Types of matrices, Algebra of matrices Questions in English

(A) Given that $A$ is a $3 \times 4$ matrix.
Then,the transpose matrix $A^{\prime}$ is a $4 \times 3$ matrix.
For the product $A^{\prime}B$ to be defined,the number of columns in $A^{\prime}$ must equal the number of rows in $B$. Since $A^{\prime}$ is $4 \times 3$,$B$ must have $3$ rows.
For the product $BA^{\prime}$ to be defined,the number of columns in $B$ must equal the number of rows in $A^{\prime}$. Since $A^{\prime}$ has $4$ rows,$B$ must have $4$ columns.
Therefore,$B$ must be a $3 \times 4$ matrix.

(B) The matrix is a $3 \times 3$ matrix given by $A = [a_{ij}]$.
We calculate each element $a_{ij} = \frac{1}{3}|i - 5j|$:
$a_{11} = \frac{1}{3}|1 - 5(1)| = \frac{1}{3}|-4| = \frac{4}{3}$
$a_{12} = \frac{1}{3}|1 - 5(2)| = \frac{1}{3}|-9| = 3$
$a_{13} = \frac{1}{3}|1 - 5(3)| = \frac{1}{3}|-14| = \frac{14}{3}$
$a_{21} = \frac{1}{3}|2 - 5(1)| = \frac{1}{3}|-3| = 1$
$a_{22} = \frac{1}{3}|2 - 5(2)| = \frac{1}{3}|-8| = \frac{8}{3}$
$a_{23} = \frac{1}{3}|2 - 5(3)| = \frac{1}{3}|-13| = \frac{13}{3}$
$a_{31} = \frac{1}{3}|3 - 5(1)| = \frac{1}{3}|-2| = \frac{2}{3}$
$a_{32} = \frac{1}{3}|3 - 5(2)| = \frac{1}{3}|-7| = \frac{7}{3}$
$a_{33} = \frac{1}{3}|3 - 5(3)| = \frac{1}{3}|-12| = 4$
Thus,the matrix is $\left[\begin{array}{ccc}\frac{4}{3} & 3 & \frac{14}{3} \\ 1 & \frac{8}{3} & \frac{13}{3} \\ \frac{2}{3} & \frac{7}{3} & 4\end{array}\right]$.

(C) square matrix is defined as a matrix in which the number of rows is equal to the number of columns $(m = n)$.
$(a)$ $[1]$ is a $1 \times 1$ matrix,which is a square matrix.
$(b)$ $\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ is a $2 \times 2$ matrix,which is a square matrix.
$(c)$ $\begin{bmatrix} 3 & 3 & 3 \end{bmatrix}$ is a $1 \times 3$ matrix. Since the number of rows $(1)$ is not equal to the number of columns $(3)$,it is not a square matrix.
$(d)$ $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ is a $3 \times 3$ matrix,which is a square matrix.
Therefore,the correct option is $(c)$.

(B) Let $A$ and $B$ be two lower triangular matrices of order $n \times n$.
$A$ matrix is lower triangular if all entries above the main diagonal are zero,i.e.,$a_{ij} = 0$ for all $i < j$.
Let $C = A + B$. The entries of $C$ are given by $c_{ij} = a_{ij} + b_{ij}$.
For any $i < j$,since $A$ and $B$ are lower triangular,we have $a_{ij} = 0$ and $b_{ij} = 0$.
Therefore,$c_{ij} = 0 + 0 = 0$ for all $i < j$.
This implies that the sum $C = A + B$ is also a lower triangular matrix.

(D) Given the equation $2A + 3B - 5C = 0$.
Rearranging for $C$,we get $5C = 2A + 3B$.
Substitute the matrices $A$ and $B$:
$5C = 2\left[\begin{array}{cccc}2 & 1 & 3 & -1 \\ 1 & -2 & 2 & -3\end{array}\right] + 3\left[\begin{array}{cccc}2 & 1 & 0 & 3 \\ 1 & -1 & 2 & 3\end{array}\right]$
$5C = \left[\begin{array}{cccc}4 & 2 & 6 & -2 \\ 2 & -4 & 4 & -6\end{array}\right] + \left[\begin{array}{cccc}6 & 3 & 0 & 9 \\ 3 & -3 & 6 & 9\end{array}\right]$
$5C = \left[\begin{array}{cccc}4+6 & 2+3 & 6+0 & -2+9 \\ 2+3 & -4-3 & 4+6 & -6+9\end{array}\right]$
$5C = \left[\begin{array}{cccc}10 & 5 & 6 & 7 \\ 5 & -7 & 10 & 3\end{array}\right]$
Dividing by $5$,we get $C = \left[\begin{array}{cccc}10/5 & 5/5 & 6/5 & 7/5 \\ 5/5 & -7/5 & 10/5 & 3/5\end{array}\right] = \left[\begin{array}{cccc}2 & 1 & 6/5 & 7/5 \\ 1 & -7/5 & 2 & 3/5\end{array}\right]$.

(D) Given the equation: $(2 A+B)^2+(A-3 B)^2=5 A^2-2 A B+10 B^2$.
Expanding the squares,we get: $(4 A^2+2 A B+2 B A+B^2)+(A^2-3 A B-3 B A+9 B^2)=5 A^2-2 A B+10 B^2$.
Combining like terms: $(4 A^2+A^2)+(B^2+9 B^2)+(2 A B-3 A B)+(2 B A-3 B A)=5 A^2-2 A B+10 B^2$.
This simplifies to: $5 A^2+10 B^2-A B-B A=5 A^2-2 A B+10 B^2$.
Subtracting $5 A^2+10 B^2$ from both sides,we get: $-A B-B A=-2 A B$.
Adding $A B$ to both sides,we get: $-B A=-A B$,which implies $A B=B A$.
Since $A$ and $B$ commute,$A B A B=A(B A) B=A(A B) B=(A A)(B B)=A^2 B^2$.

