Types of matrices, Algebra of matrices Questions in English

(B) Given $A = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = A \times A = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} = \begin{bmatrix} (0)(0) + (i)(-i) & (0)(i) + (i)(0) \\ (-i)(0) + (0)(-i) & (-i)(i) + (0)(0) \end{bmatrix}$
$A^2 = \begin{bmatrix} -i^2 & 0 \\ 0 & -i^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$,where $I$ is the identity matrix.
Now,$A^{40} = (A^2)^{20} = I^{20} = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}$.
First,calculate the product $AB$:
$AB = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}$
$AB = \begin{bmatrix} (1)(2) + (-2)(3) + (1)(1) & (1)(1) + (-2)(2) + (1)(1) \\ (2)(2) + (1)(3) + (3)(1) & (2)(1) + (1)(2) + (3)(1) \end{bmatrix}$
$AB = \begin{bmatrix} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{bmatrix} = \begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}$
Now,find the transpose $(AB)^T$ by interchanging rows and columns:
$(AB)^T = \begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}$
Thus,the correct option is $B$.

(B) Given matrices are $A_{1 \times 3} = [1, 2, 3]$,$B_{3 \times 1} = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}$,and $C_{2 \times 2} = \begin{bmatrix} 1 & 5 \\ 0 & 2 \end{bmatrix}$.
For the product of two matrices $X$ and $Y$ to be defined,the number of columns in $X$ must be equal to the number of rows in $Y$.
$1$. For $AC$: $A$ is $1 \times 3$ and $C$ is $2 \times 2$. Since $3 \neq 2$,$AC$ is not defined.
$2$. For $BA$: $B$ is $3 \times 1$ and $A$ is $1 \times 3$. Since $1 = 1$,$BA$ is defined. $BA = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} [1, 2, 3] = \begin{bmatrix} 2 & 4 & 6 \\ 3 & 6 & 9 \\ 4 & 8 & 12 \end{bmatrix}_{3 \times 3}$.
$3$. For $(AB)C$: $AB$ is defined ($1 \times 3$ and $3 \times 1$ results in $1 \times 1$). Let $P = AB = [2(1) + 3(2) + 4(3)] = [20]_{1 \times 1}$. Now,$PC$ requires $P$ to have $2$ columns,but it has $1$. Thus,$(AB)C$ is not defined.
$4$. For $(AC)B$: Since $AC$ is not defined,$(AC)B$ is not defined.
Therefore,$BA$ is the defined product.

(D) The product of two non-zero matrices can result in a null matrix. For example,consider $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$.
Calculating the product: $AB = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
Here,$A \neq O$ and $B \neq O$,yet $AB = O$.
Therefore,none of the given statements ($A=O$ and $B=O$,$A=O$ or $B=O$,$A$ is null) are necessarily true.

(D) Given that $A$ and $B$ are square matrices of order $n \times n$.
We know that the square of a matrix expression is defined as the product of the expression with itself.
Therefore,${(A - B)^2} = (A - B)(A - B)$.
Using the distributive property of matrix multiplication:
$(A - B)(A - B) = A(A - B) - B(A - B)$
$= A^2 - AB - BA + B^2$.
Since matrix multiplication is not commutative in general,$AB \neq BA$,so we cannot simplify $-AB - BA$ to $-2AB$.
Thus,the correct expression is $A^2 - AB - BA + B^2$.

(A) An identity matrix $I$ is a square matrix in which all the elements of the principal diagonal are $1$ and all other elements are $0$.
$A$ scalar matrix is a diagonal matrix in which all the diagonal elements are equal to some constant $k$.
Since an identity matrix is a diagonal matrix where all diagonal elements are equal to $1$ (a constant),it satisfies the definition of a scalar matrix.
Therefore,every identity matrix is a scalar matrix.

(B) First,calculate the product $AB$:
$AB = \begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 1 & 12 \end{bmatrix} = \begin{bmatrix} (1)(0) + (0)(1) & (1)(0) + (0)(12) \\ (2)(0) + (0)(1) & (2)(0) + (0)(12) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
Next,calculate the product $BA$:
$BA = \begin{bmatrix} 0 & 0 \\ 1 & 12 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} (0)(1) + (0)(2) & (0)(0) + (0)(0) \\ (1)(1) + (12)(2) & (1)(0) + (12)(0) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 25 & 0 \end{bmatrix} \neq O$.
Thus,$AB = O$ and $BA \neq O$.

