Solution

The possible orders that a matrix having 28 elements are ${28 \times 1,1 \times 28,2 \times 14,14 \times 2,4 \times 7,7 \times 4}$. The possible orders of a matrix having 13 elements are ${1 \times 13,13 \times 1}$.

Solution

(i) The order of the given matrix $A$ is $3 \times 3$

(ii) The number of elements in matrix $A=3 \times 3=9$

(iii) $a _{i j}=$ the elements of $i^{\text{th }}$ row and $j^{\text{th }}$ column.

So, $a _{23}=x^{2}-y, a _{31}=0, a _{12}=1$.

Solution

Let $A= \begin{bmatrix} a _{11} & a _{12} \\ a _{21} & a _{22} \end{bmatrix} _{2 \times 2}$

(i) Given that $\quad a _{i j}=\frac{(i-2 j)^{2}}{2}$

$a _{11}=\frac{(1-2 \times 1)^{2}}{2}=\frac{1}{2} ; a _{12}=\frac{(1-2 \times 2)^{2}}{2}=\frac{9}{2}$

$a _{21}=\frac{(2-2 \times 1)^{2}}{2}=0 ; a _{22}=\frac{(2-2 \times 2)^{2}}{2}=2$

Hence, the matrix $A= \begin{bmatrix} \frac{1}{2} & \frac{9}{2} \\ 0 & 2 \end{bmatrix} $ (ii) Given that $a _{ij}=|-2 i+3 j|$

$ \begin{aligned} & a _{11}=|-2 \times 1+3 \times 1|=1 ; a _{12}=|-2 \times 1+3 \times 2|=4 \\ & a _{21}=|-2 \times 2+3 \times 1|=-1 ; a _{22}=|-2 \times 2+3 \times 2|=2 \end{aligned} $

Hence, the matrix $A= \begin{bmatrix} 1 & 4 \\ -1 & 2 \end{bmatrix} $

Solution

Let

$ A= \begin{bmatrix} a _{11} & a _{12} \\ a _{21} & a _{22} \\ a _{31} & a _{32} \end{bmatrix} _{3 \times 2} $

Given that $a _{i j}=e^{i x} \sin j x$

$ \begin{matrix} a _{11}=e^{x} \sin x & a _{12}=e^{x} \sin 2 x \\ a _{21}=e^{2 x} \sin x & a _{22}=e^{2 x} \sin 2 x \\ a _{31}=e^{3 x} \sin x & a _{32}=e^{3 x} \sin 2 x \end{matrix} $

Hence, the matrix $A= \begin{bmatrix} e^{x} \sin x & e^{x} \sin 2 x \\ e^{2 x} \sin x & e^{2 x} \sin 2 x \\ e^{3 x} \sin x & e^{3 x} \sin 2 x \end{bmatrix} $

Solution

Given that $A=B$

$ \Rightarrow \quad \begin{bmatrix} a+4 & 3 b \\ 8 & -6 \end{bmatrix} = \begin{bmatrix} 2 a+2 & b^{2}+2 \\ 8 & b^{2}-5 b \end{bmatrix} $

Equating the corresponding elements, we get

$ \begin{aligned} & a+4=2 a+2, \quad 3 b=b^{2}+2 \quad \text{ and } \quad b^{2}-5 b=-6 \\ & \Rightarrow \quad 2 a-a=2, \quad b^{2}-3 b+2=0, \quad b^{2}-5 b+6=0 \\ & \therefore \quad a=2 \\ & \therefore \quad b^{2}-3 b+2=0 \\ & \therefore \quad b^{2}-5 b+6=0 \\ & \Rightarrow \quad b^{2}-2 b-b+2=0 \text{, } \\ & b^{2}-3 b-2 b+6=0 \\ & \Rightarrow b(b-2)-1(b-2)=0, \quad \Rightarrow \quad b(b-3)-2(b-3)=0 \\ & \Rightarrow \quad(b-1)(b-2)=0, \quad \Rightarrow \quad(b-2)(b-3)=0 \\ & \therefore \quad b=1,2 \Rightarrow \quad b=2,3 \end{aligned} $

but here 2 is common.

Hence, the value of $a=2$ and $b=2$.

Solution

The order of matrix $A=2 \times 2$ and the order of matrix $B=2 \times 3$ Addition of matrices is only possible when they have same order. So, $A+B$ is not possible.

Solution

Given that $X= \begin{bmatrix} 3 & 1 & -1 \\ 5 & -2 & -3\end{bmatrix} $ and $Y= \begin{bmatrix} 2 & 1 & -1 \\ 7 & 2 & 4 \end{bmatrix} $

(i) $X+Y= \begin{bmatrix} 3 & 1 & -1 \\ 5 & -2 & -3\end{bmatrix} + \begin{bmatrix} 2 & 1 & -1 \\ 7 & 2 & 4 \end{bmatrix} $

$ = \begin{bmatrix} 3+2 & 1+1 & -1-1 \\ 5+7 & -2+2 & -3+4 \end{bmatrix} = \begin{bmatrix} 5 & 2 & -2 \\ 12 & 0 & 1 \end{bmatrix} $

(ii) $2 X-3 Y=2 \begin{bmatrix} 3 & 1 & -1 \\ 5 & -2 & -3\end{bmatrix} -3 \begin{bmatrix} 2 & 1 & -1 \\ 7 & 2 & 4 \end{bmatrix} $

$ \begin{aligned} & = \begin{bmatrix} 2 \times 3 & 2 \times 1 & -2 \times 1 \\ 2 \times 5 & -2 \times 2 & -2 \times 3 \end{bmatrix} - \begin{bmatrix} 3 \times 2 & 1 \times 3 & -1 \times 3 \\ 3 \times 7 & 3 \times 2 & 3 \times 4 \end{bmatrix} \\ & = \begin{bmatrix} 6 & 2 & -2 \\ 10 & -4 & -6 \end{bmatrix} - \begin{bmatrix} 6 & 3 & -3 \\ 21 & 6 & 12 \end{bmatrix} \\ & = \begin{bmatrix} 6-6 & 2-3 & -2+3 \\ 10-21 & -4-6 & -6-12 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 1 \\ -11 & -10 & -18 \end{bmatrix} \end{aligned} $

(iii) $X+Y+Z=0$

$\Rightarrow \begin{bmatrix} 3 & 1 & -1 \\ 5 & -2 & -3\end{bmatrix} + \begin{bmatrix} 2 & 1 & -1 \\ 7 & 2 & 4\end{bmatrix} + \begin{bmatrix} a & b & c \\ d & e & f\end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} $ where $\mathbf{Z}= \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix} $

$ \begin{matrix} \Rightarrow & { \begin{bmatrix} 3+2+a & 1+1+b & -1-1+c \\ 5+7+d & -2+2+e & -3+4+f \end{bmatrix} } & = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ \Rightarrow & { \begin{bmatrix} 5+a & 2+b & -2+c \\ 12+d & e & 1+f \end{bmatrix} } & = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{matrix} $

Equating the corresponding elements, we get

$5+a=0 \Rightarrow a=-5, \quad 2+b=0 \Rightarrow b=-2, \quad-2+c=0 \Rightarrow c=2$

$12+d=0 \Rightarrow d=-12, \quad e=0, \quad 1+f=0 \Rightarrow f=-1$

Hence, the matrix $Z= \begin{bmatrix} -5 & -2 & 2 \\ -12 & 0 & -1 \end{bmatrix} $

Solution

The given equation can be written as

$ \begin{aligned} & { \begin{bmatrix} 2 x^{2} & 2 x \\ 3 x & x^{2} \end{bmatrix} + \begin{bmatrix} 16 & 10 x \\ 8 & 8 x \end{bmatrix} = \begin{bmatrix} (2 x^{2}+16) & 48 \\ 20 & 12 x \end{bmatrix} } \\ & \Rightarrow \quad \begin{bmatrix} 2 x^{2}+16 & 12 x \\ 3 x+8 & x^{2}+8 x \end{bmatrix} = \begin{bmatrix} 2 x^{2}+16 & 48 \\ 20 & 12 x \end{bmatrix} \end{aligned} $

Equating the corresponding elements we get

$ \begin{aligned} & 12 x=48, \quad 3 x+8=20, \\ & x^{2}+8 x=12 x \\ & \therefore x=\frac{48}{12}=4, \quad 3 x=20-8=12, \quad \Rightarrow \quad x^{2}=12 x-8 x=4 x \\ & \therefore \quad x=4, \quad \Rightarrow \quad x^{2}-4 x=0 \\ & x=0, x=4 \end{aligned} $

Hence, the non-zero values of $x$ is 4 .

Solution

Given that $A= \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix} $ and $B= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $

$ \begin{aligned} & A+B= \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \\ & \Rightarrow \quad A+B= \begin{bmatrix} 0+0 & 1-1 \\ 1+1 & 1+0 \end{bmatrix} \Rightarrow A+B= \begin{bmatrix} 0 & 0 \\ 2 & 1 \end{bmatrix} \\ & A-B= \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \\ & \Rightarrow \quad A-B= \begin{bmatrix} 0-0 & 1+1 \\ 1-1 & 1-0 \end{bmatrix} \Rightarrow A-B= \begin{bmatrix} 0 & 2 \\ 0 & 1 \end{bmatrix} \end{aligned} $

$ \begin{aligned} \therefore(A+B) \cdot(A-B)= \begin{bmatrix} 0 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0+0 & 0+0 \\ 0+0 & 4+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix} \\ \begin{aligned} \text{ Now, R.H.S. } & =A^{2}-B^{2} \\ & =A \cdot A-B \cdot B \\ & = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0+1 & 0+1 \\ 0+1 & 1+1 \end{bmatrix} - \begin{bmatrix} 0-1 & 0+0 \\ 0+0 & -1+0 \end{bmatrix} \\ & = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1+1 & 1-0 \\ 1-0 & 2+1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \end{aligned} \end{aligned} $

Hence, $ \begin{bmatrix} 0 & 0 \\ 0 & 5\end{bmatrix} \neq \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} $

Hence, $(A+B) \cdot(A-B) \neq A^{2}-B^{2}$

Solution

Given that $ \begin{bmatrix} 1 & x & 1\end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} =O$

$ \begin{aligned} & \Rightarrow \begin{bmatrix} 1+2 x+15 & 3+5 x+3 & 2+x+2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} =O \\ & \Rightarrow \quad \begin{bmatrix} 2 x+16 & 5 x+6 & x+4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} =O \\ & \Rightarrow[2 x+16+10 x+12+x^{2}+4 x]=0 ; \Rightarrow x^{2}+16 x+28=0 \\ & \Rightarrow \quad x^{2}+14 x+2 x+28=0 ; \Rightarrow x(x+14)+2(x+14)=0 \\ & \Rightarrow \quad(x+2)(x+14)=0 ; x+2=0 \quad \text{ or } \quad x+14=0 \\ & \therefore \quad x=-2 \text{ or } x=-14 \end{aligned} $

Hence, the values of $x$ are -2 and -14 .

Solution

Given that $A= \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} $

$ \begin{aligned} A^{2} & =A \cdot A \\ & = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 25-3 & 15-6 \\ -5+2 & -3+4 \end{bmatrix} = \begin{bmatrix} 22 & 9 \\ -3 & 1 \end{bmatrix} \end{aligned} $

$A^{2}-3 A-7 I=O$

L.H.S. $ \begin{bmatrix} 22 & 9 \\ -3 & 1\end{bmatrix} -3 \begin{bmatrix} 5 & 3 \\ -1 & -2\end{bmatrix} -7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $

$\Rightarrow \begin{bmatrix} 22 & 9 \\ -3 & 1\end{bmatrix} - \begin{bmatrix} 15 & 9 \\ -3 & -6\end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} $

$\Rightarrow \begin{bmatrix} 22-15-7 & 9-9-0 \\ -3+3-0 & 1+6-7\end{bmatrix} \Rightarrow \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \quad$ R.H.S.

We are given $A^{2}-3 A-7 I=O$

$\Rightarrow \quad A^{-1}[A^{2}-3 A-7 I]=A^{-1} O$

[Pre-multiplying both sides by $A^{-1}$ ]

$\Rightarrow \quad A^{-1} A \cdot A-3 A^{-1} \cdot A-7 A^{-1} I=O$

$[A^{-1} O=O]$

$\Rightarrow \quad I \cdot A-3 I-7 A^{-1} I=O$

$\Rightarrow A-3 I-7 A^{-1}=O$

$\Rightarrow \quad-7 A^{-1}=3 I-A$

$\Rightarrow \quad A^{-1}=\frac{1}{-7}[3 I-A]$

$\Rightarrow \quad A^{-1}=\frac{1}{-7}[3(\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix} )-(\begin{matrix} 5 & 3 \\ -1 & -2\end{matrix} )]$

$=\frac{1}{-7}[(\begin{matrix} 3 & 0 \\ 0 & 3\end{matrix} )-(\begin{matrix} 5 & 3 \\ -1 & -2\end{matrix} )]$

$=\frac{1}{-7} \begin{bmatrix} 3-5 & 0-3 \\ 0+1 & 3+2\end{bmatrix} =\frac{1}{-7} \begin{bmatrix} -2 & -3 \\ 1 & 5 \end{bmatrix} $

Hence, $A^{-1}=-\frac{1}{7} \begin{bmatrix} -2 & -3 \\ 1 & 5 \end{bmatrix} $

Solution

Let $A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} _{2 \times 2}$

$ \begin{aligned} { \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} _{2 \times 2} \begin{bmatrix} a & b \\ c & d \end{bmatrix} _{2 \times 2} \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} _{2 \times 2} } & = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} _{2 \times 2} \\ \Rightarrow \quad \begin{bmatrix} 2 a+c & 2 b+d \\ 3 a+2 c & 3 b+2 d \end{bmatrix} _{2 \times 2} \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} _{2 \times 2} & = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} _{2 \times 2} \\ \Rightarrow \quad \begin{bmatrix} -6 a-3 c+10 b+5 d & 4 a+2 c-6 b-3 d \\ -9 a-6 c+15 b+10 d & 6 a+4 c-9 b-6 d \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{aligned} $

Equating the corresponding elements, we get,

$-6 a-3 c+10 b+5 d =1\qquad$……(1)