(B) Given the matrix equation: $3\begin{bmatrix} x & y \\ z & t \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2t \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+t & 3 \end{bmatrix}$
Multiplying the scalar $3$ into the first matrix,we get:
$\begin{bmatrix} 3x & 3y \\ 3z & 3t \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+t & 2t+3 \end{bmatrix}$
Since the two matrices are equal,their corresponding elements must be equal:
$1) \ 3x = x + 4 \implies 2x = 4 \implies x = 2$
$2) \ 3t = 2t + 3 \implies t = 3$
$3) \ 3z = -1 + z + t \implies 2z = -1 + 3 \implies 2z = 2 \implies z = 1$
$4) \ 3y = 6 + x + y \implies 2y = 6 + 2 \implies 2y = 8 \implies y = 4$
Thus,the values are $(x, y, z, t) = (2, 4, 1, 3)$.

(B) Given $A = \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} (1)(1)+(0)(1) & (1)(0)+(0)(2) \\ (1)(1)+(2)(1) & (1)(0)+(2)(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 4 \end{bmatrix}$.
Now,calculate $2I$:
$2I = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$.
Finally,add $A^2$ and $2I$:
$A^2 + 2I = \begin{bmatrix} 1 & 0 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 3 & 6 \end{bmatrix}$.
Since $3A = 3 \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 3 & 6 \end{bmatrix}$,we conclude that $A^2 + 2I = 3A$.

(D) Given $A=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$. Note that $A^T = A$ and $A^2 = I$,so $A^T A = A A = I$.
We are given $X = A P A^T$.
Then $X^2 = (A P A^T)(A P A^T) = A P (A^T A) P A^T = A P I P A^T = A P^2 A^T$.
By induction,$X^n = A P^n A^T$.
Therefore,$X^{50} = A P^{50} A^T$.
Now,$A^T X^{50} A = A^T (A P^{50} A^T) A = (A^T A) P^{50} (A^T A) = I P^{50} I = P^{50}$.
Given $P = \left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$,we observe $P^2 = \left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$,$P^3 = \left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$,and in general $P^n = \left[\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right]$.
Thus,$P^{50} = \left[\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}\right]$.
Hence,$A^T X^{50} A = \left[\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}\right]$.

(D) Given $A = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{bmatrix}$. Let $B = \begin{bmatrix} x & y & z \\ a & b & c \\ u & v & w \end{bmatrix}$.
Since $AB = BA$,we perform matrix multiplication on both sides.
Calculating $AB = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{bmatrix} \begin{bmatrix} x & y & z \\ a & b & c \\ u & v & w \end{bmatrix} = \begin{bmatrix} a+2u & b+2v & c+2w \\ 2x+3a & 2y+3b & 2z+3c \\ 4x+3u & 4y+3v & 4z+3w \end{bmatrix}$.
Calculating $BA = \begin{bmatrix} x & y & z \\ a & b & c \\ u & v & w \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 2y+4z & x+3y & 2x+3z \\ 2b+4c & a+3b & 2a+3c \\ 2v+4w & u+3v & 2u+3w \end{bmatrix}$.
By comparing the elements of $AB$ and $BA$,we check the given options.
Testing option $D$: $B = \begin{bmatrix} 9 & -3 & -6 \\ -6 & -8 & 4 \\ -12 & 4 & -2 \end{bmatrix}$.
Substituting these values into the matrix multiplication confirms that $AB = BA$ holds true for this matrix.

(A) Given the matrix equation: $\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}$
Multiplying the matrices,we get:
$\alpha = x \cos \theta - y \sin \theta$ $(i)$
$\beta = x \sin \theta + y \cos \theta$ $(ii)$
$\gamma = z$ $(iii)$
Squaring and adding equations $(i)$ and $(ii)$:
$\alpha^2 + \beta^2 = (x \cos \theta - y \sin \theta)^2 + (x \sin \theta + y \cos \theta)^2$
$\alpha^2 + \beta^2 = x^2 \cos^2 \theta + y^2 \sin^2 \theta - 2xy \sin \theta \cos \theta + x^2 \sin^2 \theta + y^2 \cos^2 \theta + 2xy \sin \theta \cos \theta$
$\alpha^2 + \beta^2 = x^2(\cos^2 \theta + \sin^2 \theta) + y^2(\sin^2 \theta + \cos^2 \theta) = x^2 + y^2$
Now,consider the expression $\frac{x^2+y^2+z^2}{\gamma}$.
Substituting $x^2+y^2 = \alpha^2+\beta^2$ and $z = \gamma$:
$\frac{x^2+y^2+z^2}{\gamma} = \frac{\alpha^2+\beta^2+\gamma^2}{\gamma}$
Since $\gamma = z$,this is equal to $\frac{\alpha^2+\beta^2+\gamma^2}{z}$.

(B) Given $A = \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix}$.
We are given that $A^2 = I$.
$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Calculating the product:
Row $1$: $(1)(1) + (0)(a) + (0)(b) = 1$,$(1)(0) + (0)(-1) + (0)(c) = 0$,$(1)(0) + (0)(0) + (0)(1) = 0$.
Row $2$: $(a)(1) + (-1)(a) + (0)(b) = 0$,$(a)(0) + (-1)(-1) + (0)(c) = 1$,$(a)(0) + (-1)(0) + (0)(1) = 0$.
Row $3$: $(b)(1) + (c)(a) + (1)(b) = 2b + ac$,$(b)(0) + (c)(-1) + (1)(c) = 0$,$(b)(0) + (c)(0) + (1)(1) = 1$.
Thus,$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2b + ac & 0 & 1 \end{bmatrix}$.
Equating this to the identity matrix $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$,we get $2b + ac = 0$.
Therefore,$b = -\frac{ac}{2}$.

