Types of matrices, Algebra of matrices Questions in English

(B) Given $A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $(A - 2I)$:
$A - 2I = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -1 & -1 \end{bmatrix}$.
Next,calculate $(A - 3I)$:
$A - 3I = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix}$.
Now,multiply the two matrices:
$(A - 2I)(A - 3I) = \begin{bmatrix} 2 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} (2)(1) + (2)(-1) & (2)(2) + (2)(-2) \\ (-1)(1) + (-1)(-1) & (-1)(2) + (-1)(-2) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.

(C) To determine which expression is undefined,we check the dimensions of the matrices:
$A$ is a $3 \times 3$ matrix.
$B$ is a $3 \times 2$ matrix.
$C$ is a $1 \times 3$ matrix.
$1$. For $B'B$: $B'$ is $2 \times 3$ and $B$ is $3 \times 2$. The product is defined ($2 \times 2$ matrix).
$2$. For $CAB$: $C$ is $1 \times 3$,$A$ is $3 \times 3$,and $B$ is $3 \times 2$. The product $CA$ is $1 \times 3$,and $(CA)B$ is $1 \times 2$. This is defined.
$3$. For $A + B'$: $A$ is $3 \times 3$ and $B'$ is $2 \times 3$. Matrix addition requires matrices to have the same dimensions. Since $3 \times 3 \neq 2 \times 3$,this expression is not defined.
$4$. For $A^2 + A$: Since $A$ is a square matrix $(3 \times 3)$,$A^2$ and $A$ are both $3 \times 3$. This is defined.
Therefore,the expression $A + B'$ is not defined.

(C) Given $A = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$.
Calculate $A^2 = A \times A = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} = \begin{bmatrix} a^2 & 0 & 0 \\ 0 & b^2 & 0 \\ 0 & 0 & c^2 \end{bmatrix}$.
Similarly,$A^3 = A^2 \times A = \begin{bmatrix} a^2 & 0 & 0 \\ 0 & b^2 & 0 \\ 0 & 0 & c^2 \end{bmatrix} \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} = \begin{bmatrix} a^3 & 0 & 0 \\ 0 & b^3 & 0 \\ 0 & 0 & c^3 \end{bmatrix}$.
By the principle of mathematical induction,for any positive integer $n$,we have $A^n = \begin{bmatrix} a^n & 0 & 0 \\ 0 & b^n & 0 \\ 0 & 0 & c^n \end{bmatrix}$.
Thus,the correct option is $C$.

(A) matrix $A$ is skew-symmetric if $A^T = -A$,which implies $a_{ij} = -a_{ji}$ for all $i, j$.
For diagonal elements where $i = j$,we have $a_{ii} = -a_{ii}$,which implies $2a_{ii} = 0$,so $a_{ii} = 0$.
Checking option $A$:
Let $A = \begin{bmatrix} 0 & 4 & 5 \\ -4 & 0 & -6 \\ -5 & 6 & 0 \end{bmatrix}$.
$A^T = \begin{bmatrix} 0 & -4 & -5 \\ 4 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & 4 & 5 \\ -4 & 0 & -6 \\ -5 & 6 & 0 \end{bmatrix} = -A$.
Since $A^T = -A$,the matrix is skew-symmetric.

(A) matrix $A$ is said to be orthogonal if $AA^T = A^TA = I$,where $I$ is the identity matrix.
Given $A = \frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ -2 & 2 & -1 \end{bmatrix}$.
Then $A^T = \frac{1}{3}\begin{bmatrix} 1 & 2 & -2 \\ 2 & 1 & 2 \\ 2 & -2 & -1 \end{bmatrix}$.
Now,$AA^T = \frac{1}{9} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ -2 & 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 & -2 \\ 2 & 1 & 2 \\ 2 & -2 & -1 \end{bmatrix}$
$= \frac{1}{9} \begin{bmatrix} 1+4+4 & 2+2-4 & -2+4-2 \\ 2+2-4 & 4+1+4 & -4+2+2 \\ -2+4-2 & -4+2+2 & 4+4+1 \end{bmatrix}$
$= \frac{1}{9} \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Since $AA^T = I$,the matrix $A$ is an orthogonal matrix.

(A) An upper triangular matrix $A = [a_{ij}]$ is a square matrix where all elements below the main diagonal are zero,i.e.,$a_{ij} = 0$ for all $i > j$.
For an $n \times n$ matrix,the number of elements below the main diagonal is given by the sum of the first $(n-1)$ integers:
$1 + 2 + 3 + ... + (n - 1) = \frac{(n - 1)n}{2} = \frac{n(n - 1)}{2}$.
For example,in a $4 \times 4$ matrix,the number of zeros is $\frac{4(4 - 1)}{2} = \frac{12}{2} = 6$.
Thus,the minimum number of zeros is $\frac{n(n - 1)}{2}$.

(B) Given the equation: $A^2 - B^2 = (A - B)(A + B)$.
Expanding the right side using matrix multiplication properties:
$(A - B)(A + B) = A(A + B) - B(A + B)$
$= A^2 + AB - BA - B^2$.
Now,substitute this back into the original equation:
$A^2 - B^2 = A^2 + AB - BA - B^2$.
Subtracting $A^2$ and adding $B^2$ to both sides,we get:
$0 = AB - BA$.
Therefore,$AB = BA$.