$-9 a-6 c+15 b+10 d =0\qquad$……(2)

$4 a+2 c-6 b-3 d =0\qquad$……(3)

$6 a+4 c-9 b-6 d =1\qquad$……(4)

Multiplying eq. (1) by 2 and subtracting eq. (2), we get,

$$ \begin{matrix} -12 a-6 d+20 b+10 d=2 \\ 9 a-6 c+15 b+10 d=0 & \\ (+)\quad(+)\quad(-)\quad(-)\quad(-) \\ \hline-3a\qquad +5b\qquad =2 \end{matrix} $$

$ \begin{aligned} & -3 a+5 b=2 \end{aligned} $

Now, multiplying eq. (3) by 2 and subtracting eq. (4), we get

$$ \begin{matrix} 8 a+4 c /-12 b-6 d /=0 \\ 6 a+4 d-9 b-6 d=1 \\ (-) \quad(-) \quad(+) \quad(+)\quad (-) \\ \hline 2 a \qquad-3 b\qquad=-1 \tag{6}\\ 2 a-3 b=-1 & \end{matrix} $$

Solving eq. (5) and (6) i.e.,

$ \begin{aligned} -3 a+5 b & =2 \\ 2 a-3 b & =-1 \\ \Rightarrow \quad-6 a+10 b & =4 \\ \Rightarrow \quad 6 a-9 b & =-3 \\ \text{ Adding } \quad b & =1 \end{aligned} $

$ \begin{matrix} 2 \times(-3 a+5 b=2) & \Rightarrow & -6 a+10 b= & 4 \\ 3 \times(2 a-3 b=-1) & \Rightarrow & 6 a-9 b=-3 \end{matrix} $

Putting the value of $b$ in eq. (6), we get,

$ 2 a-3 \times 1=-1 $

$\Rightarrow 2 a-3=-1 \Rightarrow 2 a=3-1 \Rightarrow 2 a=2$

$\therefore \quad a=1$

Now, putting the values of $a$ and $b$ in equations (1) and (3)

$ \begin{aligned} & -6 \times 1-3 c+10 \times 1+5 d=1 \\ & \Rightarrow \quad-6-3 c+10+5 d=1 \end{aligned} $

$$ \begin{equation*} \Rightarrow \quad-3 c+5 d+4=1 \quad \Rightarrow-3 c+5 d=-3 \tag{7} \end{equation*} $$

and from eq. (3)

$$ \begin{align*} 4 \times 1+2 c-6 \times 1-3 d & =0 ; \\ \Rightarrow \quad & \Rightarrow 4+2 c-6-3 d=0 \tag{8}\\ -2+2 c-3 d & =0 \quad \Rightarrow \quad 2 c-3 d=2 \end{align*} $$

Solving eq. (7) and (8) we get,

$ \begin{matrix} 2 \times(-3 c+5 d=-3) & \Rightarrow & -6 c+10 d=-6 \\ 3 \times(2 c-3 d=2) & \Rightarrow & \text{ Adding } \end{matrix} $

Putting the value of $d$ in eq. (8) we get,

$ 2 c-3 \times 0=2 \Rightarrow 2 c=2 \Rightarrow c=1 $

Hence, $A= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} $.

Solution

Order of $ \begin{bmatrix} 4 \\ 1 \\ 3\end{bmatrix} $ is $3 \times 1$ and order of $ \begin{bmatrix} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{bmatrix} $ is $3 \times 3$.

So, the order of matrix A must be $1 \times 3$.

Let $A= \begin{bmatrix} a & b & c \end{bmatrix} _{1 \times 3}$

$ \begin{aligned} { \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} _{3 \times 1} \begin{bmatrix} a & b & c \end{bmatrix} _{1 \times 3} } & = \begin{bmatrix} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{bmatrix} _{3 \times 3} \\ \Rightarrow \quad \begin{bmatrix} 4 a & 4 b & 4 c \\ a & b & c \\ 3 a & 3 b & 3 c \end{bmatrix} & = \begin{bmatrix} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{bmatrix} \end{aligned} $

Equating the corresponding elements to get the values of $a, b$ and $c$

$ \begin{aligned} & 4 a=-4, \quad 4 b=8, \quad 4 c=4 \\ & \therefore \quad a=-1 \quad \therefore \quad b=2 \quad \therefore c=1 \end{aligned} $

Solution

Here, $\quad B= \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4\end{bmatrix} _{2 \times 3}$ and $A= \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix} _{3 \times 2}$

$\therefore \quad BA= \begin{bmatrix} 6+1+4 & -8+1+0 \\ 3+2+8 & -4+2+0\end{bmatrix} _{2 \times 2} \Rightarrow BA= \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix} $

L.H.S. $\quad(BA)^{2}=(BA) \cdot(BA)= \begin{bmatrix} 11 & -7 \\ 13 & -2\end{bmatrix} \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix} $

$\Rightarrow \quad \begin{bmatrix} 121-91 & -77+14 \\ 143-26 & -91+4\end{bmatrix} \Rightarrow \begin{bmatrix} 30 & -63 \\ 117 & -87 \end{bmatrix} $

R.H.S. $B^{2}=B \cdot B= \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4\end{bmatrix} _{2 \times 3} \cdot \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix} _{2 \times 3}$

Here, number of columns of first i.e., 3 is not equal to the number of rows of second matrix i.e., 2.

So, $B^{2}$ is not possible. Similarly, $A^{2}$ is also not possible.

Hence, $(BA)^{2} \neq B^{2} A^{2}$

Solution

$\quad \begin{aligned} B A & = \begin{bmatrix} 4 & 1 \\ 2 & 3 \\ 1 & 2\end{bmatrix} _{3 \times 2} \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4\end{bmatrix} _{2 \times 3} \\ B A & = \begin{bmatrix} 8+1 & 4+2 & 8+4 \\ 4+3 & 2+6 & 4+12 \\ 2+2 & 1+4 & 2+8\end{bmatrix} _{3 \times 3}= \begin{bmatrix} 9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10 \end{bmatrix} _{3 \times 3}\end{aligned}$

Now $A B= \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4\end{bmatrix} _{2 \times 3} \begin{bmatrix} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{bmatrix} _{3 \times 2}$

$ = \begin{bmatrix} 8+2+2 & 2+3+4 \\ 4+4+4 & 1+6+8 \end{bmatrix} _{2 \times 2}= \begin{bmatrix} 12 & 9 \\ 12 & 15 \end{bmatrix} _{2 \times 2} $

Hence, $\quad BA= \begin{bmatrix} 9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10\end{bmatrix} $ and $AB= \begin{bmatrix} 12 & 9 \\ 12 & 15 \end{bmatrix} $.

Solution

Let $A= \begin{bmatrix} 1 & -1 \\ -1 & 1\end{bmatrix} $ and $B= \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} $

$ \begin{aligned} & AB= \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \\ & \Rightarrow \quad AB= \begin{bmatrix} 1-1 & 1-1 \\ -1+1 & -1+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} =O \end{aligned} $

Hence, $\quad A= \begin{bmatrix} 1 & -1 \\ -1 & 1\end{bmatrix} $ and $B= \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} $.

Solution

Here, $A= \begin{bmatrix} 2 & 4 & 0 \\ 3 & 9 & 6\end{bmatrix} , B= \begin{bmatrix} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{bmatrix} $

$ \begin{aligned} AB & = \begin{bmatrix} 2 & 4 & 0 \\ 3 & 9 & 6 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 2+8+0 & 8+32+0 \\ 3+18+6 & 12+72+18 \end{bmatrix} = \begin{bmatrix} 10 & 40 \\ 27 & 102 \end{bmatrix} \end{aligned} $

L.H.S. $(A B)^{\prime}= \begin{bmatrix} 10 & 27 \\ 40 & 102 \end{bmatrix} $

Now $B= \begin{bmatrix} 1 & 4 \\ 2 & 8 \\ 1 & 3\end{bmatrix} \quad \Rightarrow \quad B^{\prime}= \begin{bmatrix} 1 & 2 & 1 \\ 4 & 8 & 3 \end{bmatrix} $

$ A= \begin{bmatrix} 2 & 4 & 0 \\ 3 & 9 & 6 \end{bmatrix} \Rightarrow A^{\prime}= \begin{bmatrix} 2 & 3 \\ 4 & 9 \\ 0 & 6 \end{bmatrix} $

R.H.S. $B^{\prime} A^{\prime}= \begin{bmatrix} 1 & 2 & 1 \\ 4 & 8 & 3\end{bmatrix} \begin{bmatrix} 2 & 3 \\ 4 & 9 \\ 0 & 6 \end{bmatrix} $

$ = \begin{bmatrix} 2+8+0 & 3+18+6 \\ 8+32+0 & 12+72+18 \end{bmatrix} = \begin{bmatrix} 10 & 27 \\ 40 & 102 \end{bmatrix} =\text{ L.H.S. } $

Hence, L.H.S. = R.H.S.

Solution

Given that: $x \begin{bmatrix} 2 \\ 1\end{bmatrix} +y \begin{bmatrix} 3 \\ 5\end{bmatrix} + \begin{bmatrix} -8 \\ 11 \end{bmatrix} =$

L.H.S. $x \begin{bmatrix} 2 \\ 1\end{bmatrix} +y \begin{bmatrix} 3 \\ 5\end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} =0$

$\Rightarrow \quad \begin{bmatrix} 2 x \\ x\end{bmatrix} + \begin{bmatrix} 3 y \\ 5 y\end{bmatrix} + \begin{bmatrix} -8 \\ -11\end{bmatrix} =O \Rightarrow \begin{bmatrix} 2 x+3 y-8 \\ x+5 y-11\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

Comparing the corresponding elements of both sides, we get,

$ \begin{aligned} & 2 x+3 y-8=0 \quad \Rightarrow \quad 2 x+3 y=8 \\ & x+5 y-11=0 \quad \Rightarrow \quad x+5 y=11 \end{aligned} $

Multiplying eq. (1) by 1 and eq. (2) by 2, and then on subtracting, we get,

$$ \begin{aligned} 2 x+3 y=8 \\ 2 x+10 y=22 \\ (-) \quad(-) \quad(-) \\ \hline -7y =-14 \end{aligned} $$

$\therefore \quad y=2$

Putting $y=2$ in eq. (2) we get,

$ x+5 \times 2=11 \Rightarrow x+10=11 $

$ \therefore \quad x=11-10=1 $

Hence, the values of $x$ and $y$ are 1 and 2 respectively.

Solution

Given that:

$$ \begin{align*} & 2 X+3 Y= \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \tag{1}\\ & 3 X+2 Y= \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix} \tag{2} \end{align*} $$

Multiplying eq. (1) by 3 and eq. (2) by 2, we get,

$ \begin{aligned} & 3[2 X+3 Y]=3 \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \quad \Rightarrow \quad 6 X+9 Y= \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix} \\ & 2[3 X+2 Y]=2 \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix} \Rightarrow 6 X+4 Y= \begin{bmatrix} -4 & 4 \\ 2 & -10 \end{bmatrix} \end{aligned} $

On subtracting eq. (4) from eq. (3) we get

$ \begin{aligned} & 5 Y= \begin{bmatrix} 6+4 & 9-4 \\ 12-2 & 0+10 \end{bmatrix} \\ & 5 Y= \begin{bmatrix} 10 & 5 \\ 10 & 10 \end{bmatrix} \Rightarrow Y= \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix} \end{aligned} $

Now, putting the value of $Y$ in equation (1) we get,

$ \begin{aligned} & 2 X+3 \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \\ & \Rightarrow 2 X+ \begin{bmatrix} 6 & 3 \\ 6 & 6 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \quad \Rightarrow 2 X= \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} 6 & 3 \\ 6 & 6 \end{bmatrix} \\ & \Rightarrow \quad 2 X= \begin{bmatrix} 2-6 & 3-3 \\ 4-6 & 0-6 \end{bmatrix} \Rightarrow 2 X= \begin{bmatrix} -4 & 0 \\ -2 & -6 \end{bmatrix} \\ & \Rightarrow \quad X=\frac{1}{2} \begin{bmatrix} -4 & 0 \\ -2 & -6 \end{bmatrix} \quad \Rightarrow \quad X= \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix} \end{aligned} $

Hence, $X= \begin{bmatrix} -2 & 0 \\ -1 & -3\end{bmatrix} $ and $Y= \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix} $.

Solution

Given that: $A= \begin{bmatrix} 3 & 5\end{bmatrix} _{1 \times 2}, B= \begin{bmatrix} 7 & 3 \end{bmatrix} _{1 \times 2}$

Let

$ \begin{aligned} C & = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} _{2 \times 1} \\ AC & = \begin{bmatrix} 3 & 5 \end{bmatrix} _{1 \times 2} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} _{2 \times 1}=[3 \alpha+5 \beta] \\ BC & = \begin{bmatrix} 7 & 3 \end{bmatrix} _{1 \times 2} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} _{2 \times 1}=[7 \alpha+3 \beta] \\ AC & =BC \end{aligned} $

$ \begin{matrix} \Rightarrow & {[3 \alpha+5 \beta]=[7 \alpha+3 \beta]} \\ \Rightarrow & 3 \alpha+5 \beta=7 \alpha+3 \beta \end{matrix} $

$ \begin{matrix} \Rightarrow & 3 \alpha-7 \alpha & =3 \beta-5 \beta \\ \Rightarrow & -4 \alpha & =-2 \beta \\ & \therefore & \frac{\alpha}{\beta} & =\frac{1}{2} \end{matrix} $

So, let $\alpha=K$ and $\beta=2 K$, $K$ is some real number.

Hence, matrix $C= \begin{bmatrix} K \\ 2 K\end{bmatrix} _{2 \times 1}$ or $ \begin{bmatrix} K & K \\ 2 K & 2 K \end{bmatrix} _{2 \times 2}$ etc.

Solution

Let $A= \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix} , B= \begin{bmatrix} 1 & 2 \\ 2 & 0\end{bmatrix} $ and $C= \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} $

$ \begin{aligned} & AB= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 0 \end{bmatrix} \Rightarrow AB= \begin{bmatrix} 1+0 & 2+0 \\ 0+0 & 0+0 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \\ & AC= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 1+0 & 2+0 \\ 0+0 & 0+0 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \end{aligned} $

Hence, $A B=A C$ for matrix $A$ is non-zero and $B \neq C$.