(C) Given $G(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Since $x+y=0$,we have $y = -x$.
Thus,$G(y) = G(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \\ \sin(-x) & \cos(-x) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Now,$G(x)G(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Performing matrix multiplication:
Row $1$,Col $1$: $(\cos x)(\cos x) + (-\sin x)(-\sin x) + (0)(0) = \cos^2 x + \sin^2 x = 1$.
Row $1$,Col $2$: $(\cos x)(\sin x) + (-\sin x)(\cos x) + (0)(0) = 0$.
Row $2$,Col $1$: $(\sin x)(\cos x) + (\cos x)(-\sin x) + (0)(0) = 0$.
Row $2$,Col $2$: $(\sin x)(\sin x) + (\cos x)(\cos x) + (0)(0) = \sin^2 x + \cos^2 x = 1$.
Thus,$G(x)G(y) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$,which is the identity matrix.

(D) Given $A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \cdot A$:
$A^2 = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} x^2 + 0 & 0 + 0 \\ x + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} x^2 & 0 \\ x + 1 & 1 \end{bmatrix}$.
Next,calculate $A^3 = A^2 \cdot A$:
$A^3 = \begin{bmatrix} x^2 & 0 \\ x + 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} x^3 + 0 & 0 + 0 \\ x(x + 1) + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} x^3 & 0 \\ x^2 + x + 1 & 1 \end{bmatrix}$.
Given $A^3 = B$,we have:
$\begin{bmatrix} x^3 & 0 \\ x^2 + x + 1 & 1 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix}$.
Comparing the corresponding elements:
$x^3 = 8 \implies x = 2$.
Also,$x^2 + x + 1 = 7 \implies x^2 + x - 6 = 0$.
Factoring the quadratic equation: $(x + 3)(x - 2) = 0$,which gives $x = 2$ or $x = -3$.
Since $x$ must satisfy both conditions,we take the common value,$x = 2$.

(B) Given $A = nI$,where $I$ is the identity matrix of order $3 \times 3$.
Then $A^2 = (nI)^2 = n^2 I^2 = n^2 I = \begin{bmatrix} n^2 & 0 & 0 \\ 0 & n^2 & 0 \\ 0 & 0 & n^2 \end{bmatrix}$.
Next,$B^2 = \begin{bmatrix} 0 & 0 & n \\ 0 & n & 0 \\ n & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & n \\ 0 & n & 0 \\ n & 0 & 0 \end{bmatrix} = \begin{bmatrix} n^2 & 0 & 0 \\ 0 & n^2 & 0 \\ 0 & 0 & n^2 \end{bmatrix} = n^2 I$.
Also,$AB = (nI)B = n(IB) = nB = \begin{bmatrix} 0 & 0 & n^2 \\ 0 & n^2 & 0 \\ n^2 & 0 & 0 \end{bmatrix}$.
Now,$A^2 + B^2 + AB = n^2 I + n^2 I + nB = 2n^2 I + nB$.
Factoring out $n$,we get $n(2nI + B)$.

(B) $A = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$
We know that $A = \omega I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Then,$A^{50} = (\omega I)^{50} = \omega^{50} I^{50}$.
Since $I^n = I$ for any positive integer $n$,we have $A^{50} = \omega^{50} I$.
We know that $\omega^3 = 1$. Therefore,$\omega^{50} = (\omega^3)^{16} \cdot \omega^2 = (1)^{16} \cdot \omega^2 = \omega^2$.
Thus,$A^{50} = \omega^2 I = \begin{bmatrix} \omega^2 & 0 \\ 0 & \omega^2 \end{bmatrix}$.
Since $A = \omega I$,we have $I = \frac{1}{\omega} A = \omega^2 A$ (as $\frac{1}{\omega} = \omega^2$).
Therefore,$A^{50} = \omega^2 (\omega^2 A) = \omega^4 A = \omega A$ (since $\omega^3 = 1$).
Alternatively,$A^{50} = \omega^{50} I = \omega^2 I$. Since $A = \omega I$,then $\omega A = \omega^2 I$.
Hence,$A^{50} = \omega A$.

(A) Given matrices are $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 3 & 4\end{array}\right]$ and $B=\left[\begin{array}{ccc}4 & 0 & -3 \\ -1 & -2 & -3\end{array}\right]$.
First,we find the transpose of matrix $A$,denoted as $A^T$:
$A^T = \left[\begin{array}{cc}1 & 0 \\ -1 & 3 \\ 2 & 4\end{array}\right]$.
Now,we compute the product $A^T B$:
$A^T B = \left[\begin{array}{cc}1 & 0 \\ -1 & 3 \\ 2 & 4\end{array}\right] \left[\begin{array}{ccc}4 & 0 & -3 \\ -1 & -2 & -3\end{array}\right]$.
Performing matrix multiplication:
$A^T B = \left[\begin{array}{ccc}(1)(4)+(0)(-1) & (1)(0)+(0)(-2) & (1)(-3)+(0)(-3) \\ (-1)(4)+(3)(-1) & (-1)(0)+(3)(-2) & (-1)(-3)+(3)(-3) \\ (2)(4)+(4)(-1) & (2)(0)+(4)(-2) & (2)(-3)+(4)(-3)\end{array}\right]$.
$A^T B = \left[\begin{array}{ccc}4+0 & 0+0 & -3+0 \\ -4-3 & 0-6 & 3-9 \\ 8-4 & 0-8 & -6-12\end{array}\right]$.
$A^T B = \left[\begin{array}{ccc}4 & 0 & -3 \\ -7 & -6 & -6 \\ 4 & -8 & -18\end{array}\right]$.

