Miscellaneous Exercise (Revised) - Chapter 3 - Matrices - Ncert Solutions class 12 - Maths

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NCERT Solutions Class 12 Maths: Chapter 3 - Matrices | Comprehensive Guide

Miscellaneous Exercise  Question 1

If $A$ and $B$ are symmetric matrices, prove that $A B-B A$ is a skew symmetric matrix.
Answer
It is given that $A$ and $B$ are symmetric matrices. Therefore, we have:
$
A^{\prime}=A \text { and } B^{\prime}=B
$

Now,
$
\begin{array}{rlrl}
(A B-B A)^{\prime} & =(A B)^{\prime}-(B A)^{\prime} & & {\left[(A-B)^{\prime}=A^{\prime}-B^{\prime}\right]} \\
& =B^{\prime} A^{\prime}-A^{\prime} B^{\prime} & & {\left[(A B)^{\prime}=B^{\prime} A^{\prime}\right]} \\
& =B A-A B & & {[\text { Using (1) }]} \\
& =-(A B-B A) &
\end{array}
$
[Using (1)]
$
\therefore(A B-B A)^{\prime}=-(A B-B A)
$

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

Miscellaneous Exercise  Question 2:
Show that the matrix $B^{\prime} A B$ is symmetric or skew symmetric according as $A$ is symmetric or skew symmetric.
Answer
We suppose that $A$ is a symmetric matrix, then $A^{\prime}=A$... (1)
Consider

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

$
\begin{aligned}
& =B^{\prime}\left(A^{\prime} B\right) \\
& =B^{\prime}(A B) \quad[\text { Using (1)] } \\
\therefore\left(B^{\prime} A B\right)^{\prime} & =B^{\prime} A B
\end{aligned}
$

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

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

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

Miscellaneous Exercise  Question 3

Find the values of $x, y, z$ if the matrix $A=\left[\begin{array}{rrr}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right]$ satisfies the equation $A^{\prime} A=I$.

Answer

Given: $\mathrm{A}=\left[\begin{array}{rrr}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right]$.
Therefore, $\mathrm{A}^{\prime}=\left[\begin{array}{rrr}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right]^{\prime}=\left[\begin{array}{rrr}0 & x & x \\ 2 y & y & -y \\ z & -z & z\end{array}\right]$
$
\begin{aligned}
& \therefore \mathrm{A}^{\prime} \mathrm{A}=\mathrm{I} \text { (given) } \\
& \Rightarrow\left[\begin{array}{rrr}
0 & x & x \\
2 y & y & -y \\
z & -z & z
\end{array}\right]\left[\begin{array}{rrr}
0 & 2 y & z \\
x & y & -z \\
x & -y & z
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$

(Here $\mathrm{I}$ is $\mathrm{I}_3$ because) matrices $\mathrm{A}$ and $\mathrm{A}^{\prime}$ are matrices of order $3 \times 3$ )
$
\begin{aligned}
& \Rightarrow\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 y z-y z-y z \\
0-z x+z x & 2 y z-y z-y z & 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] \\
& \Rightarrow \quad\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}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$

Equating corresponding entries, we have

$\begin{aligned}
& 2 x^2=1, \\
& \Rightarrow x^2=\frac{1}{2} \text {, } \\
& \Rightarrow \quad x= \pm \sqrt{\frac{1}{2}}, \\
& \therefore \quad x= \pm \frac{1}{\sqrt{2}} \text {, } \\
&
\end{aligned}$

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

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

Miscellaneous Exercise  Question 4:

For what values of
$
x,\left[\begin{array}{lll}
1 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 0 & 1 \\
1 & 0 & 2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]=O \text { ? }
$

Answer
We have:

$
\begin{aligned}
& {\left[\begin{array}{lll}
1 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 0 & 1 \\
1 & 0 & 2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]=0} \\
& \Rightarrow\left[\begin{array}{lll}
1+4+1 & 2+0+0 & 0+2+2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]=0 \\
& \Rightarrow\left[\begin{array}{lll}
6 & 2 & 4
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]=0 \\
& \Rightarrow[6(0)+2(2)+4(x)]=0 \\
& \Rightarrow[4+4 \pi]=[0] \\
& \therefore 4+4 x=0 \\
& \Rightarrow x=-1 \\
&
\end{aligned}
$

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

Miscellaneous Exercise Question 5:
If $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$, show that $A^2-5 A+7 I=O$
Answer
It is given that $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$
$
\begin{aligned}
\therefore A^2=A \cdot A & =\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right] \\
& =\left[\begin{array}{ll}
3(3)+1(-1) & 3(1)+1(2) \\
-1(3)+2(-1) & -1(1)+2(2)
\end{array}\right] \\
& =\left[\begin{array}{ll}
9-1 & 3+2 \\
-3-2 & -1+4
\end{array}\right]=\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]
\end{aligned}
$

$\begin{aligned}
&\begin{aligned}
& \therefore \text { L.H.S. }=A^2-5 A+71 \\
& =\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]-5\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]+7\left[\begin{array}{rr}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]-\left[\begin{array}{rr}
15 & 5 \\
-5 & 10
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right] \\
& =\left[\begin{array}{rr}
-7 & 0 \\
0 & -7
\end{array}\right]+\left[\begin{array}{rr}
7 & 0 \\
0 & 7
\end{array}\right] \\
& =\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}\\
&\begin{aligned}
& =O=\text { R.H.S. } \\
& \therefore A^2-5 A+7 I=O
\end{aligned}
\end{aligned}$