Solution

Given that $A= \begin{bmatrix} 1 & 2 \\ -2 & 1\end{bmatrix} , B= \begin{bmatrix} 2 & 3 \\ 3 & -4\end{bmatrix} $ and $C= \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} $

(i) To verify: $(AB) C=A(BC)$

$ AB= \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix} = \begin{bmatrix} 2+6 & 3-8 \\ -4+3 & -6-4 \end{bmatrix} = \begin{bmatrix} 8 & -5 \\ -1 & -10 \end{bmatrix} $

L.H.S.

$ \begin{aligned} (A B) C & = \begin{bmatrix} 8 & -5 \\ -1 & -10 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 8+5 & 0+0 \\ -1+10 & 0+0 \end{bmatrix} = \begin{bmatrix} 13 & 0 \\ 9 & 0 \end{bmatrix} \\ BC & = \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 2-3 & 0+0 \\ 3+4 & 0+0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 7 & 0 \end{bmatrix} \end{aligned} $

R.H.S.

$ A(BC)= \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 7 & 0 \end{bmatrix} = \begin{bmatrix} -1+14 & 0+0 \\ 2+7 & 0+0 \end{bmatrix} = \begin{bmatrix} 13 & 0 \\ 9 & 0 \end{bmatrix} $

L.H.S. $=$ R.H.S.

So, $(AB) C=A(BC)$

(ii) To verify: $A(B+C)=AB+AC$

L.H.S. $B+C= \begin{bmatrix} 2 & 3 \\ 3 & -4\end{bmatrix} + \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} =$

$ = \begin{bmatrix} 2+1 & 3+0 \\ 3-1 & -4+0 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 2 & -4 \end{bmatrix} $

L.H.S.A $(B+C)= \begin{bmatrix} 1 & 2 \\ -2 & 1\end{bmatrix} \begin{bmatrix} 3 & 3 \\ 2 & -4 \end{bmatrix} $

$ = \begin{bmatrix} 3+4 & 3-8 \\ -6+2 & -6-4 \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ -4 & -10 \end{bmatrix} $

R.H.S. $AB= \begin{bmatrix} 1 & 2 \\ -2 & 1\end{bmatrix} \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix} $

$ = \begin{bmatrix} 2+6 & 3-8 \\ -4+3 & -6-4 \end{bmatrix} = \begin{bmatrix} 8 & -5 \\ -1 & -10 \end{bmatrix} $

$ \begin{aligned} AC & = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 1-2 & 0+0 \\ -2-1 & 0+0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ -3 & 0 \end{bmatrix} \end{aligned} $

R.H.S. $AB+AC= \begin{bmatrix} 8 & -5 \\ -1 & -10\end{bmatrix} + \begin{bmatrix} -1 & 0 \\ -3 & 0\end{bmatrix} = \begin{bmatrix} 8-1 & -5+0 \\ -1-3 & -10+0 \end{bmatrix} $

$ = \begin{bmatrix} 7 & -5 \\ -4 & -10 \end{bmatrix} $

L.H.S. $=$ R.H.S.

Hence, $A(B+C)=AB+AC$

Solution

Given that: $P= \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{bmatrix} $ and $Q= \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} $

$PQ= \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{bmatrix} \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} $

$PQ= \begin{bmatrix} x a+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+y b+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+z c \end{bmatrix} $

$PQ= \begin{bmatrix} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{bmatrix} $

Now QP $= \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{bmatrix} \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} $

$QP= \begin{bmatrix} x a+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+y b+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+z c \end{bmatrix} $

$QP= \begin{bmatrix} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{bmatrix} $

Hence, $PQ=QP$.

Solution

Given that: $ \begin{bmatrix} 2 & 1 & 3\end{bmatrix} \begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} =A$

L.H.S. $\quad \begin{bmatrix} 2 & 1 & 3\end{bmatrix} _{1 \times 3} \begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix} _{3 \times 3} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} _{3 \times 1}$

$ \begin{matrix} \Rightarrow & { \begin{bmatrix} -2-1+0 & 0+1+3 & -2+0+3 \end{bmatrix} _{1 \times 3} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} _{3 \times 1}} \\ \Rightarrow & { \begin{bmatrix} -3 & 4 & 1 \end{bmatrix} _{1 \times 3} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} _{3 \times 1}} \\ \Rightarrow \quad & {[-3+0-1] _{1 \times 1}=[-4] _{1 \times 1}} \end{matrix} $

Hence, matrix $A=[-4]$

Solution

Given that: $A= \begin{bmatrix} 2 & 1\end{bmatrix} , B= \begin{bmatrix} 5 & 3 & 4 \\ 8 & 7 & 6\end{bmatrix} $ and $C= \begin{bmatrix} -1 & 2 & 1 \\ 1 & 0 & 2 \end{bmatrix} $.

L.H.S. $\quad(B+C)= \begin{bmatrix} 5 & 3 & 4 \\ 8 & 7 & 6\end{bmatrix} + \begin{bmatrix} -1 & 2 & 1 \\ 1 & 0 & 2 \end{bmatrix} $

$ = \begin{bmatrix} 5-1 & 3+2 & 4+1 \\ 8+1 & 7+0 & 2+6 \end{bmatrix} = \begin{bmatrix} 4 & 5 & 5 \\ 9 & 7 & 8 \end{bmatrix} $

$ \begin{aligned} A(B+C) & = \begin{bmatrix} 2 & 1 \end{bmatrix} _{1 \times 2} \begin{bmatrix} 4 & 5 & 5 \\ 9 & 7 & 8 \end{bmatrix} _{2 \times 3} \\ & = \begin{bmatrix} 8+9 & 10+7 & 10+8 \end{bmatrix} _{1 \times 3} \\ A(B+C) & = \begin{bmatrix} 17 & 17 & 18 \end{bmatrix} \end{aligned} $

R.H.S. $A B= \begin{bmatrix} 2 & 1\end{bmatrix} _{1 \times 2} \begin{bmatrix} 5 & 3 & 4 \\ 8 & 7 & 6 \end{bmatrix} _{2 \times 3}$

$ \begin{aligned} & = \begin{bmatrix} 10+8 & 6+7 & 8+6 \end{bmatrix} _{1 \times 3}= \begin{bmatrix} 18 & 13 & 14 \end{bmatrix} _{1 \times 3} \\ & AC= \begin{bmatrix} 2 & 1 \end{bmatrix} _{1 \times 2} \begin{bmatrix} -1 & 2 & 1 \\ 1 & 0 & 2 \end{bmatrix} _{2 \times 3} \\ & = \begin{bmatrix} -2+1 & 4+0 & 2+2 \end{bmatrix} _{1 \times 3}= \begin{bmatrix} -1 & 4 & 4 \end{bmatrix} _{1 \times 3} \\ & AB+AC= \begin{bmatrix} 18 & 13 & 14 \end{bmatrix} _{1 \times 3}+ \begin{bmatrix} -1 & 4 & 4 \end{bmatrix} _{1 \times 3} \\ & = \begin{bmatrix} 18-1 & 13+4 & 14+4 \end{bmatrix} _{1 \times 3} \\ & AB+AC= \begin{bmatrix} 17 & 17 & 18 \end{bmatrix} _{1 \times 3} \\ & \text{ L.H.S. = R.H.S. } \end{aligned} $

Solution

Given that: $A= \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} $

$ \begin{aligned} A^{2} & =A \cdot A \\ & = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 1+0+0 & 0+0-1 & -1+0-1 \\ 2+2+0 & 0+1+3 & -2+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{bmatrix} \end{aligned} $

L.H.S. $\quad A^{2}+A= \begin{bmatrix} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4\end{bmatrix} \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} $

$= \begin{bmatrix} 1+1 & -1+0 & -2-1 \\ 4+2 & 4+1 & 4+3 \\ 2+0 & 2+1 & 4+1\end{bmatrix} = \begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix} $

R.H.S. $A(A+I)= \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} [(\begin{matrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{matrix} )+(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix} )]$

$= \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix} \begin{bmatrix} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix} $

$= \begin{bmatrix} 2+0+0 & 0+0-1 & -1+0-2 \\ 4+2+0 & 0+2+3 & -2+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2\end{bmatrix} = \begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix} $

L.H.S. $=$ R.H.S.

$A^{2}+A=A(A+I)$. Hence verified.

Solution

Given that: $A= \begin{bmatrix} 0 & -1 & 2 \\ 4 & 3 & -4\end{bmatrix} , B= \begin{bmatrix} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{bmatrix} $

(i)

$ \begin{aligned} A^{\prime} & = \begin{bmatrix} 0 & -1 & 2 \\ 4 & 3 & -4 \end{bmatrix} _{2 \times 3}^{\prime}= \begin{bmatrix} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{bmatrix} _{3 \times 2} \\ (A^{\prime})^{\prime} & = \begin{bmatrix} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{bmatrix} _{3 \times 2}^{\prime}= \begin{bmatrix} 0 & -1 & 2 \\ 4 & 3 & -4 \end{bmatrix} _{2 \times 3}=A \end{aligned} $

Hence, $(A^{\prime})^{\prime}=A$

(ii) L.H.S. $\quad AB= \begin{bmatrix} 0 & -1 & 2 \\ 4 & 3 & -4\end{bmatrix} _{2 \times 3} \begin{bmatrix} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{bmatrix} _{3 \times 2}$

$= \begin{bmatrix} 0-1+4 & 0-3+12 \\ 16+3-8 & 0+9-24\end{bmatrix} _{2 \times 2}= \begin{bmatrix} 3 & 9 \\ 11 & -15 \end{bmatrix} _{2 \times 2}$

$(A B)^{\prime}= \begin{bmatrix} 3 & 11 \\ 9 & -15 \end{bmatrix} _{2 \times 2}$

R.H.S. $B^{\prime}= \begin{bmatrix} 4 & 0 \\ 1 & 3 \\ 2 & 6\end{bmatrix} ^{\prime}= \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 6 \end{bmatrix} $

$A^{\prime}= \begin{bmatrix} 0 & -1 & 2 \\ 4 & 3 & -4\end{bmatrix} ^{\prime}= \begin{bmatrix} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{bmatrix} $

$B^{\prime} A^{\prime}= \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 6\end{bmatrix} _{2 \times 3} \begin{bmatrix} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{bmatrix} _{3 \times 2}$

$= \begin{bmatrix} 0-1+4 & 16+3-8 \\ 0-3+12 & 0+9-24\end{bmatrix} _{2 \times 2}= \begin{bmatrix} 3 & 11 \\ 9 & -15 \end{bmatrix} _{2 \times 2}$

L.H.S. $=$ R.H.S.

Hence, $(A B)^{\prime}=B^{\prime} A^{\prime}$ is verified.

(iii) L.H.S. $\quad k A=k \begin{bmatrix} 0 & -1 & 2 \\ 4 & 3 & -4\end{bmatrix} = \begin{bmatrix} 0 & -k & 2 k \\ 4 k & 3 k & -4 k \end{bmatrix} $

$ (k A)^{\prime}= \begin{bmatrix} 0 & 4 k \\ -k & 3 k \\ 2 k & -4 k \end{bmatrix} $

R.H.S. $k A^{\prime}=k \begin{bmatrix} 0 & -1 & 2 \\ 4 & 3 & -4\end{bmatrix} ^{\prime}=k \begin{bmatrix} 0 & 4 \\ -1 & 3 \\ 2 & -4\end{bmatrix} = \begin{bmatrix} 0 & 4 k \\ -k & 3 k \\ 2 k & -4 k \end{bmatrix} $

Hence, L.H.S. $=$ R.H.S.

$(k A)^{\prime}=(k A^{\prime})$ is verified.

Solution

Given that: $A= \begin{bmatrix} 1 & 2 \\ 4 & 1 \\ 5 & 6\end{bmatrix} $ and $B= \begin{bmatrix} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{bmatrix} $

(i) To verify that: $(2 A+B)^{\prime}=2 A^{\prime}+B^{\prime}$

$ \begin{aligned} & \text{ L.H.S. }(2 A+B)^{\prime}=[2(\begin{matrix} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{matrix} )+(\begin{matrix} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{matrix} )]^{\prime}=[(\begin{matrix} 2 & 4 \\ 8 & 2 \\ 10 & 12 \end{matrix} )+(\begin{matrix} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{matrix} )]^{\prime} \\ & = \begin{bmatrix} 2+1 & 4+2 \\ 8+6 & 2+4 \\ 10+7 & 12+3 \end{bmatrix} ^{\prime}= \begin{bmatrix} 3 & 6 \\ 14 & 6 \\ 17 & 15 \end{bmatrix} ^{\prime}= \begin{bmatrix} 3 & 14 & 17 \\ 6 & 6 & 15 \end{bmatrix} \\ & \text{ R.H.S. } 2 A^{\prime}+B^{\prime}=2 \begin{bmatrix} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{bmatrix} ^{\prime}+ \begin{bmatrix} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{bmatrix} ^{\prime} \\ & =2 \begin{bmatrix} 1 & 4 & 5 \\ 2 & 1 & 6 \end{bmatrix} + \begin{bmatrix} 1 & 6 & 7 \\ 2 & 4 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 2 & 8 & 10 \\ 4 & 2 & 12 \end{bmatrix} + \begin{bmatrix} 1 & 6 & 7 \\ 2 & 4 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 2+1 & 8+6 & 10+7 \\ 4+2 & 2+4 & 12+3 \end{bmatrix} = \begin{bmatrix} 3 & 14 & 17 \\ 6 & 6 & 15 \end{bmatrix} \end{aligned} $

Hence, $\quad$ L.H.S. $=$ R.H.S.

