Solution

It is given that $A$ and $B$ are symmetric matrices. Therefore, we have:

$$ \begin{equation*} A^{\prime}=A \text{ and } B^{\prime}=B \tag{1} \end{equation*} $$

Now, $(A B-B A)^{\prime}=(A B)^{\prime}-(B A)^{\prime}$

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

$=B^{\prime} A^{\prime}-A^{\prime} B^{\prime}$

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

$=B A-A B$

[Using (1)]

$=-(A B-B A)$

$\therefore(A B-B A)^{\prime}=-(A B-B A)$

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

Solution

We suppose that $A$ is a symmetric matrix, then $A^{\prime}=A$

Consider

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

Thus, if $A$ is a symmetric matrix, then $B^{\prime} A B$ is a symmetric matrix.

Now, we suppose that $A$ is a skew-symmetric matrix.

Then, $A^{\prime}=-A$

Consider

$ \begin{aligned} & (B^{\prime} A B)^{\prime}=[B^{\prime}(A B)]^{\prime}=(A B)^{\prime}(B^{\prime})^{\prime} \\ & =(B^{\prime} A^{\prime}) B=B^{\prime}(-A) B \\ & =-B^{\prime} A B \\ & \therefore(B^{\prime} A B)^{\prime}=-B^{\prime} A B \end{aligned} $

Thus, if $A$ is a skew-symmetric matrix, then $B^{\prime} A B$ is a skew-symmetric matrix.

Hence, if $A$ is a symmetric or skew-symmetric matrix, then $B^{\prime} A B$ is a symmetric or skewsymmetric matrix accordingly.

Solution

Given,

$A=\begin{bmatrix}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{bmatrix}$

$A=\begin{bmatrix}0 & x & x \\ 2y & y & -y \\ z & -z & z\end{bmatrix}$

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

Now,A’A=I

Putting values

$\left[\begin{array}{ccc} 0 & x & x \\ 2 y & y & -y \\ z & -z & z \end{array}\right]\left[\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]$

$\left[\begin{array}{ccc} o(0)+x(x)+x(x) & o(2 y)+x(y)+x(-y) & o(z)+x(-z)+x(z) \\ 2 y(0)+y(x)-y(x) & 2 y(2 y)+y(y)-y(-y) & 2 y(z)+y(-z)-y(z) \\ z(0)-z(x)+z(x) & z(2 y)-z(y)+z(-y) & z(z)-z(-z)+z(z) \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]$

$\left[\begin{array}{ccc} 0+x^2+x^2 & 0+x y-x y & 0-x z+x z \\ 0+x y-x y & 4 y^2+y^2+y^2 & 2 z y-z y-z y \\ 0-x z+x z & 2 z y-z y-z y & z^2+z^2+z^2 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]$

$\left[\begin{array}{ccc} 2 x^2 & 0 & 0 \\ 0 & 6 y^2 & 0 \\ 0 & 0 & 3 z^2 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]$

Since matrices are equal, corresponding elements are equal, $$ \begin{aligned} & 2 \mathrm{x}^2=1 \\ & \mathrm{x}^2=\frac{1}{2} \\ & \mathrm{x}= \pm \sqrt{\frac{1}{2}} \\ & \mathrm{x}= \pm \frac{1}{\sqrt{2}} \end{aligned} $$

$$ \begin{aligned} & 6 y^2=1 \\ & y^2=\frac{1}{6} \\ & y= \pm \sqrt{\frac{1}{6}} \\ & y= \pm \frac{1}{\sqrt{6}} \\ \end{aligned} $$

$$ \begin{aligned} & 3 \mathrm{z}^2=1 \\ & \mathrm{z}^2=\frac{1}{3} \\ & \mathrm{z}= \pm \sqrt{\frac{1}{3}} \\ & \mathrm{z}= \pm \frac{1}{\sqrt{3}} \ \end{aligned} $$

