = \begin{bmatrix} 0 & i \\ i & 1 \end{bmatrix} \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} = iI$.
Then $A^6 = (A^3)^2 = (iI)^2 = i^2 I = -I$.
Now,$B = A^{2029} = A^{6 \times 338 + 1} = (A^6)^{338} \times A = (-I)^{338} \times A = I \times A = A$.
Since $B = A$,we need $B^{-1} = A^{-1}$.
$|A| = (i)(0) - (1)(1) = -1$.
$A^{-1} = \frac{1}{|A|} \operatorname{adj} A = \frac{1}{-1} \operatorname{adj} A = -\operatorname{adj} A$.
Therefore,$B^{-1} = -\operatorname{adj} A$.

(B) Given $A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$.
First,we find the determinant $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.
Next,we find the inverse $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{11} \begin{bmatrix} 1 & -2 \\ 5 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{bmatrix}$.
Given $A^{-1} = xA + yI$,we have:
$\begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{bmatrix} = x \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix} + y \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} x+y & 2x \\ -5x & x+y \end{bmatrix}$.
Comparing the elements,we get $2x = \frac{-2}{11} \Rightarrow x = \frac{-1}{11}$.
Also,$x+y = \frac{1}{11} \Rightarrow \frac{-1}{11} + y = \frac{1}{11} \Rightarrow y = \frac{2}{11}$.
Finally,$2x + 3y = 2(\frac{-1}{11}) + 3(\frac{2}{11}) = \frac{-2}{11} + \frac{6}{11} = \frac{4}{11}$.

(B) Given $A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}$.
The determinant $|A| = (1)(1) - (\tan x)(-\tan x) = 1 + \tan^2 x = \sec^2 x$.
The transpose $A^T = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$.
The inverse $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$.
Now,$A^T \cdot A^{-1} = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \cdot \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$
$= \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 - \tan^2 x & -\tan x - \tan x \\ \tan x + \tan x & -\tan^2 x + 1 \end{bmatrix}$
$= \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 - \tan^2 x & -2 \tan x \\ 2 \tan x & 1 - \tan^2 x \end{bmatrix}$
$= \begin{bmatrix} \frac{1 - \tan^2 x}{1 + \tan^2 x} & \frac{-2 \tan x}{1 + \tan^2 x} \\ \frac{2 \tan x}{1 + \tan^2 x} & \frac{1 - \tan^2 x}{1 + \tan^2 x} \end{bmatrix}$
Using trigonometric identities $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$ and $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$,we get:
$A^T \cdot A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}$.

(B) First,calculate the sum of matrices $A$ and $B$:
$A+B = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 1 \\ 4 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 6 & -2 \end{bmatrix}$
Next,find the determinant of $(A+B)$:
$|A+B| = (2)(-2) - (0)(6) = -4 - 0 = -4$
Since $|A+B| \neq 0$,the inverse exists.
For a matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,$M^{-1} = \frac{1}{|M|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Applying this to $A+B$:
$(A+B)^{-1} = \frac{1}{-4} \begin{bmatrix} -2 & 0 \\ -6 & 2 \end{bmatrix} = \begin{bmatrix} \frac{-2}{-4} & \frac{0}{-4} \\ \frac{-6}{-4} & \frac{2}{-4} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$

(B) Let $A = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$. The inverse of a matrix $A$ is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
First,we calculate the determinant $|A|$:
$|A| = 3(1 \times 5 - 2 \times 2) - 2(1 \times 5 - 2 \times 2) + 6(1 \times 2 - 1 \times 2)$
$|A| = 3(5 - 4) - 2(5 - 4) + 6(2 - 2) = 3(1) - 2(1) + 6(0) = 3 - 2 = 1$.
The element in the third row and second column of $A^{-1}$ is the entry $(A^{-1})_{32} = \frac{C_{23}}{|A|}$,where $C_{23}$ is the cofactor of the element at the second row and third column of matrix $A$.
The cofactor $C_{23} = (-1)^{2+3} \times M_{23}$,where $M_{23}$ is the minor of the element at $(2,3)$.
$M_{23} = \det \begin{bmatrix} 3 & 2 \\ 2 & 2 \end{bmatrix} = (3 \times 2) - (2 \times 2) = 6 - 4 = 2$.
Thus,$C_{23} = (-1)^5 \times 2 = -2$.
Therefore,$(A^{-1})_{32} = \frac{-2}{1} = -2$.

