Types of matrices, Algebra of matrices Questions in English

(A) Given the matrix equation: $\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}$
Since the two matrices are equal,their corresponding elements must be equal. Comparing the corresponding elements,we get:
$x+y+z = 9$ $(1)$
$x+z = 5$ $(2)$
$y+z = 7$ $(3)$
Substituting equation $(2)$ into equation $(1)$:
$y + (x+z) = 9$
$y + 5 = 9 \Rightarrow y = 4$
Now,substitute $y=4$ into equation $(3)$:
$4 + z = 7 \Rightarrow z = 3$
Finally,substitute $z=3$ into equation $(2)$:
$x + 3 = 5 \Rightarrow x = 2$
Thus,the values are $x=2, y=4, z=3$.

(B) Given the matrix equation: $\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$
Since the two matrices are equal,their corresponding elements must be equal.
Comparing the corresponding elements,we get:
$a-b = -1$ $(1)$
$2a-b = 0$ $(2)$
$2a+c = 5$ $(3)$
$3c+d = 13$ $(4)$
From equation $(2)$,we have $b = 2a$.
Substituting $b = 2a$ into equation $(1)$:
$a - 2a = -1$
$-a = -1 \Rightarrow a = 1$
Now,find $b$ using $b = 2a$:
$b = 2(1) = 2$
Substitute $a = 1$ into equation $(3)$:
$2(1) + c = 5$
$2 + c = 5 \Rightarrow c = 3$
Substitute $c = 3$ into equation $(4)$:
$3(3) + d = 13$
$9 + d = 13 \Rightarrow d = 4$
Therefore,the values are $a=1, b=2, c=3,$ and $d=4$.

(C) matrix is defined as a square matrix if the number of rows is equal to the number of columns.
In the given matrix $A = [a_{ij}]_{m \times n}$,the number of rows is $m$ and the number of columns is $n$.
Therefore,for the matrix to be a square matrix,the condition $m = n$ must be satisfied.

(D) Given the matrix equation: $\left[\begin{array}{cc}3x+7 & 5 \\ y+1 & 2-3x\end{array}\right]=\left[\begin{array}{cc}0 & y-2 \\ 8 & 4\end{array}\right]$
By equating the corresponding elements,we obtain the following equations:
$1$. $3x+7=0 \Rightarrow 3x=-7 \Rightarrow x=-\frac{7}{3}$
$2$. $5=y-2 \Rightarrow y=7$
$3$. $y+1=8 \Rightarrow y=7$
$4$. $2-3x=4 \Rightarrow -3x=2 \Rightarrow x=-\frac{2}{3}$
Comparing the results,we observe that the value of $x$ obtained from the first element is $-\frac{7}{3}$,while the value of $x$ obtained from the fourth element is $-\frac{2}{3}$.
Since a single variable $x$ cannot take two different values simultaneously,it is not possible to find values of $x$ and $y$ that satisfy the equality of these two matrices.

(A) matrix of order $3 \times 3$ contains $3 \times 3 = 9$ elements.
Each of these $9$ elements can be filled in $2$ ways,either with $0$ or $1$.
By the fundamental principle of counting,the total number of possible matrices is the product of the number of choices for each position.
Therefore,the total number of matrices is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^9$.
Calculating the value,we get $2^9 = 512$.

(B) Since $A$ and $B$ are of the same order $2 \times 3$,their addition is defined by adding corresponding elements.
$A + B = \begin{bmatrix} \sqrt{3} + 2 & 1 + \sqrt{5} & -1 + 1 \\ 2 + (-2) & 3 + 3 & 0 + \frac{1}{2} \end{bmatrix}$
$A + B = \begin{bmatrix} 2 + \sqrt{3} & 1 + \sqrt{5} & 0 \\ 0 & 6 & \frac{1}{2} \end{bmatrix}$

(C) We have $2A - B = 2 \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} - \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}$.
First,multiply matrix $A$ by the scalar $2$:
$2A = \begin{bmatrix} 2(1) & 2(2) & 2(3) \\ 2(2) & 2(3) & 2(1) \end{bmatrix} = \begin{bmatrix} 2 & 4 & 6 \\ 4 & 6 & 2 \end{bmatrix}$.
Now,subtract matrix $B$ from $2A$:
$2A - B = \begin{bmatrix} 2 & 4 & 6 \\ 4 & 6 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}$.
Perform element-wise subtraction:
$2A - B = \begin{bmatrix} 2 - 3 & 4 - (-1) & 6 - 3 \\ 4 - (-1) & 6 - 0 & 2 - 2 \end{bmatrix} = \begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}$.