Thus, $\mathrm{x}= \pm \frac{1}{\sqrt{2}}, \mathrm{y}= \pm \frac{1}{\sqrt{6}}, \mathrm{z}= \pm \frac{1}{\sqrt{3}}$

Solution

We have:

$ \begin{aligned} & { \begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} =0} \\ & \Rightarrow \begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} =0 \\ & \Rightarrow \begin{bmatrix} 6 & 2 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} =0 \\ & \Rightarrow[6(0)+2(2)+4(x)]=0 \\ & \Rightarrow[4+4 x]=[0] \\ & \therefore 4+4 x=0 \\ & \Rightarrow x=-1 \end{aligned} $

Thus, the required value of $x$ is -1 .

Solution

It is given that $A= \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} $

$ \begin{aligned} \therefore A^{2}=A \cdot A & = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\ & = \begin{bmatrix} 3(3)+1(-1) & 3(1)+1(2) \\ -1(3)+2(-1) & -1(1)+2(2) \end{bmatrix} \\ & = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \end{aligned} $

$\therefore$ L.H.S. $=A^{2}-5 A+7 I$

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

$= \begin{bmatrix} 8 & 5 \\ -5 & 3\end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10\end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} $

$= \begin{bmatrix} -7 & 0 \\ 0 & -7\end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} $

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

$=O=$ R.H.S.

$\therefore A^{2}-5 A+7 I=O$

Solution

We have:

$ \begin{aligned} & { \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} =O} \\ & \Rightarrow \begin{bmatrix} x+0-2 & 0-10+0 & 2 x-5-3] \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} =O \\ & \Rightarrow \begin{bmatrix} x-2 & -10 & 2 x-8 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} =O \\ & \Rightarrow[x(x-2)-40+2 x-8]=O \\ & \Rightarrow[x^{2}-2 x-40+2 x-8]=[0] \\ & \Rightarrow[x^{2}-48]=[0] \\ & \therefore x^{2}-48=0 \\ & \Rightarrow x^{2}=48 \\ & \Rightarrow x= \pm 4 \sqrt{3} \end{aligned} $

Solution

(a) The unit sale prices of $x, y$, and $z$ are respectively given as $Rs 2.50$, Rs 1.50, and Rs 1.00 .

Consequently, the total revenue in market $\mathbf{I}$ can be represented in the form of a matrix as:

$ \begin{bmatrix} 10000 & 2000 & 18000\end{bmatrix} \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} $

$=10000 \times 2.50+2000 \times 1.50+18000 \times 1.00$

$=25000+3000+18000$

$=46000$

The total revenue in market II can be represented in the form of a matrix as:

$ \begin{aligned} & { \begin{bmatrix} 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} } \\ & =6000 \times 2.50+20000 \times 1.50+8000 \times 1.00 \\ & =15000+30000+8000 \\ & =53000 \end{aligned} $

Therefore, the total revenue in market $\mathbf{I}$ isRs 46000 and the same in market II isRs 53000.

(b) The unit cost prices of $x, y$, and $z$ are respectively given as $Rs 2.00, Rs 1.00$, and 50 paise.

Consequently, the total cost prices of all the products in market $\mathbf{I}$ can be represented in the form of a matrix as:

$ \begin{aligned} & { \begin{bmatrix} 10000 & 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} } \\ & =10000 \times 2.00+2000 \times 1.00+18000 \times 0.50 \\ & =20000+2000+9000 \\ & =31000 \end{aligned} $

Since the total revenue in market $\mathbf{I}$ isRs 46000 , the gross profit in this marketis (Rs 46000 - Rs 31000) Rs 15000.

The total cost prices of all the products in market II can be represented in the form of a matrix as:

$ \begin{aligned} & { \begin{bmatrix} 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} } \\ & =6000 \times 2.00+20000 \times 1.00+8000 \times 0.50 \\ & =12000+20000+4000 \\ & =\text{ Rs } 36000 \end{aligned} $

Since the total revenue in market II isRs 53000, the gross profit in this market is (Rs 53000 - Rs 36000) Rs 17000.