(C) For a matrix $A$ to be skew-symmetric,it must satisfy $A^T = -A$.
Given the matrix $A = \begin{bmatrix} 0 & 1+2i & i-2 \\ -1-2i & 0 & K \\ 2-i & -7 & 0 \end{bmatrix}$.
Comparing the elements $A_{ij} = -A_{ji}$,we observe:
$A_{12} = 1+2i$ and $A_{21} = -1-2i = -(1+2i)$,which is consistent.
$A_{13} = i-2$ and $A_{31} = 2-i = -(i-2)$,which is consistent.
$A_{23} = K$ and $A_{32} = -7$.
For skew-symmetry,$A_{23} = -A_{32}$,so $K = -(-7) = 7$.
However,the problem states $A^{-1}$ does not exist,which implies $\det(A) = 0$.
For a skew-symmetric matrix of odd order $(3 \times 3)$,the determinant is always $0$.
Thus,the condition $A^{-1}$ does not exist is satisfied for any $K$ that makes the matrix skew-symmetric.
Therefore,$K = 7$.

(D) First,calculate the determinant $|A|$:
$|A| = 1(3 \times 3 - 4 \times 4) - 2(1 \times 3 - 4 \times 3) + 3(1 \times 4 - 3 \times 3)$
$|A| = 1(9 - 16) - 2(3 - 12) + 3(4 - 9)$
$|A| = 1(-7) - 2(-9) + 3(-5) = -7 + 18 - 15 = -4$.
Next,find the matrix of cofactors $C_{ij}$:
$C_{11} = +(9-16) = -7, C_{12} = -(3-12) = 9, C_{13} = +(4-9) = -5$
$C_{21} = -(6-12) = 6, C_{22} = +(3-9) = -6, C_{23} = -(4-6) = 2$
$C_{31} = +(8-9) = -1, C_{32} = -(4-3) = -1, C_{33} = +(3-2) = 1$
Thus,$\text{adj}(A) = \begin{bmatrix} -7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1 \end{bmatrix}$.
Finally,$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{-4} \begin{bmatrix} -7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1 \end{bmatrix} = -\frac{1}{4} \begin{bmatrix} -7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1 \end{bmatrix}$.

(A) We know that $A \times A^{-1} = I$,where $I$ is the identity matrix.
Calculating the determinant $|A|$:
$|A| = 1(7 - 20) - a(7 - 10) + 3(4 - 2) = -13 + 3a + 6 = 3a - 7$.
Using the property $A^{-1} = \frac{1}{|A|} \text{adj}(A)$,we look at the element at position $(2, 1)$ of $A^{-1}$,which is $-3$.
The cofactor $C_{12}$ of $A$ is $-(1 \times 7 - 5 \times 2) = -(7 - 10) = 3$.
Since $(A^{-1})_{21} = \frac{C_{12}}{|A|}$,we have $-3 = \frac{3}{3a - 7}$.
$-3(3a - 7) = 3 \Rightarrow -9a + 21 = 3 \Rightarrow 9a = 18 \Rightarrow a = 2$.
Now,for $b$,which is the element at $(2, 2)$ of $A^{-1}$:
$(A^{-1})_{22} = \frac{C_{22}}{|A|}$.
The cofactor $C_{22} = (1 \times 7 - 3 \times 2) = 7 - 6 = 1$.
$b = \frac{1}{3(2) - 7} = \frac{1}{6 - 7} = -1$.
Thus,$a = 2$ and $b = -1$.