(C) Given $A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$. We check the validity of the statement $A^n = \begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}$ for various values of $n \in N$.
For $n = 1$: $A^1 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$,which matches the formula $\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$.
For $n = 2$: $A^2 = A \cdot A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+0(1) & 1(0)+0(1) \\ 1(1)+1(1) & 1(0)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$. This matches the formula for $n = 2$.
For $n = 3$: $A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+0(1) & 1(0)+0(1) \\ 2(1)+1(1) & 2(0)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}$. This matches the formula for $n = 3$.
Therefore,the statement is true for $n = 3$.

(D) Let $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $N = \begin{bmatrix} p & q \\ r & s \end{bmatrix}$.
Then,$MN = \begin{bmatrix} ap+br & aq+bs \\ cp+dr & cq+ds \end{bmatrix}$ and $NM = \begin{bmatrix} ap+qc & pb+qd \\ ar+cs & br+ds \end{bmatrix}$.
For $MN = NM$,we must have $br = qc$,$aq+bs = pb+qd$,$cp+dr = ar+cs$,and $cq+ds = br+ds$.
These conditions do not hold for all matrices $M$ and $N$. For example,if $M = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $N = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$,then $MN = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ while $NM = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$,so $MN \neq NM$.
Since none of the specific conditions $(A)$,$(B)$,or $(C)$ are necessary or sufficient for $MN = NM$ to hold for any arbitrary matrices $M$ and $N$,the correct answer is $(D)$.

(D) The rotation matrix $R_{\theta}$ is given by $\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$.
For matrix $A$,$\theta = \frac{\pi}{4}$. Thus,$A^n = \left[\begin{array}{ccc}\cos \frac{n\pi}{4} & \sin \frac{n\pi}{4} & 0 \\ -\sin \frac{n\pi}{4} & \cos \frac{n\pi}{4} & 0 \\ 0 & 0 & 1\end{array}\right]$.
For $A^{2020}$,$n=2020$,so $\frac{2020\pi}{4} = 505\pi$. Since $\cos(505\pi) = -1$ and $\sin(505\pi) = 0$,$A^{2020} \neq I$.
For matrix $D$,$\theta = \frac{\pi}{2}$. Thus,$D^n = \left[\begin{array}{ccc}\cos \frac{n\pi}{2} & \sin \frac{n\pi}{2} & 0 \\ -\sin \frac{n\pi}{2} & \cos \frac{n\pi}{2} & 0 \\ 0 & 0 & 1\end{array}\right]$.
For $D^{2019}$,$n=2019$,$\frac{2019\pi}{2} = 1009.5\pi$,so $D^{2019} \neq I$.
For matrix $B$,$\theta = \frac{\pi}{3}$. Thus,$B^n = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \frac{n\pi}{3} & \sin \frac{n\pi}{3} \\ 0 & -\sin \frac{n\pi}{3} & \cos \frac{n\pi}{3}\end{array}\right]$.
For $B^{2022}$,$n=2022$,$\frac{2022\pi}{3} = 674\pi$. Since $\cos(674\pi) = 1$ and $\sin(674\pi) = 0$,$B^{2022} = I$.

(D) Given $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.
We know that for a rotation matrix $R(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$,the property $R(\theta)^n = R(n\theta)$ holds.
Thus,$A^3 = \begin{bmatrix} \cos 3 \theta & \sin 3 \theta \\ -\sin 3 \theta & \cos 3 \theta \end{bmatrix}$.
Comparing this with the given matrix $A^3 = \begin{bmatrix} \cos 3 \theta & m \\ n & \cos 3 \theta \end{bmatrix}$,we get $m = \sin 3 \theta$ and $n = -\sin 3 \theta$.
Therefore,the correct option is $D$.

(D) Given the matrix $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
First,calculate $A^2$: $A^2 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$.
This implies $A^2 + I = 0$,or $A^2 = -I$.
From this,$A^3 = A^2 \cdot A = -I \cdot A = -A$.
Also,$A^4 = (A^2)^2 = (-I)^2 = I$.
Now,let us check each option:
Option $A$: $A^3 - I = -A - I$ and $A(A - I) = A^2 - A = -I - A$. Thus,$A^3 - I = A(A - I)$ is correct.
Option $B$: $A^3 + I = -A + I$ and $A(A^3 - I) = A(-A - I) = -A^2 - A = I - A$. Thus,$A^3 + I = A(A^3 - I)$ is correct.
Option $C$: $A^4 - I = I - I = 0$ and $A^2 + I = -I + I = 0$. Thus,$A^4 - I = A^2 + I$ is correct.
Option $D$: $A^2 + I = -I + I = 0$ and $A(A^2 - I) = A(-I - I) = A(-2I) = -2A$. Since $0 \neq -2A$,this option is incorrect.

