Types of matrices, Algebra of matrices Questions in English

(D) Given the equation $M^2 - \lambda M - I_2 = 0$,where $M = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$.
First,calculate $M^2$:
$M^2 = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(2) & (1)(2) + (2)(3) \\ (2)(1) + (3)(2) & (2)(2) + (3)(3) \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}$.
Substitute $M^2$,$M$,and $I_2$ into the equation:
$\begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
This simplifies to:
$\begin{bmatrix} 5 - \lambda - 1 & 8 - 2\lambda - 0 \\ 8 - 2\lambda - 0 & 13 - 3\lambda - 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Equating the elements:
$4 - \lambda = 0 \Rightarrow \lambda = 4$.
$8 - 2\lambda = 0 \Rightarrow 2\lambda = 8 \Rightarrow \lambda = 4$.
$12 - 3\lambda = 0 \Rightarrow 3\lambda = 12 \Rightarrow \lambda = 4$.
Since $\lambda = 4$ satisfies all equations,the correct option is $D$.

(C) Given $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix}$.
Calculating $AB$:
$AB = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix}$
$= \begin{bmatrix} \cos \alpha \cos \beta - \sin \alpha \sin \beta & -\cos \alpha \sin \beta - \sin \alpha \cos \beta \\ \sin \alpha \cos \beta + \cos \alpha \sin \beta & -\sin \alpha \sin \beta + \cos \alpha \cos \beta \end{bmatrix}$
$= \begin{bmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) \end{bmatrix}$.
Similarly,calculating $BA$:
$BA = \begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$
$= \begin{bmatrix} \cos \beta \cos \alpha - \sin \beta \sin \alpha & -\cos \beta \sin \alpha - \sin \beta \cos \alpha \\ \sin \beta \cos \alpha + \cos \beta \sin \alpha & -\sin \beta \sin \alpha + \cos \beta \cos \alpha \end{bmatrix}$
$= \begin{bmatrix} \cos(\beta + \alpha) & -\sin(\beta + \alpha) \\ \sin(\beta + \alpha) & \cos(\beta + \alpha) \end{bmatrix}$.
Since $\cos(\alpha + \beta) = \cos(\beta + \alpha)$ and $\sin(\alpha + \beta) = \sin(\beta + \alpha)$,we have $AB = BA$.

(C) To determine the nature of matrix $A$,we calculate its determinant,denoted as $|A|$ or $\Delta$.
$|A| = \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{vmatrix}$
Expanding along the first row:
$|A| = 1(1 \times 0 - 1 \times 0) - 0(0 \times 0 - 1 \times 1) + 1(0 \times 0 - 1 \times 1)$
$|A| = 1(0) - 0(-1) + 1(-1)$
$|A| = 0 - 0 - 1 = -1$
Since $|A| = -1 \neq 0$,the matrix $A$ is a non-singular matrix.

(A) Given $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}$.
We need to find ${A^2} = A \times A$.
${A^2} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}$.
Calculating the product:
Row $1$: $(1 \times 1 + 0 \times 0 + 0 \times a, 1 \times 0 + 0 \times 1 + 0 \times b, 1 \times 0 + 0 \times 0 + 0 \times -1) = (1, 0, 0)$.
Row $2$: $(0 \times 1 + 1 \times 0 + 0 \times a, 0 \times 0 + 1 \times 1 + 0 \times b, 0 \times 0 + 1 \times 0 + 0 \times -1) = (0, 1, 0)$.
Row $3$: $(a \times 1 + b \times 0 + -1 \times a, a \times 0 + b \times 1 + -1 \times b, a \times 0 + b \times 0 + -1 \times -1) = (a - a, b - b, 1) = (0, 0, 1)$.
Thus,${A^2} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$,which is the Unit matrix.

(A) We are given $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.
Calculate ${A^2}$:
${A^2} = A \times A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+1(0) & 1(1)+1(1) \\ 0(1)+1(0) & 0(1)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.
Calculate ${A^3}$:
${A^3} = {A^2} \times A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+2(0) & 1(1)+2(1) \\ 0(1)+1(0) & 0(1)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.
By observing the pattern,we can generalize for any positive integer $n$:
${A^n} = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.
This can be proven using the principle of mathematical induction.