(A) Step $1$: Calculate the product $AB$.
$AB = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} (1)(2) + (2)(1) + (1)(1) & (1)(3) + (2)(2) + (1)(2) \\ (3)(2) + (1)(1) + (3)(1) & (3)(3) + (1)(2) + (3)(2) \end{bmatrix} = \begin{bmatrix} 5 & 9 \\ 10 & 17 \end{bmatrix}$.
Step $2$: Find the determinant of $AB$.
$|AB| = (5)(17) - (9)(10) = 85 - 90 = -5$.
Step $3$: Find the inverse $(AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB)$.
$(AB)^{-1} = \frac{1}{-5} \begin{bmatrix} 17 & -9 \\ -10 & 5 \end{bmatrix} = \begin{bmatrix} \frac{-17}{5} & \frac{9}{5} \\ 2 & -1 \end{bmatrix}$.

(C) Given $A = \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix}$.
First,we calculate $A^2 = \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 8 \\ 0 & 1 \end{bmatrix}$.
Then,$A^3 = A^2 \cdot A = \begin{bmatrix} 9 & 8 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 27 & 26 \\ 0 & 1 \end{bmatrix}$.
We know that $(A^{-1})^3 = (A^3)^{-1}$.
For a matrix $M = \begin{bmatrix} a & b \\ 0 & d \end{bmatrix}$,$M^{-1} = \frac{1}{ad} \begin{bmatrix} d & -b \\ 0 & a \end{bmatrix}$.
Here,$A^3 = \begin{bmatrix} 27 & 26 \\ 0 & 1 \end{bmatrix}$,so $a=27, b=26, d=1$.
$(A^3)^{-1} = \frac{1}{27 \times 1} \begin{bmatrix} 1 & -26 \\ 0 & 27 \end{bmatrix} = \frac{1}{27} \begin{bmatrix} 1 & -26 \\ 0 & 27 \end{bmatrix}$.

(C) Given the equation $A^2 - 4A + 3I = 0$.
Subtract $3I$ from both sides: $A^2 - 4A = -3I$.
Multiply by $A^{-1}$ on both sides: $A^{-1}(A^2 - 4A) = A^{-1}(-3I)$.
This simplifies to $A - 4I = -3A^{-1}$.
Therefore,$A^{-1} = -\frac{1}{3}(A - 4I) = \frac{1}{3}(4I - A)$.
Substitute $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$:
$A^{-1} = \frac{1}{3} \left( \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \right) = \frac{1}{3} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$.

(C) Given $A = \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix}$.
First,find the determinant $|A| = (2)(-7) - (-3)(5) = -14 + 15 = 1$.
The inverse of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
So,$A^{-1} = \frac{1}{1} \begin{bmatrix} -7 & 3 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 3 \\ -5 & 2 \end{bmatrix}$.
Now,calculate $A - A^{-1} = \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix} - \begin{bmatrix} -7 & 3 \\ -5 & 2 \end{bmatrix}$.
$A - A^{-1} = \begin{bmatrix} 2 - (-7) & -3 - 3 \\ 5 - (-5) & -7 - 2 \end{bmatrix} = \begin{bmatrix} 9 & -6 \\ 10 & -9 \end{bmatrix}$.
Factoring out $3$,we get $3 \begin{bmatrix} 3 & -2 \\ \frac{10}{3} & -3 \end{bmatrix}$.

(D) Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}$. The determinant $|A| = (1)(5) - (2)(3) = 5 - 6 = -1$.
Since $|A| \neq 0$,$A^{-1}$ exists.
$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{-1} \begin{bmatrix} 5 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix}$.
Given $A^{-1} = \alpha I + \beta A$,we have:
$\begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix} = \alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\beta \\ 3\beta & \alpha + 5\beta \end{bmatrix}$.
Comparing the elements:
$2\beta = 2 \Rightarrow \beta = 1$.
$\alpha + \beta = -5 \Rightarrow \alpha + 1 = -5 \Rightarrow \alpha = -6$.
Check: $3\beta = 3(1) = 3$ (Correct) and $\alpha + 5\beta = -6 + 5(1) = -1$ (Correct).
Thus,$\alpha + \beta + \alpha\beta = -6 + 1 + (-6)(1) = -5 - 6 = -11$.