(C) The given matrix $A$ is a rotation matrix of the form $R(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$ where $\theta = \frac{2 \pi}{33}$.
By the property of rotation matrices,$A^n = R(n \theta) = \begin{bmatrix} \cos(n \theta) & \sin(n \theta) \\ -\sin(n \theta) & \cos(n \theta) \end{bmatrix}$.
We need to find $A^{2017}$,so $n = 2017$.
The angle becomes $n \theta = 2017 \times \frac{2 \pi}{33} = \frac{4034 \pi}{33}$.
Dividing $4034$ by $33$: $4034 = 33 \times 122 + 8$.
So,$\frac{4034 \pi}{33} = 122 \pi + \frac{8 \pi}{33}$.
Since $\cos(122 \pi + \alpha) = \cos \alpha$ and $\sin(122 \pi + \alpha) = \sin \alpha$,we have $A^{2017} = \begin{bmatrix} \cos \frac{8 \pi}{33} & \sin \frac{8 \pi}{33} \\ -\sin \frac{8 \pi}{33} & \cos \frac{8 \pi}{33} \end{bmatrix}$.
Note that $A^4 = R(4 \times \frac{2 \pi}{33}) = R(\frac{8 \pi}{33})$.
Therefore,$A^{2017} = A^4$.

(C) Given $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}$.
Then $A^T = \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix}$.
The condition is $A A^T = 9 I = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$.
Calculating the product $A A^T$:
Row $1 \times$ Column $1$: $1(1) + 2(2) + 2(2) = 1 + 4 + 4 = 9$.
Row $2 \times$ Column $2$: $2(2) + 1(1) + (-2)(-2) = 4 + 1 + 4 = 9$.
Row $3 \times$ Column $3$: $a(a) + 2(2) + b(b) = a^2 + 4 + b^2$.
Since $A A^T = 9 I$,the diagonal elements must be $9$. Thus,$a^2 + 4 + b^2 = 9$,which implies $a^2 + b^2 = 5$.
Also,checking off-diagonal elements:
Row $1 \times$ Column $2$: $1(2) + 2(1) + 2(-2) = 2 + 2 - 4 = 0$.
Row $1 \times$ Column $3$: $1(a) + 2(2) + 2(b) = a + 4 + 2b = 0$.
Row $2 \times$ Column $3$: $2(a) + 1(2) + (-2)(b) = 2a + 2 - 2b = 0 \implies a - b = -1$.
From $a + 2b = -4$ and $a - b = -1$,subtracting gives $3b = -3 \implies b = -1$. Then $a = -2$.
Checking $a^2 + b^2 = (-2)^2 + (-1)^2 = 4 + 1 = 5$.

(B) Given matrices are $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$ and $B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{bmatrix}$.
Both matrices $A$ and $B$ have order $2 \times 3$.
Therefore,the order of $A^T$ and $B^T$ is $3 \times 2$.
Now,let us check the order of the matrices in option $B$:
The order of $A^T B^T A$ is $(3 \times 2) \times (3 \times 2) \times (2 \times 3)$,which is not defined because the inner dimensions do not match.
However,checking the order of $A^T B B^T A$:
Order of $A^T B B^T A = (3 \times 2) \times (2 \times 3) \times (3 \times 2) \times (2 \times 3) = 3 \times 3$.
Order of $B^T A A^T B = (3 \times 2) \times (2 \times 3) \times (3 \times 2) \times (2 \times 3) = 3 \times 3$.
Since both expressions $A^T B B^T A$ and $B^T A A^T B$ result in a $3 \times 3$ matrix,their orders are equal.

(C) The principal diagonal elements of matrix $A$ are $a^2, b^2, c^2$ and the principal diagonal elements of matrix $B$ are $2a, 2b, 2c-3$.
According to the given condition,the sum of the principal diagonal elements of $A$ is equal to the sum of the principal diagonal elements of $B$:
$a^2+b^2+c^2 = 2a+2b+2c-3$
Rearranging the terms,we get:
$(a^2-2a+1) + (b^2-2b+1) + (c^2-2c+1) = 0$
$(a-1)^2 + (b-1)^2 + (c-1)^2 = 0$
Since the sum of squares of real numbers is zero only if each term is zero,we have:
$a-1=0, b-1=0, c-1=0$
Therefore,$a=1, b=1, c=1$.
The product of the principal diagonal elements of $B$ is $(2a)(2b)(2c-3)$.
Substituting the values $a=1, b=1, c=1$:
$= (2 \times 1)(2 \times 1)(2 \times 1 - 3)$
$= 2 \times 2 \times (-1) = -4$.

(A) Any square matrix $A$ can be expressed as the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$,where $P = \frac{1}{2}(A + A^T)$ and $Q = \frac{1}{2}(A - A^T)$.
Given $A = \begin{bmatrix} -2 & 1 \\ 3 & 4 \end{bmatrix}$.
Then $A^T = \begin{bmatrix} -2 & 3 \\ 1 & 4 \end{bmatrix}$.
Now,$A - A^T = \begin{bmatrix} -2 & 1 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} -2 & 3 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$.
Therefore,$Q = \frac{1}{2}(A - A^T) = \frac{1}{2} \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.

(A) Given that $A$ is a non-singular matrix,its determinant $|A| \neq 0$,which implies that the inverse matrix $A^{-1}$ exists.
Given the equation $AB = O$,where $O$ is the zero matrix.
Multiplying both sides by $A^{-1}$ on the left:
$A^{-1}(AB) = A^{-1}O$
$(A^{-1}A)B = O$
$IB = O$
$B = O$
Therefore,$B$ must be a null matrix.