(C) square matrix $A = [a_{ij}]$ is called an upper triangular matrix if all elements below the main diagonal are zero,i.e.,$a_{ij} = 0$ for all $i > j$.
In the given matrix $\begin{bmatrix} 2 & 5 & -7 \\ 0 & 3 & 11 \\ 0 & 0 & 9 \end{bmatrix}$,the elements below the main diagonal are $a_{21} = 0$,$a_{31} = 0$,and $a_{32} = 0$.
Since all elements below the main diagonal are zero,the matrix is an upper triangular matrix.

(A) Given $A = \begin{bmatrix} i & 1 \\ 0 & i \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} i & 1 \\ 0 & i \end{bmatrix} \begin{bmatrix} i & 1 \\ 0 & i \end{bmatrix} = \begin{bmatrix} i^2 & i+i \\ 0 & i^2 \end{bmatrix} = \begin{bmatrix} -1 & 2i \\ 0 & -1 \end{bmatrix}$.
Next,calculate $A^4 = A^2 \times A^2 = \begin{bmatrix} -1 & 2i \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 2i \\ 0 & -1 \end{bmatrix}$.
$A^4 = \begin{bmatrix} (-1)(-1) + (2i)(0) & (-1)(2i) + (2i)(-1) \\ (0)(-1) + (-1)(0) & (0)(2i) + (-1)(-1) \end{bmatrix} = \begin{bmatrix} 1 & -2i - 2i \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -4i \\ 0 & 1 \end{bmatrix}$.

(C) To multiply a column matrix of order $3 \times 1$ by a row matrix of order $1 \times 3$,we perform matrix multiplication where each element of the resulting $3 \times 3$ matrix is the product of the corresponding row element of the first matrix and column element of the second matrix.
$\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \begin{bmatrix} 2 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 1(2) & 1(1) & 1(-1) \\ -1(2) & -1(1) & -1(-1) \\ 2(2) & 2(1) & 2(-1) \end{bmatrix} = \begin{bmatrix} 2 & 1 & -1 \\ -2 & -1 & 1 \\ 4 & 2 & -2 \end{bmatrix}$.

(C) For the subtraction of two matrices $A$ and $B$,the condition is that they must be of the same order.
Given that matrix $A$ has order $p \times q$ and matrix $B$ has order $r \times s$.
Therefore,for $A - B$ to be defined,the number of rows must be equal $(p = r)$ and the number of columns must be equal $(q = s)$.
Thus,the correct condition is $p = r$ and $q = s$.

(C) If $AB = O$,it does not necessarily imply that $A = O$ or $B = O$.
For example,consider the matrices $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$.
Calculating the product $AB$:
$AB = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} (1)(0) + (0)(1) & (1)(0) + (0)(0) \\ (0)(0) + (0)(1) & (0)(0) + (0)(0) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
Here,$A \ne O$ and $B \ne O$,yet their product $AB = O$.
Therefore,it is not necessary that either $A = O$ or $B = O$.

(B) Given the formula for elements of the matrix: ${a_{ij}} = \frac{1}{2}(3i - 2j)$.
For a $2 \times 2$ matrix $A = {[{a_{ij}}]_{2 \times 2}}$,we calculate the elements as follows:
For $i=1, j=1$: ${a_{11}} = \frac{1}{2}(3(1) - 2(1)) = \frac{1}{2}(3 - 2) = \frac{1}{2}$.
For $i=1, j=2$: ${a_{12}} = \frac{1}{2}(3(1) - 2(2)) = \frac{1}{2}(3 - 4) = -\frac{1}{2}$.
For $i=2, j=1$: ${a_{21}} = \frac{1}{2}(3(2) - 2(1)) = \frac{1}{2}(6 - 2) = \frac{4}{2} = 2$.
For $i=2, j=2$: ${a_{22}} = \frac{1}{2}(3(2) - 2(2)) = \frac{1}{2}(6 - 4) = \frac{2}{2} = 1$.
Thus,the matrix $A$ is given by:
$A = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1/2}&{-1/2}\\2&1\end{array}} \right]$.