$ (2 A+B)^{\prime}=2 A^{\prime}+B^{\prime} \text{ is verified. } $

(ii) To verify that: $(A-B)^{\prime}=A^{\prime}-B^{\prime}$

L.H.S. $(A-B)^{\prime}=[(\begin{matrix} 1 & 2 \\ 4 & 1 \\ 5 & 6\end{matrix} )-(\begin{matrix} 1 & 2 \\ 6 & 4 \\ 7 & 3\end{matrix} )]^{\prime}$

$= \begin{bmatrix} 1-1 & 2-2 \\ 4-6 & 1-4 \\ 5-7 & 6-3\end{bmatrix} ^{\prime}= \begin{bmatrix} 0 & 0 \\ -2 & -3 \\ -2 & 3\end{bmatrix} ^{\prime}= \begin{bmatrix} 0 & -2 & -2 \\ 0 & -3 & 3 \end{bmatrix} $

R.H.S. $A^{\prime}-B^{\prime}= \begin{bmatrix} 1 & 2 \\ 4 & 1 \\ 5 & 6\end{bmatrix} ^{\prime}- \begin{bmatrix} 1 & 2 \\ 6 & 4 \\ 7 & 3\end{bmatrix} ^{\prime}= \begin{bmatrix} 1 & 4 & 5 \\ 2 & 1 & 6\end{bmatrix} - \begin{bmatrix} 1 & 6 & 7 \\ 2 & 4 & 3 \end{bmatrix} $

$= \begin{bmatrix} 1-1 & 4-6 & 5-7 \\ 2-2 & 1-4 & 6-3\end{bmatrix} = \begin{bmatrix} 0 & -2 & -2 \\ 0 & -3 & 3 \end{bmatrix} $

Hence, $\quad$ L.H.S. $=$ R.H.S.

$(A-B)^{\prime}=A^{\prime}-B^{\prime}$ is verified.

Solution

Let

$ P=A^{\prime} A $

$ \begin{matrix} \Rightarrow & P^{\prime}=(A^{\prime} A)^{\prime} \\ \Rightarrow & P^{\prime}=A^{\prime}(A^{\prime})^{\prime} \\ \Rightarrow & P^{\prime}=A^{\prime} A \\ \Rightarrow & P^{\prime}=P \end{matrix} $

Hence, $A^{\prime} A$ is a symmetric matrix.

Now, Let

$Q=AA^{\prime}$

$\Rightarrow \quad Q^{\prime}=(AA^{\prime})^{\prime}$

$\begin{matrix} \Rightarrow & Q^{\prime}=(A^{\prime})^{\prime} A^{\prime} \\ \Rightarrow & Q^{\prime}=AA^{\prime} \\ \Rightarrow & Q^{\prime}=Q\end{matrix} $

Hence, $AA^{\prime}$ is also a symmetric matrix.

Solution

Given that A and B are the matrices of the order $3 \times 3$.

$ \begin{aligned} (AB)^{2} & =AB \cdot AB \\ & =AA \cdot BB \\ & =A^{2} \cdot B^{2} \end{aligned} $

Hence, $\quad(A B)^{2}=A^{2} B^{2}$

Solution

To prove that $(A+B)^{2}=A^{2}+2 A B+B^{2}$

$\begin{bmatrix} \text{ L.H.S. } \quad(A+B)^{2} & =(A+B) \cdot(A+B) & {\because \quad} & A^{2}=A \cdot A\end{bmatrix} $

So, $\quad$ L.H.S. $=$ R.H.S.

Solution

(a) To prove that: $A+(B+C)=(A+B)+C$

L.H.S. $A+(B+C)=(\begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} )+[(\begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} )+(\begin{bmatrix} 2 & 0 \\ 1 & -2\end{bmatrix} )]$

$= \begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} + \begin{bmatrix} 4+2 & 0+0 \\ 1+1 & 5-2 \end{bmatrix} $

$= \begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 2 & 3 \end{bmatrix} $

$= \begin{bmatrix} 1+6 & 2+0 \\ -1+2 & 3+3\end{bmatrix} = \begin{bmatrix} 7 & 2 \\ 1 & 6 \end{bmatrix} $

R.H.S. $(A+B)+C=[(\begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} )+(\begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} )]+(\begin{bmatrix} 2 & 0 \\ 1 & -2\end{bmatrix} )$

$= \begin{bmatrix} 1+4 & 2+0 \\ -1+1 & 3+5\end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} $

$= \begin{bmatrix} 5 & 2 \\ 0 & 8\end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} $

$= \begin{bmatrix} 5+2 & 2+0 \\ 0+1 & 8-2\end{bmatrix} = \begin{bmatrix} 7 & 2 \\ 1 & 6 \end{bmatrix} $

Hence, $\quad$ L.H.S. $=$ R.H.S.

$A+(B+C)=(A+B)+C$ Hence proved. (b) To prove that: $A(BC)=(AB) C$

$ \begin{aligned} \text{ L.H.S. } A(B C) & = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} [(\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix} )(\begin{matrix} 2 & 0 \\ 1 & -2 \end{matrix} )] \\ & = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 8+0 & 0+0 \\ 2+5 & 0-10 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 8 & 0 \\ 7 & -10 \end{bmatrix} \\ & = \begin{bmatrix} 8+14 & 0-20 \\ -8+21 & 0-30 \end{bmatrix} = \begin{bmatrix} 22 & -20 \\ 13 & -30 \end{bmatrix} \end{aligned} $

R.H.S. $(AB) C=[(\begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} )(\begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} )] \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} $

$ = \begin{bmatrix} 4+2 & 0+10 \\ -4+3 & 0+15 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} $

$ \begin{aligned} & = \begin{bmatrix} 6 & 10 \\ -1 & 15 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} \\ & = \begin{bmatrix} 12+10 & 0-20 \\ -2+15 & 0-30 \end{bmatrix} = \begin{bmatrix} 22 & -20 \\ 13 & -30 \end{bmatrix} \end{aligned} $

Hence, $\quad$ L.H.S. $=$ R.H.S.

$ A(B C)=(A B) C \text{ Hence proved. } $

(c) To prove that: $(a+b) B=a B+b B$

Here, $a=4$ and $b=-2$

L.H.S. $\quad(a+b) B=(4-2) \begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} =2 \begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 2 & 10 \end{bmatrix} $

R.H.S. $a B+b B=4 \begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} -2 \begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} = \begin{bmatrix} 16 & 0 \\ 4 & 20\end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 2 & 10 \end{bmatrix} $

$ = \begin{bmatrix} 16-8 & 0-0 \\ 4-2 & 20-10 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 2 & 10 \end{bmatrix} $

Hence, L.H.S. = R.H.S.

$ (a+b) B=a B+b B \text{ Hence proved. } $

(d) To prove that: $a(C-A)=a C-a A$

L.H.S. $\quad a(C-A)=4[(\begin{bmatrix} 2 & 0 \\ 1 & -2\end{bmatrix} )-(\begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} )]$

$ =4 \begin{bmatrix} 2-1 & 0-2 \\ 1+1 & -2-3 \end{bmatrix} =4 \begin{bmatrix} 1 & -2 \\ 2 & -5 \end{bmatrix} = \begin{bmatrix} 4 & -8 \\ 8 & -20 \end{bmatrix} $

R.H.S. $a C-a A=4 \begin{bmatrix} 2 & 0 \\ 1 & -2\end{bmatrix} -4 \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} $

$ \begin{aligned} & = \begin{bmatrix} 8 & 0 \\ 4 & -8 \end{bmatrix} - \begin{bmatrix} 4 & 8 \\ -4 & 12 \end{bmatrix} \\ & = \begin{bmatrix} 8-4 & 0-8 \\ 4+4 & -8-12 \end{bmatrix} = \begin{bmatrix} 4 & -8 \\ 8 & -20 \end{bmatrix} \end{aligned} $

Hence, $\quad$ L.H.S. = R.H.S.

$ a(C-A)=a C-a A \text{ Hence proved. } $

(e) To prove that: $(A^{T})^{T}=A$

L.H.S.

$ \begin{aligned} A^{T} & = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} ^{T}= \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \\ (A^{T})^{T} & = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} ^{T}= \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} =A \quad \text{ R.H.S. } \end{aligned} $

Hence, $\quad(A^{T})^{T}=A$

(f) To prove that: $(b A)^{T}=b A^{T}$ L.H.S. $(b A)^{T}=[-2(\begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} )]^{T}= \begin{bmatrix} -2 & -4 \\ 2 & -6\end{bmatrix} ^{T}= \begin{bmatrix} -2 & 2 \\ -4 & -6 \end{bmatrix} $ R.H.S. $b A^{T}=-2 \begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} ^{T}=-2 \begin{bmatrix} 1 & -1 \\ 2 & 3\end{bmatrix} = \begin{bmatrix} -2 & 2 \\ -4 & -6 \end{bmatrix} $

Hence, L.H.S. = R.H.S.

$ (b A)^{T}=b A^{T} \text{ Hence proved. } $

(g) To prove that: $(A B)^{T}=B^{T} A^{T}$

L.H.S. $(A B)^{T}=[(\begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} )(\begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} )]^{T}$

$ = \begin{bmatrix} 4+2 & 0+10 \\ -4+3 & 0+15 \end{bmatrix} ^{T}= \begin{bmatrix} 6 & 10 \\ -1 & 15 \end{bmatrix} ^{T}= \begin{bmatrix} 6 & -1 \\ 10 & 15 \end{bmatrix} $

R.H.S. $B^{T} A^{T}= \begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} ^{T} \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} ^{T}$

$ = \begin{bmatrix} 4 & 1 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 4+2 & -4+3 \\ 0+10 & 0+15 \end{bmatrix} = \begin{bmatrix} 6 & -1 \\ 10 & 15 \end{bmatrix} $

Hence, $\quad$ L.H.S. = R.H.S.

$(A B)^{T}=B^{T} A^{T}$ Hence proved. (h) To prove that: $(A-B) C=AC-BC$

L.H.S. $\quad(A-B) C=[(\begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} )-(\begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} )] \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} $

$= \begin{bmatrix} 1-4 & 2-0 \\ -1-1 & 3-5\end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} $

$= \begin{bmatrix} -3 & 2 \\ -2 & -2\end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} $

$= \begin{bmatrix} -6+2 & 0-4 \\ -4-2 & 0+4\end{bmatrix} = \begin{bmatrix} -4 & -4 \\ -6 & 4 \end{bmatrix} $

R.H.S. $AC-BC= \begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & -2\end{bmatrix} - \begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} $

$= \begin{bmatrix} 2+2 & 0-4 \\ -2+3 & 0-6\end{bmatrix} - \begin{bmatrix} 8+0 & 0+0 \\ 2+5 & 0-10 \end{bmatrix} $

$= \begin{bmatrix} 4 & -4 \\ 1 & -6\end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 7 & -10 \end{bmatrix} $

$= \begin{bmatrix} 4-8 & -4-0 \\ 1-7 & -6+10\end{bmatrix} = \begin{bmatrix} -4 & -4 \\ -6 & 4 \end{bmatrix} $

Hence, $\quad$ L.H.S. $=$ R.H.S.

$ (A-B) C=A C-B C $

(i) To prove that: $(A-B)^{T}=A^{T}-B^{T}$

L.H.S. $\quad(A-B)^{T}=[(\begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} )-(\begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} )]^{T}$

$ = \begin{bmatrix} 1-4 & 2-0 \\ -1-1 & 3-5 \end{bmatrix} ^{T}= \begin{bmatrix} -3 & 2 \\ -2 & -2 \end{bmatrix} ^{T}= \begin{bmatrix} -3 & -2 \\ 2 & -2 \end{bmatrix} $

R.H.S. $\quad A^{T}-B^{T}= \begin{bmatrix} 1 & 2 \\ -1 & 3\end{bmatrix} ^{T}- \begin{bmatrix} 4 & 0 \\ 1 & 5\end{bmatrix} ^{T}= \begin{bmatrix} 1 & -1 \\ 2 & 3\end{bmatrix} - \begin{bmatrix} 4 & 1 \\ 0 & 5 \end{bmatrix} $

$ = \begin{bmatrix} 1-4 & -1-1 \\ 2-0 & 3-5 \end{bmatrix} = \begin{bmatrix} -3 & -2 \\ 2 & -2 \end{bmatrix} $

Hence, $\quad$ L.H.S. $=$ R.H.S.

$ (A-B)^{T}=A^{T}-B^{T} \text{ Hence proved. } $

Solution

Given that

$ \begin{aligned} A & = \begin{bmatrix} \cos q & \sin q \\ -\sin q & \cos q \end{bmatrix} \\ A=A \cdot A & = \begin{bmatrix} \cos q & \sin q \\ -\sin q & \cos q \end{bmatrix} \begin{bmatrix} \cos q & \sin q \\ -\sin q & \cos q \end{bmatrix} \\ & = \begin{bmatrix} \cos ^{2} q-\sin ^{2} q & \cos q \sin q+\sin q \cos q \\ \sin q \cos q-\cos q \sin q & -\sin ^{2} q+\cos ^{2} q \end{bmatrix} \\ & = \begin{bmatrix} \cos 2 q & \sin 2 q \\ -\sin 2 q & \cos 2 q \end{bmatrix} \begin{bmatrix} \because \cos ^{2} A-\sin ^{2} A=\cos 2 A \\ 2 \sin A \cos A=\sin 2 A \end{bmatrix} \end{aligned} $

Hence proved.

Solution

Given that: $A= \begin{bmatrix} 0 & -x \\ x & 0\end{bmatrix} $ and $B= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $

L.H.S. $\quad(A+B)^{2}=(A+B) \cdot(A+B)$

$ \begin{aligned} & =[(\begin{matrix} 0 & -x \\ x & 0 \end{matrix} )+(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} )] \cdot[(\begin{matrix} 0 & -x \\ x & 0 \end{matrix} )+(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} )] \\ & = \begin{bmatrix} 0+0 & -x+1 \\ x+1 & 0+0 \end{bmatrix} \cdot \begin{bmatrix} 0+0 & -x+1 \\ x+1 & 0+0 \end{bmatrix} \\ & = \begin{bmatrix} 0 & -x+1 \\ x+1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0 & -x+1 \\ x+1 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0+(-x+1)(x+1) & 0+0 \\ 0+0 & (x+1)(-x+1)+0 \end{bmatrix} \\ & = \begin{bmatrix} 1-x^{2} & 0 \\ 0 & 1-x^{2} \end{bmatrix} \end{aligned} $

Put $x^{2}=-1$

R.H.S.