(D) Given: $A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$
First,find the transpose of $A$: $A^T = \begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix}$
Now,calculate $(A^T)^2$:
$(A^T)^2 = \begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} (2)(2) + (-4)(-3) & (2)(-4) + (-4)(1) \\ (-3)(2) + (1)(-3) & (-3)(-4) + (1)(1) \end{bmatrix} = \begin{bmatrix} 16 & -12 \\ -9 & 13 \end{bmatrix}$
Next,calculate $(12A)^T$:
$12A = \begin{bmatrix} 24 & -36 \\ -48 & 12 \end{bmatrix} \Rightarrow (12A)^T = \begin{bmatrix} 24 & -48 \\ -36 & 12 \end{bmatrix}$
Finally,add the two matrices:
$(A^T)^2 + (12A)^T = \begin{bmatrix} 16 & -12 \\ -9 & 13 \end{bmatrix} + \begin{bmatrix} 24 & -48 \\ -36 & 12 \end{bmatrix} = \begin{bmatrix} 40 & -60 \\ -45 & 25 \end{bmatrix}$

(B) Given the equation $ABCDE = I$.
Multiply both sides by $A^{-1}$ from the left: $A^{-1}(ABCDE) = A^{-1}I \Rightarrow BCDE = A^{-1}$.
Multiply both sides by $B^{-1}$ from the left: $B^{-1}(BCDE) = B^{-1}A^{-1} \Rightarrow CDE = B^{-1}A^{-1}$.
Multiply both sides by $E^{-1}$ from the right: $(CDE)E^{-1} = B^{-1}A^{-1}E^{-1} \Rightarrow CD = B^{-1}A^{-1}E^{-1}$.
Multiply both sides by $D^{-1}$ from the right: $(CD)D^{-1} = B^{-1}A^{-1}E^{-1}D^{-1} \Rightarrow C = B^{-1}A^{-1}E^{-1}D^{-1}$.
Taking the inverse of both sides: $C^{-1} = (B^{-1}A^{-1}E^{-1}D^{-1})^{-1}$.
Using the property $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$,we get $C^{-1} = (D^{-1})^{-1}(E^{-1})^{-1}(A^{-1})^{-1}(B^{-1})^{-1} = DEAB$.

(A) Given $M = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$.
First,calculate $M^2 = M \times M$:
$M^2 = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{bmatrix} = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$.
Now,calculate $4M = 4 \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix}$.
Finally,calculate $M^2 - 4M = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} = 5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 5I$.

(D) Given,$a = \sin \frac{\pi}{6} = \frac{1}{2}$,$b = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,and $c = \cot \frac{\pi}{2} = 0$.
Substituting these values into the matrix $A$:
$A = \begin{bmatrix} (\frac{1}{2})^2 + 0^2 & (\frac{1}{2})^2 & (\frac{1}{2})^2 \\ (\frac{1}{\sqrt{2}})^2 & 0^2 + (\frac{1}{2})^2 & (\frac{1}{\sqrt{2}})^2 \\ 0^2 & 0^2 & (\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{2} \\ 0 & 0 & \frac{3}{4} \end{bmatrix}$.
Now,calculate the determinant $|A|$:
$|A| = \frac{1}{4} \left( \frac{1}{4} \cdot \frac{3}{4} - 0 \right) - \frac{1}{4} \left( \frac{1}{2} \cdot \frac{3}{4} - 0 \right) + \frac{1}{4} (0 - 0) = \frac{1}{4} \cdot \frac{3}{16} - \frac{1}{4} \cdot \frac{3}{8} = \frac{3}{64} - \frac{6}{64} = -\frac{3}{64}$.
Wait,let us re-evaluate the matrix $A$ construction:
$A = \begin{bmatrix} b^2+c^2 & a^2 & a^2 \\ b^2 & c^2+a^2 & b^2 \\ c^2 & c^2 & a^2+b^2 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/4 & 1/4 \\ 1/2 & 1/4 & 1/2 \\ 0 & 0 & 3/4 \end{bmatrix}$.
Expanding along the third row: $|A| = 0 - 0 + \frac{3}{4} \left( \frac{1}{4} \cdot \frac{1}{4} - \frac{1}{4} \cdot \frac{1}{2} \right) = \frac{3}{4} \left( \frac{1}{16} - \frac{2}{16} \right) = \frac{3}{4} \left( -\frac{1}{16} \right) = -\frac{3}{64} \neq 0$.
Re-checking the question values: $a^2 = 1/4$,$b^2 = 1/2$,$c^2 = 0$.
$A = \begin{bmatrix} 1/2 & 1/4 & 1/4 \\ 1/2 & 1/4 & 1/2 \\ 0 & 0 & 3/4 \end{bmatrix}$.
Since $|A| \neq 0$,it is a non-singular matrix.

(D) Given $A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 1+0+6 & 0+0+4 & 2+0+8 \\ 2+2+9 & 0+1+6 & 4+3+12 \\ 3+4+12 & 0+2+8 & 6+6+16 \end{bmatrix} = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix}$.
Now,calculate $5A = 5 \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix}$.
And $6I = 6 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$.
Finally,compute $A^2 - 5A + 6I$:
$A^2 - 5A + 6I = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = \begin{bmatrix} 7-5+6 & 4-0+0 & 10-10+0 \\ 13-10+0 & 7-5+6 & 19-15+0 \\ 19-15+0 & 10-10+0 & 28-20+6 \end{bmatrix} = \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix}$.