(D) Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$.
Calculate $AB$:
$AB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} = \begin{bmatrix} a & 2b \\ 3a & 4b \end{bmatrix}$.
Calculate $BA$:
$BA = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} a & 2a \\ 3b & 4b \end{bmatrix}$.
For $AB = BA$,we must have:
$\begin{bmatrix} a & 2b \\ 3a & 4b \end{bmatrix} = \begin{bmatrix} a & 2a \\ 3b & 4b \end{bmatrix}$.
Comparing the elements,we get $2b = 2a$ and $3a = 3b$,which both imply $a = b$.
Since $a, b \in \mathbb{N}$,there are infinitely many pairs $(a, b)$ such that $a = b$ (e.g.,$(1, 1), (2, 2), (3, 3), \dots$).
Therefore,there exist infinitely many matrices $B$ such that $AB = BA$.

(C) Given $H = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$.
We calculate the powers of $H$:
$H^2 = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} = \begin{bmatrix} \omega^2 & 0 \\ 0 & \omega^2 \end{bmatrix}$.
By induction,$H^n = \begin{bmatrix} \omega^n & 0 \\ 0 & \omega^n \end{bmatrix}$.
For $n = 70$,$H^{70} = \begin{bmatrix} \omega^{70} & 0 \\ 0 & \omega^{70} \end{bmatrix}$.
Since $\omega^3 = 1$,we have $\omega^{70} = (\omega^3)^{23} \cdot \omega = (1)^{23} \cdot \omega = \omega$.
Therefore,$H^{70} = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} = H$.

(A) Given $AA^T = 9I$,where $I$ is the $3 \times 3$ identity matrix.
$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$
Multiplying the matrices,we look at the elements in the third row:
For the element at $(3, 1)$: $a(1) + 2(2) + b(2) = 0 \Rightarrow a + 2b + 4 = 0 \Rightarrow a + 2b = -4$ ... $(i)$
For the element at $(3, 2)$: $a(2) + 2(1) + b(-2) = 0 \Rightarrow 2a - 2b + 2 = 0 \Rightarrow a - b = -1$ ... $(ii)$
Solving the system of equations:
From $(ii)$,$a = b - 1$.
Substitute into $(i)$: $(b - 1) + 2b = -4 \Rightarrow 3b = -3 \Rightarrow b = -1$.
Then,$a = -1 - 1 = -2$.
Thus,the ordered pair $(a, b)$ is $(-2, -1)$.

(A) Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix}$,$B = \begin{bmatrix} -1 & -2 \\ -2 & 0 \\ 0 & -4 \end{bmatrix}$,and $C = \begin{bmatrix} -4 & -5 & -6 \\ 0 & 0 & 1 \end{bmatrix}$.
First,calculate $D = A \times B$:
$D = \begin{bmatrix} (1)(-1)+(2)(-2)+(3)(0) & (1)(-2)+(2)(0)+(3)(-4) \\ (2)(-1)+(3)(-2)+(4)(0) & (2)(-2)+(3)(0)+(4)(-4) \\ (3)(-1)+(4)(-2)+(5)(0) & (3)(-2)+(4)(0)+(5)(-4) \end{bmatrix} = \begin{bmatrix} -5 & -14 \\ -8 & -20 \\ -11 & -26 \end{bmatrix}$.
Now,calculate $P = D \times C$:
$P = \begin{bmatrix} -5 & -14 \\ -8 & -20 \\ -11 & -26 \end{bmatrix} \begin{bmatrix} -4 & -5 & -6 \\ 0 & 0 & 1 \end{bmatrix}$.
The element $P_{22}$ is the product of the second row of $D$ and the second column of $C$:
$P_{22} = (-8)(-5) + (-20)(0) = 40 + 0 = 40$.

(C) The dimensions of the matrices are: $A$ is $3 \times 3$,$B$ is $3 \times 2$,and $C$ is $3 \times 1$.
$(i) (AB)^T C$: $AB$ is $(3 \times 3)(3 \times 2) = 3 \times 2$. Thus,$(AB)^T$ is $2 \times 3$. The product $(2 \times 3)(3 \times 1)$ is defined.
$(ii) C^T C (AB)^T$: $C^T$ is $1 \times 3$ and $C$ is $3 \times 1$,so $C^T C$ is $1 \times 1$. $(AB)^T$ is $2 \times 3$. The product $(1 \times 1)(2 \times 3)$ is not defined because the number of columns in the first matrix $(1)$ does not equal the number of rows in the second matrix $(2)$.
$(iii) C^T AB$: $C^T$ is $1 \times 3$ and $AB$ is $3 \times 2$. The product $(1 \times 3)(3 \times 2)$ is defined.
$(iv) A^T AB B^T C$: $A^T$ is $3 \times 3$,$AB$ is $3 \times 2$,$B^T$ is $2 \times 3$,and $C$ is $3 \times 1$. The product $(3 \times 3)(3 \times 2)(2 \times 3)(3 \times 1)$ is defined.
Thus,$(i), (iii),$ and $(iv)$ are defined. Exactly three are defined.

(A) We have $A^2 = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix}$.
$A^3 = A^2 A = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3a \\ 0 & 1 \end{bmatrix}$.
By observing the pattern,we can generalize this using mathematical induction. For any $n \in N$,$A^n = \begin{bmatrix} 1 & na \\ 0 & 1 \end{bmatrix}$.

(D) Given the equation $A + 2X = B$.
Subtracting $A$ from both sides,we get $2X = B - A$.
First,calculate $B - A$:
$B - A = \begin{bmatrix} -2 & 5 \\ 6 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 4 \\ 1 & -6 \end{bmatrix} = \begin{bmatrix} -2 - 3 & 5 - 4 \\ 6 - 1 & 1 - (-6) \end{bmatrix} = \begin{bmatrix} -5 & 1 \\ 5 & 7 \end{bmatrix}$.
Now,$X = \frac{1}{2} (B - A) = \frac{1}{2} \begin{bmatrix} -5 & 1 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix} -2.5 & 0.5 \\ 2.5 & 3.5 \end{bmatrix}$.
Comparing this with the given options,the correct answer is none of these.

