Types of matrices, Algebra of matrices Questions in English

(D) Given $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$.
First,calculate $A^{2} = A \times A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 9+2 & 6+2 \\ 3+1 & 2+1 \end{bmatrix} = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}$.
Substitute $A^{2}$,$A$,and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ into the equation $A^{2} + aA + bI = 0$:
$\begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix} + a \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} + b \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
This gives the matrix equation:
$\begin{bmatrix} 11 + 3a + b & 8 + 2a \\ 4 + a & 3 + a + b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Comparing the elements:
$8 + 2a = 0 \Rightarrow 2a = -8 \Rightarrow a = -4$.
$4 + a = 0 \Rightarrow 4 + (-4) = 0$ (Consistent).
$11 + 3a + b = 0 \Rightarrow 11 + 3(-4) + b = 0 \Rightarrow 11 - 12 + b = 0 \Rightarrow -1 + b = 0 \Rightarrow b = 1$.
Thus,$a = -4$ and $b = 1$.

(A) We know that $A = (A^{\prime})^{\prime}$.
Therefore,$A = \begin{bmatrix}-2 & 1 \\ 3 & 2\end{bmatrix}$.
Now,$A + 2B = \begin{bmatrix}-2 & 1 \\ 3 & 2\end{bmatrix} + 2\begin{bmatrix}-1 & 0 \\ 1 & 2\end{bmatrix}$.
$A + 2B = \begin{bmatrix}-2 & 1 \\ 3 & 2\end{bmatrix} + \begin{bmatrix}-2 & 0 \\ 2 & 4\end{bmatrix} = \begin{bmatrix}-4 & 1 \\ 5 & 6\end{bmatrix}$.
Finally,$(A + 2B)^{\prime} = \begin{bmatrix}-4 & 5 \\ 1 & 6\end{bmatrix}$.

(A) First,calculate the product $AB$:
$AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}$
Now,find the transpose $(AB)^{\prime}$:
$(AB)^{\prime} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$
Next,find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$
$B^{\prime} = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$
Now,calculate the product $B^{\prime}A^{\prime}$:
$B^{\prime}A^{\prime} = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$
Since $(AB)^{\prime} = B^{\prime}A^{\prime}$,the property is verified.

(A) Given $A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
The transpose of matrix $A$,denoted by $A^{\prime}$,is obtained by interchanging its rows and columns:
$A^{\prime} = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.
Now,calculate the product $A^{\prime} A$:
$A^{\prime} A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
Performing matrix multiplication:
$A^{\prime} A = \begin{bmatrix} (\sin \alpha)(\sin \alpha) + (-\cos \alpha)(-\cos \alpha) & (\sin \alpha)(\cos \alpha) + (-\cos \alpha)(\sin \alpha) \\ (\cos \alpha)(\sin \alpha) + (\sin \alpha)(-\cos \alpha) & (\cos \alpha)(\cos \alpha) + (\sin \alpha)(\sin \alpha) \end{bmatrix}$.
Simplifying the terms using trigonometric identities $\sin^2 \alpha + \cos^2 \alpha = 1$:
$A^{\prime} A = \begin{bmatrix} \sin^2 \alpha + \cos^2 \alpha & \sin \alpha \cos \alpha - \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha - \sin \alpha \cos \alpha & \cos^2 \alpha + \sin^2 \alpha \end{bmatrix}$.
$A^{\prime} A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Thus,it is verified that $A^{\prime} A = I$.

Let $A = \left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right]$.
Then,the transpose $A^{\prime} = \left[\begin{array}{cc}3 & 1 \\ 5 & -1\end{array}\right]$.
Any square matrix $A$ can be written as $A = P + Q$,where $P = \frac{1}{2}(A + A^{\prime})$ is a symmetric matrix and $Q = \frac{1}{2}(A - A^{\prime})$ is a skew-symmetric matrix.
First,calculate $P = \frac{1}{2}(A + A^{\prime})$:
$A + A^{\prime} = \left[\begin{array}{cc}3+3 & 5+1 \\ 1+5 & -1-1\end{array}\right] = \left[\begin{array}{cc}6 & 6 \\ 6 & -2\end{array}\right]$
$P = \frac{1}{2} \left[\begin{array}{cc}6 & 6 \\ 6 & -2\end{array}\right] = \left[\begin{array}{cc}3 & 3 \\ 3 & -1\end{array}\right]$
Since $P^{\prime} = P$,$P$ is symmetric.
Next,calculate $Q = \frac{1}{2}(A - A^{\prime})$:
$A - A^{\prime} = \left[\begin{array}{cc}3-3 & 5-1 \\ 1-5 & -1-(-1)\end{array}\right] = \left[\begin{array}{cc}0 & 4 \\ -4 & 0\end{array}\right]$
$Q = \frac{1}{2} \left[\begin{array}{cc}0 & 4 \\ -4 & 0\end{array}\right] = \left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]$
Since $Q^{\prime} = -Q$,$Q$ is skew-symmetric.
Thus,$A = P + Q = \left[\begin{array}{cc}3 & 3 \\ 3 & -1\end{array}\right] + \left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]$.

Let $A = \left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]$. Then,the transpose $A^{\prime} = \left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]$.
To express $A$ as the sum of a symmetric and a skew-symmetric matrix,we use the formula $A = P + Q$,where $P = \frac{1}{2}(A + A^{\prime})$ is symmetric and $Q = \frac{1}{2}(A - A^{\prime})$ is skew-symmetric.
First,calculate $A + A^{\prime} = \left[\begin{array}{ccc}3+3 & 3-2 & -1-4 \\ -2+3 & -2-2 & 1-5 \\ -4-1 & -5+1 & 2+2\end{array}\right] = \left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]$.
Thus,$P = \frac{1}{2}(A + A^{\prime}) = \left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]$. Since $P^{\prime} = P$,$P$ is symmetric.
Next,calculate $A - A^{\prime} = \left[\begin{array}{ccc}3-3 & 3-(-2) & -1-(-4) \\ -2-3 & -2-(-2) & 1-(-5) \\ -4-(-1) & -5-1 & 2-2\end{array}\right] = \left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]$.
Thus,$Q = \frac{1}{2}(A - A^{\prime}) = \left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{array}\right]$. Since $Q^{\prime} = -Q$,$Q$ is skew-symmetric.
Therefore,$A = \left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right] + \left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{array}\right]$.