(D) The condition $AB = O$ for matrices does not necessarily imply that $A = O$ or $B = O$.
In matrix algebra,there exist non-zero matrices $A$ and $B$ such that their product $AB$ is the zero matrix $O$.
For example,if $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$,then $AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$,even though $A \neq O$ and $B \neq O$.
Therefore,none of the given options $A \neq O, B = O$,$A = O, B \neq O$,or $A = O$ or $B = O$ are necessary conditions for $AB = O$.
Thus,the correct option is $(d)$.

(A) Given $A = \begin{bmatrix} 2 & 2 \\ a & b \end{bmatrix}$ and $A^2 = O$,where $O$ is the zero matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Calculate $A^2$:
$A^2 = \begin{bmatrix} 2 & 2 \\ a & b \end{bmatrix} \begin{bmatrix} 2 & 2 \\ a & b \end{bmatrix} = \begin{bmatrix} 4 + 2a & 4 + 2b \\ 2a + ab & 2a + b^2 \end{bmatrix}$.
Equating $A^2$ to the zero matrix:
$\begin{bmatrix} 4 + 2a & 4 + 2b \\ 2a + ab & 2a + b^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
From the first row:
$4 + 2a = 0 \Rightarrow 2a = -4 \Rightarrow a = -2$.
$4 + 2b = 0 \Rightarrow 2b = -4 \Rightarrow b = -2$.
Check consistency with the second row:
$2a + ab = 2(-2) + (-2)(-2) = -4 + 4 = 0$.
$2a + b^2 = 2(-2) + (-2)^2 = -4 + 4 = 0$.
Since all equations are satisfied,$(a, b) = (-2, -2)$.

(B) Given the matrix equation: $[m \ n] \begin{bmatrix} m \\ n \end{bmatrix} = [25]$.
Performing matrix multiplication: $[m^2 + n^2] = [25]$.
This implies $m^2 + n^2 = 25$.
We are given the condition $m < n$ and $m, n$ are typically assumed to be positive integers in such problems.
Testing integer pairs $(m, n)$ such that $m^2 + n^2 = 25$:
For $m=3, n=4$: $3^2 + 4^2 = 9 + 16 = 25$.
Since $3 < 4$,the condition $m < n$ is satisfied.
Thus,$(m, n) = (3, 4)$.

(A) To add or subtract matrices,they must have the same order.
Matrix $A$ is of order $3 \times 3$.
Matrix $B$ is of order $3 \times 2$.
In the expression $A^2 + 2B - 2A$,we are attempting to add matrix $A^2$ (which is $3 \times 3$) to matrix $2B$ (which is $3 \times 2$).
Since the orders of $A^2$ and $2B$ are different,the addition is not defined.
Therefore,the expression $A^2 + 2B - 2A$ is not defined.

(B) Given $A = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$ and $B = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} = \begin{bmatrix} i^2 & 0 \\ 0 & (-i)^2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$.
Next,calculate $B^2$:
$B^2 = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} = \begin{bmatrix} i^2 & 0 \\ 0 & i^2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$.
Since $A^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ and $B^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$,we observe that $A^2 = B^2$.

(A) Given $A = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix}$.
We calculate $A^2$:
$A^2 = A \cdot A = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix}$
$A^2 = \begin{bmatrix} (ab)(ab) + (b^2)(-a^2) & (ab)(b^2) + (b^2)(-ab) \\ (-a^2)(ab) + (-ab)(-a^2) & (-a^2)(b^2) + (-ab)(-ab) \end{bmatrix}$
$A^2 = \begin{bmatrix} a^2b^2 - a^2b^2 & ab^3 - ab^3 \\ -a^3b + a^3b & -a^2b^2 + a^2b^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
Since $A^2 = O$,it follows that $A^n = O$ for all $n \ge 2$.
Therefore,the minimum value of $n$ is $2$.