(D) Given the equation $2A + 3X = 5B$.
Subtract $2A$ from both sides: $3X = 5B - 2A$.
Thus,$X = \frac{1}{3}(5B - 2A)$.
Substitute the matrices $A$ and $B$:
$X = \frac{1}{3} \left( 5 \begin{bmatrix} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{bmatrix} - 2 \begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix} \right)$
$X = \frac{1}{3} \left( \begin{bmatrix} 10 & -10 \\ 20 & 10 \\ -25 & 5 \end{bmatrix} - \begin{bmatrix} 16 & 0 \\ 8 & -4 \\ 6 & 12 \end{bmatrix} \right)$
$X = \frac{1}{3} \begin{bmatrix} 10-16 & -10-0 \\ 20-8 & 10-(-4) \\ -25-6 & 5-12 \end{bmatrix}$
$X = \frac{1}{3} \begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix}$
$X = \begin{bmatrix} -2 & -\frac{10}{3} \\ 4 & \frac{14}{3} \\ -\frac{31}{3} & -\frac{7}{3} \end{bmatrix}$

(A) Given equations are:
$X + Y = \left[\begin{array}{ll}5 & 2 \\ 0 & 9\end{array}\right]$ --- $(1)$
$X - Y = \left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$ --- $(2)$
Adding equations $(1)$ and $(2)$:
$(X + Y) + (X - Y) = \left[\begin{array}{ll}5 & 2 \\ 0 & 9\end{array}\right] + \left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$
$2X = \left[\begin{array}{ll}5+3 & 2+6 \\ 0+0 & 9-1\end{array}\right] = \left[\begin{array}{ll}8 & 8 \\ 0 & 8\end{array}\right]$
$X = \frac{1}{2} \left[\begin{array}{ll}8 & 8 \\ 0 & 8\end{array}\right] = \left[\begin{array}{ll}4 & 4 \\ 0 & 4\end{array}\right]$
Subtracting equation $(2)$ from $(1)$:
$(X + Y) - (X - Y) = \left[\begin{array}{ll}5 & 2 \\ 0 & 9\end{array}\right] - \left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$
$2Y = \left[\begin{array}{ll}5-3 & 2-6 \\ 0-0 & 9-(-1)\end{array}\right] = \left[\begin{array}{ll}2 & -4 \\ 0 & 10\end{array}\right]$
$Y = \frac{1}{2} \left[\begin{array}{ll}2 & -4 \\ 0 & 10\end{array}\right] = \left[\begin{array}{ll}1 & -2 \\ 0 & 5\end{array}\right]$

(A) Given the equation:
$2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$
Multiply the first matrix by $2$:
$\begin{bmatrix} 2x & 10 \\ 14 & 2y-6 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$
Add the two matrices on the left side:
$\begin{bmatrix} 2x+3 & 10-4 \\ 14+1 & 2y-6+2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$
$\begin{bmatrix} 2x+3 & 6 \\ 15 & 2y-4 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$
By comparing the corresponding elements:
$2x+3 = 7 \Rightarrow 2x = 4 \Rightarrow x = 2$
$2y-4 = 14 \Rightarrow 2y = 18 \Rightarrow y = 9$
Thus,$x=2$ and $y=9$.

(N/A) $(i)$ The combined sales in September and October for each farmer in each variety is given by the sum of matrices $A$ and $B$:
$A+B = \begin{bmatrix} \text{Basmati} & \text{Permal} & \text{Naura} \\ 15,000 & 30,000 & 36,000 \\ 70,000 & 40,000 & 20,000 \end{bmatrix} \begin{matrix} \\ \text{Ramkishan} \\ \text{Gurcharan Singh} \end{matrix}$
$(ii)$ The decrease in sales from September to October is given by the difference of matrices $A$ and $B$:
$A-B = \begin{bmatrix} \text{Basmati} & \text{Permal} & \text{Naura} \\ 5,000 & 10,000 & 24,000 \\ 30,000 & 20,000 & 0 \end{bmatrix} \begin{matrix} \\ \text{Ramkishan} \\ \text{Gurcharan Singh} \end{matrix}$
$(iii)$ The profit is $2\%$ of the October sales $(B)$:
$0.02 \times B = 0.02 \times \begin{bmatrix} 5,000 & 10,000 & 6,000 \\ 20,000 & 10,000 & 10,000 \end{bmatrix} = \begin{bmatrix} 100 & 200 & 120 \\ 400 & 200 & 200 \end{bmatrix}$
Thus,in October,Ramkishan receives Rs. $100$,Rs. $200$,and Rs. $120$ as profit for the three varieties,and Gurcharan Singh receives Rs. $400$,Rs. $200$,and Rs. $200$ as profit for the three varieties,respectively.