(C) Given that $A$ and $B$ are symmetric matrices,we have:
$A^{\prime} = A$ and $B^{\prime} = B$ .......... $(1)$
Consider the transpose of the matrix $(AB - BA)$:
$(AB - BA)^{\prime} = (AB)^{\prime} - (BA)^{\prime}$
Using the property $(XY)^{\prime} = Y^{\prime}X^{\prime}$,we get:
$(AB - BA)^{\prime} = B^{\prime}A^{\prime} - A^{\prime}B^{\prime}$
Substituting the values from $(1)$:
$(AB - BA)^{\prime} = BA - AB$
Factoring out the negative sign:
$(AB - BA)^{\prime} = -(AB - BA)$
Since the transpose of $(AB - BA)$ is equal to its negative,$(AB - BA)$ is a skew-symmetric matrix.

(A) We shall prove the result by using the principle of mathematical induction.
Let $P(n)$ be the statement: If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$,then $A^{n} = \begin{bmatrix} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{bmatrix}$ for $n \in N$.
Step $1$: For $n = 1$,$A^{1} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$. This is true.
Step $2$: Assume the result is true for $n = k$. That is,$A^{k} = \begin{bmatrix} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{bmatrix}$.
Step $3$: We prove the result for $n = k + 1$.
$A^{k+1} = A \cdot A^{k} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{bmatrix}$
$= \begin{bmatrix} \cos \theta \cos k \theta - \sin \theta \sin k \theta & \cos \theta \sin k \theta + \sin \theta \cos k \theta \\ -\sin \theta \cos k \theta - \cos \theta \sin k \theta & -\sin \theta \sin k \theta + \cos \theta \cos k \theta \end{bmatrix}$
Using trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$A^{k+1} = \begin{bmatrix} \cos(k+1)\theta & \sin(k+1)\theta \\ -\sin(k+1)\theta & \cos(k+1)\theta \end{bmatrix}$.
Thus,the result holds for $n = k+1$. By the principle of mathematical induction,the statement is true for all $n \in N$.

(N/A) It is given that $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$.
To prove: $P(n): A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$ for all $n \in \mathbb{N}$.
We shall prove the result by using the principle of mathematical induction.
For $n=1$,we have:
$P(1): A^1 = \begin{bmatrix} 1+2(1) & -4(1) \\ 1 & 1-2(1) \end{bmatrix} = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = A$.
Therefore,the result is true for $n=1$.
Assume the result is true for $n=k$:
$P(k): A^k = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix}$.
Now,we prove that the result is true for $n=k+1$:
$A^{k+1} = A^k \cdot A = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$.
$A^{k+1} = \begin{bmatrix} 3(1+2k) - 4k & -4(1+2k) + 4k \\ 3k + 1 - 2k & -4k - (1-2k) \end{bmatrix}$.
$A^{k+1} = \begin{bmatrix} 3 + 6k - 4k & -4 - 8k + 4k \\ k + 1 & -4k - 1 + 2k \end{bmatrix}$.
$A^{k+1} = \begin{bmatrix} 3 + 2k & -4 - 4k \\ k + 1 & -1 - 2k \end{bmatrix}$.
$A^{k+1} = \begin{bmatrix} 1 + 2(k+1) & -4(k+1) \\ k+1 & 1 - 2(k+1) \end{bmatrix}$.
Therefore,the result is true for $n=k+1$.
Thus,by the principle of mathematical induction,the result holds for all $n \in \mathbb{N}$.

(A) Given $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$.
The transpose $A^{\prime} = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}$.
Given $A^{\prime} A = I$,where $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Calculating $A^{\prime} A$:
$\begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix} \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Performing matrix multiplication:
$\begin{bmatrix} 0+x^2+x^2 & 0+xy-xy & 0-xz+xz \\ 0+xy-xy & 4y^2+y^2+y^2 & 2zy-zy-zy \\ 0-xz+xz & 2zy-zy-zy & z^2+z^2+z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Simplifying the matrix:
$\begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Equating corresponding elements:
$2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$
$6y^2 = 1 \implies y^2 = \frac{1}{6} \implies y = \pm \frac{1}{\sqrt{6}}$
$3z^2 = 1 \implies z^2 = \frac{1}{3} \implies z = \pm \frac{1}{\sqrt{3}}$
Thus,$x = \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{6}}, z = \pm \frac{1}{\sqrt{3}}$.

(A) We have $\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$.
First,multiply the first two matrices:
$\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right] = \left[\begin{array}{lll}1(1)+2(2)+1(1) & 1(2)+2(0)+1(0) & 1(0)+2(1)+1(2)\end{array}\right] = \left[\begin{array}{lll}6 & 2 & 4\end{array}\right]$.
Now,multiply the result by the third matrix:
$\left[\begin{array}{lll}6 & 2 & 4\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right] = 6(0) + 2(2) + 4(x) = 0 + 4 + 4x = 4 + 4x$.
Setting the result to $0$:
$4 + 4x = 0
\Rightarrow 4x = -4
\Rightarrow x = -1$.
Thus,the required value of $x$ is $-1$.

(A) Given $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.
First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.
$A^{2} = \begin{bmatrix} 3(3) + 1(-1) & 3(1) + 1(2) \\ -1(3) + 2(-1) & -1(1) + 2(2) \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$.
Now,evaluate the expression $L.H.S. = A^{2} - 5A + 7I$.
$L.H.S. = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
$L.H.S. = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$.
$L.H.S. = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5-(-5)+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$.
Thus,$A^{2} - 5A + 7I = 0$ is proved.

(A) Let the sales matrix be $A = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}$.
Let the unit sale price matrix be $S = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$ and unit cost price matrix be $C = \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix}$.
Profit per unit is $P = S - C = \begin{bmatrix} 2.50-2.00 \\ 1.50-1.00 \\ 1.00-0.50 \end{bmatrix} = \begin{bmatrix} 0.50 \\ 0.50 \\ 0.50 \end{bmatrix}$.
Total profit is given by the matrix product $A \times P$:
$\begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 0.50 \\ 0.50 \\ 0.50 \end{bmatrix} = \begin{bmatrix} 10000(0.50) + 2000(0.50) + 18000(0.50) \\ 6000(0.50) + 20000(0.50) + 8000(0.50) \end{bmatrix}$
$= \begin{bmatrix} 5000 + 1000 + 9000 \\ 3000 + 10000 + 4000 \end{bmatrix} = \begin{bmatrix} 15000 \\ 17000 \end{bmatrix}$.
Total gross profit = $15000 + 17000 = 32000$.