(C) Given the equation: $2X - \begin{bmatrix} 1 & 2 \\ 7 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 0 & -2 \end{bmatrix}$
Adding the matrix $\begin{bmatrix} 1 & 2 \\ 7 & 4 \end{bmatrix}$ to both sides,we get:
$2X = \begin{bmatrix} 3 & 2 \\ 0 & -2 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 7 & 4 \end{bmatrix}$
Performing matrix addition:
$2X = \begin{bmatrix} 3+1 & 2+2 \\ 0+7 & -2+4 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 7 & 2 \end{bmatrix}$
Now,divide every element by $2$ to find $X$:
$X = \frac{1}{2} \begin{bmatrix} 4 & 4 \\ 7 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 7/2 & 1 \end{bmatrix}$
Thus,the correct option is $C$.

(A) Given $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.
Calculate ${A^2} = A \times A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+2(0) & 1(2)+2(1) \\ 0(1)+1(0) & 0(2)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix}$.
Calculate ${A^3} = {A^2} \times A = \begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+4(0) & 1(2)+4(1) \\ 0(1)+1(0) & 0(2)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 6 \\ 0 & 1 \end{bmatrix}$.
Observing the pattern,we see that for any positive integer $n$,the matrix follows the form ${A^n} = \begin{bmatrix} 1 & 2n \\ 0 & 1 \end{bmatrix}$.
Thus,the correct option is $A$.

(C) Given,$kA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$.
Substituting $A$,we get $k \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix} = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$.
Multiplying $k$ inside the matrix: $\begin{bmatrix} 0 & 2k \\ 3k & -4k \end{bmatrix} = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$.
Comparing the corresponding elements:
$1) -4k = 24 \implies k = -6$.
$2) 2k = 3a \implies 2(-6) = 3a \implies -12 = 3a \implies a = -4$.
$3) 3k = 2b \implies 3(-6) = 2b \implies -18 = 2b \implies b = -9$.
Thus,the values are $k = -6, a = -4, b = -9$.

(B) First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 1 \\ 5 & 6 & 2 \\ 3 & 4 & 1 \end{bmatrix}$
Next,calculate $A^3 = A^2 \times A$:
$A^3 = \begin{bmatrix} 2 & 3 & 1 \\ 5 & 6 & 2 \\ 3 & 4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 7 & 9 & 3 \\ 15 & 19 & 6 \\ 9 & 12 & 4 \end{bmatrix}$
Now,calculate $A^3 - 3A^2$:
$A^3 - 3A^2 = \begin{bmatrix} 7 & 9 & 3 \\ 15 & 19 & 6 \\ 9 & 12 & 4 \end{bmatrix} - 3 \begin{bmatrix} 2 & 3 & 1 \\ 5 & 6 & 2 \\ 3 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 9 & 3 \\ 15 & 19 & 6 \\ 9 & 12 & 4 \end{bmatrix} - \begin{bmatrix} 6 & 9 & 3 \\ 15 & 18 & 6 \\ 9 & 12 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$
Therefore,$A^3 - 3A^2 = I$,which implies $A^3 - 3A^2 - I = O$.

(B) Given $A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} (3)(3) + (-5)(-4) & (3)(-5) + (-5)(2) \\ (-4)(3) + (2)(-4) & (-4)(-5) + (2)(2) \end{bmatrix} = \begin{bmatrix} 9 + 20 & -15 - 10 \\ -12 - 8 & 20 + 4 \end{bmatrix} = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}$.
Next,calculate $5A$:
$5A = 5 \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix}$.
Now,find $A^2 - 5A$:
$A^2 - 5A = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} - \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} = \begin{bmatrix} 29-15 & -25-(-25) \\ -20-(-20) & 24-10 \end{bmatrix} = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$.
This can be written as $14 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 14I$.
Thus,the correct option is $B$.

