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 $(a)$.

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 $(d)$.

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).

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 $(d)$.

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 $(d)$.

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 $(d)$.

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).

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).

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).

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 $(d)$.

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 $(a)$.

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).

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 $(d)$.

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 $(d)$.

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).

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.