(N/A) Given that $A$ and $B$ are square matrices of the same order such that $AB = BA$.
Part $1$: To prove $P(n): AB^{n} = B^{n}A$ for all $n \in N$ by induction.
For $n = 1$,$AB^{1} = B^{1}A$,which is true as $AB = BA$ is given.
Assume the result is true for $n = k$,i.e.,$AB^{k} = B^{k}A$ $(1)$.
For $n = k + 1$,we have $AB^{k+1} = (AB^{k})B = (B^{k}A)B = B^{k}(AB) = B^{k}(BA) = (B^{k}B)A = B^{k+1}A$.
Thus,by the principle of mathematical induction,$AB^{n} = B^{n}A$ for all $n \in N$.
Part $2$: To prove $Q(n): (AB)^{n} = A^{n}B^{n}$ for all $n \in N$ by induction.
For $n = 1$,$(AB)^{1} = A^{1}B^{1} = AB$,which is true.
Assume the result is true for $n = k$,i.e.,$(AB)^{k} = A^{k}B^{k}$ $(2)$.
For $n = k + 1$,we have $(AB)^{k+1} = (AB)^{k}(AB) = (A^{k}B^{k})(AB) = A^{k}(B^{k}A)B$.
Using the result from Part $1$,$B^{k}A = AB^{k}$,so $(AB)^{k+1} = A^{k}(AB^{k})B = (A^{k}A)(B^{k}B) = A^{k+1}B^{k+1}$.
Thus,by the principle of mathematical induction,$(AB)^{n} = A^{n}B^{n}$ for all $n \in N$.

(B) Given: $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$.
Then,$A^{2} = A \cdot A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$.
Calculating the product:
$A^{2} = \begin{bmatrix} \alpha^{2} + \beta \gamma & \alpha \beta - \alpha \beta \\ \alpha \gamma - \alpha \gamma & \beta \gamma + \alpha^{2} \end{bmatrix} = \begin{bmatrix} \alpha^{2} + \beta \gamma & 0 \\ 0 & \alpha^{2} + \beta \gamma \end{bmatrix}$.
Given $A^{2} = I$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we have:
$\begin{bmatrix} \alpha^{2} + \beta \gamma & 0 \\ 0 & \alpha^{2} + \beta \gamma \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Comparing the corresponding elements:
$\alpha^{2} + \beta \gamma = 1$.
Rearranging the equation:
$1 - \alpha^{2} - \beta \gamma = 0$.
Thus,the correct option is $B$.

(A) Since it is a $2 \times 2$ matrix,it has $2$ rows and $2$ columns.
Let the matrix be $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$.
Given $a_{ij} = \frac{(i+j)^2}{2}$,we calculate each element:

$a_{11} = \frac{(1+1)^2}{2} = \frac{4}{2} = 2$	$a_{12} = \frac{(1+2)^2}{2} = \frac{9}{2}$
$a_{21} = \frac{(2+1)^2}{2} = \frac{9}{2}$	$a_{22} = \frac{(2+2)^2}{2} = \frac{16}{2} = 8$

Thus,the required matrix $A$ is $A = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}$.

(A) $2 \times 2$ matrix has $2$ rows and $2$ columns.
Let the matrix be $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$.
Given $a_{ij} = \frac{i}{j}$,we calculate each element:

$a_{11} = \frac{1}{1} = 1$	$a_{12} = \frac{1}{2}$
$a_{21} = \frac{2}{1} = 2$	$a_{22} = \frac{2}{2} = 1$

Thus,the required matrix is $A = \begin{bmatrix} 1 & \frac{1}{2} \\ 2 & 1 \end{bmatrix}$.

(N/A) Since it is a $2 \times 2$ matrix,it has $2$ rows and $2$ columns. Let the matrix be $A$.
Where $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$.
Now,it is given that $a_{ij} = \frac{(i + 2j)^2}{2}$.

Element	Calculation
$a_{11}$	$a_{11} = \frac{(1 + 2(1))^2}{2} = \frac{(1 + 2)^2}{2} = \frac{3^2}{2} = \frac{9}{2}$
$a_{12}$	$a_{12} = \frac{(1 + 2(2))^2}{2} = \frac{(1 + 4)^2}{2} = \frac{5^2}{2} = \frac{25}{2}$
$a_{21}$	$a_{21} = \frac{(2 + 2(1))^2}{2} = \frac{(2 + 2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8$
$a_{22}$	$a_{22} = \frac{(2 + 2(2))^2}{2} = \frac{(2 + 4)^2}{2} = \frac{6^2}{2} = \frac{36}{2} = 18$

Hence,the required matrix $A$ is:
$A = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}$

(D) Given $A$ is a $2 \times 2$ matrix with entries from $\{0, 1\}$ and $|A| \neq 0$.
For statement $(P)$: If $A \neq I_{2}$,then $|A| = -1$.
Consider $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. Here $A \neq I_{2}$ and $|A| = (1)(1) - (1)(0) = 1$.
Since we found a case where $A \neq I_{2}$ but $|A| = 1$,statement $(P)$ is false.
For statement $(Q)$: If $|A| = 1$,then $\operatorname{tr}(A) = 2$.
The possible $2 \times 2$ matrices with entries from $\{0, 1\}$ such that $|A| = 1$ are:
$A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \implies \operatorname{tr}(A_1) = 1+1 = 2$
$A_2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \implies \operatorname{tr}(A_2) = 1+1 = 2$
$A_3 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \implies \operatorname{tr}(A_3) = 1+1 = 2$
In all cases where $|A| = 1$,the trace is $2$. Thus,statement $(Q)$ is true.
Therefore,$(P)$ is false and $(Q)$ is true.