(D) Given,matrix $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
First,we calculate $A^2$:
$A^2 = A \cdot A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} (0)(0) + (-1)(1) & (0)(-1) + (-1)(0) \\ (1)(0) + (0)(1) & (1)(-1) + (0)(0) \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$,where $I$ is the identity matrix.
Now,we calculate $A^{16}$:
$A^{16} = (A^2)^8 = (-I)^8 = (-1)^8 \cdot I^8 = 1 \cdot I = I$.
Thus,$A^{16} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.
Calculate $A^2$:
$A^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = 2^1 A$.
Calculate $A^3$:
$A^3 = A^2 \cdot A = (2A) \cdot A = 2 A^2 = 2(2A) = 4A = 2^2 A$.
By induction,we can generalize for $A^n$:
$A^n = 2^{n-1} A$.
Therefore,for $n = 100$:
$A^{100} = 2^{100-1} A = 2^{99} A$.
Thus,the correct option is $B$.

(B) For any three matrices $A$,$B$,and $C$,matrix multiplication is associative,meaning $(AB)C = A(BC)$.
Matrix multiplication is generally not commutative,meaning $AB \neq BA$ in most cases.
Therefore,the correct statement is that matrix multiplication is associative.

(A) Given the matrix equation: $\begin{bmatrix} x + y & 2x + z \\ x - y & 2z + w \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ 0 & 10 \end{bmatrix}$.
By equating the corresponding elements of the two matrices,we get the following system of linear equations:
$1) x + y = 4$
$2) x - y = 0$
$3) 2x + z = 7$
$4) 2z + w = 10$
From equation $(2)$,we have $x = y$. Substituting this into equation $(1)$,we get $x + x = 4$,which implies $2x = 4$,so $x = 2$. Since $x = y$,we have $y = 2$.
Now,substitute $x = 2$ into equation $(3)$: $2(2) + z = 7 \implies 4 + z = 7 \implies z = 3$.
Finally,substitute $z = 3$ into equation $(4)$: $2(3) + w = 10 \implies 6 + w = 10 \implies w = 4$.
Thus,the values are $x=2, y=2, z=3, w=4$.

(A) Given $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$.
First,we calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} (2)(2) + (-1)(-1) & (2)(-1) + (-1)(2) \\ (-1)(2) + (2)(-1) & (-1)(-1) + (2)(2) \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$.
Now,we evaluate the options. Let's check option $A$ $(4A - 3I)$:
$4A - 3I = 4 \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$.
Since $A^2 = 4A - 3I$,the correct option is $A$.

(B) Given $P = \begin{bmatrix} i & 0 & -i \\ 0 & -i & i \\ -i & i & 0 \end{bmatrix}$ and $Q = \begin{bmatrix} -i & i \\ 0 & 0 \\ i & -i \end{bmatrix}$.
To find $PQ$,we perform matrix multiplication:
$PQ = \begin{bmatrix} i & 0 & -i \\ 0 & -i & i \\ -i & i & 0 \end{bmatrix} \begin{bmatrix} -i & i \\ 0 & 0 \\ i & -i \end{bmatrix}$
$= \begin{bmatrix} i(-i) + 0(0) + (-i)(i) & i(i) + 0(0) + (-i)(-i) \\ 0(-i) + (-i)(0) + i(i) & 0(i) + (-i)(0) + i(-i) \\ (-i)(-i) + i(0) + 0(i) & (-i)(i) + i(0) + 0(-i) \end{bmatrix}$
Since $i^2 = -1$,we have:
$PQ = \begin{bmatrix} -i^2 + 0 - i^2 & i^2 + 0 + i^2 \\ 0 + 0 + i^2 & 0 + 0 - i^2 \\ i^2 + 0 + 0 & -i^2 + 0 + 0 \end{bmatrix} = \begin{bmatrix} -(-1) - (-1) & -1 + (-1) \\ -1 & -(-1) \\ -1 & -(-1) \end{bmatrix} = \begin{bmatrix} 1+1 & -1-1 \\ -1 & 1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 \\ -1 & 1 \\ -1 & 1 \end{bmatrix}$.

(B) unit matrix (or identity matrix) $I$ of order $n$ is a square matrix with $1$s on the main diagonal and $0$s elsewhere.
For any identity matrix $I_n$,the determinant is given by $|I_n| = 1^n = 1$.
Since $I$ is a unit matrix of order $10$,its determinant is $|I| = 1^{10} = 1$.