(B) Given that the order of $A$ is $3 \times 4$,therefore the order of $A'$ is $4 \times 3$.
Since $A'B$ is defined,let the order of $B$ be $m \times n$. For $A'B$ to be defined,the number of columns in $A'$ must equal the number of rows in $B$. Thus,$m = 3$.
Now,consider $BA'$. For $BA'$ to be defined,the number of columns in $B$ must equal the number of rows in $A'$. Since $A'$ is $4 \times 3$,the number of rows is $4$. Thus,$n = 4$.
Therefore,the order of $B$ is $3 \times 4$.
Consequently,the order of $B'$ is $4 \times 3$.
Finally,the order of $B'A$ is $(4 \times 3) \times (3 \times 4) = 4 \times 4$.

(B) Let $D = \begin{bmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{bmatrix}$.
Clearly,$D' = D$ because a diagonal matrix is symmetric.
Now,consider $AD$ and $DA$ for a general matrix $A = [a_{ij}]$.
$AD = \begin{bmatrix} d_1 a_{11} & d_2 a_{12} & d_3 a_{13} \\ d_1 a_{21} & d_2 a_{22} & d_3 a_{23} \\ d_1 a_{31} & d_2 a_{32} & d_3 a_{33} \end{bmatrix}$ and $DA = \begin{bmatrix} d_1 a_{11} & d_1 a_{12} & d_1 a_{13} \\ d_2 a_{21} & d_2 a_{22} & d_2 a_{23} \\ d_3 a_{31} & d_3 a_{32} & d_3 a_{33} \end{bmatrix}$.
Comparing these,$AD = DA$ only if $d_1 = d_2 = d_3$ (i.e.,$D$ is a scalar matrix). Thus,the statement $AD = DA$ for every matrix $A$ is false.
Also,if $D^{-1}$ exists,$D^{-1} = \text{diag}(d_1^{-1}, d_2^{-1}, d_3^{-1})$,which is not necessarily a scalar matrix unless $d_1 = d_2 = d_3$. Therefore,both $B$ and $C$ are technically false,but $B$ is the most standard counter-example in matrix theory.

(B) Given that $A_k$ is a skew-symmetric matrix,we have $A_k^T = -A_k$.
For any odd power $m$,$(A_k^m)^T = (A_k^T)^m = (-A_k)^m = (-1)^m A_k^m$.
Since $m = 2r-1$ is always odd,$(A_{2r-1})^{2r-1}$ is skew-symmetric.
Thus,$B = \sum_{r=1}^n (2r-1)(A_{2r-1})^{2r-1}$ is a sum of skew-symmetric matrices.
Taking the transpose: $B^T = \sum_{r=1}^n (2r-1)((A_{2r-1})^{2r-1})^T = \sum_{r=1}^n (2r-1)(-(A_{2r-1})^{2r-1}) = -B$.
Therefore,$B$ is skew-symmetric.

(D) Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
Given $A \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$,we get:
$a - b = -1$ $(1)$
$c - d = 2$ $(2)$
Given $A^2 \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$,we can write this as $A \left( A \begin{bmatrix} 1 \\ -1 \end{bmatrix} \right) = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
Substituting $(1)$,we get $A \begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
This gives the system:
$-a + 2b = 1$ $(3)$
$-c + 2d = 0$ $(4)$
Solving $(1)$ and $(3)$: From $(1)$,$a = b - 1$. Substituting into $(3)$: $-(b - 1) + 2b = 1 \Rightarrow -b + 1 + 2b = 1 \Rightarrow b = 0$. Then $a = -1$.
Solving $(2)$ and $(4)$: From $(2)$,$c = d + 2$. Substituting into $(4)$: $-(d + 2) + 2d = 0 \Rightarrow -d - 2 + 2d = 0 \Rightarrow d = 2$. Then $c = 4$.
The matrix $A = \begin{bmatrix} -1 & 0 \\ 4 & 2 \end{bmatrix}$.
The sum of the elements is $a + b + c + d = -1 + 0 + 4 + 2 = 5$.

(D) Since $A$ is an idempotent matrix,$A^2 = A$.
This implies $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$,and $A^4 = A$.
Using the binomial expansion for $(I + A)^4$:
$(I + A)^4 = {^4C_0}I^4 + {^4C_1}I^3A + {^4C_2}I^2A^2 + {^4C_3}IA^3 + {^4C_4}A^4$
Since $I^n = I$ and $A^n = A$ for all $n \geq 1$:
$(I + A)^4 = I + 4A + 6A + 4A + A$
$(I + A)^4 = I + (4 + 6 + 4 + 1)A$
$(I + A)^4 = I + 15A$

(D) Given $A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$.
We calculate the powers of $A$:
$A^2 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}$
By induction,we can hypothesize that $A^n = \begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}$.
Now,let us check option $(D)$: $nA - (n-1)I = n \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} - (n-1) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} n & 0 \\ n & n \end{bmatrix} - \begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}$.
This matches our derived form for $A^n$. Thus,option $(D)$ is correct.