(A) Given $A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$.
First,calculate $A^{2}$:
$A^{2} = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} x^{2} + 1 & x \\ x & 1 \end{bmatrix}$.
Next,calculate $A^{4} = A^{2} \times A^{2}$:
$A^{4} = \begin{bmatrix} x^{2} + 1 & x \\ x & 1 \end{bmatrix} \begin{bmatrix} x^{2} + 1 & x \\ x & 1 \end{bmatrix} = \begin{bmatrix} (x^{2} + 1)^{2} + x^{2} & x(x^{2} + 1) + x \\ x(x^{2} + 1) + x & x^{2} + 1 \end{bmatrix}$.
We are given $a_{11} = 109$,so:
$(x^{2} + 1)^{2} + x^{2} = 109$.
Let $y = x^{2}$. Then $(y + 1)^{2} + y = 109$.
$y^{2} + 2y + 1 + y = 109 \Rightarrow y^{2} + 3y - 108 = 0$.
$(y + 12)(y - 9) = 0$.
Since $y = x^{2} \geq 0$,we have $y = 9$,so $x^{2} = 9$.
Now,find $a_{22}$:
From the matrix $A^{4}$,$a_{22} = x^{2} + 1$.
Substituting $x^{2} = 9$,we get $a_{22} = 9 + 1 = 10$.

(B) Let $t = \tan(\frac{\theta}{2})$. Then $A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}$.
$I_{2} + A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$ and $I_{2} - A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$.
$(I_{2} - A)^{-1} = \frac{1}{1 + t^{2}} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$.
$(I_{2} + A)(I_{2} - A)^{-1} = \frac{1}{1 + t^{2}} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = \frac{1}{1 + t^{2}} \begin{bmatrix} 1 - t^{2} & -2t \\ 2t & 1 - t^{2} \end{bmatrix} = \begin{bmatrix} \frac{1 - t^{2}}{1 + t^{2}} & -\frac{2t}{1 + t^{2}} \\ \frac{2t}{1 + t^{2}} & \frac{1 - t^{2}}{1 + t^{2}} \end{bmatrix}$.
Comparing with $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$,we get $a = \frac{1 - t^{2}}{1 + t^{2}} = \cos \theta$ and $b = \frac{2t}{1 + t^{2}} = \sin \theta$.
Thus,$a^{2} + b^{2} = \cos^{2} \theta + \sin^{2} \theta = 1$.
Therefore,$13(a^{2} + b^{2}) = 13(1) = 13$.

(A) Let $A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$,where $a, b, c \in \mathbb{Z}$.
Then $A^{2} = \begin{bmatrix} a & b \\ b & c \end{bmatrix} \begin{bmatrix} a & b \\ b & c \end{bmatrix} = \begin{bmatrix} a^{2} + b^{2} & ab + bc \\ ab + bc & b^{2} + c^{2} \end{bmatrix}$.
The sum of the diagonal elements of $A^{2}$ is given by $\text{tr}(A^{2}) = a^{2} + b^{2} + b^{2} + c^{2} = a^{2} + 2b^{2} + c^{2}$.
We are given that $a^{2} + 2b^{2} + c^{2} = 1$,where $a, b, c \in \mathbb{Z}$.
Since $a^{2}, b^{2}, c^{2} \ge 0$,for the sum to be $1$,we must have $b^{2} = 0$,which implies $b = 0$.
Then the equation reduces to $a^{2} + c^{2} = 1$.
Since $a, c \in \mathbb{Z}$,the possible integer solutions for $(a, c)$ are:
$1$. If $a = 0$,then $c^{2} = 1 \Rightarrow c = \pm 1$. This gives two matrices: $\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$ and $\begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix}$.
$2$. If $c = 0$,then $a^{2} = 1 \Rightarrow a = \pm 1$. This gives two matrices: $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix}$.
Thus,the total number of such matrices is $2 + 2 = 4$.

(B) Given $A = I + C$,where $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix}$.
Since $I$ and $C$ commute,we use the binomial expansion $(I+C)^n = I + nC + \frac{n(n-1)}{2}C^2$.
Note that $C^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and $C^3 = O$.
Thus,$A^n = (I+C)^n = I + nC + \frac{n(n-1)}{2}C^2$.
The element $b_{13}$ is the entry in the first row and third column of $B = 7A^{20} - 20A^7 + 2I$.
The $(1,3)$ entry of $I$ is $0$,the $(1,3)$ entry of $C$ is $0$,and the $(1,3)$ entry of $C^2$ is $1$.
Therefore,the $(1,3)$ entry of $A^n$ is $\frac{n(n-1)}{2}$.
$b_{13} = 7 \times \left( \frac{20 \times 19}{2} \right) - 20 \times \left( \frac{7 \times 6}{2} \right) + 2(0)$.
$b_{13} = 7 \times 190 - 20 \times 21 = 1330 - 420 = 910$.

(C) Given $A = [a_{ij}]$ is a $3 \times 3$ matrix such that the sum of elements in each row is $1$.
Let $X = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$.
Then $AX = \begin{bmatrix} a_{11} + a_{12} + a_{13} \\ a_{21} + a_{22} + a_{23} \\ a_{31} + a_{32} + a_{33} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = X$.
Thus,$AX = X$.
Multiplying both sides by $A$,we get $A^2X = A(AX) = AX = X$.
Multiplying by $A$ again,we get $A^3X = A(A^2X) = AX = X$.
Let $A^3 = [b_{ij}]$. Then $A^3X = \begin{bmatrix} b_{11} + b_{12} + b_{13} \\ b_{21} + b_{22} + b_{23} \\ b_{31} + b_{32} + b_{33} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$.
The sum of all entries of $A^3$ is $(b_{11} + b_{12} + b_{13}) + (b_{21} + b_{22} + b_{23}) + (b_{31} + b_{32} + b_{33}) = 1 + 1 + 1 = 3$.

(B) Let matrix $B = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$.
Given $AB = BA$,we have:
$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Performing matrix multiplication:
$\begin{bmatrix} d & e & f \\ a & b & c \\ g & h & i \end{bmatrix} = \begin{bmatrix} b & a & c \\ e & d & f \\ h & g & i \end{bmatrix}$
Comparing the corresponding elements:
$d = b, e = a, f = c, g = h$
Thus,matrix $B$ takes the form:
$B = \begin{bmatrix} a & b & c \\ b & a & c \\ g & g & i \end{bmatrix}$
Since each entry $a, b, c, g, i$ can be chosen from the set $\{1, 2, 3, 4, 5\}$,there are $5$ choices for each of the $5$ independent variables.
Total number of matrices $B = 5 \times 5 \times 5 \times 5 \times 5 = 5^5 = 3125$.