(C) For matrices,the property $AB = O$ does not necessarily imply that $A = O$ or $B = O$.
For example,consider $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$.
Then $AB = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
However,$A \neq O$ and $B \neq O$.
Therefore,the statement in option $C$ is not true.

(A) Given matrices $A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 2 \\ 4 & 5 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}$.
To find $AB$,we perform matrix multiplication:
$AB = \begin{bmatrix} (1)(1)+(2)(2)+(-1)(0) & (1)(0)+(2)(1)+(-1)(1) & (1)(0)+(2)(0)+(-1)(3) \\ (3)(1)+(0)(2)+(2)(0) & (3)(0)+(0)(1)+(2)(1) & (3)(0)+(0)(0)+(2)(3) \\ (4)(1)+(5)(2)+(0)(0) & (4)(0)+(5)(1)+(0)(1) & (4)(0)+(5)(0)+(0)(3) \end{bmatrix}$
Calculating each element:
Row $1$: $(1+4+0) = 5$,$(0+2-1) = 1$,$(0+0-3) = -3$
Row $2$: $(3+0+0) = 3$,$(0+0+2) = 2$,$(0+0+6) = 6$
Row $3$: $(4+10+0) = 14$,$(0+5+0) = 5$,$(0+0+0) = 0$
Thus,$AB = \begin{bmatrix} 5 & 1 & -3 \\ 3 & 2 & 6 \\ 14 & 5 & 0 \end{bmatrix}$.

(D) Given $A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$.
We calculate $A^2$ as follows:
$A^2 = A \times A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \alpha^2 + 0 & 0 + 0 \\ \alpha + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} \alpha^2 & 0 \\ \alpha + 1 & 1 \end{bmatrix}$.
We are given that $A^2 = B$,so:
$\begin{bmatrix} \alpha^2 & 0 \\ \alpha + 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$.
Comparing the corresponding elements,we get:
$1$) $\alpha^2 = 1 \implies \alpha = \pm 1$
$2$) $\alpha + 1 = 5 \implies \alpha = 4$
Since there is no value of $\alpha$ that satisfies both equations simultaneously,there are no real values of $\alpha$ for which $A^2 = B$.

(A) First,calculate the product of the matrix and the column vector:
$\begin{bmatrix} 7 & 1 & 2 \\ 9 & 2 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} = \begin{bmatrix} (7 \times 3) + (1 \times 4) + (2 \times 5) \\ (9 \times 3) + (2 \times 4) + (1 \times 5) \end{bmatrix} = \begin{bmatrix} 21 + 4 + 10 \\ 27 + 8 + 5 \end{bmatrix} = \begin{bmatrix} 35 \\ 40 \end{bmatrix}$.
Next,calculate the scalar multiplication:
$2 \begin{bmatrix} 4 \\ 2 \end{bmatrix} = \begin{bmatrix} 8 \\ 4 \end{bmatrix}$.
Finally,add the two resulting matrices:
$\begin{bmatrix} 35 \\ 40 \end{bmatrix} + \begin{bmatrix} 8 \\ 4 \end{bmatrix} = \begin{bmatrix} 35 + 8 \\ 40 + 4 \end{bmatrix} = \begin{bmatrix} 43 \\ 44 \end{bmatrix}$.
Thus,the correct option is $A$.

(A) Given $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$.
We evaluate the options:
$(i)$ Calculate $A^2$:
$A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Thus,option $A$ is correct.
$(ii)$ $(-1)I = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \neq A$.
$(iii)$ Calculate the determinant $|A|$:
$|A| = 0(0 - 0) - 0(0 - 0) - 1(0 - 1) = 1$.
Since $|A| \neq 0$,$A^{-1}$ exists.
$(iv)$ $A$ is clearly not a zero matrix as it contains non-zero elements.