$ = \begin{bmatrix} 1+1 & 0 \\ 0 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} $

(given)

$ \begin{aligned} A^{2}+B^{2} & =A \cdot A+B \cdot B \\ & = \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix} \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0-x^{2} & 0+0 \\ 0+0 & -x^{2}+0 \end{bmatrix} + \begin{bmatrix} 0+1 & 0+0 \\ 0+0 & 1+0 \end{bmatrix} \\ & = \begin{bmatrix} -x^{2} & 0 \\ 0 & -x^{2} \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -x^{2}+1 & 0+0 \\ 0+0 & -x^{2}+1 \end{bmatrix} \\ & = \begin{bmatrix} -x^{2}+1 & 0 \\ 0 & -x^{2}+1 \end{bmatrix} = \begin{bmatrix} 1+1 & 0 \\ 0 & 1+1 \end{bmatrix} [\because x^{2}=-1] \\ & = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \end{aligned} $

Hence,

L.H.S. $=$ R.H.S.

$ (A+B)^{2}=A^{2}+B^{2} $

Solution

Given that: $A= \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} $

L.H.S. $\quad A^{2}=A \cdot A= \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{bmatrix} \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} $

$ \begin{aligned} & = \begin{bmatrix} 0+4-3 & 0-3+3 & 0+4-4 \\ 0-12+12 & 4+9-12 & -4-12+16 \\ 0-12+12 & 3+9-12 & -3-12+16 \end{bmatrix} \\ & = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =I \quad \text{ R.H.S. } \end{aligned} $

Hence, $A^{2}=I$ is verified.

Solution

To prove that $(A^{\prime})^{n}=(A^{n})^{\prime}$

Let $P(n):(A^{\prime})^{n}=(A^{n})^{\prime}$

Step 1: Put $n=1, P(1): A^{\prime}=A^{\prime} \quad$ which is true for $n=1$

Step 2: $\quad$ Put $n=K, P(K):(A^{\prime})^{K}=(A^{K})^{\prime} \quad$ Let it be true for $n=K$

Step 3: $\quad$ Put $n=K+1, P(K+1):(A^{\prime})^{K+1}=(A^{K+1})^{\prime}$

L.H.S.

$ \begin{aligned} (A^{\prime})^{K+1} & =(A^{\prime})^{K} \cdot(A^{\prime}) \\ & =(A^{K})^{\prime} \cdot(A^{\prime}) \\ & =(A^{K} \cdot A)^{\prime} \\ & =(A^{K+1})^{\prime} \quad \text{ R.H.S. } \end{aligned} $

(From step 2)

The given statement is true for $P(K+1)$ whenever it is true for $P(K)$, where $K \in N$.

Solution

(i) Let

$ \begin{aligned} A & = \begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix} \\ |A| & =1 \times 7-(-5) \times 3=7+15=22 \neq 0 \end{aligned} $

So, $A$ is invertible.

Let

$A=IA$

$ \begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A $

$R_2 \to R_2+5 R_1$

$\Rightarrow \quad \begin{bmatrix} 1 & 3 \\ 0 & 22\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix} A$

$R_2 \to \frac{1}{22} R_2$

$\Rightarrow \quad \begin{bmatrix} 1 & 3 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{5}{22} & \frac{1}{22} \end{bmatrix} A$

$R_1 \to R_1-3 R_2$

$\Rightarrow \quad \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} \frac{7}{22} & \frac{-3}{22} \\ \frac{5}{22} & \frac{1}{22} \end{bmatrix} A$

So

$ A^{-1}= \begin{bmatrix} \frac{7}{22} & \frac{-3}{22} \\ \frac{5}{22} & \frac{1}{22} \end{bmatrix} \Rightarrow \frac{1}{22} \begin{bmatrix} 7 & -3 \\ 5 & 1 \end{bmatrix} $

Hence, inverse of $ \begin{bmatrix} 1 & 3 \\ -5 & 7\end{bmatrix} $ is $\frac{1}{22} \begin{bmatrix} 7 & -3 \\ 5 & 1 \end{bmatrix} $

(ii) Let

$ \begin{aligned} A & = \begin{bmatrix} 1 & -3 \\ -2 & 6 \end{bmatrix} \\ |A| & =1 \times 6-(-3)(-2)=6-6=0 \\ |A| & =0 \text{ so } A \text{ is not invertible. } \end{aligned} $

Hence, inverse of $ \begin{bmatrix} 1 & -3 \\ -2 & 6 \end{bmatrix} $ is not possible.

Solution

Given that: $ \begin{bmatrix} x y & 4 \\ z+6 & x+y\end{bmatrix} = \begin{bmatrix} 8 & w \\ 0 & 6 \end{bmatrix} $

Equating the corresponding elements,

$x y=8, w=4, z+6=0 \quad \Rightarrow \quad z=-6, x+y=6$

Now, solving $x+y=6$

and

$x y=8$

From eqn. (i), $\quad y=6-x$

Putting the value of $y$ in eqn. (ii) we get,

$ \begin{aligned} & x(6-x)=8 \Rightarrow 6 x-x^{2}=8 \\ & \Rightarrow x^{2}-6 x+8=0 \Rightarrow x^{2}-4 x-2 x+8=0 \\ & \Rightarrow x(x-4)-2(x-4)=0 \Rightarrow(x-4)(x-2)=0 \\ & \therefore \quad x=4,2 \end{aligned} $

Solution

Order of matrices A and B is $2 \times 2$.

$\therefore$ Order of matrix C must be $2 \times 2$.

$ \begin{aligned} & \text{ Let } C= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \\ & \therefore \\ & \Rightarrow 3 \begin{bmatrix} 1 & 5 \\ 7 & 12 \end{bmatrix} +5 \begin{bmatrix} 9 & 1 \\ 7 & 8 \end{bmatrix} +2 \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 3 & 15 \\ 21 & 36 \end{bmatrix} + \begin{bmatrix} 45 & 5 \\ 35 & 40 \end{bmatrix} + \begin{bmatrix} 2 a & 2 b \\ 2 c & 2 d \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 3+45+2 a & 15+5+2 b \\ 21+35+2 c & 36+40+2 d \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 48+2 a & 20+2 b \\ 56+2 c & 76+2 d \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \end{aligned} $

Equating the corresponding elements, we get,

$ \begin{aligned} & 48+2 a=0 \Rightarrow 2 a=-48 \Rightarrow a=-24 \\ & 20+2 b=0 \Rightarrow 2 b=-20 \Rightarrow b=-10 \\ & 56+2 c=0 \Rightarrow 2 c=-56 \Rightarrow c=-28 \\ & 76+2 d=0 \Rightarrow 2 d=-76 \Rightarrow d=-38 \\ & C= \begin{bmatrix} -24 & -10 \\ -28 & -38 \end{bmatrix} \end{aligned} $

Solution

Given that: $\quad A= \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} $

$ \begin{aligned} A^{2}=A \cdot A & = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \\ & = \begin{bmatrix} 9+20 & -15-10 \\ -12-8 & 20+4 \end{bmatrix} = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} \\ \therefore \quad A^{2}-5 A-14 I & = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} -5 \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} -14 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} - \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} - \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} \end{aligned} $

$ \begin{aligned} & = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} - \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} \\ & = \begin{bmatrix} 29-29 & -25+25 \\ -20+20 & 24-24 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \end{aligned} $

Hence, $A^{2}-5 A-14 I=O$

Now, multiplying both sides by A, we get,

$ \begin{aligned} & A^{2} \cdot A-5 A \cdot A-14 IA=OA \\ & \Rightarrow \quad A^{3}-5 A^{2}-14 A=0 \\ & \Rightarrow \quad A^{3}=5 A^{2}+14 A \\ & \Rightarrow \quad A^{3}=5 \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} +14 \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \\ & = \begin{bmatrix} 145 & -125 \\ -100 & 120 \end{bmatrix} + \begin{bmatrix} 42 & -70 \\ -56 & 28 \end{bmatrix} \\ & = \begin{bmatrix} 145+42 & -125-70 \\ -100-56 & 120+28 \end{bmatrix} = \begin{bmatrix} 187 & -195 \\ -156 & 148 \end{bmatrix} \end{aligned} $

Hence, $A^{3}= \begin{bmatrix} 187 & -195 \\ -156 & 148 \end{bmatrix}$

Solution

Given that: $3 \begin{bmatrix} a & b \\ c & d\end{bmatrix} = \begin{bmatrix} a & 6 \\ -1 & 2 d\end{bmatrix} + \begin{bmatrix} 4 & a+b \\ c+d & 3 \end{bmatrix} $

$ \begin{bmatrix} 3 a & 3 b \\ 3 c & 3 d \end{bmatrix} = \begin{bmatrix} a+4 & 6+a+b \\ -1+c+d & 2 d+3 \end{bmatrix} $

Equating the corresponding elements, we get,

$ \begin{aligned} & 3 a=a+4 \quad \Rightarrow 3 a-a=4 \quad \Rightarrow 2 a=4 \quad \Rightarrow a=2 \\ & 3 b=6+a+b \quad \Rightarrow 3 b-b-a=6 \quad \Rightarrow 2 b-a=6 \quad \Rightarrow 2 b-2=6 \\ & \Rightarrow 2 b=8 \\ & \Rightarrow b=4 \\ & 3 c=-1+c+d \Rightarrow 3 c-c-d=-1 \Rightarrow 2 c-d=-1 \\ & \text{ and } 3 d=2 d+3 \Rightarrow 3 d-2 d=3 \quad \Rightarrow d=3 \\ & \text{ Now } 2 c-d=-1 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 2 c-3=-1 \Rightarrow 2 c=3-1 \Rightarrow 2 c=2 \\ & \therefore \quad c=1 \\ & \therefore a=2, b=4, c=1 \text{ and } d=3 . \end{aligned} $

Solution

Order of matrix $ \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix} $ is $3 \times 2$ and the matrix

$ \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix} \text{ is } 3 \times 3 $

$\therefore$ Order of matrix A must be $2 \times 3$

Let $A= \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix} _{2 \times 3}$

So,$\begin{aligned} { \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4\end{bmatrix} \begin{bmatrix} a & b & c \\ d & e & f\end{bmatrix} } & = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{bmatrix} \\ { \begin{bmatrix} 2 a-d & 2 b-e & 2 c-f \\ a+0 & b+0 & c+0 \\ -3 a+4 d & -3 b+4 e & -3 c+4 f\end{bmatrix} } & = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix} \end{aligned}$

Equating the corresponding elements, we get,

$ \begin{aligned} & 2 a-d=-1 \text{ and } a=1 \Rightarrow 2 \times 1-d=-1 \Rightarrow d=2+1 \Rightarrow d=3 \\ & 2 b-e=-8 \text{ and } b=-2 \Rightarrow 2(-2)-e=-8 \Rightarrow-4-e=-8 \\ & \Rightarrow e=4 \\ & 2 c-f=-10 \text{ and } c=-5 \Rightarrow 2(-5)-f=-10 \Rightarrow-10-f=-10 \\ & \Rightarrow f=0 \\ & a=1, b=-2, c=-5, d=3, e=4 \text{ and } f=0 \\ & A= \begin{bmatrix} 1 & -2 & -5 \\ 3 & 4 & 0 \end{bmatrix} \end{aligned} $

Solution

Given that: $\quad A= \begin{bmatrix} 1 & 2 \\ 4 & 1 \end{bmatrix} $

$ \begin{aligned} A^{2}=A \cdot A & = \begin{bmatrix} 1 & 2 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & 1 \end{bmatrix} = \begin{bmatrix} 1+8 & 2+2 \\ 4+4 & 8+1 \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 8 & 9 \end{bmatrix} \\ A^{2}+2 A+7 I & = \begin{bmatrix} 9 & 4 \\ 8 & 9 \end{bmatrix} +2 \begin{bmatrix} 1 & 2 \\ 4 & 1 \end{bmatrix} +7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 9 & 4 \\ 8 & 9 \end{bmatrix} + \begin{bmatrix} 2 & 4 \\ 8 & 2 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \\ & = \begin{bmatrix} 9+2+7 & 4+4+0 \\ 8+8+0 & 9+2+7 \end{bmatrix} = \begin{bmatrix} 18 & 8 \\ 16 & 18 \end{bmatrix} \end{aligned} $

Hence, $\quad A^{2}+2 A+7 I= \begin{bmatrix} 18 & 8 \\ 16 & 18 \end{bmatrix} $.

Solution

Here,

$ A= \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} $

Given that: $\quad A^{-1}=A^{\prime}$

Pre-multiplying both sides by $A$

$ AA^{-1}=AA^{\prime} $

$\Rightarrow \quad I=AA^{\prime}$

$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} $

$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} \cos ^{2} \alpha+\sin ^{2} \alpha & -\sin \alpha \cos \alpha+\sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha+\cos \alpha \sin \alpha & \sin ^{2} \alpha+\cos ^{2} \alpha \end{bmatrix} $

$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $

Hence, it is true for all values of $\alpha$.

Solution

Let $\quad A= \begin{bmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0\end{bmatrix} A^{\prime}= \begin{bmatrix} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{bmatrix} $

For skew symmetric matrix, $A^{\prime}=-A$.

$ \begin{aligned} & { \begin{bmatrix} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{bmatrix} =- \begin{bmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{bmatrix} } \\ & \Rightarrow \quad \begin{bmatrix} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -a & -3 \\ -2 & -b & 1 \\ -c & -1 & 0 \end{bmatrix} \end{aligned} $

Equating the corresponding elements, we get

$a=-2, b=-b \Rightarrow 2 b=0 \Rightarrow b=0$ and $c=-3$

Hence, $a=-2, b=0$ and $c=-3$.

Solution

Given that:

$ \begin{aligned} P(x) & = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} \\ P(y) & = \begin{bmatrix} \cos y & \sin y \\ -\sin y & \cos y \end{bmatrix} \\ P(x) \cdot P(y) & = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} \begin{bmatrix} \cos y & \sin y \\ -\sin y & \cos y \end{bmatrix} \\ & = \begin{bmatrix} \cos x \cos y-\sin x \sin y & \cos x \sin y+\sin x \cos y \\ -\sin x \cos y-\cos x \sin y & -\sin x \sin y+\cos x \cos y \end{bmatrix} \\ & = \begin{bmatrix} \cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y) \end{bmatrix} \\ & =P(x+y) \end{aligned} $

Now

$ \begin{aligned} P(y) \cdot P(x) & = \begin{bmatrix} \cos y & \sin y \\ -\sin y & \cos y \end{bmatrix} \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} \\ & = \begin{bmatrix} \cos x \cos y-\sin x \sin y & \sin x \cos y+\cos x \sin y \\ -\cos x \sin y-\cos y \sin x & -\sin x \sin y+\cos x \cos y \end{bmatrix} \end{aligned} $

$ \begin{aligned} & = \begin{bmatrix} \cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y) \end{bmatrix} \\ & =P(x+y) \end{aligned} $

Hence, $P(x) \cdot P(y)=P(x+y)=P(y) \cdot P(x)$.