(A) Performing the matrix multiplication on the left side:
$\begin{bmatrix} 7(1) + 5(3) + \alpha(2) \\ \beta(1) + 2(3) + 11(2) \\ 3(1) + \gamma(3) + 1(2) \end{bmatrix} = \begin{bmatrix} 22 + 2\alpha \\ \beta + 28 \\ 5 + 3\gamma \end{bmatrix}$.
Equating this to the right side matrix:
$1) \ 22 + 2\alpha = \alpha + \beta \implies \alpha - \beta = -22$
$2) \ \beta + 28 = -2\alpha + \beta - 2\gamma \implies 2\alpha + 2\gamma = -28 \implies \alpha + \gamma = -14$
$3) \ 5 + 3\gamma = \alpha + 2\beta + 3\gamma \implies \alpha + 2\beta = 5$
From $(1)$,$\beta = \alpha + 22$. Substituting into $(3)$:
$\alpha + 2(\alpha + 22) = 5 \implies 3\alpha + 44 = 5 \implies 3\alpha = -39 \implies \alpha = -13$.
Then $\beta = -13 + 22 = 9$.
From $(2)$,$\gamma = -14 - \alpha = -14 - (-13) = -1$.
Now,calculate $100 + \frac{2\alpha + 11\beta}{\gamma} = 100 + \frac{2(-13) + 11(9)}{-1} = 100 + \frac{-26 + 99}{-1} = 100 + \frac{73}{-1} = 100 - 73 = 27$.

(C) Let the order of matrix $A$ be $m \times n$. Since $A$ and $B$ are added to form $P = A + B$,$B$ must also have the order $m \times n$.
For $Q = A^T B$,the order of $A^T$ is $n \times m$ and $B$ is $m \times n$. Thus,the product $Q$ has order $n \times n$.
For $R = A B^T$,the order of $A$ is $m \times n$ and $B^T$ is $n \times m$. Thus,the product $R$ has order $m \times m$.
Now,let us check the options:
$P$ is $m \times n$,$Q$ is $n \times n$,$R$ is $m \times m$.
$PQ$ has order $(m \times n) \times (n \times n) = m \times n$.
$RP$ has order $(m \times m) \times (m \times n) = m \times n$.
Both $PQ$ and $RP$ have the same order as $A$,which is $m \times n$.

(A) First,we calculate $AA^T$:
$AA^T = \left[\begin{array}{lll}9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2\end{array}\right] \left[\begin{array}{lll}9 & 1 & 7 \\ 3 & 5 & 6 \\ 0 & 8 & 2\end{array}\right] = \left[\begin{array}{lll}90 & 24 & 81 \\ 24 & 90 & 53 \\ 81 & 53 & 89\end{array}\right]$
Next,we calculate $A^2$:
$A^2 = \left[\begin{array}{lll}9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2\end{array}\right] \left[\begin{array}{lll}9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2\end{array}\right] = \left[\begin{array}{lll}84 & 42 & 24 \\ 70 & 76 & 56 \\ 83 & 63 & 52\end{array}\right]$
Now,find $AA^T - A^2$:
$AA^T - A^2 = \left[\begin{array}{ccc}90-84 & 24-42 & 81-24 \\ 24-70 & 90-76 & 53-56 \\ 81-83 & 53-63 & 89-52\end{array}\right] = \left[\begin{array}{ccc}6 & -18 & 57 \\ -46 & 14 & -3 \\ -2 & -10 & 37\end{array}\right]$
The sum of all elements $a_{ij}$ is:
$\sum a_{ij} = 6 - 18 + 57 - 46 + 14 - 3 - 2 - 10 + 37 = 35$.

(C) Given $A = \begin{bmatrix} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -3 & -2 & 4 \\ 2 & 2 & -1 \\ -2 & 0 & 3 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} (1-2-1) & (2+0-2) & (-1+4+0) \\ (-1+0+2) & (-2+0+4) & (1+0+0) \\ (1-2+0) & (2+0+0) & (-1+4+0) \end{bmatrix} = \begin{bmatrix} -2 & 0 & 3 \\ 1 & 2 & 1 \\ -1 & 2 & 3 \end{bmatrix}$.
Next,calculate $A+B$:
$A+B = \begin{bmatrix} 1-3 & 2-2 & -1+4 \\ -1+2 & 0+2 & 2-1 \\ 1-2 & 2+0 & 0+3 \end{bmatrix} = \begin{bmatrix} -2 & 0 & 3 \\ 1 & 2 & 1 \\ -1 & 2 & 3 \end{bmatrix}$.
Comparing both results,we see that $A^2 = A+B$.

(B) Given $A = \begin{bmatrix} 0 & 1 \\ 1 & k \end{bmatrix}$.
First,calculate $A^2 = \begin{bmatrix} 0 & 1 \\ 1 & k \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & k \end{bmatrix} = \begin{bmatrix} 1 & k \\ k & 1+k^2 \end{bmatrix}$.
Next,calculate $A^3 = A^2 \cdot A = \begin{bmatrix} 1 & k \\ k & 1+k^2 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & k \end{bmatrix} = \begin{bmatrix} k & 1+k^2 \\ 1+k^2 & k+k(1+k^2) \end{bmatrix} = \begin{bmatrix} k & 1+k^2 \\ 1+k^2 & k+k+k^3 \end{bmatrix} = \begin{bmatrix} k & 1+k^2 \\ 1+k^2 & 2k+k^3 \end{bmatrix}$.
We are given $d = 228$,so $2k + k^3 = 228$.
By testing values,if $k = 6$,then $2(6) + 6^3 = 12 + 216 = 228$. Thus,$k = 6$.
Now,$b = 1 + k^2 = 1 + 6^2 = 1 + 36 = 37$.
Also,$c = 1 + k^2 = 1 + 36 = 37$.
Therefore,$b + c = 37 + 37 = 74$.

(A) Given $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$.
We calculate the powers of $A$:
$A^2 = A \cdot A = \left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] \left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] = \left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] = 2A - I = 2A - (2-1)I$.
$A^3 = A^2 \cdot A = \left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] \left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] = \left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] = 3A - 2I = 3A - (3-1)I$.
By mathematical induction,we can generalize this pattern for any $n \in N$:
$A^n = nA - (n-1)I$.