(A) The matrix $A$ has $2$ columns,which is equal to the number of rows of matrix $B$. Therefore,the product $AB$ is defined.
$AB = \left[\begin{array}{ll}6 & 9 \\ 2 & 3\end{array}\right] \left[\begin{array}{lll}2 & 6 & 0 \\ 7 & 9 & 8\end{array}\right]$
$AB = \left[\begin{array}{lll}6(2)+9(7) & 6(6)+9(9) & 6(0)+9(8) \\ 2(2)+3(7) & 2(6)+3(9) & 2(0)+3(8)\end{array}\right]$
$AB = \left[\begin{array}{lll}12+63 & 36+81 & 0+72 \\ 4+21 & 12+27 & 0+24\end{array}\right]$
$AB = \left[\begin{array}{ccc}75 & 117 & 72 \\ 25 & 39 & 24\end{array}\right]$

Since $A$ is a $2 \times 3$ matrix and $B$ is a $3 \times 2$ matrix,both $AB$ and $BA$ are defined. The order of $AB$ is $2 \times 2$ and the order of $BA$ is $3 \times 3$.
$AB = \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} (1)(2) + (-2)(4) + (3)(2) & (1)(3) + (-2)(5) + (3)(1) \\ (-4)(2) + (2)(4) + (5)(2) & (-4)(3) + (2)(5) + (5)(1) \end{bmatrix} = \begin{bmatrix} 2 - 8 + 6 & 3 - 10 + 3 \\ -8 + 8 + 10 & -12 + 10 + 5 \end{bmatrix} = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}$.
$BA = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix} = \begin{bmatrix} (2)(1) + (3)(-4) & (2)(-2) + (3)(2) & (2)(3) + (3)(5) \\ (4)(1) + (5)(-4) & (4)(-2) + (5)(2) & (4)(3) + (5)(5) \\ (2)(1) + (1)(-4) & (2)(-2) + (1)(2) & (2)(3) + (1)(5) \end{bmatrix} = \begin{bmatrix} 2 - 12 & -4 + 6 & 6 + 15 \\ 4 - 20 & -8 + 10 & 12 + 25 \\ 2 - 4 & -4 + 2 & 6 + 5 \end{bmatrix} = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}$.
Since $AB$ is a $2 \times 2$ matrix and $BA$ is a $3 \times 3$ matrix,clearly $AB \neq BA$.

(B) The statement that $AB \neq BA$ does not hold for every pair of matrices $A$ and $B$ for which $AB$ and $BA$ are defined.
For instance,if $A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}$,then:
$AB = \begin{bmatrix} 1 \times 3 + 0 \times 0 & 1 \times 0 + 0 \times 4 \\ 0 \times 3 + 2 \times 0 & 0 \times 0 + 2 \times 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 8 \end{bmatrix}$
$BA = \begin{bmatrix} 3 \times 1 + 0 \times 0 & 3 \times 0 + 0 \times 2 \\ 0 \times 1 + 4 \times 0 & 0 \times 0 + 4 \times 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 8 \end{bmatrix}$
Since $AB = BA$,matrix multiplication is commutative in this case.
Thus,the multiplication of diagonal matrices of the same order is commutative.

(A) Given $A = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}$.
To find $AB$,we perform matrix multiplication:
$AB = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}$
$AB = \begin{bmatrix} (0 \times 3) + (-1 \times 0) & (0 \times 5) + (-1 \times 0) \\ (0 \times 3) + (2 \times 0) & (0 \times 5) + (2 \times 0) \end{bmatrix}$
$AB = \begin{bmatrix} 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Thus,the product $AB$ is the zero matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.

(N/A) First,we calculate $AB$:
$AB = \left[\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2\end{array}\right] \left[\begin{array}{rr}1 & 3 \\ 0 & 2 \\ -1 & 4\end{array}\right] = \left[\begin{array}{rr}1+0+1 & 3+2-4 \\ 2+0-3 & 6+0+12 \\ 3+0-2 & 9-2+8\end{array}\right] = \left[\begin{array}{rr}2 & 1 \\ -1 & 18 \\ 1 & 15\end{array}\right]$
Next,we calculate $(AB)C$:
$(AB)C = \left[\begin{array}{rr}2 & 1 \\ -1 & 18 \\ 1 & 15\end{array}\right] \left[\begin{array}{rrrr}1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1\end{array}\right] = \left[\begin{array}{rrrr}2+2 & 4+0 & 6-2 & -8+1 \\ -1+36 & -2+0 & -3-36 & 4+18 \\ 1+30 & 2+0 & 3-30 & -4+15\end{array}\right] = \left[\begin{array}{rrrr}4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11\end{array}\right]$
Now,we calculate $BC$:
$BC = \left[\begin{array}{rr}1 & 3 \\ 0 & 2 \\ -1 & 4\end{array}\right] \left[\begin{array}{rrrr}1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1\end{array}\right] = \left[\begin{array}{rrrr}1+6 & 2+0 & 3-6 & -4+3 \\ 0+4 & 0+0 & 0-4 & 0+2 \\ -1+8 & -2+0 & -3-8 & 4+4\end{array}\right] = \left[\begin{array}{rrrr}7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8\end{array}\right]$
Finally,we calculate $A(BC)$:
$A(BC) = \left[\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2\end{array}\right] \left[\begin{array}{rrrr}7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8\end{array}\right] = \left[\begin{array}{rrrr}7+4-7 & 2+0+2 & -3-4+11 & -1+2-8 \\ 14+0+21 & 4+0-6 & -6+0-33 & -2+0+24 \\ 21-4+14 & 6+0-4 & -9+4-22 & -3-2+16\end{array}\right] = \left[\begin{array}{rrrr}4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11\end{array}\right]$
Since $(AB)C = A(BC)$,the associative property of matrix multiplication is verified.