(B) Given $P = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix}$.
Calculate $P^2 = P \times P = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1/2 + 1/2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$.
Calculate $P^3 = P^2 \times P = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 + 1/2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3/2 & 1 \end{bmatrix}$.
Calculate $P^4 = P^2 \times P^2 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 + 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$.
Observing the pattern,$P^n = \begin{bmatrix} 1 & 0 \\ n/2 & 1 \end{bmatrix}$.
Therefore,for $n = 50$,$P^{50} = \begin{bmatrix} 1 & 0 \\ 50/2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 25 & 1 \end{bmatrix}$.

(A) First,calculate $A^2$:
$A^2 = \begin{bmatrix} 2 & -2 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4-2 & -4+2 \\ 2-1 & -2+1 \end{bmatrix} = \begin{bmatrix} 2 & -2 \\ 1 & -1 \end{bmatrix} = A$.
Since $A^2 = A$,it follows that $A^n = A$ for all $n \geq 1$.
Next,calculate $B^2$:
$B^2 = \begin{bmatrix} -1 & 2 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} -1 & 2 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1-2 & -2+4 \\ 1-2 & -2+4 \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ -1 & 2 \end{bmatrix} = B$.
Since $B^2 = B$,it follows that $B^m = B$ for all $m \geq 1$.
The given equation $nA^n + mB^m = I$ becomes $nA + mB = I$.
Substituting the matrices:
$n \begin{bmatrix} 2 & -2 \\ 1 & -1 \end{bmatrix} + m \begin{bmatrix} -1 & 2 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This results in the system of equations:
$2n - m = 1$
$-2n + 2m = 0 \implies n = m$.
Substituting $n = m$ into $2n - m = 1$,we get $2n - n = 1$,so $n = 1$.
Thus,$n = 1$ and $m = 1$.
The only pair $(n, m)$ is $(1, 1)$,so there is $1$ element in the set.

(B) Let the number of $1$s be $x$,the number of $-1$s be $y$,and the number of $0$s be $z$. Since the matrix is $3 \times 3$,we have $x + y + z = 9$.
The sum of entries is $1(x) + (-1)(y) + 0(z) = 5$,which simplifies to $x - y = 5$,or $x = y + 5$.
Substituting $x$ in the first equation: $(y + 5) + y + z = 9 \implies 2y + z = 4$.
We analyze the possible non-negative integer solutions for $(x, y, z)$:
Case-$I$: If $y = 0$,then $z = 4$ and $x = 5$. Number of matrices $= \frac{9!}{5!0!4!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.
Case-$II$: If $y = 1$,then $z = 2$ and $x = 6$. Number of matrices $= \frac{9!}{6!1!2!} = \frac{9 \times 8 \times 7}{2 \times 1} = 252$.
Case-$III$: If $y = 2$,then $z = 0$ and $x = 7$. Number of matrices $= \frac{9!}{7!2!0!} = \frac{9 \times 8}{2 \times 1} = 36$.
Total number of matrices $= 126 + 252 + 36 = 414$.

(A) Given $A = [a_{ij}]$ of order $3 \times 3$ with $a_{ij} = 2^{j-i}$.
$A = \begin{bmatrix} 2^{1-1} & 2^{2-1} & 2^{3-1} \\ 2^{1-2} & 2^{2-2} & 2^{3-2} \\ 2^{1-3} & 2^{2-3} & 2^{3-3} \end{bmatrix} = \begin{bmatrix} 1 & 2 & 4 \\ 1/2 & 1 & 2 \\ 1/4 & 1/2 & 1 \end{bmatrix}$.
Calculate $A^2$:
$A^2 = \begin{bmatrix} 1 & 2 & 4 \\ 1/2 & 1 & 2 \\ 1/4 & 1/2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 4 \\ 1/2 & 1 & 2 \\ 1/4 & 1/2 & 1 \end{bmatrix} = \begin{bmatrix} 1+1+1 & 2+2+2 & 4+4+4 \\ 1/2+1/2+1/2 & 1+1+1 & 2+2+2 \\ 1/4+1/4+1/4 & 1/2+1/2+1/2 & 1+1+1 \end{bmatrix} = \begin{bmatrix} 3 & 6 & 12 \\ 3/2 & 3 & 6 \\ 3/4 & 3/2 & 3 \end{bmatrix} = 3A$.
Since $A^2 = 3A$,we have $A^3 = A^2 \cdot A = (3A) \cdot A = 3A^2 = 3(3A) = 3^2 A$.
By induction,$A^n = 3^{n-1} A$.
We need to find $S = A^2 + A^3 + \ldots + A^{10}$.
$S = 3A + 3^2 A + \ldots + 3^9 A = (3 + 3^2 + \ldots + 3^9) A$.
This is a geometric progression with $n=9$ terms,first term $a=3$,and common ratio $r=3$.
Sum $= \frac{a(r^n - 1)}{r - 1} = \frac{3(3^9 - 1)}{3 - 1} = \frac{3^{10} - 3}{2}$.
Thus,$S = \left(\frac{3^{10} - 3}{2}\right) A$.

(D) Given $A = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$,so $A^{\prime} = \begin{bmatrix} 1 & 1 & 1 \end{bmatrix}$.
We need to calculate $A^{\prime} BA = \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$.
First,compute $A^{\prime} B = \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2 \end{bmatrix} = \begin{bmatrix} 9^2+12^2-15^2 & -10^2+13^2+16^2 & 11^2-14^2+17^2 \end{bmatrix}$.
Calculating the values:
$9^2+12^2-15^2 = 81+144-225 = 0$.
$-10^2+13^2+16^2 = -100+169+256 = 325$.
$11^2-14^2+17^2 = 121-196+289 = 214$.
So,$A^{\prime} B = \begin{bmatrix} 0 & 325 & 214 \end{bmatrix}$.
Now,$(A^{\prime} B) A = \begin{bmatrix} 0 & 325 & 214 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 0(1) + 325(1) + 214(1) = 539$.