(D) Given $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -5 & 7 & 1 \\ 1 & -5 & 7 \\ 7 & 1 & -5 \end{bmatrix}$.
Calculating the product $AB$:
$AB = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} -5 & 7 & 1 \\ 1 & -5 & 7 \\ 7 & 1 & -5 \end{bmatrix}$
Performing matrix multiplication:
Row $1$: $(1)(-5) + (2)(1) + (3)(7) = -5 + 2 + 21 = 18$; $(1)(7) + (2)(-5) + (3)(1) = 7 - 10 + 3 = 0$; $(1)(1) + (2)(7) + (3)(-5) = 1 + 14 - 15 = 0$.
Row $2$: $(3)(-5) + (1)(1) + (2)(7) = -15 + 1 + 14 = 0$; $(3)(7) + (1)(-5) + (2)(1) = 21 - 5 + 2 = 18$; $(3)(1) + (1)(7) + (2)(-5) = 3 + 7 - 10 = 0$.
Row $3$: $(2)(-5) + (3)(1) + (1)(7) = -10 + 3 + 7 = 0$; $(2)(7) + (3)(-5) + (1)(1) = 14 - 15 + 1 = 0$; $(2)(1) + (3)(7) + (1)(-5) = 2 + 21 - 5 = 18$.
Thus,$AB = \begin{bmatrix} 18 & 0 & 0 \\ 0 & 18 & 0 \\ 0 & 0 & 18 \end{bmatrix} = 18 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 18I_3$.

(A) Given the equation: $2X + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix}$
Subtract the matrix $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ from both sides:
$2X = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
$2X = \begin{bmatrix} 3-1 & 8-2 \\ 7-3 & 2-4 \end{bmatrix}$
$2X = \begin{bmatrix} 2 & 6 \\ 4 & -2 \end{bmatrix}$
Divide the entire matrix by $2$:
$X = \frac{1}{2} \begin{bmatrix} 2 & 6 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix}$

(C) Given equations are:
$(1)$ $A + B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$
$(2)$ $A - 2B = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$
To eliminate $B$,multiply equation $(1)$ by $2$:
$2A + 2B = 2 \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix}$
Now,add this to equation $(2)$:
$(2A + 2B) + (A - 2B) = \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$
$3A = \begin{bmatrix} 2-1 & 0+1 \\ 2+0 & 2-1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}$
Dividing by $3$,we get:
$A = \frac{1}{3} \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 1/3 & 1/3 \\ 2/3 & 1/3 \end{bmatrix}$

(B) Let $E = \begin{bmatrix} a & a \\ a & a \end{bmatrix}$ be the identity element in the set $M$ such that for any matrix $A = \begin{bmatrix} x & x \\ x & x \end{bmatrix} \in M$,we have $A \cdot E = A$.
Performing the matrix multiplication:
$\begin{bmatrix} x & x \\ x & x \end{bmatrix} \begin{bmatrix} a & a \\ a & a \end{bmatrix} = \begin{bmatrix} xa+xa & xa+xa \\ xa+xa & xa+xa \end{bmatrix} = \begin{bmatrix} 2ax & 2ax \\ 2ax & 2ax \end{bmatrix}$.
Equating this to $A$:
$\begin{bmatrix} 2ax & 2ax \\ 2ax & 2ax \end{bmatrix} = \begin{bmatrix} x & x \\ x & x \end{bmatrix}$.
This implies $2ax = x$. Since $x \neq 0$,we can divide by $x$ to get $2a = 1$,which means $a = \frac{1}{2}$.
Thus,the identity element is $\begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.

(B) The property of the transpose of a product of two matrices $A$ and $B$ is given by the reversal law of transpose: $(AB)' = B'A'$.
Option $A$ is incorrect as the order is not reversed.
Option $C$ is incorrect because the denominator should be the determinant $|A|$,not the matrix $A$.
Option $D$ is incorrect as the reversal law for the inverse of a product is $(AB)^{-1} = B^{-1}A^{-1}$.
Therefore,the correct relation is $(AB)' = B'A'$.

(A) Given that $A$ is an involutory matrix,by definition,$A^2 = I$.
We need to evaluate the expression $(I - A)(I + A)$.
Expanding the product:
$(I - A)(I + A) = I(I) + I(A) - A(I) - A(A)$
$= I^2 + IA - AI - A^2$
$= I + A - A - A^2$
$= I - A^2$
Since $A^2 = I$,substituting this into the expression:
$I - A^2 = I - I = O$,where $O$ is the zero matrix.