(N/A) First,we calculate $A+B$:
$A+B = \left[\begin{array}{rrr}0+0 & 6+1 & 7+1 \\ -6+1 & 0+0 & 8+2 \\ 7+1 & -8+2 & 0+0\end{array}\right] = \left[\begin{array}{rrr}0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0\end{array}\right]$.
Now,calculate $(A+B)C$:
$(A+B)C = \left[\begin{array}{rrr}0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0\end{array}\right] \left[\begin{array}{r}2 \\ -2 \\ 3\end{array}\right] = \left[\begin{array}{r}0(2) + 7(-2) + 8(3) \\ -5(2) + 0(-2) + 10(3) \\ 8(2) + (-6)(-2) + 0(3)\end{array}\right] = \left[\begin{array}{r}0 - 14 + 24 \\ -10 + 0 + 30 \\ 16 + 12 + 0\end{array}\right] = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$.
Next,calculate $AC$:
$AC = \left[\begin{array}{rrr}0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0\end{array}\right] \left[\begin{array}{r}2 \\ -2 \\ 3\end{array}\right] = \left[\begin{array}{r}0(2) + 6(-2) + 7(3) \\ -6(2) + 0(-2) + 8(3) \\ 7(2) + (-8)(-2) + 0(3)\end{array}\right] = \left[\begin{array}{r}0 - 12 + 21 \\ -12 + 0 + 24 \\ 14 + 16 + 0\end{array}\right] = \left[\begin{array}{r}9 \\ 12 \\ 30\end{array}\right]$.
Next,calculate $BC$:
$BC = \left[\begin{array}{rrr}0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0\end{array}\right] \left[\begin{array}{r}2 \\ -2 \\ 3\end{array}\right] = \left[\begin{array}{r}0(2) + 1(-2) + 1(3) \\ 1(2) + 0(-2) + 2(3) \\ 1(2) + 2(-2) + 0(3)\end{array}\right] = \left[\begin{array}{r}0 - 2 + 3 \\ 2 + 0 + 6 \\ 2 - 4 + 0\end{array}\right] = \left[\begin{array}{r}1 \\ 8 \\ -2\end{array}\right]$.
Finally,calculate $AC + BC$:
$AC + BC = \left[\begin{array}{r}9 \\ 12 \\ 30\end{array}\right] + \left[\begin{array}{r}1 \\ 8 \\ -2\end{array}\right] = \left[\begin{array}{r}9+1 \\ 12+8 \\ 30-2\end{array}\right] = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$.
Since $(A+B)C = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$ and $AC+BC = \left[\begin{array}{r}10 \\ 20 \\ 28\end{array}\right]$,we have verified that $(A+B)C = AC + BC$.

(A) To find $A + B$,we add the corresponding elements of matrices $A$ and $B$:
$A + B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$
$= \begin{bmatrix} 2 + 1 & 4 + 3 \\ 3 + (-2) & 2 + 5 \end{bmatrix}$
$= \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}$

(A) To find $A - B$,we subtract the corresponding elements of matrix $B$ from matrix $A$:
$A - B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$
$= \begin{bmatrix} 2 - 1 & 4 - 3 \\ 3 - (-2) & 2 - 5 \end{bmatrix}$
$= \begin{bmatrix} 1 & 1 \\ 3 + 2 & -3 \end{bmatrix}$
$= \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}$

(A) Given $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ and $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$.
First,calculate $3A$:
$3A = 3 \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 3 \times 2 & 3 \times 4 \\ 3 \times 3 & 3 \times 2 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}$.
Now,subtract $C$ from $3A$:
$3A - C = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$.
Perform element-wise subtraction:
$3A - C = \begin{bmatrix} 6 - (-2) & 12 - 5 \\ 9 - 3 & 6 - 4 \end{bmatrix} = \begin{bmatrix} 6 + 2 & 7 \\ 6 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}$.