(D) Given that $A$ is a nilpotent matrix of index $2$,we have $A^{2} = O$,where $O$ is the null matrix.
Since $A^{2} = O$,it follows that $A^{n} = O$ for all $n \geq 2$.
Using the binomial expansion for $(I_2+A)^{51}$:
$(I_2+A)^{51} = \binom{51}{0} I_2^{51} + \binom{51}{1} I_2^{50} A + \binom{51}{2} I_2^{49} A^{2} + \dots + \binom{51}{51} A^{51}$.
Since $A^{2} = O$,all terms containing $A^{k}$ for $k \geq 2$ become $O$.
Thus,$(I_2+A)^{51} = I_2 + 51A$.
Now,multiply by $A$:
$A(I_2+A)^{51} = A(I_2 + 51A) = AI_2 + 51A^{2}$.
Since $AI_2 = A$ and $A^{2} = O$,we get:
$A(I_2+A)^{51} = A + 51(O) = A$.

(B) Let $A_k = \begin{bmatrix} 1 & 3k + \frac{1}{3} \\ 0 & 1 \end{bmatrix}$.
We know that $\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a+b \\ 0 & 1 \end{bmatrix}$.
Therefore,the product $\prod_{k=1}^{36} A_k = \begin{bmatrix} 1 & \sum_{k=1}^{36} (3k + \frac{1}{3}) \\ 0 & 1 \end{bmatrix}$.
Calculate the sum: $\sum_{k=1}^{36} (3k + \frac{1}{3}) = 3 \sum_{k=1}^{36} k + \sum_{k=1}^{36} \frac{1}{3}$.
$= 3 \times \frac{36 \times 37}{2} + 36 \times \frac{1}{3}$.
$= 3 \times 18 \times 37 + 12$.
$= 54 \times 37 + 12 = 1998 + 12 = 2010$.
Thus,the product is $\begin{bmatrix} 1 & 2010 \\ 0 & 1 \end{bmatrix}$.

(C) First,we calculate $BB^T$. Given $B = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$,its transpose is $B^T = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$.
$BB^T = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} \frac{3}{4} + \frac{1}{4} & -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \\ -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} & \frac{1}{4} + \frac{3}{4} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Now,$(BB^TA)^5 = (IA)^5 = A^5$.
We have $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.
$A^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.
$A^3 = A^2 A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.
By induction,$A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.
Therefore,$A^5 = \begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} 2 & -1 \\ -7 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 1 \\ 7 & 2 \end{bmatrix}$.
First,calculate $AB$:
$AB = \begin{bmatrix} 2 & -1 \\ -7 & 4 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 7 & 2 \end{bmatrix} = \begin{bmatrix} (2)(4) + (-1)(7) & (2)(1) + (-1)(2) \\ (-7)(4) + (4)(7) & (-7)(1) + (4)(2) \end{bmatrix} = \begin{bmatrix} 8-7 & 2-2 \\ -28+28 & -7+8 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Since $AB = I$,then $B = A^{-1}$.
Also,$BA = I$ because $A$ and $B$ are inverses of each other.
Now check the options:
Option $A$: $A^T = \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix}$,so $AA^T = \begin{bmatrix} 2 & -1 \\ -7 & 4 \end{bmatrix} \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 5 & -18 \\ -18 & 65 \end{bmatrix} \neq I$.
Option $B$: $(AB)^T = I^T = I$. This is correct.
Option $C$: $B^T = \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix}$,so $BB^T = \begin{bmatrix} 4 & 1 \\ 7 & 2 \end{bmatrix} \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 17 & 30 \\ 30 & 53 \end{bmatrix} \neq I$.
Option $D$: Since $AB = I$ and $BA = I$,then $AB = BA$. Thus,$AB \neq BA$ is false.
Therefore,the correct option is $B$.

(B) Let the given matrices be $A = [x \, y \, z]$,$B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$,and $C = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$.
The order of matrix $A$ is $1 \times 3$.
The order of matrix $B$ is $3 \times 3$.
The order of matrix $C$ is $3 \times 1$.
First,we calculate the product $AB$. Since the number of columns in $A$ $(3)$ equals the number of rows in $B$ $(3)$,the product $AB$ is defined and its order is $1 \times 3$.
Next,we calculate the product $(AB)C$. Since the number of columns in $(AB)$ $(3)$ equals the number of rows in $C$ $(3)$,the product is defined.
The resulting order is $(1 \times 3) \times (3 \times 1) = 1 \times 1$.

(D) square matrix $A$ with exactly one non-zero entry in each row and each column,where the non-zero entry is $\pm 1$,is known as a generalized permutation matrix.
Let $A$ be an $n \times n$ matrix. Since each row and column has exactly one non-zero entry equal to $1$ or $-1$,the product $AA^T$ will result in the identity matrix $I_n$.
Specifically,the $(i, j)$-th entry of $AA^T$ is the dot product of the $i$-th row and $j$-th row of $A$. If $i \neq j$,the rows are orthogonal,so the dot product is $0$. If $i = j$,the dot product is $(\pm 1)^2 = 1$.
Thus,$AA^T = I_n$,which implies that $A$ is an orthogonal matrix.
Since $A$ is orthogonal,$\det(A) = \pm 1$,so $A$ is always non-singular.

(B) Given $A = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix}$.
Calculate $A^2 = A \times A = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1(1) + 0(\frac{1}{2}) & 1(0) + 0(1) \\ \frac{1}{2}(1) + 1(\frac{1}{2}) & \frac{1}{2}(0) + 1(1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2(\frac{1}{2}) & 1 \end{bmatrix}$.
Calculate $A^3 = A^2 \times A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 + \frac{1}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{3}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3(\frac{1}{2}) & 1 \end{bmatrix}$.
By induction,we can generalize that $A^n = \begin{bmatrix} 1 & 0 \\ n(\frac{1}{2}) & 1 \end{bmatrix}$.
Therefore,for $n = 50$,$A^{50} = \begin{bmatrix} 1 & 0 \\ 50(\frac{1}{2}) & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 25 & 1 \end{bmatrix}$.