Miscellaneous Exercise  Question 6:
Find $x$, if
$
\left[\begin{array}{lll}
x & -5 & -1
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
4 \\
1
\end{array}\right]=O
$

Answer
We have:
$
\begin{aligned}
& {\left[\begin{array}{lll}
x & -5 & -1
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
4 \\
1
\end{array}\right]=O} \\
& \Rightarrow\left[\begin{array}{lll}
x+0-2 & 0-10+0 & 2 x-5-3]
\end{array}\right]\left[\begin{array}{l}
x \\
4 \\
1
\end{array}\right]=O \\
& \Rightarrow\left[\begin{array}{lll}
x-2 & -10 & 2 x-8
\end{array}\right]\left[\begin{array}{l}
x \\
4 \\
1
\end{array}\right]=O \\
& \Rightarrow[x(x-2)-40+2 x-8]=O \\
& \Rightarrow\left[x^2-2 x-40+2 x-8\right]=[0] \\
& \Rightarrow\left[x^2-48\right]=[0] \\
& \therefore x^2-48=0 \\
& \Rightarrow x^2=48 \\
& \Rightarrow x= \pm 4 \sqrt{3} \\
&
\end{aligned}
$

Miscellaneous Exercise  Question 7:
A manufacturer produces three products $x, y, z$ which he sells in two markets. Annual sales are indicated below:

(a) If unit sale prices of $x, y$ and $z$ are Rs 2.50 , Rs 1.50 and Rs 1.00 , respectively, find the total revenue in each market with the help of matrix algebra.
(b) If the unit costs of the above three commodities are Rs 2.00 , Rs 1.00 and 50 paise respectively. Find the gross profit.

Answer

(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{aligned}
& {\left[\begin{array}{lll}
10000 & 2000 & 18000
\end{array}\right]\left[\begin{array}{l}
2.50 \\
1.50 \\
1.00
\end{array}\right]} \\
& =10000 \times 2.50+2000 \times 1.50+18000 \times 1.00 \\
& =25000+3000+18000 \\
& =46000
\end{aligned}
$

The total revenue in market II can be represented in the form of a matrix as:
$
\begin{aligned}
& {\left[\begin{array}{lll}
6000 & 20000 & 8000
\end{array}\right]\left[\begin{array}{l}
2.50 \\
1.50 \\
1.00
\end{array}\right]} \\
& =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}
& {\left[\begin{array}{lll}
10000 & 2000 & 18000
\end{array}\right]\left[\begin{array}{l}
2.00 \\
1.00 \\
0.50
\end{array}\right]} \\
& =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}
& {\left[\begin{array}{lll}
6000 & 20000 & 8000
\end{array}\right]\left[\begin{array}{l}
2.00 \\
1.00 \\
0.50
\end{array}\right]} \\
& =6000 \times 2.00+20000 \times 1.00+8000 \times 0.50 \\
& =12000+20000+4000 \\
& =\operatorname{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.

Miscellaneous Exercise Question 8:
Find the matrix $X$ so that $X\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$
Answer
It is given that:
$
X\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]=\left[\begin{array}{rrr}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]
$

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=\left[\begin{array}{ll}
a & c \\
b & d
\end{array}\right]
$

Therefore, we have:

$
\begin{aligned}
& {\left[\begin{array}{ll}
a & c \\
b & d
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]=\left[\begin{array}{rrr}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{lll}
\alpha+4 c & 2 a+5 c & 3 a+6 c \\
b+4 d & 2 b+5 d & 3 b+6 d
\end{array}\right]=\left[\begin{array}{rrr}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]
\end{aligned}
$

Equating the corresponding elements of the two matrices, we have:
$
\begin{array}{lll}
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{array}
$

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 $\left[\begin{array}{rr}1 & -2 \\ 2 & 0\end{array}\right]$.

Miscellaneous Exercise Question 9:
Choose the correct answer in the following questions:
If $A=\left[\begin{array}{rr}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]_{\text {is such that } A^2=I \text { then }}$
A. $1+\alpha^2+\beta \gamma=0$
B. $1-\alpha^2+\beta \gamma=0$
c. $1-\alpha^2-\beta \gamma=0$
D. $1+\alpha^2-\beta \gamma=0$

Answer

Answer: C
$
\begin{aligned}
& A=\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right] \\
& \therefore A^2=A \cdot A=\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right] \\
&=\left[\begin{array}{cc}
\alpha^2+\beta \gamma & \alpha \beta-\alpha \beta \\
\alpha \gamma-\alpha \gamma & \beta \gamma+\alpha^2
\end{array}\right] \\
&=\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \beta \gamma+\alpha^2
\end{array}\right]
\end{aligned}
$

Now, $A^2=I \Rightarrow\left[\begin{array}{cc}\alpha^2+\beta \gamma & 0 \\ 0 & \beta \gamma+\alpha^2\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
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}
$

Miscellaneous Exercise Question 10:
If the matrix $A$ is both symmetric and skew symmetric, then
A. $A$ is a diagonal matrix
B. $A$ is a zero matrix
C. $A$ is a square matrix
D. None of these

Answer
Answer: B
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.

Miscellaneous Exercise Question 11:
If $A$ is square matrix such that $A^2=A$, then $(I+A)^3-7 A$ is equal to
A. $A$
B. $I-A$
C. $I$
D. $3 A$

Answer
Answer: C
$
\begin{aligned}
(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 \quad\left[A^2=A\right] \\
& =I+A \cdot A-A \\
& =I+A^2-A \\
& =I+A-A \\
& =I \\
\therefore(I+A)^3-7 A & =I
\end{aligned}
$