(B) Given $(BA)^{-1} = C$.
Using the property of inverse matrices $(BA)^{-1} = A^{-1}B^{-1}$,we have $A^{-1}B^{-1} = C$.
Multiplying both sides by $B$ on the right,we get $A^{-1}B^{-1}B = CB$.
Since $B^{-1}B = I$,it follows that $A^{-1} = CB$.
Now,calculate the product $CB$:
$A^{-1} = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$
$A^{-1} = \begin{bmatrix} (-1)(2) + (0)(1) + (1)(-1) & (-1)(6) + (0)(0) + (1)(1) & (-1)(4) + (0)(1) + (1)(-1) \\ (1)(2) + (1)(1) + (3)(-1) & (1)(6) + (1)(0) + (3)(1) & (1)(4) + (1)(1) + (3)(-1) \\ (2)(2) + (0)(1) + (2)(-1) & (2)(6) + (0)(0) + (2)(1) & (2)(4) + (0)(1) + (2)(-1) \end{bmatrix}$
$A^{-1} = \begin{bmatrix} -3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{bmatrix}$

(A) The inverse of a matrix $A$ does not exist if its determinant is zero,i.e.,$|A| = 0$.
Given matrix $A = \begin{bmatrix} 1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{bmatrix}$.
Setting the determinant to zero:
$|A| = 1((-1)(-6) - (7)(4)) - 2((4)(-6) - (7)(2)) + x((4)(4) - (-1)(2)) = 0$
$|A| = 1(6 - 28) - 2(-24 - 14) + x(16 + 2) = 0$
$|A| = 1(-22) - 2(-38) + x(18) = 0$
$-22 + 76 + 18x = 0$
$54 + 18x = 0$
$18x = -54$
$x = -3$.

(C) We know that for any invertible matrices $A$ and $B$,the property $(AB)^{-1} = B^{-1} A^{-1}$ holds true.
Given $A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix}$.
Substituting these values,we get:
$(AB)^{-1} = B^{-1} A^{-1} = \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$
Performing matrix multiplication:
$= \begin{bmatrix} (1)(2) + (0)(-1) & (1)(-3) + (0)(2) \\ (-3)(2) + (1)(-1) & (-3)(-3) + (1)(2) \end{bmatrix}$
$= \begin{bmatrix} 2 + 0 & -3 + 0 \\ -6 - 1 & 9 + 2 \end{bmatrix}$
$= \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix}$

(A) Given $F(\alpha) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
First,we find the determinant $|F(\alpha)|$:
$|F(\alpha)| = \cos \alpha(\cos \alpha - 0) - (-\sin \alpha)(\sin \alpha - 0) + 0 = \cos^2 \alpha + \sin^2 \alpha = 1$.
Since $|F(\alpha)| = 1 \neq 0$,the inverse exists.
The inverse is given by $[F(\alpha)]^{-1} = \frac{1}{|F(\alpha)|} \text{adj}(F(\alpha))$.
The cofactor matrix $C$ is calculated by finding the cofactor of each element.
The adjoint is the transpose of the cofactor matrix:
$\text{adj}(F(\alpha)) = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
We know that $\cos(-\alpha) = \cos \alpha$ and $\sin(-\alpha) = -\sin \alpha$.
Substituting these,we get $\text{adj}(F(\alpha)) = \begin{bmatrix} \cos(-\alpha) & -\sin(-\alpha) & 0 \\ \sin(-\alpha) & \cos(-\alpha) & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(-\alpha)$.
Thus,$[F(\alpha)]^{-1} = F(-\alpha)$.