(C) Given $A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix}$.
Calculating powers of $A$:
$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$
$A^3 = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 3 & 1 \end{bmatrix}$
$A^4 = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 10 & 4 & 1 \end{bmatrix}$
By observing the pattern,for $A^n$,the first column elements are $1$,$n$,and $\frac{n(n+1)}{2}$.
Thus,$A^n = \begin{bmatrix} 1 & 0 & 0 \\ n & 1 & 0 \\ \frac{n(n+1)}{2} & n & 1 \end{bmatrix}$.
For $n = 20$:
$A^{20} = \begin{bmatrix} 1 & 0 & 0 \\ 20 & 1 & 0 \\ \frac{20(21)}{2} & 20 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 20 & 1 & 0 \\ 210 & 20 & 1 \end{bmatrix}$.
The sum of the elements of the first column is $1 + 20 + 210 = 231$.

(B) Given equations are:
$(1) \ A + B = 2B^T$
$(2) \ 3A + 2B = I_3$
Taking the transpose of equation $(1)$:
$(A + B)^T = (2B^T)^T \Rightarrow A^T + B^T = 2B$
From $(1)$,$A = 2B^T - B$. Substitute this into $(2)$:
$3(2B^T - B) + 2B = I_3 \Rightarrow 6B^T - 3B + 2B = I_3 \Rightarrow 6B^T - B = I_3 \Rightarrow B = 6B^T - I_3$
Substitute $B$ into $(1)$:
$A + (6B^T - I_3) = 2B^T \Rightarrow A = I_3 - 4B^T$
Taking transpose of $A = I_3 - 4B^T$:
$A^T = I_3 - 4B$
From $A^T + B^T = 2B$,we have $B^T = 2B - A^T$. Substitute $A^T = I_3 - 4B$:
$B^T = 2B - (I_3 - 4B) = 6B - I_3$
Substitute $B^T$ back into $B = 6B^T - I_3$:
$B = 6(6B - I_3) - I_3 = 36B - 6I_3 - I_3 = 36B - 7I_3$
$35B = 7I_3 \Rightarrow B = \frac{1}{5}I_3$
Now find $A$:
$A = I_3 - 4B^T = I_3 - 4(6B - I_3) = I_3 - 24B + 4I_3 = 5I_3 - 24(\frac{1}{5}I_3) = \frac{25I_3 - 24I_3}{5} = \frac{1}{5}I_3$
Check the options:
$10A + 5B = 10(\frac{1}{5}I_3) + 5(\frac{1}{5}I_3) = 2I_3 + I_3 = 3I_3$
Thus,the correct option is $B$.

(C) Given $P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$.
Since $P$ is an orthogonal matrix,$P^T P = P P^T = I$.
Given $Q = PAP^T$,we need to find $P^T Q^{2015} P$.
$Q^2 = (PAP^T)(PAP^T) = PA(P^T P)AP^T = PA(I)AP^T = PA^2 P^T$.
By induction,$Q^n = PA^n P^T$.
Therefore,$Q^{2015} = PA^{2015} P^T$.
Now,$P^T Q^{2015} P = P^T (PA^{2015} P^T) P = (P^T P) A^{2015} (P^T P) = I A^{2015} I = A^{2015}$.
Given $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$,we observe the pattern:
$A^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.
$A^3 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.
By induction,$A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.
Thus,$A^{2015} = \begin{bmatrix} 1 & 2015 \\ 0 & 1 \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} 1 & 2 & x \\ 3 & -1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} y \\ x \\ 1 \end{bmatrix}$.
Calculating the product $AB$:
$AB = \begin{bmatrix} 1 & 2 & x \\ 3 & -1 & 2 \end{bmatrix} \begin{bmatrix} y \\ x \\ 1 \end{bmatrix} = \begin{bmatrix} 1(y) + 2(x) + x(1) \\ 3(y) - 1(x) + 2(1) \end{bmatrix} = \begin{bmatrix} y + 3x \\ 3y - x + 2 \end{bmatrix}$.
Equating this to the given matrix $\begin{bmatrix} 6 \\ 8 \end{bmatrix}$:
$y + 3x = 6$ (Equation $1$)
$3y - x + 2 = 8 \Rightarrow 3y - x = 6$ (Equation $2$)
From Equation $1$,$y = 6 - 3x$. Substituting this into Equation $2$:
$3(6 - 3x) - x = 6$
$18 - 9x - x = 6$
$18 - 10x = 6$
$10x = 12 \Rightarrow x = \frac{12}{10} = \frac{6}{5}$.
Substituting $x = \frac{6}{5}$ into $y = 6 - 3x$:
$y = 6 - 3(\frac{6}{5}) = 6 - \frac{18}{5} = \frac{30 - 18}{5} = \frac{12}{5}$.
Comparing $x$ and $y$,we see that $y = 2x$ (since $\frac{12}{5} = 2 \times \frac{6}{5}$).
Thus,the correct option is $A$.