(B) Given $A = \begin{bmatrix} 1/3 & 2 \\ 0 & 2x - 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$.
We are given that $AB = I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Calculating the product $AB$:
$AB = \begin{bmatrix} 1/3 & 2 \\ 0 & 2x - 3 \end{bmatrix} \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} (1/3 \times 3) + (2 \times 0) & (1/3 \times 6) + (2 \times -1) \\ (0 \times 3) + ((2x - 3) \times 0) & (0 \times 6) + ((2x - 3) \times -1) \end{bmatrix}$
$AB = \begin{bmatrix} 1 + 0 & 2 - 2 \\ 0 + 0 & 0 - (2x - 3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 3 - 2x \end{bmatrix}$.
Since $AB = I$,we have $\begin{bmatrix} 1 & 0 \\ 0 & 3 - 2x \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Comparing the elements,we get $3 - 2x = 1$.
$2x = 3 - 1 = 2$.
$x = 1$.

(B) For the matrix multiplication $AB = C$ to be defined,the number of columns in matrix $A$ must be equal to the number of rows in matrix $B$.
If $A$ is an $m \times n$ matrix and $B$ is an $n \times p$ matrix,then the resulting matrix $C$ will have the dimensions $m \times p$.
Looking at option $B$: If $A$ is $3 \times 2$ and $B$ is $2 \times 3$,then $C$ must be a $3 \times 3$ matrix.
Therefore,the correct dimensions are $A_{3 \times 2}, B_{2 \times 3}, C_{3 \times 3}$.

(B) Given $A = \begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix}$.
We need to find $\lambda$ such that $A^2 = O$,where $O$ is the zero matrix.
$A^2 = A \cdot A = \begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix} \begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix}$
$A^2 = \begin{bmatrix} (\lambda)(\lambda) + (1)(-1) & (\lambda)(1) + (1)(-\lambda) \\ (-1)(\lambda) + (-\lambda)(-1) & (-1)(1) + (-\lambda)(-\lambda) \end{bmatrix}$
$A^2 = \begin{bmatrix} \lambda^2 - 1 & \lambda - \lambda \\ -\lambda + \lambda & -1 + \lambda^2 \end{bmatrix} = \begin{bmatrix} \lambda^2 - 1 & 0 \\ 0 & \lambda^2 - 1 \end{bmatrix}$.
Since $A^2 = O$,we have $\begin{bmatrix} \lambda^2 - 1 & 0 \\ 0 & \lambda^2 - 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Comparing the elements,we get $\lambda^2 - 1 = 0$.
$\lambda^2 = 1 \Rightarrow \lambda = \pm 1$.

(C) Given $A = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}$.
$A^2 = \begin{bmatrix} (4 \times 4 + 1 \times 3) & (4 \times 1 + 1 \times 2) \\ (3 \times 4 + 2 \times 3) & (3 \times 1 + 2 \times 2) \end{bmatrix} = \begin{bmatrix} 19 & 6 \\ 18 & 7 \end{bmatrix}$.
Now,calculate $6A = 6 \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 24 & 6 \\ 18 & 12 \end{bmatrix}$.
Finally,compute $A^2 - 6A = \begin{bmatrix} 19 & 6 \\ 18 & 7 \end{bmatrix} - \begin{bmatrix} 24 & 6 \\ 18 & 12 \end{bmatrix} = \begin{bmatrix} -5 & 0 \\ 0 & -5 \end{bmatrix}$.
Since $\begin{bmatrix} -5 & 0 \\ 0 & -5 \end{bmatrix} = -5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -5I$,the correct option is $C$.

(C) The order of matrix $A$ is $3 \times 3$ and the order of matrix $B$ is $2 \times 3$.
For matrix addition $A+B$,the orders must be identical,which is not the case here.
For matrix multiplication $AB$,the number of columns in $A$ $(3)$ must equal the number of rows in $B$ $(2)$,which is not true.
Now,consider the transpose matrices:
$A'$ is of order $3 \times 3$.
$B'$ is of order $3 \times 2$.
For the product $A'B'$,the number of columns in $A'$ $(3)$ equals the number of rows in $B'$ $(3)$. Thus,$A'B'$ is defined.
For the product $B'A'$,the number of columns in $B'$ $(2)$ must equal the number of rows in $A'$ $(3)$,which is not true.
Therefore,the correct option is $C$.