(A) Matrix $A$ has $2$ columns,and matrix $B$ has $2$ rows. Since the number of columns in $A$ equals the number of rows in $B$,the product $AB$ is defined.
$AB = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$
$AB = \begin{bmatrix} (2 \times 1) + (4 \times -2) & (2 \times 3) + (4 \times 5) \\ (3 \times 1) + (2 \times -2) & (3 \times 3) + (2 \times 5) \end{bmatrix}$
$AB = \begin{bmatrix} 2 - 8 & 6 + 20 \\ 3 - 4 & 9 + 10 \end{bmatrix}$
$AB = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}$

(A) To find the product $BA$,we multiply matrix $B$ by matrix $A$ as follows:
$BA = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$
Performing matrix multiplication:
$BA = \begin{bmatrix} (1 \times 2) + (3 \times 3) & (1 \times 4) + (3 \times 2) \\ (-2 \times 2) + (5 \times 3) & (-2 \times 4) + (5 \times 2) \end{bmatrix}$
$BA = \begin{bmatrix} 2 + 9 & 4 + 6 \\ -4 + 15 & -8 + 10 \end{bmatrix}$
$BA = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}$

(A) To add two matrices,we add their corresponding elements:
$\begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} a+a & b+b \\ -b+b & a+a \end{bmatrix}$
$= \begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}$

(A) To compute the sum of the two matrices,we add the corresponding elements of each matrix:
$\begin{bmatrix} a^2 + b^2 & b^2 + c^2 \\ a^2 + c^2 & a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix}$
$= \begin{bmatrix} a^2 + b^2 + 2ab & b^2 + c^2 + 2bc \\ a^2 + c^2 - 2ac & a^2 + b^2 - 2ab \end{bmatrix}$
Using the algebraic identities $(x+y)^2 = x^2 + y^2 + 2xy$ and $(x-y)^2 = x^2 + y^2 - 2xy$,we get:
$= \begin{bmatrix} (a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2 \end{bmatrix}$

(A) To add two matrices,we add their corresponding elements:
$\begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix}$
$= \begin{bmatrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{bmatrix}$
$= \begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}$

(A) To compute the sum of the two matrices,we add the corresponding elements of each matrix:
$\left[\begin{array}{cc}\cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x\end{array}\right]+\left[\begin{array}{cc}\sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x\end{array}\right]$
$= \left[\begin{array}{cc}\cos ^{2} x+\sin ^{2} x & \sin ^{2} x+\cos ^{2} x \\ \sin ^{2} x+\cos ^{2} x & \cos ^{2} x+\sin ^{2} x\end{array}\right]$
Using the trigonometric identity $\sin^2 x + \cos^2 x = 1$,we substitute the value into the matrix:
$= \left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right]$

(A) To multiply the two matrices,we use the row-by-column multiplication rule:
$\begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix} = \begin{bmatrix} a(a) + b(b) & a(-b) + b(a) \\ -b(a) + a(b) & -b(-b) + a(a) \end{bmatrix}$
$= \begin{bmatrix} a^2 + b^2 & -ab + ab \\ -ab + ab & b^2 + a^2 \end{bmatrix}$
$= \begin{bmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{bmatrix}$

(A) To find the product of the matrices $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} 2 & 3 & 4 \end{bmatrix}$,we multiply each element of the column matrix by the row matrix.
The product is given by:
$\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 1(2) & 1(3) & 1(4) \\ 2(2) & 2(3) & 2(4) \\ 3(2) & 3(3) & 3(4) \end{bmatrix}$
Calculating each element:
$= \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}$

(A) To multiply the two matrices,we perform row-by-column multiplication:
$\left[\begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$
$= \left[\begin{array}{ccc}1(1) + (-2)(2) & 1(2) + (-2)(3) & 1(3) + (-2)(1) \\ 2(1) + 3(2) & 2(2) + 3(3) & 2(3) + 3(1)\end{array}\right]$
$= \left[\begin{array}{ccc}1 - 4 & 2 - 6 & 3 - 2 \\ 2 + 6 & 4 + 9 & 6 + 3\end{array}\right]$
$= \left[\begin{array}{ccc}-3 & -4 & 1 \\ 8 & 13 & 9\end{array}\right]$

(A) To compute the product of the two matrices,we multiply the rows of the first matrix by the columns of the second matrix:
$\left[\begin{array}{lll}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]$
$= \left[\begin{array}{lll}2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \\ 3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \\ 4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5)\end{array}\right]$
$= \left[\begin{array}{lll}2+0+12 & -6+6+0 & 10+12+20 \\ 3+0+15 & -9+8+0 & 15+16+25 \\ 4+0+18 & -12+10+0 & 20+20+30\end{array}\right]$
$= \left[\begin{array}{ccc}14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70\end{array}\right]$