Solution

To show that: $(I+A)^{3}=7 A+I$

L.H.S. $\quad(I+A)^{3}=I^{3}+A^{3}+3 I^{2} A+3 I A^{2}$

$ \begin{aligned} & \Rightarrow I+A^{2} \cdot A+3 IA+3 IA^{2} \\ & \Rightarrow I+A \cdot A+3 IA+3 IA \\ & {[\because A^{2}=A]} \\ & \Rightarrow I+A^{2}+3 IA+3 IA \\ & \Rightarrow I+A+3 IA+3 IA \quad[\because A^{2}=A] \\ & \Rightarrow I+A+3 A+3 A \Rightarrow 7 A+I \quad \text{ R.H.S. } \end{aligned} $

L.H.S. $=$ R.H.S. Hence, Proved.

Solution

Given that $B$ is a skew symmetric matrix $\therefore B^{\prime}=-B$

Let

$ \begin{aligned} P & =A^{\prime} BA \\ P^{\prime} & =(A^{\prime} BA)^{\prime} \\ & =A^{\prime} B^{\prime}(A^{\prime})^{\prime} \\ & =A^{\prime}(-B) A \\ & =-A^{\prime} BA=-P \\ P^{\prime} & =-P \end{aligned} $

$ \Rightarrow $

So

Hence, $A^{\prime} BA$ is a skew symmetric matrix.

Solution

Let $P(n):(AB)^{n}=A^{n} B^{n}$

Step 1:

Put $n=1, P(1): AB=AB$

Step 2:

Put $n=k, P(k):(AB)^{k}=A^{k} B^{k} \quad$ (Let it be true for any $k \in N$ )

Step 3:

Put $n=k+1, P(k+1):(AB)^{k+1}=A^{k+1} B^{k+1}$

L.H.S.

$ \begin{aligned} (AB)^{k+1} & =(AB)^{k} \cdot AB \\ & =A^{k} B^{k} \cdot AB \\ & =A^{k+1} B^{k+1} \quad \text{ R.H.S. } \end{aligned} $

$ =A^{k} B^{k} \cdot AB \quad[\text{ from Step 2] } $

L.H.S. $=$ R.H.S.

Hence, if $P(n)$ is true for $P(k)$ then it is true for $P(k+1)$.

Solution

Given that: $\quad A= \begin{bmatrix} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} $ and $A^{\prime}=A^{-1}$

Pre-multiplying both sides by $A$ we get,

$ \begin{gathered} AA^{\prime}=AA^{-1} \\ \Rightarrow \\ \Rightarrow \begin{bmatrix} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} \begin{bmatrix} 0 & x & x \\ 2 y & y & -y \\ z & -z & z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ \Rightarrow \begin{bmatrix} 0+4 y^{2}+z^{2} & 0+2 y^{2}-z^{2} & 0-2 y^{2}+z^{2} \\ 0+2 y^{2}-z^{2} & x^{2}+y^{2}+z^{2} & x^{2}-y^{2}-z^{2} \\ 0-2 y^{2}+z^{2} & x^{2}-y^{2}-z^{2} & x^{2}+y^{2}+z^{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{gathered} $

Equating the corresponding elements, we get,

$$ \begin{align*} & 4 y^{2}+z^{2}=1 \tag{i}\\ & 2 y^{2}-z^{2}=0 \tag{ii} \end{align*} $$

Adding ( $i$ ) and (ii) we get, $6 y^{2}=1 \Rightarrow y^{2}=\frac{1}{6} \Rightarrow y= \pm \frac{1}{\sqrt{6}}$

From eqn. (i), we get,

$$ \begin{align*} 4 y^{2}+z^{2} & =1 \\ \Rightarrow 4(\frac{1}{\sqrt{6}})^{2}+z^{2} & =1 \Rightarrow \frac{2}{3}+z^{2}=1 \Rightarrow z^{2}=1-\frac{2}{3}=\frac{1}{3} \\ \therefore \quad z & = \pm \frac{1}{\sqrt{3}} \\ x^{2}+y^{2}+z^{2} & =1 \tag{iii}\\ x^{2}-y^{2}-z^{2} & =0 \tag{iv} \end{align*} $$

Adding (iii) and (iv) we get,

$ 2 x^{2}=1 \Rightarrow x^{2}=\frac{1}{2} \Rightarrow x= \pm \frac{1}{\sqrt{2}} $

Hence, $x= \pm \frac{1}{\sqrt{ }}, y= \pm \frac{1}{\sqrt{ }} \underset{6}{and} z= \pm \frac{1}{\sqrt{ }}$.

Solution

(i) Here, $A= \begin{bmatrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{bmatrix} $ for elementary row transformation we put

$A=IA$

$ \begin{bmatrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A$

$R_2 \to R_2+R_1$

$ \begin{bmatrix} 2 & -1 & 3 \\ -3 & 2 & 4 \\ -3 & 2 & 3\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A$

$R_3 \to R_3-R_2$

$ \begin{bmatrix} 2 & -1 & 3 \\ -3 & 2 & 4 \\ 0 & 0 & -1\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{bmatrix} A$

$R_1 \to R_1+R_2$

$ \begin{bmatrix} -1 & 1 & 7 \\ -3 & 2 & 4 \\ 0 & 0 & -1\end{bmatrix} = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{bmatrix} A$

$R_2 \to R_2-3 R_1$

$ \begin{bmatrix} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & -1\end{bmatrix} = \begin{bmatrix} 2 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1 \end{bmatrix} A$

$R_1 \to R_1+R_2$ and $R_3 \to-1 . R_3$

$ \begin{bmatrix} -1 & 0 & -10 \\ 0 & -1 & -17 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} -3 & -1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{bmatrix} A$

$ \begin{aligned} & R_1 \to R_1+10 R_3 \text{ and } R_2 \to R_2+17 R_3 \\ & { \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ 1 & 1 & -1 \end{bmatrix} A} \\ & R_1 \to-1 \cdot R_1 \text{ and } R_2 \to-1 . R_2 \\ & { \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{bmatrix} A} \\ & A^{-1}= \begin{bmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{bmatrix} \\ & A= \begin{bmatrix} 2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1 \end{bmatrix} \\ & \text{ Put } \quad A=IA \\ & { \begin{bmatrix} 2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A} \\ & R_1 \to R_1-2 R_3 \text{ and } R_2 \to R_2+R_1 \\ & { \begin{bmatrix} 0 & 1 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} A} \\ & R_1 \to R_1+R_2 \\ & { \begin{bmatrix} 0 & 0 & 0 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} A} \end{aligned} $

First row on L.H.S. contains all zeros, so the inverse of the given matrix A does not exist.

Hence, matrix A has no inverse.

(iii) Here,

$ A= \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} $

$ \begin{aligned} & \text{ Put } \quad A=I A \\ & { \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A} \\ & R_1 \to 3 R_1-R_2 \\ & { \begin{bmatrix} 1 & -1 & -3 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A} \\ & R_2 \to R_2-5 R_1 \\ & { \begin{bmatrix} 1 & -1 & -3 \\ 0 & 6 & 15 \\ 0 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -1 & 0 \\ -15 & 6 & 0 \\ 0 & 0 & 1 \end{bmatrix} A} \\ & R_2 \to R_2-5 R_3 \\ & { \begin{bmatrix} 1 & -1 & -3 \\ 0 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -1 & 0 \\ -15 & 6 & -5 \\ 0 & 0 & 1 \end{bmatrix} A} \\ & R_3 \to R_3-R_2 \\ & { \begin{bmatrix} 1 & -1 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -1 & 0 \\ -15 & 6 & -5 \\ 15 & -6 & 6 \end{bmatrix} A} \\ & R_1 \to R_1+R_2 \\ & { \begin{bmatrix} 1 & 0 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} -12 & 5 & -5 \\ -15 & 6 & -5 \\ 15 & -6 & 6 \end{bmatrix} A} \\ & R_3 \to \frac{1}{3} R_3 \\ & { \begin{bmatrix} 1 & 0 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -12 & 5 & -5 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} A} \\ & R_1 \to R_1+3 R_3 \\ & { \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} A} \end{aligned} $

Hence,

$ A^{-1}= \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} $

Solution

We know that any square matrix can be expressed as the sum of symmetric and skew symmetric matrix i.e. $A=\frac{1}{2}[A+A^{\prime}]+\frac{1}{2}[A-A^{\prime}]$.

Let $P=\frac{1}{2}[A+A^{\prime}]$ and $Q=\frac{1}{2}[A-A^{\prime}]$

So

$ \begin{aligned} P & =\frac{1}{2}[(\begin{matrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{matrix} )+(\begin{matrix} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{matrix} )] \\ & =\frac{1}{2} \begin{bmatrix} 2+2 & 3+1 & 1+4 \\ 1+3 & -1-1 & 2+1 \\ 4+1 & 1+2 & 2+2 \end{bmatrix} =\frac{1}{2} \begin{bmatrix} 4 & 4 & 5 \\ 4 & -2 & 3 \\ 5 & 3 & 4 \end{bmatrix} \\ & = \begin{bmatrix} 2 & 2 & 5 / 2 \\ 2 & -1 & 3 / 2 \\ 5 / 2 & 3 / 2 & 2 \end{bmatrix} \\ P^{\prime} & = \begin{bmatrix} 2 & 2 & 5 / 2 \\ 2 & -1 & 3 / 2 \\ 5 / 2 & 3 / 2 & 2 \end{bmatrix} =P \end{aligned} $

As $P^{\prime}=P \quad \therefore P$ is a symmetric matrix.

Now $Q=\frac{1}{2}[A-A^{\prime}]$

$ \begin{aligned} & =\frac{1}{2}[(\begin{matrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{matrix} )-(\begin{matrix} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{matrix} )] \\ & =\frac{1}{2}[(\begin{matrix} 2-2 & 3-1 & 1-4 \\ 1-3 & -1+1 & 2-1 \\ 4-1 & 1-2 & 2-2 \end{matrix} )]=\frac{1}{2}[(\begin{matrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{matrix} )] \end{aligned} $

$ = \begin{bmatrix} 0 & 1 & -3 / 2 \\ -1 & 0 & 1 / 2 \\ 3 / 2 & -1 / 2 & 0 \end{bmatrix} =- \begin{bmatrix} 0 & -1 & 3 / 2 \\ 1 & 0 & -1 / 2 \\ -3 / 2 & 1 / 2 & 0 \end{bmatrix} =-Q . $

As $Q=-Q \quad \therefore Q$ is a skew symmetric matrix.

So

$ \begin{aligned} A & =P+Q \\ A & = \begin{bmatrix} 2 & 2 & 5 / 2 \\ 2 & -1 & 3 / 2 \\ 5 / 2 & 3 / 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 1 & -3 / 2 \\ -1 & 0 & 1 / 2 \\ 3 / 2 & -1 / 2 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 2+0 & 2+1 & \frac{5}{2}-\frac{3}{2} \\ 2-1 & -1+0 & \frac{3}{2}+\frac{1}{2} \\ \frac{5}{2}+\frac{3}{2} & \frac{3}{2}-\frac{1}{2} & 2+0 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{bmatrix} =A \end{aligned} $

Hence, the required relation is

$ A= \begin{bmatrix} 2 & 2 & 5 / 2 \\ 2 & -1 & 3 / 2 \\ 5 / 2 & 3 / 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 1 & -3 / 2 \\ -1 & 0 & 1 / 2 \\ 3 / 2 & -1 / 2 & 0 \end{bmatrix} $

Solution

Given that $A= \begin{bmatrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{bmatrix} $

Here number of columns and the number of rows are equal i.e., 3. So, $A$ is a square matrix.

Hence, the correct option is $(a)$.

• Option (b) Diagonal matrix: A diagonal matrix is a square matrix in which all the elements outside the main diagonal are zero. In the given matrix ( P ), the elements outside the main diagonal are not zero (e.g., ( P_{1,3} = 4 ), ( P_{3,1} = 4 )), so it is not a diagonal matrix.

• Option (c) Unit matrix: A unit matrix, also known as an identity matrix, is a square matrix in which all the elements on the main diagonal are 1 and all other elements are 0. In the given matrix ( P ), the elements on the main diagonal are not all 1 (e.g., ( P_{2,2} = 4 )), so it is not a unit matrix.

• Option (d) None: This option is incorrect because the matrix ( P ) is indeed a square matrix, as established in the given answer.

Solution

Total number of possible matrices of order $3 \times 3$ with each entry 0 or $2=2^{3 \times 3}=2^{9}=512$.

Hence, the correct option is $(d)$.

• Option (a) 9 is incorrect because it does not account for the exponential growth of combinations. The number of possible matrices is not simply the number of entries (9), but rather the number of combinations of 0s and 2s for each entry.

• Option (b) 27 is incorrect because it suggests that each entry has only 3 possible values (which would be 3^3 for a 3x3 matrix), but in this problem, each entry has only 2 possible values (0 or 2).

• Option (c) 81 is incorrect because it implies that each entry has 4 possible values (which would be 4^2 for a 3x3 matrix), but in this problem, each entry has only 2 possible values (0 or 2).

Solution

Given that: $ \begin{bmatrix} 2 x+y & 4 x \\ 5 x-7 & 4 x\end{bmatrix} = \begin{bmatrix} 7 & 7 y-13 \\ y & x+6 \end{bmatrix} $

Equating the corresponding elements, we get,

$\qquad 2 x+y=7 \qquad ……(1)$

$ \begin{aligned} & \text{ and } \quad 4 x=x+6 \qquad ……(2) \\ & \text{ from eqn. (ii) } \quad 4 x-x=6 \\ & 3 x=6 \\ & \therefore \quad x=2 \\ & \text{ from eqn. (i) } 2 \times 2+y=7 \\ & 4+y=7 \quad \therefore y=7-4=3 \end{aligned} $

Hence, the correct option is (b).