(C) Given $A = [1\, 2\, 3]$ (a $1 \times 3$ matrix) and $B = \begin{bmatrix} -5 & 4 & 0 \\ 0 & 2 & -1 \\ 1 & -3 & 2 \end{bmatrix}$ (a $3 \times 3$ matrix).
To find $AB$,we multiply the row of $A$ by each column of $B$:
$AB = [1 \times (-5) + 2 \times 0 + 3 \times 1, \quad 1 \times 4 + 2 \times 2 + 3 \times (-3), \quad 1 \times 0 + 2 \times (-1) + 3 \times 2]$
$AB = [-5 + 0 + 3, \quad 4 + 4 - 9, \quad 0 - 2 + 6]$
$AB = [-2, \quad -1, \quad 4]$
Thus,$AB = [-2\, -1\, 4]$.

(B) Let the order of matrix $A$ be $m \times n$.
Let the order of matrix $B$ be $p \times q$.
For the product $AB$ to be defined,the number of columns in $A$ must be equal to the number of rows in $B$. Thus,$n = p$.
So,the order of $B$ is $n \times q$.
For the product $BA$ to be defined,the number of columns in $B$ must be equal to the number of rows in $A$. Thus,$q = m$.
Substituting these values,the order of $B$ is $n \times m$.

(C) Given the matrix equation: $\begin{bmatrix} 2 & -3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} a & c \\ b & d \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 2 & -5 \end{bmatrix}$
Rearranging the terms to solve for the matrix $\begin{bmatrix} a & c \\ b & d \end{bmatrix}$:
$\begin{bmatrix} a & c \\ b & d \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 4 \\ 2 & -5 \end{bmatrix}$
Performing the subtraction element-wise:
$a = 2 - 1 = 1$
$c = -3 - 4 = -7$
$b = 4 - 2 = 2$
$d = 0 - (-5) = 5$
Thus,$(a, b, c, d) = (1, 2, -7, 5)$.

(B) Given $A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$.
To find $A^2$,we multiply $A$ by itself:
$A^2 = A \times A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
Performing matrix multiplication:
$A^2 = \begin{bmatrix} (-1)(-1) + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + (-1)(-1) + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + (-1)(-1) \end{bmatrix}$
$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$
Thus,$A^2$ is the unit matrix (identity matrix) $I$.

(B) To find the element in the $3^{rd}$ row and $3^{rd}$ column of the product matrix $AB$,we denote it as $C_{33}$.
According to the rule of matrix multiplication,$C_{33}$ is obtained by multiplying the elements of the $3^{rd}$ row of matrix $A$ with the corresponding elements of the $3^{rd}$ column of matrix $B$ and summing them up.
$C_{33} = (A_{31} \times B_{13}) + (A_{32} \times B_{23}) + (A_{33} \times B_{33})$
$C_{33} = (-2 \times 3) + (2 \times 5) + (0 \times 0)$
$C_{33} = -6 + 10 + 0$
$C_{33} = 4$.
Thus,the correct option is $B$.

(C) Given $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} = 2I$,where $I$ is the identity matrix of order $3 \times 3$.
We need to find $A^5$.
$A^5 = (2I)^5 = 2^5 I^5$.
Since $I^n = I$ for any positive integer $n$,we have $I^5 = I$.
$A^5 = 32I$.
We can rewrite this as $A^5 = 16 \times 2I = 16A$.

(D) Let $B = \begin{bmatrix} x & y \\ z & w \end{bmatrix}$.
Given $AB = O$,we have $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Performing matrix multiplication: $\begin{bmatrix} 0(x) + 1(z) & 0(y) + 1(w) \\ 0(x) + 0(z) & 0(y) + 0(w) \end{bmatrix} = \begin{bmatrix} z & w \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Comparing the elements,we get $z = 0$ and $w = 0$,while $x$ and $y$ can be any real numbers.
Checking the options,option $D$ gives $B = \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix}$,which satisfies the condition $AB = O$ because $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.

(B) For any two square matrices $A$ and $B$ of the same order,the square of their sum is defined as:
$(A + B)^2 = (A + B)(A + B)$
Expanding the product 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 generally not commutative (i.e.,$AB \neq BA$ in general),we cannot simplify $AB + BA$ to $2AB$. Therefore,the correct expression is $A^2 + AB + BA + B^2$.