(D) Given $A^{-1} = \frac{-1}{2} \begin{bmatrix} 5 & 8 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -2.5 & -4 \\ 0.5 & -1 \end{bmatrix}$.
To find $A$,we use the formula $A = (A^{-1})^{-1}$.
The determinant $|A^{-1}| = (-2.5)(-1) - (-4)(0.5) = 2.5 + 2 = 4.5 = \frac{9}{2}$.
$A = \frac{1}{|A^{-1}|} \text{adj}(A^{-1}) = \frac{1}{9/2} \begin{bmatrix} -1 & 4 \\ -0.5 & -2.5 \end{bmatrix} = \frac{2}{9} \begin{bmatrix} -1 & 4 \\ -0.5 & -2.5 \end{bmatrix} = \begin{bmatrix} -2/9 & 8/9 \\ -1/9 & -5/9 \end{bmatrix}$.
Alternatively,using the property $A = \frac{1}{|A^{-1}|} \text{adj}(A^{-1})$ on the original matrix:
$|A^{-1}| = \frac{1}{4} (10 - (-8)) = \frac{18}{4} = 4.5$.
$A = \frac{1}{4.5} \cdot \frac{-1}{2} \begin{bmatrix} 2 & -8 \\ 1 & 5 \end{bmatrix} = \frac{-1}{9} \begin{bmatrix} 2 & -8 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} -2/9 & 8/9 \\ -1/9 & -5/9 \end{bmatrix}$.
Calculating $2A + I_2 = 2 \begin{bmatrix} -2/9 & 8/9 \\ -1/9 & -5/9 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -4/9+1 & 16/9 \\ -2/9 & -10/9+1 \end{bmatrix} = \begin{bmatrix} 5/9 & 16/9 \\ -2/9 & -1/9 \end{bmatrix}$.
Note: Based on the provided options,there is a discrepancy in the question's matrix values. Assuming the intended question was $A = \frac{-1}{2} \begin{bmatrix} 5 & 8 \\ -1 & 2 \end{bmatrix}$,then $2A+I_2 = \begin{bmatrix} -5 & -8 \\ 1 & -2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -4 & -8 \\ 1 & -1 \end{bmatrix}$. Given the options,$D$ is the intended answer if $A = \begin{bmatrix} 2 & 4 \\ 1 & 1 \end{bmatrix}$.

(D) We know that $AA^{-1} = I$.
Multiplying the given matrices:
$\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix} \times \frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ -8 & 6 & 2c \\ 5 & -3 & 1 \end{bmatrix} = I$
$\frac{1}{2} \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ -8 & 6 & 2c \\ 5 & -3 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Performing matrix multiplication for the first row and third column:
$0(1) + 1(2c) + 2(1) = 0 \implies 2c + 2 = 0 \implies 2c = -2 \implies c = -1$.
Performing matrix multiplication for the third row and first column:
$3(1) + a(-8) + 1(5) = 0 \implies 3 - 8a + 5 = 0 \implies 8 - 8a = 0 \implies 8a = 8 \implies a = 1$.
Thus,the values are $a = 1$ and $c = -1$.

(B) To find $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
First,calculate the determinant $|A|$:
$|A| = 0(2-3) - 1(1-9) + 2(1-6) = 0 - 1(-8) + 2(-5) = 8 - 10 = -2$.
Next,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,$\text{adj}(A) = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}^T = \begin{bmatrix} -1 & 8 & -5 \\ 1 & -6 & 3 \\ -1 & 2 & -1 \end{bmatrix}$.
Finally,$A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & 8 & -5 \\ 1 & -6 & 3 \\ -1 & 2 & -1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & -8 & 5 \\ -1 & 6 & -3 \\ 1 & -2 & 1 \end{bmatrix}$.
Comparing with the given options,option $D$ is $\frac{1}{2} \begin{bmatrix} 1 & -1 & -1 \\ -8 & 6 & -2 \\ 5 & -3 & 1 \end{bmatrix}$,which is incorrect. Re-evaluating the calculation,the correct inverse is $\begin{bmatrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{bmatrix}$,which matches option $B$.

(B) The inverse of a matrix $A$,denoted as $A^{-1}$,does not exist if and only if the determinant of the matrix is zero,i.e.,$|A| = 0$.
Given $A = \begin{bmatrix} k & 2 \\ -2 & -k \end{bmatrix}$.
The determinant $|A|$ is calculated as:
$|A| = (k)(-k) - (2)(-2)$
$|A| = -k^2 + 4$
For $A^{-1}$ to not exist,we set $|A| = 0$:
$-k^2 + 4 = 0$
$k^2 = 4$
$k = \pm 2$
Therefore,$A^{-1}$ does not exist when $k = \pm 2$.