(B) Given $A = \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix}$ and $f(t) = t^2 - 3t + 7$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 1(1) + (-2)(4) & 1(-2) + (-2)(5) \\ 4(1) + 5(4) & 4(-2) + 5(5) \end{bmatrix} = \begin{bmatrix} -7 & -12 \\ 24 & 17 \end{bmatrix}$.
Now,calculate $f(A) = A^2 - 3A + 7I$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$:
$f(A) = \begin{bmatrix} -7 & -12 \\ 24 & 17 \end{bmatrix} - 3 \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$f(A) = \begin{bmatrix} -7 & -12 \\ 24 & 17 \end{bmatrix} - \begin{bmatrix} 3 & -6 \\ 12 & 15 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$
$f(A) = \begin{bmatrix} -7-3+7 & -12+6+0 \\ 24-12+0 & 17-15+7 \end{bmatrix} = \begin{bmatrix} -3 & -6 \\ 12 & 9 \end{bmatrix}$.
Finally,calculate $f(A) + \begin{bmatrix} 3 & 6 \\ -12 & -9 \end{bmatrix}$:
$\begin{bmatrix} -3 & -6 \\ 12 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ -12 & -9 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.

(D) Given the matrix equation:
$m[-3, 4] + n[4, -3] = [10, -11]$
Multiplying the scalars $m$ and $n$ into the matrices,we get:
$[-3m, 4m] + [4n, -3n] = [10, -11]$
Adding the matrices,we have:
$[-3m + 4n, 4m - 3n] = [10, -11]$
By equating the corresponding elements,we obtain two linear equations:
$-3m + 4n = 10$ $\dots(i)$
$4m - 3n = -11$ $\dots(ii)$
To solve for $m$ and $n$,multiply equation $(i)$ by $3$ and equation $(ii)$ by $4$:
$-9m + 12n = 30$ $\dots(iii)$
$16m - 12n = -44$ $\dots(iv)$
Adding equations $(iii)$ and $(iv)$:
$7m = -14 \Rightarrow m = -2$
Substituting $m = -2$ into equation $(i)$:
$-3(-2) + 4n = 10 \Rightarrow 6 + 4n = 10 \Rightarrow 4n = 4 \Rightarrow n = 1$
Finally,calculating $3m + 7n$:
$3(-2) + 7(1) = -6 + 7 = 1$

(B) square matrix is defined as a scalar matrix if all its non-diagonal elements are zero and all its diagonal elements are equal to a constant $k$.
Given that $a_{ij} = 0$ for $i \neq j$ (non-diagonal elements are zero) and $a_{ij} = k$ for $i = j$ (diagonal elements are equal to a constant $k$),the matrix satisfies the definition of a scalar matrix.
Therefore,the correct option is $B$.

(C) Given that $A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$.
Multiplying matrix $A$ by a scalar $h$,we get:
$hA = \begin{bmatrix} 0 & 2h \\ 3h & -4h \end{bmatrix}$.
We are given that $hA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$.
Comparing the corresponding elements of the two matrices:
$1$) $-4h = 24 \implies h = -6$.
$2$) $2h = 3a \implies 2(-6) = 3a \implies -12 = 3a \implies a = -4$.
$3$) $3h = 2b \implies 3(-6) = 2b \implies -18 = 2b \implies b = -9$.
Thus,the values are $h = -6, a = -4, b = -9$.

(A) Given $A = \begin{bmatrix} b & a & 0 \\ c & 0 & b \\ a & a & b \end{bmatrix}$ and $B = \begin{bmatrix} 0 & a & b \\ b & 0 & c \\ b & a & a \end{bmatrix}$.
Performing matrix multiplication $AB$:
$AB = \begin{bmatrix} b(0)+a(b)+0(b) & b(a)+a(0)+0(a) & b(b)+a(c)+0(a) \\ c(0)+0(b)+b(b) & c(a)+0(0)+b(a) & c(b)+0(c)+b(a) \\ a(0)+a(b)+b(b) & a(a)+a(0)+b(a) & a(b)+a(c)+b(a) \end{bmatrix} = \begin{bmatrix} ab & ab & b^2+ac \\ b^2 & ac+ab & bc+ab \\ ab+b^2 & a^2+ab & 2ab+ac \end{bmatrix}$.
Equating this to the given matrix $\begin{bmatrix} 2 & 2 & 7 \\ 1 & 8 & 5 \\ 3 & 6 & 10 \end{bmatrix}$:
From the element at $(2,1)$,$b^2 = 1$.
From the element at $(1,1)$,$ab = 2$.
From the element at $(1,3)$,$b^2+ac = 7 \implies 1+ac = 7 \implies ac = 6$.
Since $ab=2$ and $b^2=1$,we have $a = 2/b$. If $b=1$,$a=2$. If $b=-1$,$a=-2$.
Case $1$: $b=1, a=2$. Then $ac=6 \implies 2c=6 \implies c=3$.
Case $2$: $b=-1, a=-2$. Then $ac=6 \implies -2c=6 \implies c=-3$.
In both cases,$a^2+b^2+c^2 = (\pm 2)^2 + (\pm 1)^2 + (\pm 3)^2 = 4+1+9 = 14$.

(C) Given $A = \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Since $A^2 = 0$,all higher powers $A^n = 0$ for $n \geq 2$.
Given $f(x) = x + x^2 + x^3 + \ldots + x^{2023}$,we have $f(A) = A + A^2 + A^3 + \ldots + A^{2023}$.
Substituting $A^n = 0$ for $n \geq 2$,we get $f(A) = A + 0 + 0 + \ldots + 0 = A$.
Therefore,$f(A) + I = A + I$.
$f(A) + I = \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.