(C) square matrix $A = [a_{ij}]$ is called a lower triangular matrix if all elements above the main diagonal are zero,i.e.,$a_{ij} = 0$ for all $i < j$.
In the given matrix $A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 2 & 3 & 0 & 0 \\ 4 & 5 & 6 & 0 \\ 7 & 8 & 9 & 10 \end{bmatrix}$,the elements above the main diagonal are $a_{12}=0, a_{13}=0, a_{14}=0, a_{23}=0, a_{24}=0, a_{34}=0$.
Since all elements $a_{ij} = 0$ for $i < j$,the matrix $A$ is a lower triangular matrix.
Therefore,the correct option is $C$.

(B) An upper triangular matrix is a square matrix in which all the entries below the main diagonal are zero.
For a matrix $[a_{ij}]_{n \times n}$,the elements below the main diagonal are those where the row index $i$ is greater than the column index $j$ $(i > j)$.
Therefore,the condition for an upper triangular matrix is $a_{ij} = 0$ for all $i > j$.

(B) Given $A = \text{diag}(2, -1, 3)$ and $B = \text{diag}(-1, 3, 2)$.
For a diagonal matrix $D = \text{diag}(d_1, d_2, d_3)$,$D^n = \text{diag}(d_1^n, d_2^n, d_3^n)$.
Therefore,$A^2 = \text{diag}(2^2, (-1)^2, 3^2) = \text{diag}(4, 1, 9)$.
Now,$A^2B = \text{diag}(4, 1, 9) \times \text{diag}(-1, 3, 2)$.
Multiplying two diagonal matrices involves multiplying their corresponding diagonal elements:
$A^2B = \text{diag}(4 \times (-1), 1 \times 3, 9 \times 2) = \text{diag}(-4, 3, 18)$.

(D) Multiply the scalar $\cos \theta$ into the first matrix and $\sin \theta$ into the second matrix:
$\begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ - \sin \theta \cos \theta & \cos^2 \theta \end{bmatrix} + \begin{bmatrix} \sin^2 \theta & - \sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix}$
Now,add the corresponding elements of the two matrices:
$= \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & \cos \theta \sin \theta - \sin \theta \cos \theta \\ - \sin \theta \cos \theta + \cos \theta \sin \theta & \cos^2 \theta + \sin^2 \theta \end{bmatrix}$
Using the trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

(C) unit matrix $I$ is a diagonal matrix where all diagonal elements are $1$.
Multiplying a matrix by a scalar $k$ results in a matrix where every element is multiplied by $k$.
Thus,$3I$ results in a diagonal matrix where all diagonal elements are $3$ and all non-diagonal elements are $0$.
$A$ scalar matrix is a diagonal matrix in which all diagonal elements are equal.
Since all diagonal elements of $3I$ are $3$,it is a scalar matrix.

(C) Given matrices are $A = [a\, b]$,$B = [-b\, -a]$,and $C = \begin{bmatrix} a \\ -a \end{bmatrix}$.
First,calculate $AC$:
$AC = [a\, b] \begin{bmatrix} a \\ -a \end{bmatrix} = [a(a) + b(-a)] = [a^2 - ab]$.
Next,calculate $BC$:
$BC = [-b\, -a] \begin{bmatrix} a \\ -a \end{bmatrix} = [-b(a) + (-a)(-a)] = [-ab + a^2] = [a^2 - ab]$.
Since $AC = [a^2 - ab]$ and $BC = [a^2 - ab]$,it follows that $AC = BC$.

(D) Given $A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}$.
Calculate $A^2$:
$A^2 = A \cdot A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+a(0) & 1(a)+a(1) \\ 0(1)+1(0) & 0(a)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix}$.
Calculate $A^3$:
$A^3 = A \cdot A^2 = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3a \\ 0 & 1 \end{bmatrix}$.
Calculate $A^4$:
$A^4 = A \cdot A^3 = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3a \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 4a \\ 0 & 1 \end{bmatrix}$.
Thus,$A^4 = \begin{bmatrix} 1 & 4a \\ 0 & 1 \end{bmatrix}$.