(A) Given $A = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(4) & (1)(2) + (2)(-3) \\ (4)(1) + (-3)(4) & (4)(2) + (-3)(-3) \end{bmatrix} = \begin{bmatrix} 9 & -4 \\ -8 & 17 \end{bmatrix}$.
Next,calculate $4A$:
$4A = 4 \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 16 & -12 \end{bmatrix}$.
Next,calculate $5I$:
$5I = 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$.
Now,compute $A^2 + 4A - 5I$:
$A^2 + 4A - 5I = \begin{bmatrix} 9 & -4 \\ -8 & 17 \end{bmatrix} + \begin{bmatrix} 4 & 8 \\ 16 & -12 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$
$= \begin{bmatrix} 9+4-5 & -4+8-0 \\ -8+16-0 & 17-12-5 \end{bmatrix} = \begin{bmatrix} 8 & 4 \\ 8 & 0 \end{bmatrix}$
$= 4 \begin{bmatrix} 2 & 1 \\ 2 & 0 \end{bmatrix}$.

(D) matrix $A = \begin{bmatrix} a & b \\ c & a \end{bmatrix}$ is in $S$ where $a, b, c \in \{0, 1, 2\}$.
There are $3$ choices for each of $a, b, c$,so the total number of matrices in $S$ is $3 \times 3 \times 3 = 27$.
For a matrix to be singular,its determinant must be zero: $\det(A) = a^2 - bc = 0$,which implies $a^2 = bc$.
We check the cases for $a \in \{0, 1, 2\}$:
Case $1$: $a = 0$. Then $bc = 0$. The pairs $(b, c)$ can be $(0, 0), (0, 1), (0, 2), (1, 0), (2, 0)$. There are $5$ such matrices.
Case $2$: $a = 1$. Then $bc = 1$. The only pair $(b, c)$ is $(1, 1)$. There is $1$ such matrix.
Case $3$: $a = 2$. Then $bc = 4$. The only pair $(b, c)$ is $(2, 2)$. There is $1$ such matrix.
Total singular matrices $= 5 + 1 + 1 = 7$.
Number of non-singular matrices $= \text{Total} - \text{Singular} = 27 - 7 = 20$.

(C) Given $A = \begin{bmatrix} \alpha - 1 \\ 0 \\ 0 \end{bmatrix}$ and $B = \begin{bmatrix} \alpha + 1 \\ 0 \\ 0 \end{bmatrix}$.
The transpose of matrix $B$ is $B^T = \begin{bmatrix} \alpha + 1 & 0 & 0 \end{bmatrix}$.
Now,calculate the product $AB^T$:
$AB^T = \begin{bmatrix} \alpha - 1 \\ 0 \\ 0 \end{bmatrix} \begin{bmatrix} \alpha + 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} (\alpha - 1)(\alpha + 1) & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \alpha^2 - 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
For $AB^T$ to be a non-zero matrix,at least one element must be non-zero.
Thus,$\alpha^2 - 1 \neq 0$,which implies $\alpha^2 \neq 1$.
Taking the square root on both sides,we get $|\alpha| \neq 1$.

(A) Given matrices are $A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 2 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 7 & -2 & 1 \end{bmatrix}$.
To find $AB$,we perform matrix multiplication:
$AB = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 7 & -2 & 1 \end{bmatrix}$
Calculating each element:
Row $1$: $(1)(1) + (0)(-2) + (0)(7) = 1$,$(1)(0) + (0)(1) + (0)(-2) = 0$,$(1)(0) + (0)(0) + (0)(1) = 0$
Row $2$: $(2)(1) + (1)(-2) + (0)(7) = 2 - 2 = 0$,$(2)(0) + (1)(1) + (0)(-2) = 1$,$(2)(0) + (1)(0) + (0)(1) = 0$
Row $3$: $(-3)(1) + (2)(-2) + (1)(7) = -3 - 4 + 7 = 0$,$(-3)(0) + (2)(1) + (1)(-2) = 2 - 2 = 0$,$(-3)(0) + (2)(0) + (1)(1) = 1$
Thus,$AB = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.

(A) Given $P = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix}$.
We can write $P = I + A$,where $A = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix}$.
Note that $A^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$ and $A^3 = O$.
Using the binomial expansion for $P^n = (I + A)^n = I + nA + \frac{n(n-1)}{2}A^2$ (since $A^3 = O$):
$P^5 = I + 5A + \frac{5 \times 4}{2}A^2 = I + 5A + 10A^2$.
$P^5 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + 5 \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} + 10 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 45+90 & 15 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 135 & 15 & 1 \end{bmatrix}$.
Given $Q = P^5 + I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 135 & 15 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 15 & 2 & 0 \\ 135 & 15 & 2 \end{bmatrix}$.
Thus,$q_{21} = 15$,$q_{31} = 135$,and $q_{32} = 15$.
The value of $\frac{q_{21} + q_{31}}{q_{32}} = \frac{15 + 135}{15} = \frac{150}{15} = 10$.

(B) Let $A = [a_{ij}]$ be a $3 \times 3$ matrix where $a_{ij} \in \{-1, 0, 1\}$.
The sum of the diagonal elements of $AA^{T}$ is given by $\operatorname{trace}(AA^{T})$.
We know that $\operatorname{trace}(AA^{T}) = \sum_{i=1}^{3} \sum_{j=1}^{3} a_{ij}^{2}$.
Given that $\operatorname{trace}(AA^{T}) = 3$,we have $\sum_{i=1}^{3} \sum_{j=1}^{3} a_{ij}^{2} = 3$.
Since $a_{ij} \in \{-1, 0, 1\}$,$a_{ij}^{2}$ can only be $0$ or $1$.
For the sum of nine such squares to be $3$,exactly three of the entries $a_{ij}$ must be $\pm 1$,and the remaining six entries must be $0$.
First,we choose $3$ positions out of $9$ for the non-zero entries in $\binom{9}{3}$ ways.
Then,for each of these $3$ chosen positions,the entry can be either $1$ or $-1$,which gives $2^{3}$ possibilities.
Therefore,the total number of such matrices is $\binom{9}{3} \times 2^{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 8 = 84 \times 8 = 672$.