(B) Given the matrix $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$.
First,we calculate the determinant of $A$ $(|A|)$:
$|A| = 1(2 - 6) - 0(0 - 3) + 1(0 - 2)$
$|A| = 1(-4) - 0 + 1(-2)$
$|A| = -4 - 2 = -6$.
We know the property of determinants that $|A^{-1}| = \frac{1}{|A|}$.
Therefore,$|A^{-1}| = \frac{1}{-6} = -\frac{1}{6}$.

(C) First,calculate the product $AB$:
$AB = \left[\begin{array}{ccc}1 & 2 & 1 \\ -1 & 1 & 3\end{array}\right] \left[\begin{array}{cc}1 & 2 \\ -3 & 1 \\ 0 & 2\end{array}\right]$
$AB = \left[\begin{array}{cc} (1)(1) + (2)(-3) + (1)(0) & (1)(2) + (2)(1) + (1)(2) \\ (-1)(1) + (1)(-3) + (3)(0) & (-1)(2) + (1)(1) + (3)(2) \end{array}\right]$
$AB = \left[\begin{array}{cc} 1 - 6 + 0 & 2 + 2 + 2 \\ -1 - 3 + 0 & -2 + 1 + 6 \end{array}\right] = \left[\begin{array}{cc} -5 & 6 \\ -4 & 5 \end{array}\right]$
Now,find the inverse of the matrix $M = \left[\begin{array}{cc} -5 & 6 \\ -4 & 5 \end{array}\right]$.
The determinant $|M| = (-5)(5) - (6)(-4) = -25 + 24 = -1$.
The inverse is given by $M^{-1} = \frac{1}{|M|} \text{adj}(M)$.
For a $2 \times 2$ matrix $\left[\begin{array}{cc} a & b \\ c & d \end{array}\right]$,the adjoint is $\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]$.
So,$\text{adj}(M) = \left[\begin{array}{cc} 5 & -6 \\ 4 & -5 \end{array}\right]$.
$(AB)^{-1} = \frac{1}{-1} \left[\begin{array}{cc} 5 & -6 \\ 4 & -5 \end{array}\right] = \left[\begin{array}{cc} -5 & 6 \\ -4 & 5 \end{array}\right]$.

(D) The determinant of matrix $A$ is given by $|A| = 1(7 - 4a) - 2(7 - 2a) + 3(4 - 2) = 7 - 4a - 14 + 4a + 6 = -1$.
Since $B = A^{-1}$,we know that $B = \frac{1}{|A|} \text{adj}(A)$.
Given $|A| = -1$,the element $B_{ij} = \frac{C_{ji}}{|A|} = -C_{ji}$,where $C_{ji}$ is the cofactor of the element $A_{ji}$.
For element $B_{13} = b$,we have $b = -C_{31}$.
The cofactor $C_{31} = (-1)^{3+1} \begin{vmatrix} 2 & 3 \\ 1 & a \end{vmatrix} = 2a - 3$.
Thus,$b = -(2a - 3) = 3 - 2a$,which implies $2a + b = 3$.
For element $B_{21} = -3$,we have $-3 = -C_{12}$.
The cofactor $C_{12} = (-1)^{1+2} \begin{vmatrix} 1 & a \\ 2 & 7 \end{vmatrix} = -(7 - 2a) = 2a - 7$.
Thus,$-3 = -(2a - 7) = 7 - 2a$,which implies $2a = 10$,so $a = 5$.
Substituting $a = 5$ into $2a + b = 3$,we get $2(5) + b = 3$,so $10 + b = 3$,which gives $b = -7$.