(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 $3 \times 1$.
When multiplying matrices,if the order of matrix $A$ is $m \times n$ and matrix $B$ is $n \times p$,the resulting matrix $AB$ has the order $m \times p$.
First,calculate the product $AB$: $(1 \times 3) \times (3 \times 3) = (1 \times 3)$.
Next,multiply the result by $C$: $(1 \times 3) \times (3 \times 1) = (1 \times 1)$.
Therefore,the order of the final product is $1 \times 1$.

(A) Given the equation: $(A + B)(A - B) = A^2 - B^2$
Expanding the left side using the distributive property of matrix multiplication:
$(A + B)(A - B) = A(A - B) + B(A - B)$
$= A^2 - AB + BA - B^2$
Equating this to the right side:
$A^2 - AB + BA - B^2 = A^2 - B^2$
Subtracting $A^2$ and adding $B^2$ to both sides:
$-AB + BA = 0$
Therefore:
$BA = AB$

(B) Given $A = \begin{bmatrix} 5 & -3 \\ 2 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & -4 \\ 3 & 6 \end{bmatrix}$.
To find $A - B$,we subtract the corresponding elements of matrix $B$ from matrix $A$:
$A - B = \begin{bmatrix} 5-6 & -3-(-4) \\ 2-3 & 4-6 \end{bmatrix}$
$A - B = \begin{bmatrix} -1 & -3+4 \\ -1 & -2 \end{bmatrix}$
$A - B = \begin{bmatrix} -1 & 1 \\ -1 & -2 \end{bmatrix}$
Thus,the correct option is $B$.

(D) Given $X = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$.
We calculate $X^2 = X \times X = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} (3)(3) + (-4)(1) & (3)(-4) + (-4)(-1) \\ (1)(3) + (-1)(1) & (1)(-4) + (-1)(-1) \end{bmatrix} = \begin{bmatrix} 9-4 & -12+4 \\ 3-1 & -4+1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$.
Now,let us check the options for $n=2$:
Option $(a)$ gives $\begin{bmatrix} 3(2) & -4(2) \\ 2 & -2 \end{bmatrix} = \begin{bmatrix} 6 & -8 \\ 2 & -2 \end{bmatrix} \neq X^2$.
Option $(b)$ gives $\begin{bmatrix} 2+2 & 5-2 \\ 2 & -2 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 2 & -2 \end{bmatrix} \neq X^2$.
Option $(c)$ gives $\begin{bmatrix} 3^2 & (-4)^2 \\ 1^2 & (-1)^2 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 1 & 1 \end{bmatrix} \neq X^2$.
Since none of the options match $X^2$,the correct answer is $(d)$.

(B) Given $A = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}$.
To find ${A^2}$,we calculate $A \times A$:
${A^2} = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}$
Performing matrix multiplication:
${A^2} = \begin{bmatrix} (i \times i) + (0 \times 0) & (i \times 0) + (0 \times i) \\ (0 \times i) + (i \times 0) & (0 \times 0) + (i \times i) \end{bmatrix}$
${A^2} = \begin{bmatrix} i^2 & 0 \\ 0 & i^2 \end{bmatrix}$
Since $i^2 = -1$,we substitute this value:
${A^2} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$.

(A) Matrix addition is commutative for matrices of the same order.
For any two square matrices $A$ and $B$ of the same order $n \times n$,the sum $A + B$ is equal to $B + A$.
This is a fundamental property of matrix algebra.
Therefore,the correct option is $A$.

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

(D) Given $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.
To find ${A^2}$,we perform matrix multiplication $A \times A$:
${A^2} = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$
Calculating the elements:
Row $1$,Column $1$: $(3 \times 3) + (1 \times -1) = 9 - 1 = 8$
Row $1$,Column $2$: $(3 \times 1) + (1 \times 2) = 3 + 2 = 5$
Row $2$,Column $1$: $(-1 \times 3) + (2 \times -1) = -3 - 2 = -5$
Row $2$,Column $2$: $(-1 \times 1) + (2 \times 2) = -1 + 4 = 3$
Thus,${A^2} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$.