(C) Given that $A^{T} = A$ and $B^{T} = -B$.
For option $A$: Let $C = A^{4} - B^{4}$.
$C^{T} = (A^{4} - B^{4})^{T} = (A^{T})^{4} - (B^{T})^{4} = A^{4} - (-B)^{4} = A^{4} - B^{4} = C$. Thus,$A^{4} - B^{4}$ is symmetric.
For option $B$: Let $C = AB - BA$.
$C^{T} = (AB - BA)^{T} = (AB)^{T} - (BA)^{T} = B^{T}A^{T} - A^{T}B^{T} = (-B)(A) - (A)(-B) = -BA + AB = AB - BA = C$. Thus,$AB - BA$ is symmetric.
For option $C$: Let $C = B^{5} - A^{5}$.
$C^{T} = (B^{5} - A^{5})^{T} = (B^{T})^{5} - (A^{T})^{5} = (-B)^{5} - A^{5} = -B^{5} - A^{5} = -(B^{5} + A^{5})$. This is not necessarily equal to $-C = -(B^{5} - A^{5})$. Thus,this statement is not true.
For option $D$: Let $C = AB + BA$.
$C^{T} = (AB + BA)^{T} = (AB)^{T} + (BA)^{T} = B^{T}A^{T} + A^{T}B^{T} = (-B)(A) + (A)(-B) = -BA - AB = -(AB + BA) = -C$. Thus,$AB + BA$ is skew-symmetric.
Therefore,option $C$ is not true.

(C) Let $A = \left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]$.
$1$. For option $A$: Applying $R_1 \rightarrow R_1 + R_2$,we get $\left[\begin{array}{cc}-1+1 & 2-1 \\ 1 & -1\end{array}\right] = \left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]$. This is possible.
$2$. For option $B$: Applying $R_1 \leftrightarrow R_2$,we get $\left[\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right]$. This is possible.
$3$. For option $C$: To get $\left[\begin{array}{cc}-1 & 2 \\ -2 & 7\end{array}\right]$,we would need $R_2 \rightarrow R_2 + k R_1$. Here,$-1 + k(-1) = -2 \implies k=1$,but then $2 + k(2) = 2 + 1(2) = 4 \neq 7$. Thus,this is not possible.
$4$. For option $D$: Applying $R_2 \rightarrow R_2 + 2R_1$,we get $\left[\begin{array}{cc}-1 & 2 \\ 1+2(-1) & -1+2(2)\end{array}\right] = \left[\begin{array}{cc}-1 & 2 \\ -1 & 3\end{array}\right]$. This is possible.
Therefore,the matrix in option $C$ cannot be obtained by a single elementary row operation.

(A) matrix $A$ of order $3 \times 3$ has $9$ entries,where each entry $a_{ij} \in \{0, 1\}$.
The sum of all entries $S = \sum_{i=1}^{3} \sum_{j=1}^{3} a_{ij}$ can range from $0$ to $9$.
Since the sum must be a prime number,the possible values for the sum are $2, 3, 5, 7$ (as $0$ and $1$ are not prime,and $9$ is not prime).
The number of ways to choose $k$ entries to be $1$ (and the rest $0$) is given by the combination formula $\binom{9}{k}$.
Total number of matrices = $\binom{9}{2} + \binom{9}{3} + \binom{9}{5} + \binom{9}{7}$.
Calculating each term:
$\binom{9}{2} = \frac{9 \times 8}{2} = 36$
$\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
$\binom{9}{5} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$
$\binom{9}{7} = \binom{9}{2} = 36$
Total = $36 + 84 + 126 + 36 = 282$.

(A) Given $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $A^2 = A$.
Calculating $A^2$:
$A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{bmatrix}$.
Since $A^2 = A$,we have:
$\begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
Comparing the elements,we get $ab + bd = b$,which implies $b(a + d) = b$.
Since the entries are non-zero,$b \neq 0$,so we can divide by $b$ to get $a + d = 1$.

(B) square matrix of order $5$ where each row sum is $1$ and each column sum is $1$ with entries from $\{0, 1\}$ is a permutation matrix.
In the first row,there are $5$ possible positions to place the $1$.
In the second row,there are $4$ remaining positions available to place the $1$ (since the column used in the first row cannot be used again).
In the third row,there are $3$ remaining positions.
In the fourth row,there are $2$ remaining positions.
In the fifth row,there is only $1$ remaining position.
Therefore,the total number of such matrices is $5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$.

(C) Given $A = \begin{bmatrix} \cos 60^{\circ} & \sin 60^{\circ} \\ -\sin 60^{\circ} & \cos 60^{\circ} \end{bmatrix}$.
Let $\alpha = 60^{\circ} = \frac{\pi}{3}$.
Then $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$.
By the property of rotation matrices,$A^n = \begin{bmatrix} \cos(n\alpha) & \sin(n\alpha) \\ -\sin(n\alpha) & \cos(n\alpha) \end{bmatrix}$.
For $A^{30}$,$n = 30$,so $n\alpha = 30 \times \frac{\pi}{3} = 10\pi$.
$A^{30} = \begin{bmatrix} \cos(10\pi) & \sin(10\pi) \\ -\sin(10\pi) & \cos(10\pi) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
For $A^{25}$,$n = 25$,so $n\alpha = 25 \times \frac{\pi}{3} = 8\pi + \frac{\pi}{3}$.
$A^{25} = \begin{bmatrix} \cos(8\pi + \frac{\pi}{3}) & \sin(8\pi + \frac{\pi}{3}) \\ -\sin(8\pi + \frac{\pi}{3}) & \cos(8\pi + \frac{\pi}{3}) \end{bmatrix} = \begin{bmatrix} \cos(\frac{\pi}{3}) & \sin(\frac{\pi}{3}) \\ -\sin(\frac{\pi}{3}) & \cos(\frac{\pi}{3}) \end{bmatrix} = A$.
Thus,$A^{30} = I$ and $A^{25} = A$.
Checking the options: $A^{30} + A^{25} - A = I + A - A = I$. Therefore,option $C$ is correct.