• Option (a) $x=3, y=1$:

  • Substituting $x=3$ and $y=1$ into the first equation $2x + y = 7$:
    • $2(3) + 1 = 6 + 1 = 7$ (This is correct)
  • Substituting $x=3$ into the second equation $4x = x + 6$:
    • $4(3) = 3 + 6$
    • $12 \neq 9$ (This is incorrect)
  • Therefore, $x=3, y=1$ does not satisfy both equations.

• Option (c) $x=2, y=4$:

  • Substituting $x=2$ and $y=4$ into the first equation $2x + y = 7$:
    • $2(2) + 4 = 4 + 4 = 8$ (This is incorrect)
  • Substituting $x=2$ into the second equation $4x = x + 6$:
    • $4(2) = 2 + 6$
    • $8 = 8$ (This is correct)
  • Therefore, $x=2, y=4$ does not satisfy both equations.

• Option (d) $x=3, y=3$:

  • Substituting $x=3$ and $y=3$ into the first equation $2x + y = 7$:
    • $2(3) + 3 = 6 + 3 = 9$ (This is incorrect)
  • Substituting $x=3$ into the second equation $4x = x + 6$:
    • $4(3) = 3 + 6$
    • $12 \neq 9$ (This is incorrect)
  • Therefore, $x=3, y=3$ does not satisfy both equations.

Solution

Given that: $A=\frac{1}{\pi} \begin{bmatrix} \sin ^{-1}(x \pi) & \tan ^{-1}(\frac{x}{\pi}) \\ \sin ^{-1}(\frac{x}{\pi}) & \cot ^{-1}(\pi x) \end{bmatrix} $

and

$ B=\frac{1}{\pi} \begin{bmatrix} -\cos ^{-1}(x \pi) & \tan ^{-1}(\frac{x}{\pi}) \\ \sin ^{-1}(\frac{x}{\pi}) & -\tan ^{-1}(\pi x) \end{bmatrix} $

$ \begin{aligned} & A-B=\frac{1}{\pi} \begin{bmatrix} \sin ^{-1}(x \pi) & \tan ^{-1}(\frac{x}{\pi}) \\ \sin ^{-1}(\frac{x}{\pi}) & \cot ^{-1}(\pi x) \end{bmatrix} -\frac{1}{\pi} \begin{bmatrix} -\cos ^{-1}(x \pi) & \tan ^{-1}(\frac{x}{\pi}) \\ \sin ^{-1}(\frac{x}{\pi}) & -\tan ^{-1}(\pi x) \end{bmatrix} \\ & =\frac{1}{\pi} \begin{bmatrix} \sin ^{-1}(x \pi)+\cos ^{-1}(x \pi) & \tan ^{-1}(\frac{x}{\pi})-\tan ^{-1}(\frac{x}{\pi}) \\ \sin ^{-1}(\frac{x}{\pi})-\sin ^{-1}(\frac{x}{\pi}) & \cot ^{-1}(\pi x)+\tan ^{-1}(\pi x) \end{bmatrix} \\ & =\frac{1}{\pi} \begin{bmatrix} \frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2} \end{bmatrix} \quad \begin{bmatrix} \because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2} \end{bmatrix} \\ & =\frac{1}{\pi} \times \frac{\pi}{2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =\frac{1}{2} I \end{aligned} $

Hence, the correct option is $(d)$.

• Option (a) I:

  • This option is incorrect because the resulting matrix ( A - B ) is not the identity matrix ( I ). The identity matrix ( I ) has 1s on the diagonal and 0s elsewhere, but ( A - B ) results in a matrix with (\frac{1}{2}) on the diagonal and 0s elsewhere.

• Option (b) O:

  • This option is incorrect because the resulting matrix ( A - B ) is not the zero matrix ( O ). The zero matrix ( O ) has all elements equal to 0, but ( A - B ) results in a matrix with (\frac{1}{2}) on the diagonal and 0s elsewhere.

• Option (c) 2I:

  • This option is incorrect because the resulting matrix ( A - B ) is not twice the identity matrix ( 2I ). The matrix ( 2I ) would have 2s on the diagonal and 0s elsewhere, but ( A - B ) results in a matrix with (\frac{1}{2}) on the diagonal and 0s elsewhere.

Solution

As we know that the addition and subtraction of two matrices is only possible when they have same order. It is also given that $m=n$.

$\therefore$ Order of $(5 A-2 B)$ is $3 \times n$

Hence, the correct option is $(d)$.

• Option (a) $m \times 3$: This option is incorrect because the order of the resulting matrix $(5A - 2B)$ must match the order of the original matrices $A$ and $B$, which is $3 \times n$. Since $m = n$, the resulting matrix cannot have an order of $m \times 3$.

• Option (b) $3 \times 3$: This option is incorrect because the order of the resulting matrix $(5A - 2B)$ must match the order of the original matrices $A$ and $B$, which is $3 \times n$. There is no indication that $n$ is equal to 3, so the resulting matrix cannot have an order of $3 \times 3$.

• Option (c) $m \times n$: This option is incorrect because the order of the resulting matrix $(5A - 2B)$ must match the order of the original matrices $A$ and $B$, which is $3 \times n$. Since $m = n$, the resulting matrix cannot have an order of $m \times n$; it must be $3 \times n$.

Solution

Given that $A= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $

$ A^{2}=A \cdot A= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0+1 & 0+0 \\ 0+0 & 1+0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $

Hence, the correct option is $(d)$.

• Option (a): This option suggests that ( A^2 = A ). However, squaring a matrix generally does not yield the original matrix unless it is an idempotent matrix (i.e., ( A^2 = A )). In this case, ( A ) is not idempotent, as shown by the calculation ( A^2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} ).

• Option (b): This option suggests that ( A^2 = \begin{bmatrix} 1 & 0 \ 1 & 0 \end{bmatrix} ). This is incorrect because the calculation of ( A^2 ) involves matrix multiplication, and the resulting matrix does not match this option. The correct result is the identity matrix ( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} ).

• Option (c): This option suggests that ( A^2 = \begin{bmatrix} 0 & 1 \ 0 & 1 \end{bmatrix} ). This is incorrect because the calculation of ( A^2 ) involves matrix multiplication, and the resulting matrix does not match this option. The correct result is the identity matrix ( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} ).

Solution

Given that: $\quad A=[a _{i j}] _{2 \times 2}$

Let

$ \begin{aligned} & A= \begin{bmatrix} a _{11} & a _{12} \\ a _{21} & a _{22} \end{bmatrix} _{2 \times 2} \\ & a _{11}=0 \quad[\because i=j] \\ & a _{12}=1 \quad[\because i \neq j] \\ & a _{21}=1 \quad[\because i \neq j] \\ & a _{22}=0 \quad[\because i=j] \end{aligned} $

$\therefore \quad A= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $

Now, $A^{2}=A \cdot A= \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix} 0+1 & 0+0 \\ 0+0 & 1+0\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =I$

Hence, the correct option is (a).

• Option (b) is incorrect because ( A^2 \neq A ). Specifically, ( A^2 = I ), which is the identity matrix, not the original matrix ( A ).

• Option (c) is incorrect because ( A^2 \neq 0 ). The result of ( A^2 ) is the identity matrix ( I ), not the zero matrix.

• Option (d) is incorrect because there is a correct option among the given choices, which is (a). Therefore, “None of these” is not applicable.

Solution

Let

$ \begin{aligned} A & = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix} \\ A^{\prime} & = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix} =A \end{aligned} $

$A^{\prime}=A$, so $A$ is a symmetric matrix.

Hence, the correct option is (b).

• Option (a) Identity matrix: An identity matrix is a square matrix in which all the elements of the principal diagonal are ones and all other elements are zeros. The given matrix has diagonal elements 1, 2, and 4, not all ones, so it is not an identity matrix.

• Option (c) Skew symmetric matrix: A skew symmetric matrix is a square matrix that is equal to the negative of its transpose, i.e., ( A’ = -A ). For the given matrix, ( A’ = A ), which is not equal to (-A). Therefore, it is not a skew symmetric matrix.

• Option (d) None of these: This option is incorrect because the matrix is indeed a symmetric matrix, as shown in the solution.

Solution

Let

$ \begin{aligned} & A= \begin{bmatrix} 0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0 \end{bmatrix} \\ & A^{\prime}= \begin{bmatrix} 0 & 5 & -8 \\ -5 & 0 & -12 \\ 8 & 12 & 0 \end{bmatrix} \end{aligned} $

$ \Rightarrow \quad A^{\prime}=- \begin{bmatrix} 0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0 \end{bmatrix} =-A $

$A^{\prime}=-A$, so $A$ is a skew symmetric matrix.

Hence, the correct option is (c).

• Option (a) diagonal matrix: A diagonal matrix is a square matrix in which all the elements outside the main diagonal are zero. In the given matrix, there are non-zero elements outside the main diagonal, so it is not a diagonal matrix.

• Option (b) symmetric matrix: A symmetric matrix is a square matrix that is equal to its transpose, i.e., ( A = A^{\prime} ). In the given matrix, the transpose ( A^{\prime} ) is not equal to ( A ), so it is not a symmetric matrix.

• Option (d) scalar matrix: A scalar matrix is a diagonal matrix in which all the elements on the main diagonal are equal. Since the given matrix is not even a diagonal matrix, it cannot be a scalar matrix either.

Solution

Order of matrix $A=m \times n$

Let order of matrix $B$ be $K \times P$

Order of matrix $B^{\prime}=P \times K$

If $AB^{\prime}$ is defined then the order of $AB^{\prime}$ is $m \times K$ if $n=P$

If $B^{\prime} A$ is defined then order of $B^{\prime} A$ is $P \times n$ when $K=m$

Now, order of $B^{\prime}=P \times K$

$\therefore \quad$ Order of $B=K \times P$

$ =m \times n \quad[\because K=m, P=n] $

Hence, the correct option is $(d)$.

• Option (a) $m \times m$: This option is incorrect because if $B$ were of order $m \times m$, then $B’$ would be of order $m \times m$. For $AB’$ to be defined, the number of columns of $A$ (which is $n$) must equal the number of rows of $B’$ (which is $m$), but this is not necessarily true unless $m = n$. Additionally, for $B’A$ to be defined, the number of columns of $B’$ (which is $m$) must equal the number of rows of $A$ (which is $m$), which is true, but it does not satisfy the condition for $AB’$.

• Option (b) $n \times n$: This option is incorrect because if $B$ were of order $n \times n$, then $B’$ would be of order $n \times n$. For $AB’$ to be defined, the number of columns of $A$ (which is $n$) must equal the number of rows of $B’$ (which is $n$), which is true. However, for $B’A$ to be defined, the number of columns of $B’$ (which is $n$) must equal the number of rows of $A$ (which is $m$), but this is not necessarily true unless $m = n$.

• Option (c) $n \times m$: This option is incorrect because if $B$ were of order $n \times m$, then $B’$ would be of order $m \times n$. For $AB’$ to be defined, the number of columns of $A$ (which is $n$) must equal the number of rows of $B’$ (which is $m$), but this is not necessarily true unless $m = n$. Additionally, for $B’A$ to be defined, the number of columns of $B’$ (which is $n$) must equal the number of rows of $A$ (which is $m$), which is true, but it does not satisfy the condition for $AB’$.

Solution

Let

$ \begin{aligned} P & =(A B^{\prime}-BA^{\prime}) \\ P^{\prime} & =(A B^{\prime}-BA^{\prime})^{\prime} \\ & =(A B^{\prime})^{\prime}-(BA^{\prime})^{\prime} \\ & =(B^{\prime}) A^{\prime}-(A^{\prime})^{\prime} B^{\prime} \quad[\because(AB)^{\prime}=B^{\prime} A^{\prime}] \\ & =BA^{\prime}-AB^{\prime} \\ & =-(A B^{\prime}-BA^{\prime})=-P \\ P^{\prime} & =-P, \text{ so it is a skew symmetric matrix. } \end{aligned} $

Hence, the correct option is $(a)$.

• Option (b) null matrix: A null matrix is a matrix in which all elements are zero. The expression (A B^{\prime} - B A^{\prime}) does not necessarily result in a matrix with all zero elements unless (A) and (B) are specifically chosen such that this condition holds. In general, this expression will not yield a null matrix.

• Option (c) symmetric matrix: A symmetric matrix is a matrix that is equal to its transpose, i.e., (P = P^{\prime}). However, for the given expression (P = A B^{\prime} - B A^{\prime}), we have shown that (P^{\prime} = -P), which means (P) is skew symmetric, not symmetric. Therefore, it cannot be a symmetric matrix.

• Option (d) unit matrix: A unit matrix (or identity matrix) is a square matrix with ones on the diagonal and zeros elsewhere. The expression (A B^{\prime} - B A^{\prime}) does not necessarily result in such a matrix. The result of this expression depends on the specific entries of matrices (A) and (B) and does not generally produce a unit matrix.

Solution

$(A-I)^{3}+(A+I)^{3}-7 A=A^{3}-I^{3}-3 A^{2} I+3 A I^{2}+A^{3}+I^{3}+3 A^{2} I$

$ \begin{aligned} & +3 AI^{2}-7 A \\ = & 2 A^{3}+6 AI-7 A \\ = & 2 A \cdot A^{2}+6 AI-7 A \\ = & 2 AI+6 AI-7 A \\ = & 8 AI-7 A=8 A-7 A \\ = & A \end{aligned} $

$ [A^{2}=I] $

Hence, the correct option is (a).

• Option (b) $I-A$: This option is incorrect because the expression $(A-I)^{3}+(A+I)^{3}-7 A$ simplifies to $A$, not $I-A$. The detailed simplification shows that the terms involving $I$ cancel out, and the remaining terms simplify to $A$.

• Option (c) $I+A$: This option is incorrect because the expression $(A-I)^{3}+(A+I)^{3}-7 A$ simplifies to $A$, not $I+A$. The detailed simplification shows that the terms involving $I$ cancel out, and the remaining terms simplify to $A$.