(B) For the sum $A+B$ to be defined,matrices $A$ and $B$ must have the same order. Let the order of $A$ be $m \times n$. Then the order of $B$ must also be $m \times n$.
For the product $AB$ to be defined,the number of columns in $A$ must be equal to the number of rows in $B$.
Since $A$ is $m \times n$,it has $n$ columns. Since $B$ is $m \times n$,it has $m$ rows.
Therefore,for $AB$ to be defined,we must have $n = m$.
Since $m = n$,the matrices $A$ and $B$ must be square matrices of the same order $n \times n$.

(B) To find the product $AB$,we multiply matrix $A$ and matrix $B$ row by column:
$AB = \begin{bmatrix} 1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \end{bmatrix}$
$= \begin{bmatrix} (1)(2)+(3)(1)+(0)(-1) & (1)(3)+(3)(2)+(0)(1) & (1)(4)+(3)(3)+(0)(2) \\ (-1)(2)+(2)(1)+(1)(-1) & (-1)(3)+(2)(2)+(1)(1) & (-1)(4)+(2)(3)+(1)(2) \\ (0)(2)+(0)(1)+(2)(-1) & (0)(3)+(0)(2)+(2)(1) & (0)(4)+(0)(3)+(2)(2) \end{bmatrix}$
$= \begin{bmatrix} 2+3+0 & 3+6+0 & 4+9+0 \\ -2+2-1 & -3+4+1 & -4+6+2 \\ 0+0-2 & 0+0+2 & 0+0+4 \end{bmatrix}$
$= \begin{bmatrix} 5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4 \end{bmatrix}$
Thus,the correct option is $B$.

(B) Given matrices are $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.
To find the product $AB$,we perform matrix multiplication:
$AB = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
$AB = \begin{bmatrix} (1 \times 0) + (1 \times 1) & (1 \times 1) + (1 \times 0) \\ (0 \times 0) + (1 \times 1) & (0 \times 1) + (1 \times 0) \end{bmatrix}$
$AB = \begin{bmatrix} 0 + 1 & 1 + 0 \\ 0 + 1 & 0 + 0 \end{bmatrix}$
$AB = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix}$.
First,calculate $AB$:
$AB = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix} = \begin{bmatrix} 0 & -i^2 \\ -i^2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.
Next,calculate $BA$:
$BA = \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} = \begin{bmatrix} 0 & -i^2 \\ -i^2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.
Since $AB = BA$,the matrices $A$ and $B$ commute.
For any two square matrices $A$ and $B$ that commute,the identity $(A + B)(A - B) = A^2 - AB + BA - B^2$ simplifies to $A^2 - B^2$ because $-AB + BA = 0$.
Therefore,$(A + B)(A - B) = A^2 - B^2$.

(B) Given $A = \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix}$,$B = \begin{bmatrix} -1 & 4 \\ 2 & 3 \end{bmatrix}$,and $C = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
We need to calculate $5A - 3B - 2C$.
First,calculate the scalar multiplications:
$5A = 5 \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 5 & -10 \\ 15 & 0 \end{bmatrix}$
$3B = 3 \begin{bmatrix} -1 & 4 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} -3 & 12 \\ 6 & 9 \end{bmatrix}$
$2C = 2 \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$
Now,perform the matrix subtraction:
$5A - 3B - 2C = \begin{bmatrix} 5 & -10 \\ 15 & 0 \end{bmatrix} - \begin{bmatrix} -3 & 12 \\ 6 & 9 \end{bmatrix} - \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$
$= \begin{bmatrix} 5 - (-3) - 0 & -10 - 12 - (-2) \\ 15 - 6 - 2 & 0 - 9 - 0 \end{bmatrix}$
$= \begin{bmatrix} 5 + 3 & -22 + 2 \\ 7 & -9 \end{bmatrix}$
$= \begin{bmatrix} 8 & -20 \\ 7 & -9 \end{bmatrix}$.