(D) symmetric matrix $A$ of order $3 \times 3$ is defined as $A = A^T$.
For a $3 \times 3$ matrix,it takes the form:
$A = \begin{bmatrix} a & b & c \\ b & d & e \\ c & e & f \end{bmatrix}$
Here,the independent entries are $a, b, c, d, e,$ and $f$.
There are $6$ independent positions in the matrix that can be filled.
Each of these $6$ positions can be filled by any of the $10$ digits from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
Therefore,the total number of such symmetric matrices is $10 \times 10 \times 10 \times 10 \times 10 \times 10 = 10^6$.

(A) Given $A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$.
Calculate the characteristic equation of $A$: $|A - \lambda I| = 0$.
$\begin{vmatrix} 2-\lambda & -1 \\ 1 & 1-\lambda \end{vmatrix} = (2-\lambda)(1-\lambda) + 1 = \lambda^2 - 3\lambda + 3 = 0$.
By Cayley-Hamilton theorem,$A^2 - 3A + 3I = 0$,so $A^2 = 3A - 3I$.
We observe $A^6 = -27I = -3^3 I$.
Then $A^{12} = (A^6)^2 = (-3^3 I)^2 = 3^6 I$.
$A^{13} = A^{12} \cdot A = 3^6 A = 3^6 \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 \cdot 3^6 & -3^6 \\ 3^6 & 3^6 \end{bmatrix}$.
The sum of the diagonal elements (trace) of $A^{13}$ is $2 \cdot 3^6 + 3^6 = 3 \cdot 3^6 = 3^7$.
Given the sum is $3^n$,we have $3^n = 3^7$,which implies $n = 7$.

(A) Let $M = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$.
From the first condition,$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} b \\ e \\ h \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$. Thus,$b = -1, e = 2, h = 3$.
Now,$M = \begin{bmatrix} a & -1 & c \\ d & 2 & f \\ g & 3 & i \end{bmatrix}$.
From the second condition,$\begin{bmatrix} a & -1 & c \\ d & 2 & f \\ g & 3 & i \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} a+1 \\ d-2 \\ g-3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}$.
Solving these,we get $a+1 = 1 \Rightarrow a = 0$,$d-2 = 1 \Rightarrow d = 3$,and $g-3 = -1 \Rightarrow g = 2$.
Now,$M = \begin{bmatrix} 0 & -1 & c \\ 3 & 2 & f \\ 2 & 3 & i \end{bmatrix}$.
From the third condition,$\begin{bmatrix} 0 & -1 & c \\ 3 & 2 & f \\ 2 & 3 & i \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -1+c \\ 5+f \\ 5+i \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix}$.
From the third row,$5+i = 12 \Rightarrow i = 7$.
The sum of the diagonal entries is $a + e + i = 0 + 2 + 7 = 9$.

(A) Given $M = \begin{bmatrix} \frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2} \end{bmatrix}$.
We can write $M$ as $M = I + \frac{3}{2} A$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 1-1 & 1-1 \\ -1+1 & -1+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
Since $A^2 = O$,we can use the binomial expansion for $(I + \frac{3}{2} A)^n$:
$M^n = (I + \frac{3}{2} A)^n = I^n + n(I^{n-1})(\frac{3}{2} A) + \frac{n(n-1)}{2} I^{n-2} (\frac{3}{2} A)^2 + \dots$
Since $A^2 = O$,all terms involving $A^k$ for $k \ge 2$ are zero.
Thus,$M^n = I + n \cdot \frac{3}{2} A$.
For $n = 2022$:
$M^{2022} = I + 2022 \cdot \frac{3}{2} A = I + 3033 A$.
$M^{2022} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 3033 \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 1+3033 & 0+3033 \\ 0-3033 & 1-3033 \end{bmatrix} = \begin{bmatrix} 3034 & 3033 \\ -3033 & -3032 \end{bmatrix}$.

(A) Let $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$.
From $A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$,we get the second column of $A$ as $\begin{bmatrix} a_{12} \\ a_{22} \\ a_{32} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$. Thus,$a_{12} = 0, a_{22} = 0, a_{32} = 1$.
From $A \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$,the second row equation is $4a_{21} + a_{22} + 3a_{23} = 1$. Since $a_{22} = 0$,we have $4a_{21} + 3a_{23} = 1$ (Equation $1$).
From $A \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$,the second row equation is $2a_{21} + a_{22} + 2a_{23} = 0$. Since $a_{22} = 0$,we have $2a_{21} + 2a_{23} = 0$,which implies $a_{21} = -a_{23}$ (Equation $2$).
Substituting Equation $2$ into Equation $1$: $4(-a_{23}) + 3a_{23} = 1 \Rightarrow -4a_{23} + 3a_{23} = 1 \Rightarrow -a_{23} = 1 \Rightarrow a_{23} = -1$.

(A) Given $A^n = A^{n-2} + A^2 - I$. For $n=50$,we have $A^{50} = A^{48} + A^2 - I$.
By repeating this recurrence relation,we get $A^{50} = A^{48} + (A^2 - I) = A^{46} + 2(A^2 - I) = A^{44} + 3(A^2 - I) = \dots = A^2 + 24(A^2 - I) = 25A^2 - 24I$.
First,calculate $A^2$: $A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$.
Now,$A^{50} = 25 \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} - 24 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 25-24 & 0 & 0 \\ 25 & 25-24 & 0 \\ 25 & 0 & 25-24 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$.
The sum of all elements is $1 + 0 + 0 + 25 + 1 + 0 + 25 + 0 + 1 = 53$.

(C) First,we calculate $A^{2}$:
$A^{2} = A \times A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} (1)(1) + (0)(-1) & (1)(0) + (0)(7) \\ (-1)(1) + (7)(-1) & (-1)(0) + (7)(7) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -8 & 49 \end{bmatrix}$
Given the equation $A^{2} = 8A + kI$,we substitute the matrices:
$\begin{bmatrix} 1 & 0 \\ -8 & 49 \end{bmatrix} = 8 \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} + k \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 1 & 0 \\ -8 & 49 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ -8 & 56 \end{bmatrix} + \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$
$\begin{bmatrix} 1 & 0 \\ -8 & 49 \end{bmatrix} = \begin{bmatrix} 8 + k & 0 \\ -8 & 56 + k \end{bmatrix}$
By comparing the corresponding elements,we get:
$8 + k = 1 \Rightarrow k = 1 - 8 = -7$
Also,$56 + k = 49 \Rightarrow k = 49 - 56 = -7$
Thus,the value of $k$ is $-7$.