• Option (d) $3 A$: This option is incorrect because the expression $(A-I)^{3}+(A+I)^{3}-7 A$ simplifies to $A$, not $3A$. The detailed simplification shows that the terms involving $I$ cancel out, and the remaining terms simplify to $A$.

Solution

We know that for any two matrices $A$ and $B$, we may have $AB=BA, AB \neq BA$ and $AB=0$, but it is not always true.

Hence, the correct option is $(d)$.

• Option (a) $AB=BA$: This is incorrect because it is not always true that the product of two matrices is commutative. In general, matrix multiplication is not commutative, meaning $AB$ does not necessarily equal $BA$ for all matrices $A$ and $B$.

• Option (b) $AB \neq BA$: This is incorrect because there are cases where $AB$ can equal $BA$. While it is true that matrix multiplication is generally not commutative, there exist specific pairs of matrices for which $AB = BA$.

• Option (c) $AB=O$: This is incorrect because it is not always true that the product of two matrices is the zero matrix. While it is possible for the product of two non-zero matrices to be the zero matrix, it is not a general rule that applies to all matrices.

Solution

Given that: $ \begin{bmatrix} 1 & -3 \\ 2 & 4\end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1\end{bmatrix} \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} $

Using $C_2 \to C_2-2 C_1$, we get

$ \begin{bmatrix} 1 & -5 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ 2 & 0 \end{bmatrix} $

Hence, the correct option is $(d)$.

• Option (a) is incorrect because the matrix on the right-hand side of the equation does not match the result of applying the column operation ( C_2 \to C_2 - 2C_1 ) to the given matrices. Specifically, the second column of the second matrix on the right-hand side should be (\begin{bmatrix} -5 \ 0 \end{bmatrix}) instead of (\begin{bmatrix} -5 \ 2 \end{bmatrix}).

• Option (b) is incorrect because the matrix on the right-hand side of the equation does not match the result of applying the column operation ( C_2 \to C_2 - 2C_1 ) to the given matrices. Specifically, the second column of the second matrix on the right-hand side should be (\begin{bmatrix} -5 \ 0 \end{bmatrix}) instead of (\begin{bmatrix} -5 \ 2 \end{bmatrix}).

• Option (c) is incorrect because the matrix on the left-hand side of the equation does not match the result of applying the column operation ( C_2 \to C_2 - 2C_1 ) to the given matrices. Specifically, the second row of the left-hand side matrix should be (\begin{bmatrix} 2 & 0 \end{bmatrix}) instead of (\begin{bmatrix} 2 & 4 \end{bmatrix}).

Solution

We have, $\quad \begin{bmatrix} 4 & 2 \\ 3 & 3\end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 3\end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} $

Using elementary row transformation $R_1 \to R_1-3 R_2$,

$ \begin{bmatrix} -5 & -7 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -7 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} $

Hence, the correct option is (a).

• Option (b) is incorrect because the matrix multiplication $\begin{bmatrix} 1 & 2 \\ 0 & 3\end{bmatrix} \begin{bmatrix} -1 & -3 \\ 1 & 1 \end{bmatrix}$ does not result in $\begin{bmatrix} -5 & -7 \\ 3 & 3\end{bmatrix}$. The correct multiplication would yield a different matrix.

• Option (c) is incorrect because the matrix $\begin{bmatrix} 1 & 2 \\ 1 & -7\end{bmatrix}$ does not match the required form after applying the elementary row operation $R_1 \to R_1-3 R_2$. Additionally, the matrix multiplication $\begin{bmatrix} 1 & 2 \\ 1 & -7\end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$ does not result in $\begin{bmatrix} -5 & -7 \\ 3 & 3\end{bmatrix}$.

• Option (d) is incorrect because the matrix $\begin{bmatrix} 4 & 2 \\ -5 & -7\end{bmatrix}$ does not match the result of the elementary row operation $R_1 \to R_1-3 R_2$ applied to the original matrix. Additionally, the matrix multiplication $\begin{bmatrix} 1 & 2 \\ -3 & -3\end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$ does not result in $\begin{bmatrix} 4 & 2 \\ -5 & -7\end{bmatrix}$.

Solution

Null matrix i.e. $ \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix} $ or $ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $ is both symmetric and skew symmetric matrix.

Solution

Let $A$ and $B$ be any two matrices

$\therefore$ For skew symmetric matrices

$$ \begin{align*} & A=-A^{\prime} \tag{i}\\ & B=-B^{\prime} \tag{ii} \end{align*} $$

Adding (i) and (ii) we get

$ \Rightarrow \quad \begin{aligned} & A+B=-A^{\prime}-B^{\prime} \\ & A+B=-(A^{\prime}+B^{\prime}), \text{ so } A+B \text{ is skew symmetric } \\ & \text{ matrix. } \end{aligned} $

Hence, the sum of two skew symmetric matrices is always skew symmetric matrix.

Solution

Let A be a matrix

$ \therefore \quad-A=-1 . A $

Hence, negative of a matrix is obtained by multiplying it by -1 .

Solution

Let $A$ be any matrix

$ \therefore \quad 0 \cdot A=A \cdot 0=0 $

Hence, the product of any matrix by the scalar $\mathbf{0}$ is the null matrix.

Solution

A matrix which is not a square matrix is called a rectangular matrix.

Solution

Matrix multiplication is distributive over addition. Let A, B and $C$ be any matrices.

So, (i) $A(B+C)=AB+AC$

(ii) $(A+B) C=AC+BC$

Solution

Let $A$ be a symmetric matrix

$ \therefore \quad A^{\prime}=A $

$ (A^{3})^{\prime}=(A^{\prime})^{3}=A^{3} \qquad[\because(A^{\prime})^{k}=(A^{k})^{\prime}] $

Hence, if $A$ is a symmetric matrix, then $A^{3}$ is a symmetric matrix.

Solution

If $A$ is a skew symmetric matrix,

$ \therefore \quad \begin{aligned} A^{\prime} & =-A \\ (A^{2})^{\prime} & =(A^{\prime})^{2} \\ & =(-A)^{2} \\ & =A^{2} \end{aligned} $

Hence, $A^{2}$ is a symmetric matrix.

Solution

(i) $(AB)^{\prime}=B^{\prime} A^{\prime}$

(ii) $(kA)^{\prime}=k \cdot A^{\prime}$

(iii) $[k(A-B)]^{\prime}=k(A-B)^{\prime}=k(A^{\prime}-B^{\prime})$

Solution

If $A$ is a skew symmetric matrix

$ \begin{aligned} \therefore \quad A^{\prime} & =-A \\ (k A)^{\prime} & =k A^{\prime}=k(-A)=-k A \end{aligned} $

Hence, $k A$ is a skew symmetric matrix.

Solution

(i) Let

$ \begin{matrix} P & =(AB-BA) \\ P^{\prime} & =(AB-BA)^{\prime} \\ & =(AB)^{\prime}-(BA)^{\prime} \\ & =B^{\prime} A^{\prime}-A^{\prime} B^{\prime} & \\ & =BA-AB \quad[\because(AB)^{\prime}=B^{\prime} A^{\prime}] \\ & =-(AB-BA) & {[\because A^{\prime}=A \text{ and } B^{\prime}=B]} \\ & =-P \end{matrix} $

Hence, $(A B-B A)$ is a skew symmetric matrix.

(ii) Let

$ \begin{aligned} Q & =(B A-2 A B) \\ Q^{\prime} & =(B A-2 A B)^{\prime} \\ & =(B A)^{\prime}-(2 A B)^{\prime} \\ & =A^{\prime} B^{\prime}-2(A B)^{\prime} \quad[\because(k A)^{\prime}=k A^{\prime}] \\ & =A^{\prime} B^{\prime}-2 B^{\prime} A^{\prime} \quad \\ & =A B-2 B A \quad[\because A^{\prime}=A \text{ and } B^{\prime}=B] \\ & =-(2 B A-A B) \end{aligned} $

Hence, $(B A-2 A B)$ is neither a symmetric nor a skew symmetric matrix.

Solution

If $A$ is a symmetric matrix

$ \therefore \quad A^{\prime}=A $

Let

$ \begin{matrix} P & =B^{\prime} AB \\ P^{\prime} & =(B^{\prime} AB)^{\prime} \\ & =B^{\prime} A^{\prime}(B^{\prime})^{\prime} & \\ & =B^{\prime} AB \quad[\because(AB)^{\prime}=B^{\prime} A^{\prime}] \\ & {[\because A^{\prime}=A \text{ and }(B^{\prime})^{\prime}=B]} \end{matrix} $

$ \therefore \quad P^{\prime}=P $

So, $P$ is a symmetric matrix.

Hence, $B^{\prime} AB$ is a symmetric matrix.

Solution

Given that

$A^{\prime}=A$

and

$B^{\prime}=B$

Let

$ \begin{aligned} P & =AB \\ P^{\prime} & =(AB)^{\prime} \\ & =B^{\prime} A^{\prime} \\ P^{\prime} & =BA \\ & =P \end{aligned} $

$ [\because A^{\prime}=A \text{ and } B^{\prime}=B] $

Hence, $AB$ is symmetric if and only if $\mathbf{A B}=\mathbf{B A}$.

Solution

$A^{-1}$ does not exist if we apply one or more row operations while finding $A^{-1}$ by elementary row operations, obtain all zeroes in one or more rows.

Solution

False.

A matrix is an array of elements, numbers or functions having rows and columns.

Solution

False.

The matrices having same order can only be added.

Solution

False.

The two matrices are said to be equal if their corresponding elements are same.

Solution

True.

For addition and subtraction, the order of the two matrices should be same.

Solution

True.

If $A, B$ and $C$ are the matrices of addition then

$ \begin{align*} A+(B+C) & =(A+B)+C \tag{associative}\\ A+B & =B+A \end{align*} $

(commutative)

Solution

False.

Since $A B \neq B A$ if $A B$ and $B A$ are well defined.

Solution

False.

Since, in identity matrix all the elements of principal diagonal are unity rest are zero.

e.g., $\quad A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =I_3$

Solution

True.

If $A$ and $B$ are square matrices then their addition is commutative i.e., $A+B=B+A$.

Solution

False.

Since subtraction of any two matrices of the same order is not commutative i.e., $A-B \neq B-A$.

Solution

False.

Since for any two non-zero matrices A and B, we may get $AB=0$.

Solution

False.

Transpose of a column matrix is a row matrix.

e.g., $A= \begin{bmatrix} 2 \\ 3 \\ 5\end{bmatrix} _{3 \times 1} \quad \therefore A^{\prime}= \begin{bmatrix} 2 & 3 & 5 \end{bmatrix} _{1 \times 3}$

Solution

False.

For two square matrices $A$ and $B, AB=BA$ is not always true.

Solution

True.

Let $A, B$ and $C$ be three matrices of the same order.

Given that $A^{\prime}=A, B^{\prime}=B$ and $C^{\prime}=C$

Let

$ \begin{aligned} P & =A+B+C \\ P^{\prime} & =(A+B+C)^{\prime} \\ & =A^{\prime}+B^{\prime}+C^{\prime} \\ & =A+B+C \\ & =P \end{aligned} $

$ \Rightarrow \quad P^{\prime}=(A+B+C)^{\prime} $

So, $A+B+C$ is also a symmetric matrix.

Solution

False.

Since $(A B)^{\prime}=B^{\prime} A^{\prime}$.

Solution

True.

Let $A=[a _{i j}] _{m \times n}$ and $B=[b _{i j}] _{p \times q}$

$AB$ is defined when $n=P$

$\therefore \quad$ Order of $AB=m \times q$

$\Rightarrow \quad$ Order of $(AB)^{\prime}=q \times m$

Order of $B^{\prime}$ is $q \times p$ and order of $A^{\prime}$ is $n \times m$

$\therefore B^{\prime} A^{\prime}$ is defined when $P=n$

and the order of $B^{\prime} A^{\prime}$ is $q \times m$

Hence, order of $(A B)^{\prime}=$ Order of $B^{\prime} A^{\prime}$ i.e., $q \times m$.

Solution

False.

Let $A= \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix} , B= \begin{bmatrix} 0 & 0 \\ 2 & 0\end{bmatrix} $ and $C= \begin{bmatrix} 0 & 0 \\ 3 & 4 \end{bmatrix} $

$ \begin{matrix} \therefore & AB & = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\ AC & = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \end{matrix} $

Here $A B=A C=0$ but $B \neq C$.

Solution

True.

Let

$ \begin{aligned} P & =AA^{\prime} \\ P^{\prime} & =(AA^{\prime})^{\prime} \\ & =(A^{\prime})^{\prime} \cdot A^{\prime} \\ & =AA^{\prime} \\ & =P \end{aligned} $

So, $P$ is symmetric matrix.

Hence, $AA^{\prime}$ is always a symmetric matrix.

Solution

False. $A= \begin{bmatrix} 2 & 3 & -1 \\ 1 & 4 & 2\end{bmatrix} $ and $B= \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix} $

Since $A B$ is defined

$ \begin{aligned} \therefore \quad AB & = \begin{bmatrix} 2 & 3 & -1 \\ 1 & 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 4+12-2 & 6+15-1 \\ 2+16+4 & 3+20+2 \end{bmatrix} = \begin{bmatrix} 14 & 20 \\ 22 & 25 \end{bmatrix} \end{aligned} $

BA is also defined.

$ \begin{aligned} \therefore \quad BA & = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 & -1 \\ 1 & 4 & 2 \end{bmatrix} \\ & = \begin{bmatrix} 4+3 & 6+12 & -2+6 \\ 8+5 & 12+20 & -4+10 \\ 4+1 & 6+4 & -2+2 \end{bmatrix} = \begin{bmatrix} 7 & 18 & 4 \\ 13 & 32 & 6 \\ 5 & 10 & 0 \end{bmatrix} \end{aligned} $

So $AB \neq BA$

Solution

True.

$ \begin{matrix} (A^{2})^{\prime} & =(A^{\prime})^{2} & \\ & =[-A]^{2} & {[\because A^{\prime}=-A]} \\ & =A^{2} & \end{matrix} $

So, $A^{2}$ is a symmetric matrix.

Solution

True.

If $A$ and $B$ are invertible matrices of the same order

$\therefore A$ and $B$ satisfy commutative property w.r.t. multiplication.