(A) The given matrix is $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$.
First,calculate $2A$:
$2A = 2 \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$.
Now,calculate the determinant of $2A$ $(LHS)$:
$|2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} = (2 \times 4) - (4 \times 8) = 8 - 32 = -24$.
Next,calculate the determinant of $A$:
$|A| = \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = (1 \times 2) - (2 \times 4) = 2 - 8 = -6$.
Now,calculate $4|A|$ $(RHS)$:
$4|A| = 4 \times (-6) = -24$.
Since $LHS$ = $-24$ and $RHS$ = $-24$,we have shown that $|2A| = 4|A|$.

(N/A) The information is represented in the form of a $3 \times 2$ matrix $A$ as follows:
$A = \begin{bmatrix} 30 & 25 \\ 25 & 31 \\ 27 & 26 \end{bmatrix}$
In this matrix,the rows represent the factories $(I, II, III)$ and the columns represent the gender of the workers (Men,Women).
The entry in the third row and second column is $26$.
This entry represents the number of women workers in factory $III$.

(A) We know that if a matrix is of order $m \times n$,it has $mn$ elements.
To find all possible orders of a matrix with $8$ elements,we need to find all pairs of natural numbers $(m, n)$ such that their product $mn = 8$.
The factors of $8$ are $1, 2, 4, 8$.
The possible pairs $(m, n)$ are:
$1 \times 8 = 8$
$8 \times 1 = 8$
$4 \times 2 = 8$
$2 \times 4 = 8$
Therefore,the possible orders are $1 \times 8, 8 \times 1, 4 \times 2,$ and $2 \times 4$.

(C) In general,a $3 \times 2$ matrix is given by $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}$.
Given the formula $a_{ij} = \frac{1}{2}|i - 3j|$,where $i = 1, 2, 3$ and $j = 1, 2$,we calculate each element:
$a_{11} = \frac{1}{2}|1 - 3(1)| = \frac{1}{2}|-2| = 1$
$a_{12} = \frac{1}{2}|1 - 3(2)| = \frac{1}{2}|-5| = \frac{5}{2}$
$a_{21} = \frac{1}{2}|2 - 3(1)| = \frac{1}{2}|-1| = \frac{1}{2}$
$a_{22} = \frac{1}{2}|2 - 3(2)| = \frac{1}{2}|-4| = 2$
$a_{31} = \frac{1}{2}|3 - 3(1)| = \frac{1}{2}|0| = 0$
$a_{32} = \frac{1}{2}|3 - 3(2)| = \frac{1}{2}|-3| = \frac{3}{2}$
Thus,the required matrix is $A = \begin{bmatrix} 1 & \frac{5}{2} \\ \frac{1}{2} & 2 \\ 0 & \frac{3}{2} \end{bmatrix}$.

(D) Since the given matrices are equal,their corresponding elements must be equal.
Comparing the corresponding elements,we get:
$x+3 = 0 \implies x = -3$
$z+4 = 6 \implies z = 2$
$2y-7 = 3y-2 \implies 2y-3y = -2+7 \implies -y = 5 \implies y = -5$
$a-1 = -3 \implies a = -3+1 \implies a = -2$
$0 = 2c+2 \implies 2c = -2 \implies c = -1$
$b-3 = 2b+4 \implies b-2b = 4+3 \implies -b = 7 \implies b = -7$
Thus,the values are $a=-2, b=-7, c=-1, x=-3, y=-5, z=2$.

(A) By the definition of equality of two matrices,we equate the corresponding elements:
$1) \ 2a + b = 4$
$2) \ a - 2b = -3$
$3) \ 5c - d = 11$
$4) \ 4c + 3d = 24$
Solving equations $(1)$ and $(2)$:
From $(2)$,$a = 2b - 3$. Substituting into $(1)$:
$2(2b - 3) + b = 4 \implies 4b - 6 + b = 4 \implies 5b = 10 \implies b = 2$.
Then $a = 2(2) - 3 = 1$.
Solving equations $(3)$ and $(4)$:
From $(3)$,$d = 5c - 11$. Substituting into $(4)$:
$4c + 3(5c - 11) = 24 \implies 4c + 15c - 33 = 24 \implies 19c = 57 \implies c = 3$.
Then $d = 5(3) - 11 = 4$.
Thus,$a=1, b=2, c=3, d=4$.

(N/A) $(i)$ In the given matrix,the number of rows is $3$ and the number of columns is $4$. Therefore,the order of the matrix is $3 \times 4$.
$(ii)$ Since the order of the matrix is $3 \times 4$,the total number of elements is $3 \times 4 = 12$.
$(iii)$ By identifying the position of each element in the matrix:
$a_{13} = 19$ (first row,third column)
$a_{21} = 35$ (second row,first column)
$a_{33} = -5$ (third row,third column)
$a_{24} = 12$ (second row,fourth column)
$a_{23} = \frac{5}{2}$ (second row,third column)

(N/A) We know that if a matrix is of the order $m \times n$,it has $mn$ elements.
Thus,to find all the possible orders of a matrix having $24$ elements,we need to find all the ordered pairs of natural numbers whose product is $24$.
The ordered pairs are: $(1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6),$ and $(6, 4)$.
Hence,the possible orders of a matrix having $24$ elements are: $1 \times 24, 24 \times 1, 2 \times 12, 12 \times 2, 3 \times 8, 8 \times 3, 4 \times 6,$ and $6 \times 4$.
Similarly,for $13$ elements,we find the ordered pairs of natural numbers whose product is $13$. Since $13$ is a prime number,the only pairs are $(1, 13)$ and $(13, 1)$.
Hence,the possible orders of a matrix having $13$ elements are $1 \times 13$ and $13 \times 1$.