Solution

It is given that:

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

The matrix given on the R.H.S. of the equation is a $2 \times 3$ matrix and the one given on the L.H.S. of the equation is a $2 \times 3$ matrix. Therefore, $X$ has to be a $2 \times 2$ matrix.

Now, let $X= \begin{cases} a & c \\ b & d \end{cases} $

Therefore, we have:

$ \begin{aligned} & { \begin{bmatrix} a & c \\ b & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} } \\ & \Rightarrow \begin{bmatrix} a+4 c & 2 a+5 c & 3 a+6 c \\ b+4 d & 2 b+5 d & 3 b+6 d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \end{aligned} $

Equating the corresponding elements of the two matrices, we have:

$ \begin{matrix} a+4 c=-7, & 2 a+5 c=-8, & 3 a+6 c=-9 \\ b+4 d=2, & 2 b+5 d=4, & 3 b+6 d=6 \end{matrix} $

Now, $a+4 c=-7 \Rightarrow a=-7-4 c$

$ \begin{aligned} \therefore 2 a+5 c=-8 & \Rightarrow-14-8 c+5 c=-8 \\ & \Rightarrow-3 c=6 \\ & \Rightarrow c=-2 \end{aligned} $

$\therefore a=-7-4(-2)=-7+8=1$

Now, $b+4 d=2 \Rightarrow b=2-4 d$

$ \begin{aligned} \therefore 2 b+5 d=4 & \Rightarrow 4-8 d+5 d=4 \\ & \Rightarrow-3 d=0 \\ & \Rightarrow d=0 \end{aligned} $

$\therefore b=2-4(0)=2$

Thus, $a=1, b=2, c=-2, d=0$

Hence, the required matrix $X$ is $ \begin{cases} 1 & -2 \\ 2 & 0 \end{cases} $.

Solution

$ \begin{aligned} & A= \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \\ & \therefore A^{2}=A \cdot A= \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \\ & = \begin{bmatrix} \alpha^{2}+\beta \gamma & \alpha \beta-\alpha \beta \\ \alpha \gamma-\alpha \gamma & \beta \gamma+\alpha^{2} \end{bmatrix} \\ & = \begin{bmatrix} \alpha^{2}+\beta \gamma & 0 \\ 0 & \beta \gamma+\alpha^{2} \end{bmatrix} \end{aligned} $

Now, $A^{2}=I \Rightarrow \begin{bmatrix} \alpha^{2}+\beta \gamma & 0 \\ 0 & \beta \gamma+\alpha^{2}\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $

On comparing the corresponding elements, we have:

$ \begin{aligned} & \alpha^{2}+\beta \gamma=1 \\ & \Rightarrow \alpha^{2}+\beta \gamma-1=0 \\ & \Rightarrow 1-\alpha^{2}-\beta \gamma=0 \end{aligned} $

Solution

If $A$ is both symmetric and skew-symmetric matrix, then we should have

$ \begin{aligned} & A^{\prime}=A \text{ and } A^{\prime}=-A \\ & \Rightarrow A=-A \\ & \Rightarrow A+A=O \\ & \Rightarrow 2 A=O \\ & \Rightarrow A=O \end{aligned} $

Therefore, $A$ is a zero matrix.

Solution

$ \begin{matrix} (I+A)^{3}-7 A & =I^{3}+A^{3}+3 I^{2} A+3 A^{2} I-7 A & \\ & =I+A^{3}+3 A+3 A^{2}-7 A & \\ & =I+A^{2} \cdot A+3 A+3 A-7 A & \\ & =I+A \cdot A-A \\ & =I+A^{2}-A \\ & =I+A-A \\ & =I \\ \therefore(I+A)^{3}-7 & =I \end{matrix} $