(A) To multiply the two matrices,we perform row-by-column multiplication:
$\left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ -1 & 1\end{array}\right] \times \left[\begin{array}{ccc}1 & 0 & 1 \\ -1 & 2 & 1\end{array}\right]$
$= \left[\begin{array}{ccc}2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\ 3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1) \\ -1(1)+1(-1) & -1(0)+1(2) & -1(1)+1(1)\end{array}\right]$
$= \left[\begin{array}{ccc}2-1 & 0+2 & 2+1 \\ 3-2 & 0+4 & 3+2 \\ -1-1 & 0+2 & -1+1\end{array}\right]$
$= \left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0\end{array}\right]$

(A) To compute the product of the two matrices,we multiply the rows of the first matrix by the columns of the second matrix:
$\left[\begin{array}{ccc}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right]\left[\begin{array}{cc}2 & -3 \\ 1 & 0 \\ 3 & 1\end{array}\right]$
$= \left[\begin{array}{cc} (3)(2) + (-1)(1) + (3)(3) & (3)(-3) + (-1)(0) + (3)(1) \\ (-1)(2) + (0)(1) + (2)(3) & (-1)(-3) + (0)(0) + (2)(1) \end{array}\right]$
$= \left[\begin{array}{cc} 6 - 1 + 9 & -9 - 0 + 3 \\ -2 + 0 + 6 & 3 + 0 + 2 \end{array}\right]$
$= \left[\begin{array}{cc} 14 & -6 \\ 4 & 5 \end{array}\right]$

First,we compute $A+B$:
$A+B = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{bmatrix} = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix}$
Next,we compute $B-C$:
$B-C = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3 \end{bmatrix} = \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix}$
Now,we verify $A+(B-C) = (A+B)-C$:
$A+(B-C) = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$
$(A+B)-C = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$
Since both sides are equal,the property $A+(B-C)=(A+B)-C$ is verified.

(A) Given $A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$.
First,calculate $3A$:
$3A = 3 \times \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$.
Next,calculate $5B$:
$5B = 5 \times \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$.
Finally,compute $3A - 5B$:
$3A - 5B = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.

(A) Given expression: $\cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix}$
Multiply the scalar into the matrices:
$= \begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ -\sin \theta \cos \theta & \cos^2 \theta \end{bmatrix} + \begin{bmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix}$
Add the two matrices:
$= \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & \cos \theta \sin \theta - \sin \theta \cos \theta \\ -\sin \theta \cos \theta + \sin \theta \cos \theta & \cos^2 \theta + \sin^2 \theta \end{bmatrix}$
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$:
$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

(A) Given equations are:
$X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$ ............ $(1)$
$X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$ ............ $(2)$
Adding equations $(1)$ and $(2)$,we get:
$2X = \left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right] + \left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right] = \left[\begin{array}{ll}7+3 & 0+0 \\ 2+0 & 5+3\end{array}\right] = \left[\begin{array}{ll}10 & 0 \\ 2 & 8\end{array}\right]$
$\therefore X = \frac{1}{2}\left[\begin{array}{ll}10 & 0 \\ 2 & 8\end{array}\right] = \left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right]$
Now,substituting $X$ in equation $(1)$:
$\left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right] + Y = \left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$
$Y = \left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right] - \left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right]$
$Y = \left[\begin{array}{ll}7-5 & 0-0 \\ 2-1 & 5-4\end{array}\right] = \left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]$
Thus,$X = \left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right]$ and $Y = \left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]$.

(A) Given equations are:
$2X+3Y=\left[\begin{array}{cc}2 & 3 \\ 4 & 0\end{array}\right]$ ... $(1)$
$3X+2Y=\left[\begin{array}{cc}2 & -2 \\ -1 & 5\end{array}\right]$ ... $(2)$
Multiply equation $(1)$ by $2$ and equation $(2)$ by $3$:
$4X+6Y=\left[\begin{array}{cc}4 & 6 \\ 8 & 0\end{array}\right]$ ... $(3)$
$9X+6Y=\left[\begin{array}{cc}6 & -6 \\ -3 & 15\end{array}\right]$ ... $(4)$
Subtracting $(3)$ from $(4)$:
$(9X+6Y)-(4X+6Y)=\left[\begin{array}{cc}6 & -6 \\ -3 & 15\end{array}\right]-\left[\begin{array}{cc}4 & 6 \\ 8 & 0\end{array}\right]$
$5X=\left[\begin{array}{cc}2 & -12 \\ -11 & 15\end{array}\right] \Rightarrow X=\left[\begin{array}{cc}\frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3\end{array}\right]$
Now,substitute $X$ in $(1)$:
$3Y=\left[\begin{array}{cc}2 & 3 \\ 4 & 0\end{array}\right]-2\left[\begin{array}{cc}\frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3\end{array}\right]$
$3Y=\left[\begin{array}{cc}2 & 3 \\ 4 & 0\end{array}\right]-\left[\begin{array}{cc}\frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6\end{array}\right]=\left[\begin{array}{cc}\frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6\end{array}\right]$
$Y=\left[\begin{array}{cc}\frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2\end{array}\right]$