(A) We know that $A^4 A^{-1} = A^{4-1} = A^3$.
Since $A$ is a diagonal matrix,$A^n = \begin{bmatrix} a_{11}^n & 0 & 0 \\ 0 & a_{22}^n & 0 \\ 0 & 0 & a_{33}^n \end{bmatrix}$.
Given $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -1 \end{bmatrix}$,we calculate $A^3$:
$A^3 = \begin{bmatrix} 2^3 & 0 & 0 \\ 0 & (-2)^3 & 0 \\ 0 & 0 & (-1)^3 \end{bmatrix} = \begin{bmatrix} 8 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -1 \end{bmatrix}$.
Thus,$A^4 A^{-1} = \begin{bmatrix} 8 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -1 \end{bmatrix}$.

(C) Given that $A^2 = A$ (idempotent matrix).
We need to evaluate $(I + A)^2 - 3A$.
Expanding the expression: $(I + A)^2 = I^2 + IA + AI + A^2 = I + A + A + A^2 = I + 2A + A^2$.
Since $A^2 = A$,we substitute it into the expression: $I + 2A + A = I + 3A$.
Now,subtract $3A$: $(I + 3A) - 3A = I$.
Therefore,the correct option is $C$.

(D) Given the matrix $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$.
First,we calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$
Performing matrix multiplication:
Row $1$: $(0)(0) + (0)(0) + (-1)(-1) = 1$,$(0)(0) + (0)(-1) + (-1)(0) = 0$,$(0)(-1) + (0)(0) + (-1)(0) = 0$
Row $2$: $(0)(0) + (-1)(0) + (0)(-1) = 0$,$(0)(0) + (-1)(-1) + (0)(0) = 1$,$(0)(-1) + (-1)(0) + (0)(0) = 0$
Row $3$: $(-1)(0) + (0)(0) + (0)(-1) = 0$,$(-1)(0) + (0)(-1) + (0)(0) = 0$,$(-1)(-1) + (0)(0) + (0)(0) = 1$
Thus,$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Now,we calculate $I + A^2 = I + I = 2I$.

(A) Given $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.
Then,the transpose $A^{\prime} = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
Given the condition $A+A^{\prime} = I$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Adding $A$ and $A^{\prime}$:
$A+A^{\prime} = \begin{bmatrix} \sin \alpha + \sin \alpha & -\cos \alpha + \cos \alpha \\ \cos \alpha - \cos \alpha & \sin \alpha + \sin \alpha \end{bmatrix} = \begin{bmatrix} 2\sin \alpha & 0 \\ 0 & 2\sin \alpha \end{bmatrix}$.
Equating this to the identity matrix $I$:
$\begin{bmatrix} 2\sin \alpha & 0 \\ 0 & 2\sin \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This implies $2\sin \alpha = 1$,so $\sin \alpha = \frac{1}{2}$.
Since $\sin^2 \alpha + \cos^2 \alpha = 1$,we have $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$.
Therefore,$\cos \alpha = \pm \frac{\sqrt{3}}{2}$.
Looking at the options,$\frac{\sqrt{3}}{2}$ is provided.

(A) Given $A = \begin{bmatrix} 0 & 0 & -5 \\ 0 & -5 & 0 \\ -5 & 0 & 0 \end{bmatrix}$.
We need to calculate $A^2 = A \times A$.
$A^2 = \begin{bmatrix} 0 & 0 & -5 \\ 0 & -5 & 0 \\ -5 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -5 \\ 0 & -5 & 0 \\ -5 & 0 & 0 \end{bmatrix}$.
Performing matrix multiplication:
Row $1$: $(0)(0) + (0)(0) + (-5)(-5) = 25$,$(0)(0) + (0)(-5) + (-5)(0) = 0$,$(0)(-5) + (0)(0) + (-5)(0) = 0$.
Row $2$: $(0)(0) + (-5)(0) + (0)(-5) = 0$,$(0)(0) + (-5)(-5) + (0)(0) = 25$,$(0)(-5) + (-5)(0) + (0)(0) = 0$.
Row $3$: $(-5)(0) + (0)(0) + (0)(-5) = 0$,$(-5)(0) + (0)(-5) + (0)(0) = 0$,$(-5)(-5) + (0)(0) + (0)(0) = 25$.
Thus,$A^2 = \begin{bmatrix} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{bmatrix} = 25 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 25 I$.
Therefore,the correct option is $A$.

(D) Given $A = [2]$ and $B = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$.
First,calculate the product $BA$:
$BA = \begin{bmatrix} 3 \\ 4 \end{bmatrix} [2] = \begin{bmatrix} 3 \times 2 \\ 4 \times 2 \end{bmatrix} = \begin{bmatrix} 6 \\ 8 \end{bmatrix}$.
Now,find the transpose $(BA)'$:
$(BA)' = \begin{bmatrix} 6 & 8 \end{bmatrix}$.
Note: Based on the provided options,there seems to be a typo in the original question's matrix $A$. If $A = [2]$,the result is $[6 \quad 8]$. If the question intended $B = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$ and $A = [1 \quad 2]$,then $BA = \begin{bmatrix} 3 \\ 4 \end{bmatrix} [1 \quad 2] = \begin{bmatrix} 3 & 6 \\ 4 & 8 \end{bmatrix}$.
Then $(BA)' = \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}$,which matches option $D$.

(A) Let $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} x & y \\ 0 & x \end{bmatrix}$.
First,calculate $AB$:
$AB = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x & y \\ 0 & x \end{bmatrix} = \begin{bmatrix} 1(x) + 1(0) & 1(y) + 1(x) \\ 0(x) + 1(0) & 0(y) + 1(x) \end{bmatrix} = \begin{bmatrix} x & x+y \\ 0 & x \end{bmatrix}$.
Next,calculate $BA$:
$BA = \begin{bmatrix} x & y \\ 0 & x \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} x(1) + y(0) & x(1) + y(1) \\ 0(1) + x(0) & 0(1) + x(1) \end{bmatrix} = \begin{bmatrix} x & x+y \\ 0 & x \end{bmatrix}$.
Since $AB = BA$ for any $x, y$,the matrix $B$ must be of the form $\begin{bmatrix} x & y \\ 0 & x \end{bmatrix}$.
Thus,option $A$ is the correct choice.