(C) Given $R(t) = \begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{bmatrix}$.
Then $R(s) = \begin{bmatrix} \cos s & \sin s \\ -\sin s & \cos s \end{bmatrix}$.
Calculating the product $R(s) \cdot R(t)$:
$R(s) \cdot R(t) = \begin{bmatrix} \cos s & \sin s \\ -\sin s & \cos s \end{bmatrix} \begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{bmatrix}$
$= \begin{bmatrix} \cos s \cos t - \sin s \sin t & \cos s \sin t + \sin s \cos t \\ -\sin s \cos t - \cos s \sin t & -\sin s \sin t + \cos s \cos t \end{bmatrix}$
Using trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$= \begin{bmatrix} \cos(s+t) & \sin(s+t) \\ -\sin(s+t) & \cos(s+t) \end{bmatrix}$
$= R(s+t)$.

(B) matrix $A$ is symmetric if $A = A^T$,which implies that the element at the $i$-th row and $j$-th column is equal to the element at the $j$-th row and $i$-th column,i.e.,$a_{ij} = a_{ji}$.
For the given matrix $A = \begin{bmatrix} 4 & x + 2 \\ 2x - 3 & x + 1 \end{bmatrix}$,we have $a_{12} = x + 2$ and $a_{21} = 2x - 3$.
Since the matrix is symmetric,we set $a_{12} = a_{21}$:
$x + 2 = 2x - 3$
Subtracting $x$ from both sides,we get $2 = x - 3$.
Adding $3$ to both sides,we get $x = 5$.
Therefore,the correct option is $B$.

(C) Let $A = \left[ {\begin{array}{*{20}{c}}0&{ - 4}&1\\4&0&{ - 5}\\{ - 1}&5&0\end{array}} \right]$.
To check if the matrix is skew-symmetric,we find its transpose $A^T$.
$A^T = \left[ {\begin{array}{*{20}{c}}0&4&{ - 1}\\{ - 4}&0&5\\1&{ - 5}&0\end{array}} \right]$.
We observe that $A^T = -A$.
Since $A^T = -A$,the matrix $A$ is a skew-symmetric matrix.

(D) Given that $A = [a_{ij}]$ is a square matrix where $a_{ij} = i^2 - j^2$.
For any element $a_{ji}$,we have $a_{ji} = j^2 - i^2$.
We can rewrite this as $a_{ji} = -(i^2 - j^2) = -a_{ij}$.
$A$ matrix $A$ is called a skew-symmetric matrix if $a_{ji} = -a_{ij}$ for all $i, j$.
Since $a_{ji} = -a_{ij}$ holds true for all $i, j$,the matrix $A$ is a skew-symmetric matrix.
Therefore,the correct option is $D$.

(A) The condition $AI = A$ holds for any matrix $A$ where the product is defined,as $I$ is the identity matrix.
However,the condition $AA^T = I$ is the definition of an orthogonal matrix.
An orthogonal matrix must be a square matrix.
Therefore,the given conditions imply that $A$ is an orthogonal matrix,which requires $A$ to be a square matrix.

(B) First,we calculate the product $AB$:
$AB = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}$
$AB = \begin{bmatrix} (1)(2) + (-2)(3) + (1)(1) & (1)(1) + (-2)(2) + (1)(1) \\ (2)(2) + (1)(3) + (3)(1) & (2)(1) + (1)(2) + (3)(1) \end{bmatrix}$
$AB = \begin{bmatrix} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{bmatrix} = \begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}$
Now,we find the transpose $(AB)^T$ by swapping rows and columns:
$(AB)^T = \begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix}$
Thus,the correct option is $B$.

(D) Given $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
First,calculate the determinant of $A$:
$|A| = (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta) = \cos^2 \theta + \sin^2 \theta = 1$.
Since $|A| = 1 \neq 0$,the matrix $A$ is invertible.
Next,check if $A$ is orthogonal:
$A A^T = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & \cos \theta \sin \theta - \sin \theta \cos \theta \\ \sin \theta \cos \theta - \cos \theta \sin \theta & \sin^2 \theta + \cos^2 \theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Since $A A^T = I$,$A$ is an orthogonal matrix. The transpose of an orthogonal matrix is also orthogonal,so $A'$ is also orthogonal.
Therefore,the statement '$A$ is not invertible' is incorrect.