(A) We know the property of inverse of matrices: $(XY)^{-1} = Y^{-1} X^{-1}$.
Applying this property to the given expression:
$(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1} = AB$.
Now,we calculate the product $AB$:
$AB = \left[\begin{array}{ll}2 & -2 \\ 2 & -3\end{array}\right] \left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$
$AB = \left[\begin{array}{ll} (2)(0) + (-2)(1) & (2)(-1) + (-2)(0) \\ (2)(0) + (-3)(1) & (2)(-1) + (-3)(0) \end{array}\right]$
$AB = \left[\begin{array}{ll} 0 - 2 & -2 + 0 \\ 0 - 3 & -2 + 0 \end{array}\right] = \left[\begin{array}{ll} -2 & -2 \\ -3 & -2 \end{array}\right]$.

(D) Given $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$.
First,calculate the determinant of $A$:
$|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$.
Next,find the adjoint of $A$:
$\text{adj}(A) = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$.
We know that $A^{-1} = \frac{1}{|A|} \text{adj}(A)$:
$A^{-1} = \frac{1}{-19} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} = \frac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$.
Since $A^{-1} = KA$ and $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$,we have:
$KA = \frac{1}{19} A$.
Therefore,$K = \frac{1}{19}$.

(D) matrix is invertible if and only if its determinant is non-zero $(|M| \neq 0)$.
For matrix $A$: $|A| = (2 \times 15) - (3 \times 10) = 30 - 30 = 0$. Thus,$A$ is not invertible.
For matrix $B$: Since row $R_1$ and row $R_3$ are identical,$|B| = 0$. Thus,$B$ is not invertible.
For matrix $C$: $|C| = 1(4 \times 8 - 5 \times 6) - 2(3 \times 8 - 5 \times 4) + 3(3 \times 6 - 4 \times 4) = 1(32 - 30) - 2(24 - 20) + 3(18 - 16) = 1(2) - 2(4) + 3(2) = 2 - 8 + 6 = 0$. Thus,$C$ is not invertible.
For matrix $D$: $|D| = 2(1 \times 5 - 0 \times 4) - 4(1 \times 5 - 0 \times 1) + 2(1 \times 4 - 1 \times 1) = 2(5) - 4(5) + 2(3) = 10 - 20 + 6 = -4$. Since $|D| \neq 0$,$D$ is invertible.

(D) Given the matrix $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$.
The determinant of $A$ is $|A| = (2)(2) - (-1)(-1) = 4 - 1 = 3$.
The adjoint of $A$ is $\text{adj}(A) = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$.
The inverse of a matrix is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
Substituting the values,we get $A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$.
Alternatively,using the characteristic equation $A^{2} - 4A + 3I = 0$,multiply by $A^{-1}$:
$A - 4I + 3A^{-1} = 0 \Rightarrow 3A^{-1} = 4I - A$.
$3A^{-1} = 4 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$.
Thus,$A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$.

(C) square matrix $A$ is invertible if and only if its determinant $|A| \neq 0$.
Let us calculate the determinant for each matrix:
$1$. For $A_{1} = \begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix}$,$|A_{1}| = (4 \times 1) - (2 \times 2) = 4 - 4 = 0$. Thus,$A_{1}$ is not invertible.
$2$. For $A_{2} = \begin{bmatrix} -1 & -2 & 3 \\ 4 & 5 & 7 \\ 2 & 4 & -6 \end{bmatrix}$,notice that row $3$ is $-2$ times row $1$ $(R_{3} = -2R_{1})$. Since two rows are proportional,$|A_{2}| = 0$. Thus,$A_{2}$ is not invertible.
$3$. For $A_{3} = \begin{bmatrix} 1 & 0 & 0 \\ 5 & 2 & 1 \\ 7 & 2 & 1 \end{bmatrix}$,notice that row $2$ and row $3$ are not proportional,but let's calculate: $|A_{3}| = 1(2-2) - 0 + 0 = 0$. Thus,$A_{3}$ is not invertible.
$4$. For $A_{4} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$,$|A_{4}| = 1(2-6) - 0(0-3) + 1(0-2) = 1(-4) + 1(-2) = -4 - 2 = -6$.
Since $|A_{4}| = -6 \neq 0$,the matrix $A_{4}$ is invertible.