(A) Given the equation $2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$ and $Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$.
Substitute the matrix $Y$ into the equation:
$2X + \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$
Subtract matrix $Y$ from both sides:
$2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$
$2X = \begin{bmatrix} 1-3 & 0-2 \\ -3-1 & 2-4 \end{bmatrix}$
$2X = \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}$
Divide by $2$ to find $X$:
$X = \frac{1}{2} \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}$

(A) Given equation: $2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
Multiplying the first matrix by $2$: $\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
Adding the two matrices: $\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
Comparing corresponding elements:
$2+y = 5 \Rightarrow y = 3$
$2x+2 = 8 \Rightarrow 2x = 6 \Rightarrow x = 3$
Thus,$x=3$ and $y=3$.

(A) Given the equation:
$2\begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$
Multiplying the matrices by the scalars:
$\begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$
Adding the matrices on the left side:
$\begin{bmatrix} 2x+3 & 2z-3 \\ 2y+0 & 2t+6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$
Comparing the corresponding elements:
$1) 2x+3 = 9 \Rightarrow 2x = 6 \Rightarrow x = 3$
$2) 2z-3 = 15 \Rightarrow 2z = 18 \Rightarrow z = 9$
$3) 2y = 12 \Rightarrow y = 6$
$4) 2t+6 = 18 \Rightarrow 2t = 12 \Rightarrow t = 6$
Thus,$x=3, y=6, z=9, t=6$.

(A) Given the matrix equation: $x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$
Multiplying the scalars $x$ and $y$ into the matrices,we get:
$\begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$
Adding the matrices on the left side:
$\begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$
By comparing the corresponding elements,we obtain a system of linear equations:
$1) \ 2x - y = 10$
$2) \ 3x + y = 5$
Adding equation $(1)$ and $(2)$:
$(2x - y) + (3x + y) = 10 + 5$
$5x = 15$
$x = 3$
Substituting $x = 3$ into equation $(2)$:
$3(3) + y = 5$
$9 + y = 5$
$y = 5 - 9$
$y = -4$
Thus,the values are $x = 3$ and $y = -4$.

(A) Given the matrix equation:
$3\begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix}$
Multiplying the scalar $3$ into the first matrix:
$\begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix}$
By equating the corresponding elements of the matrices:
$1) \ 3x = x + 4 \Rightarrow 2x = 4 \Rightarrow x = 2$
$2) \ 3y = 6 + x + y \Rightarrow 2y = 6 + 2 = 8 \Rightarrow y = 4$
$3) \ 3w = 2w + 3 \Rightarrow w = 3$
$4) \ 3z = -1 + z + w \Rightarrow 2z = -1 + 3 = 2 \Rightarrow z = 1$
Thus,the values are $x=2, y=4, z=1, w=3$.

(A) Given $F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $F(y) = \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
$L.H.S = F(x) F(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Performing matrix multiplication:
$= \begin{bmatrix} \cos x \cos y - \sin x \sin y & -\cos x \sin y - \sin x \cos y & 0 \\ \sin x \cos y + \cos x \sin y & -\sin x \sin y + \cos x \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Using trigonometric identities:
$\cos(A+B) = \cos A \cos B - \sin A \sin B$
$\sin(A+B) = \sin A \cos B + \cos A \sin B$
$= \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$= F(x+y) = R.H.S$
Hence,$F(x) F(y) = F(x+y)$.

Let $A = \left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]$ and $B = \left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]$.
First,calculate the product $AB$:
$AB = \left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right] \left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]$
$= \left[\begin{array}{cc}5(2) + (-1)(3) & 5(1) + (-1)(4) \\ 6(2) + 7(3) & 6(1) + 7(4)\end{array}\right]$
$= \left[\begin{array}{cc}10 - 3 & 5 - 4 \\ 12 + 21 & 6 + 28\end{array}\right] = \left[\begin{array}{cc}7 & 1 \\ 33 & 34\end{array}\right]$
Next,calculate the product $BA$:
$BA = \left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right] \left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]$
$= \left[\begin{array}{cc}2(5) + 1(6) & 2(-1) + 1(7) \\ 3(5) + 4(6) & 3(-1) + 4(7)\end{array}\right]$
$= \left[\begin{array}{cc}10 + 6 & -2 + 7 \\ 15 + 24 & -3 + 28\end{array}\right] = \left[\begin{array}{cc}16 & 5 \\ 39 & 25\end{array}\right]$
Since $\left[\begin{array}{cc}7 & 1 \\ 33 & 34\end{array}\right] \ne \left[\begin{array}{cc}16 & 5 \\ 39 & 25\end{array}\right]$,it is shown that $AB \ne BA$.