(A) Using the reversal law of inverses,we know that $(B^{-1}A^{-1})^{-1} = (A^{-1})^{-1}(B^{-1})^{-1} = AB$.
Now,we calculate the product $AB$:
$AB = \begin{bmatrix} 2 & 2 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$
$AB = \begin{bmatrix} (2 \times 0) + (2 \times 1) & (2 \times -1) + (2 \times 0) \\ (-3 \times 0) + (2 \times 1) & (-3 \times -1) + (2 \times 0) \end{bmatrix}$
$AB = \begin{bmatrix} 0 + 2 & -2 + 0 \\ 0 + 2 & 3 + 0 \end{bmatrix} = \begin{bmatrix} 2 & -2 \\ 2 & 3 \end{bmatrix}$.

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

(A) To find the index of nilpotency,we calculate $A^2$:
$A^2 = \begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix} \begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix}$
$A^2 = \begin{bmatrix} (1-1-4) & (-3+9+12) & (-4-12+16) \\ (-1+3+4) & (3-9-12) & (4+12-16) \\ (1+3-4) & (-3-9+12) & (-4-12+16) \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$
Since $A^2 = O$ and $A \neq O$,the matrix $A$ is nilpotent of index $2$.

(B) The transpose of a product of matrices is equal to the product of their transposes in reverse order.
For any two matrices $X$ and $Y$,the property is $(XY)' = Y' X'$.
Applying this property to the product of three matrices $A, B,$ and $C$:
$(ABC)' = ((AB)C)' = C'(AB)'$
Using the property again for $(AB)' = B' A'$:
$(ABC)' = C'(B' A') = C' B' A'$
Therefore,the correct option is $B$.

(B) Given $A = \begin{bmatrix} a & b \\ b & a \end{bmatrix}$.
We need to find $A^2 = A \times A$.
$A^2 = \begin{bmatrix} a & b \\ b & a \end{bmatrix} \begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & ab + ba \\ ba + ab & b^2 + a^2 \end{bmatrix}$.
$A^2 = \begin{bmatrix} a^2 + b^2 & 2ab \\ 2ab & a^2 + b^2 \end{bmatrix}$.
Comparing this with $A^2 = \begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix}$,we get $\alpha = a^2 + b^2$ and $\beta = 2ab$.

(C) Given $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$.
To find $A^2$,we calculate $A \times A$:
$A^2 = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$
Performing matrix multiplication:
$A^2 = \begin{bmatrix} (\cos \alpha)(\cos \alpha) + (\sin \alpha)(-\sin \alpha) & (\cos \alpha)(\sin \alpha) + (\sin \alpha)(\cos \alpha) \\ (-\sin \alpha)(\cos \alpha) + (\cos \alpha)(-\sin \alpha) & (-\sin \alpha)(\sin \alpha) + (\cos \alpha)(\cos \alpha) \end{bmatrix}$
$A^2 = \begin{bmatrix} \cos^2 \alpha - \sin^2 \alpha & 2\sin \alpha \cos \alpha \\ -2\sin \alpha \cos \alpha & \cos^2 \alpha - \sin^2 \alpha \end{bmatrix}$
Using trigonometric identities $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ and $\sin 2\alpha = 2\sin \alpha \cos \alpha$:
$A^2 = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{bmatrix}$.

(D) Given $A = \begin{bmatrix} 1 & 2 & 3 \\ -2 & 3 & -1 \\ 3 & 1 & 2 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 1 & 2 & 3 \\ -2 & 3 & -1 \\ 3 & 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ -2 & 3 & -1 \\ 3 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 11 & 7 \\ -11 & 4 & -11 \\ 7 & 11 & 12 \end{bmatrix}$.
Now,calculate $9I = 9 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$.
Finally,add $A^2$ and $9I$:
$A^2 + 9I = \begin{bmatrix} 6 & 11 & 7 \\ -11 & 4 & -11 \\ 7 & 11 & 12 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} = \begin{bmatrix} 15 & 11 & 7 \\ -11 & 13 & -11 \\ 7 & 11 & 21 \end{bmatrix}$.
Comparing this result with the given options,it does not match $2A$,$4A$,or $6A$. Therefore,the correct option is $(d)$.