(A) First,calculate the product $AB$:
$AB = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} (2 \times 1 + 3 \times 3) & (2 \times 0 + 3 \times 1) \\ (1 \times 1 + 2 \times 3) & (1 \times 0 + 2 \times 1) \end{bmatrix} = \begin{bmatrix} 11 & 3 \\ 7 & 2 \end{bmatrix}$
Next,find the determinant $|AB|$:
$|AB| = (11 \times 2) - (3 \times 7) = 22 - 21 = 1$
Now,find the adjoint of $AB$:
For a matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Thus,$\text{adj}(AB) = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix}$
Finally,$(AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB) = \frac{1}{1} \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix}$

(D) 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 $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} -\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
Finally,the inverse 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,$A^{-1} = A$.

(D) We know that for any invertible matrices $A$ and $B$,the property $(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1}$ holds true.
Since $(A^{-1})^{-1} = A$ and $(B^{-1})^{-1} = B$,the expression simplifies to $AB$.
Now,we calculate the product $AB$:
$AB = \left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] \left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$
$AB = \left[\begin{array}{cc} (2)(2) + (3)(-1) & (2)(-3) + (3)(2) \\ (1)(2) + (2)(-1) & (1)(-3) + (2)(2) \end{array}\right]$
$AB = \left[\begin{array}{cc} 4 - 3 & -6 + 6 \\ 2 - 2 & -3 + 4 \end{array}\right]$
$AB = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
Thus,the correct option is $D$.

(B) Given $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$.
First,we find the determinant $|A| = 0(0) - 0(0) - 1(0 - 1) = -1(-1) = 1$.
Since $|A| \neq 0$,the matrix $A$ is invertible.
Now,we calculate $A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Since $A^2 = I$,multiplying both sides by $A^{-1}$ gives $A = A^{-1}$.

(C) First,calculate the sum of matrices $A$ and $B$:
$A+B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 4 & 3 \end{bmatrix}$
Next,find the determinant of $(A+B)$:
$|A+B| = (5 \times 3) - (2 \times 4) = 15 - 8 = 7$
Now,find the adjoint of $(A+B)$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A+B) = \begin{bmatrix} 3 & -2 \\ -4 & 5 \end{bmatrix}$
Finally,use the formula $(A+B)^{-1} = \frac{1}{|A+B|} \text{adj}(A+B)$:
$(A+B)^{-1} = \frac{1}{7} \begin{bmatrix} 3 & -2 \\ -4 & 5 \end{bmatrix}$

(B) First,calculate the product $AB$:
$AB = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} (1)(1)+(2)(2)+(1)(0) & (1)(2)+(2)(1)+(1)(1) \\ (2)(1)+(1)(2)+(0)(0) & (2)(2)+(1)(1)+(0)(1) \end{bmatrix} = \begin{bmatrix} 5 & 5 \\ 4 & 5 \end{bmatrix}$
Now,find the determinant $|AB|$:
$|AB| = (5)(5) - (5)(4) = 25 - 20 = 5$
Next,find the adjoint of $AB$:
$adj(AB) = \begin{bmatrix} 5 & -5 \\ -4 & 5 \end{bmatrix}$
Finally,calculate the inverse $(AB)^{-1} = \frac{1}{|AB|} adj(AB)$:
$(AB)^{-1} = \frac{1}{5} \begin{bmatrix} 5 & -5 \\ -4 & 5 \end{bmatrix}$
Thus,the correct option is $B$.

(B) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 \times 3$.
$\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Multiplying the matrices on the left side:
Row $1$,Column $1$: $(2)(3) + (0)(\alpha) + (-1)(\beta) = 6 - \beta$
Since the result must be $I$,$6 - \beta = 1$,which gives $\beta = 5$.
Row $2$,Column $1$: $(5)(3) + (1)(\alpha) + (0)(\beta) = 15 + \alpha$
Since the result must be $I$,$15 + \alpha = 0$,which gives $\alpha = -15$.
Thus,the values are $\alpha = -15$ and $\beta = 5$.