(B) Given the matrix equation: $\begin{bmatrix} x & 0 \\ 1 & y \end{bmatrix} + \begin{bmatrix} -2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 6 & 3 \end{bmatrix} - \begin{bmatrix} 2 & 4 \\ 2 & 1 \end{bmatrix}$
First,perform the addition on the left side:
$\begin{bmatrix} x - 2 & 0 + 1 \\ 1 + 3 & y + 4 \end{bmatrix} = \begin{bmatrix} x - 2 & 1 \\ 4 & y + 4 \end{bmatrix}$
Next,perform the subtraction on the right side:
$\begin{bmatrix} 3 - 2 & 5 - 4 \\ 6 - 2 & 3 - 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$
Equating the corresponding elements of the two matrices:
$x - 2 = 1 \Rightarrow x = 3$
$y + 4 = 2 \Rightarrow y = -2$
Thus,the values are $x = 3$ and $y = -2$.

(D) Given $A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$.
Since $A^2 = I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we have:
$A^2 = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} x^2 + 1 & x \\ x & 1 \end{bmatrix}$.
Equating this to the identity matrix:
$\begin{bmatrix} x^2 + 1 & x \\ x & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Comparing the corresponding elements,we get $x = 0$ and $x^2 + 1 = 1$.
From $x^2 + 1 = 1$,we get $x^2 = 0$,which implies $x = 0$.
Thus,the value of $x$ is $0$.

(B) Given $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $aI + bA$:
$aI + bA = a \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + b \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} + \begin{bmatrix} 0 & b \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} a & b \\ 0 & a \end{bmatrix}$.
Now,calculate $(aI + bA)^2$:
$(aI + bA)^2 = \begin{bmatrix} a & b \\ 0 & a \end{bmatrix} \begin{bmatrix} a & b \\ 0 & a \end{bmatrix} = \begin{bmatrix} a^2 + 0 & ab + ba \\ 0 + 0 & 0 + a^2 \end{bmatrix} = \begin{bmatrix} a^2 & 2ab \\ 0 & a^2 \end{bmatrix}$.
This can be written as:
$a^2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 2ab \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = a^2I + 2abA$.

(B) The concept of matrix theory was formally introduced by the British mathematician $Arthur \ Cayley$ in the $19^{th}$ century. While the $Cayley-Hamilton$ theorem is a famous result in matrix algebra,the foundational development of matrix theory is attributed to $Cayley$. Therefore,the correct option is $B$.

(C) First,we calculate $A^2$:
$A^2 = \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(-3) & (1)(2) + (2)(0) \\ (-3)(1) + (0)(-3) & (-3)(2) + (0)(0) \end{bmatrix} = \begin{bmatrix} -5 & 2 \\ -3 & -6 \end{bmatrix} \neq A$.
Next,we calculate $B^2$:
$B^2 = \begin{bmatrix} -1 & 0 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} (-1)(-1) + (0)(2) & (-1)(0) + (0)(3) \\ (2)(-1) + (3)(2) & (2)(0) + (3)(3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 4 & 9 \end{bmatrix} \neq B$.
Now,we calculate $AB$:
$AB = \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} (1)(-1) + (2)(2) & (1)(0) + (2)(3) \\ (-3)(-1) + (0)(2) & (-3)(0) + (0)(3) \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 3 & 0 \end{bmatrix}$.
Finally,we calculate $BA$:
$BA = \begin{bmatrix} -1 & 0 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix} (-1)(1) + (0)(-3) & (-1)(2) + (0)(0) \\ (2)(1) + (3)(-3) & (2)(2) + (3)(0) \end{bmatrix} = \begin{bmatrix} -1 & -2 \\ -7 & 4 \end{bmatrix}$.
Since $AB = \begin{bmatrix} 3 & 6 \\ 3 & 0 \end{bmatrix}$ and $BA = \begin{bmatrix} -1 & -2 \\ -7 & 4 \end{bmatrix}$,it is clear that $AB \neq BA$.

(C) For any two matrices $A$ and $B$,matrix addition is commutative,i.e.,$A + B = B + A$.
Matrix addition is associative,i.e.,$(A + B) + C = A + (B + C)$.
Matrix multiplication is associative,i.e.,$(AB)C = A(BC)$.
However,matrix multiplication is generally not commutative,i.e.,$AB \neq BA$ in most cases.
Therefore,the statement that matrix multiplication is commutative is not true.