(N/A) We know that if a matrix is of the order $m \times n$,it has $mn$ elements.
To find all possible orders of a matrix with $18$ elements,we need to find all ordered pairs of natural numbers $(m, n)$ such that $mn = 18$.
The factors of $18$ are $1, 2, 3, 6, 9, 18$.
The possible ordered pairs $(m, n)$ are $(1, 18), (18, 1), (2, 9), (9, 2), (3, 6), (6, 3)$.
Thus,the possible orders are $1 \times 18, 18 \times 1, 2 \times 9, 9 \times 2, 3 \times 6, 6 \times 3$.
Similarly,for a matrix with $5$ elements,we need to find all ordered pairs $(m, n)$ such that $mn = 5$.
Since $5$ is a prime number,its only factors are $1$ and $5$.
The possible ordered pairs are $(1, 5)$ and $(5, 1)$.
Thus,the possible orders are $1 \times 5$ and $5 \times 1$.

(A) In general,a $3 \times 4$ matrix is given by $A=\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$.
Given: $a_{ij}=\frac{1}{2}|-3i+j|$,where $i=1, 2, 3$ and $j=1, 2, 3, 4$.
Calculating the elements:
$a_{11} = \frac{1}{2}|-3(1)+1| = \frac{1}{2}|-2| = 1$
$a_{12} = \frac{1}{2}|-3(1)+2| = \frac{1}{2}|-1| = \frac{1}{2}$
$a_{13} = \frac{1}{2}|-3(1)+3| = \frac{1}{2}|0| = 0$
$a_{14} = \frac{1}{2}|-3(1)+4| = \frac{1}{2}|1| = \frac{1}{2}$
$a_{21} = \frac{1}{2}|-3(2)+1| = \frac{1}{2}|-5| = \frac{5}{2}$
$a_{22} = \frac{1}{2}|-3(2)+2| = \frac{1}{2}|-4| = 2$
$a_{23} = \frac{1}{2}|-3(2)+3| = \frac{1}{2}|-3| = \frac{3}{2}$
$a_{24} = \frac{1}{2}|-3(2)+4| = \frac{1}{2}|-2| = 1$
$a_{31} = \frac{1}{2}|-3(3)+1| = \frac{1}{2}|-8| = 4$
$a_{32} = \frac{1}{2}|-3(3)+2| = \frac{1}{2}|-7| = \frac{7}{2}$
$a_{33} = \frac{1}{2}|-3(3)+3| = \frac{1}{2}|-6| = 3$
$a_{34} = \frac{1}{2}|-3(3)+4| = \frac{1}{2}|-5| = \frac{5}{2}$
Thus,the matrix is $A=\begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix}$.

(A) The matrix $A$ is of order $3 \times 4$,where $i \in \{1, 2, 3\}$ and $j \in \{1, 2, 3, 4\}$.
The elements are calculated using $a_{i j} = 2i - j$:
For $i=1$: $a_{11} = 2(1)-1 = 1, a_{12} = 2(1)-2 = 0, a_{13} = 2(1)-3 = -1, a_{14} = 2(1)-4 = -2$.
For $i=2$: $a_{21} = 2(2)-1 = 3, a_{22} = 2(2)-2 = 2, a_{23} = 2(2)-3 = 1, a_{24} = 2(2)-4 = 0$.
For $i=3$: $a_{31} = 2(3)-1 = 5, a_{32} = 2(3)-2 = 4, a_{33} = 2(3)-3 = 3, a_{34} = 2(3)-4 = 2$.
Thus,the required matrix is $A = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}$.

(C) Given the matrix equation: $\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$
Since the two matrices are equal,their corresponding elements must be equal.
By comparing the corresponding elements of the matrices:
$1$. For the element in the first row and first column: $y = 4$
$2$. For the element in the first row and second column: $z = 3$
$3$. For the element in the second row and first column: $x = 1$
Thus,the values are $x = 1, y = 4, z = 3$.

(D) Given the matrix equation: $\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$.
Since the matrices are equal,their corresponding elements must be equal.
Comparing the elements,we get:
$1) x+y = 6$
$2) xy = 8$
$3) 5+z = 5$
From equation $(3)$,$z = 5-5 = 0$.
Now,we have $x+y = 6$ and $xy = 8$. We can use the identity $(x-y)^2 = (x+y)^2 - 4xy$.
$(x-y)^2 = (6)^2 - 4(8) = 36 - 32 = 4$.
Taking the square root,$x-y = \pm 2$.
Case $1$: If $x-y = 2$ and $x+y = 6$,adding the equations gives $2x = 8 \Rightarrow x = 4$. Substituting $x=4$ into $x+y=6$ gives $y = 2$.
Case $2$: If $x-y = -2$ and $x+y = 6$,adding the equations gives $2x = 4 \Rightarrow x = 2$. Substituting $x=2$ into $x+y=6$ gives $y = 4$.
Thus,the values are $(x=4, y=2, z=0)$ or $(x=2, y=4, z=0)$.