(N/A) Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}$.
First,calculate $AB$:
$AB = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4) \\ 0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \\ 1(-1)+1(0)+0(2) & 1(1)+1(-1)+0(3) & 1(0)+1(1)+0(4) \end{bmatrix}$
$= \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$.
Next,calculate $BA$:
$BA = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} -1(1)+1(0)+0(1) & -1(2)+1(1)+0(1) & -1(3)+1(0)+0(0) \\ 0(1)+(-1)(0)+1(1) & 0(2)+(-1)(1)+1(1) & 0(3)+(-1)(0)+1(0) \\ 2(1)+3(0)+4(1) & 2(2)+3(1)+4(1) & 2(3)+3(0)+4(0) \end{bmatrix}$
$= \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}$.
Since $AB \ne BA$,the given statement is proved.

(N/A) We have $A^{2}=A \times A$.
$A^{2}=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$
$A^{2}=\left[\begin{array}{ccc}2(2)+0(2)+1(1) & 2(0)+0(1)+1(-1) & 2(1)+0(3)+1(0) \\ 2(2)+1(2)+3(1) & 2(0)+1(1)+3(-1) & 2(1)+1(3)+3(0) \\ 1(2)+(-1)(2)+0(1) & 1(0)+(-1)(1)+0(-1) & 1(1)+(-1)(3)+0(0)\end{array}\right]$
$A^{2}=\left[\begin{array}{ccc}4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1+0 & 1-3+0\end{array}\right] = \left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]$
Now,substitute the matrices into the expression $A^{2}-5 A+6 I$:
$= \left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right] - 5\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right] + 6\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$= \left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right] - \left[\begin{array}{ccc}10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0\end{array}\right] + \left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]$
$= \left[\begin{array}{ccc}5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+6\end{array}\right]$
$= \left[\begin{array}{ccc}1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4\end{array}\right]$

(A) Given $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$.
First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$.
$A^{2} = \begin{bmatrix} 3(3) + (-2)(4) & 3(-2) + (-2)(-2) \\ 4(3) + (-2)(4) & 4(-2) + (-2)(-2) \end{bmatrix} = \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$.
Now,substitute into the equation $A^{2} = kA - 2I$:
$\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = k \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
$\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$.
$\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}$.
Comparing the corresponding elements,we get $3k - 2 = 1$.
$3k = 3 \Rightarrow k = 1$.

(C) Matrices $P$ and $Y$ are of the orders $p \times k$ and $3 \times k$ respectively.
For the product $PY$ to be defined,the number of columns in $P$ must equal the number of rows in $Y$. Thus,$k=3$.
The resulting matrix $PY$ has the order $p \times k$.
Matrices $W$ and $Y$ are of the orders $n \times 3$ and $3 \times k$ respectively.
The product $WY$ is defined because the number of columns in $W$ $(3)$ equals the number of rows in $Y$ $(3)$.
The resulting matrix $WY$ has the order $n \times k$.
For the sum $PY + WY$ to be defined,the matrices $PY$ and $WY$ must have the same order.
Since $PY$ is $p \times k$ and $WY$ is $n \times k$,we must have $p=n$.
Therefore,the restrictions are $k=3$ and $p=n$.

(D) Matrix $X$ is of the order $2 \times n$.
Therefore,matrix $7X$ is also of the same order $2 \times n$.
Matrix $Z$ is of the order $2 \times p$. Since it is given that $n=p,$ the order of matrix $Z$ is $2 \times n$.
Therefore,matrix $5Z$ is also of the order $2 \times n$.
Since both matrices $7X$ and $5Z$ have the same order $2 \times n,$ their difference $7X - 5Z$ is well-defined.
The order of the resulting matrix $7X - 5Z$ is $2 \times n$.

(N/A) We have $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$.
Multiplying by a constant $k$,we get $kB = k \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 2k & -k & 2k \\ k & 2k & 4k \end{bmatrix}$.
Now,taking the transpose of $kB$,we get $(kB)^{\prime} = \begin{bmatrix} 2k & k \\ -k & 2k \\ 2k & 4k \end{bmatrix}$.
We can factor out $k$ from the matrix: $(kB)^{\prime} = k \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix}$.
Since $B^{\prime} = \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix}$,it follows that $(kB)^{\prime} = kB^{\prime}$.
Thus,the property is verified.