Exercise 1.8 - Chapter 1 Applications of Matrices and Determinants 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $1.8$
Choose the correct or the most suitable answer from the given four alternatives.
Question 1.
If $|\operatorname{adj}(\operatorname{adj} \mathrm{A})|=|\mathrm{A}|^9$, then the order of the square matrix $\mathrm{A}$ is
(a) 3
(b) 4
(c) 2
(d) 5
Answer:
(b) 4
Hint. We know $|\operatorname{adj}(\operatorname{adj} \mathrm{A})|=|\mathrm{A}|^{(n-1)^2}=(\mathrm{A})^9$ given
$
\begin{aligned}
\Rightarrow & (n-1)^2 & =9 \\
\Rightarrow & n & =4
\end{aligned} \quad \Rightarrow n-1=3
$
So the order of $\mathrm{A}$ is 4 .
Question 2.
If $\mathrm{A}$ is a $3 \times 3$ non-singular matrix such that $\mathrm{AA}^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$ and $\mathrm{B}=\mathrm{A}^{-1} \mathrm{~A}^{\mathrm{T}}$, then $\mathrm{BB}^{\mathrm{T}}=$
(a) $\mathrm{A}$
(b) $\mathrm{B}$
(c) $\mathrm{I}_3$
(d) $\mathrm{B}^{\mathrm{T}}$
Answer:
(c) $\mathrm{I}_3$
Hint. Given
$
\begin{aligned}
B & =A^{-1} A^T \\
B^T & =\left(A^{-1} A^T\right)^T=\left(A^T\right)^T\left(A^{-1}\right)^T=A\left(A^T\right)^{-1} \\
B B^T & =\left(A^{-1} A^T\right)\left(A A^T\right)^{-1}=A^{-1}(A)\left(A^T\right)\left(A^T\right)^{-1} \quad\left(\therefore A A^T=A^T A\right) \\
& =\left(A^{-1} A\right)\left[\left(A^T\right)\left(A^T\right)^{-1}\right] \\
& =I I=I
\end{aligned}
$
Question 3.
If $A=\left[\begin{array}{ll}3 & 5 \\ 1 & 2\end{array}\right], B=\operatorname{adj} A$ and $C=3 A$, then $\frac{|\operatorname{adj} B|}{|C|}=$.
(a) $\frac{1}{3}$
(b) $\frac{1}{9}$
(c) $\frac{1}{4}$
(d) 1

Answer:
(b) $\frac{1}{9}$
Hint. Here
Now
$
\begin{aligned}
& A=\left(\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right) \quad \Rightarrow \operatorname{adj} A=\left(\begin{array}{cc}
2 & -5 \\
-1 & 3
\end{array}\right)=B \\
& B=\left(\begin{array}{cc}
2 & -5 \\
-1 & 3
\end{array}\right) \quad \Rightarrow \operatorname{adj} B=\left(\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right) \\
& |\operatorname{adj} \mathrm{B}|=\left|\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right|=6-5=1 \\
&
\end{aligned}
$
Given
$C=3 A \quad \Rightarrow C=3\left(\begin{array}{ll}3 & 5 \\ 1 & 2\end{array}\right)=\left(\begin{array}{cc}9 & 15 \\ 3 & 6\end{array}\right)$
$|C|=\left|\begin{array}{cc}9 & 15 \\ 3 & 6\end{array}\right|=54-45=9$
So
$\frac{|\operatorname{adj} B|}{|C|}=\frac{1}{9}$
Question 4.
If $A=\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]=\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right]$, then $A=$
(a) $\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]$
(b) $\left[\begin{array}{rr}1 & 2 \\ -1 & 4\end{array}\right]$
(c) $\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]$
(d) $\left[\begin{array}{cc}4 & -1 \\ 2 & 1\end{array}\right]$
Answer:
(c) $\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]$
Hint. Given
Let $A=\left(\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right)=\left(\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right)$ $\Rightarrow \quad \begin{aligned} a+b & =6 \\ -2 a+4 b & =0 \\ c+d & =0 \\ -2 c+4 d & =6\end{aligned}$
From $(2) \Rightarrow \quad a=2 b \quad \therefore(1) \Rightarrow 3 b=6 \Rightarrow b=2$. So $a=4$
From (3) $\Rightarrow \quad c=-d \quad \therefore(4) \Rightarrow 6 d=6 \Rightarrow d=1$. So $c=-1$ $\therefore \quad \mathrm{A}=\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]$

Question 5.
If $\mathrm{A}=\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]$ then $9 \mathrm{I}_2-\mathrm{A}=$
(a) $\mathrm{A}^{-1}$
(b) $\frac{\mathrm{A}^{-1}}{2}$
(c) $3 \mathrm{~A}^{-1}$
(d) $2 \mathrm{~A}^{-1}$
Answer:
(d) $2 \mathrm{~A}^{-1}$
Hint.
$
A=\left(\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right)
$
So
Here
$
9 I-A=\left(\begin{array}{ll}
9 & 0 \\
0 & 9
\end{array}\right)-\left(\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right)=\left(\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right)
$
$
\begin{aligned}
A & =\left(\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right) \\
|A| & =\left|\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right|=14-12=2
\end{aligned}
$
$\operatorname{adj} A=\left(\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right)$
So $A^{-1}=\frac{1}{2}\left(\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right)$
$
\Rightarrow \quad\left(\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right)=2 A^{-1}
$
Question 6.
If $A=\left[\begin{array}{ll}2 & 0 \\ 1 & 5\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 4 \\ 2 & 0\end{array}\right]$ then $|\operatorname{adj}(A B)|=$
(a) $-40$
(b) $-80$
(c) $-60$
(d) $-20$
Answer:
(b) $-80$
Hint:
$
\begin{aligned}
\mathrm{AB} & =\left(\begin{array}{ll}
2 & 0 \\
1 & 5
\end{array}\right)\left(\begin{array}{ll}
1 & 4 \\
2 & 0
\end{array}\right)=\left(\begin{array}{cc}
2 & 8 \\
11 & 4
\end{array}\right) \\
(\operatorname{adj} \mathrm{AB}) & =\left(\begin{array}{cc}
4 & -8 \\
-11 & 2
\end{array}\right) \text { and }|\operatorname{adj} \mathrm{AB}|=\left|\begin{array}{cc}
4 & -8 \\
-11 & 2
\end{array}\right|=8-88=-80
\end{aligned}
$

Question 7.
If $P=\left[\begin{array}{ccc}1 & x & 0 \\ 1 & 3 & 0 \\ 2 & 4 & -2\end{array}\right]$ is the adjoint of $3 \times 3$ matrix $A$ and $|A|=4$, then $x$ is
(a) 15
(b) 12
(c) 14
(d) 11
Answer:
(d) 11
Hint: Given $|A|=4$ and $P$ is the adjoint matrix of $A$
$
\begin{aligned}
& |\mathrm{P}|=4^2=16 \\
& \Rightarrow-2(3-\mathrm{x})=16 \\
& \Rightarrow-6+2 \mathrm{x}=16 \\
& \Rightarrow 2 \mathrm{x}=22 \\
& \Rightarrow \mathrm{x}=11
\end{aligned}
$
Question 8.
If $\mathrm{A}=\left[\begin{array}{ccc}3 & 1 & -1 \\ 2 & -2 & 0 \\ 1 & 2 & -1\end{array}\right]$ and $\mathrm{A}^{-1}=\left[\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$ then the value of $a_{23}$ is
(a) 0
(b) $-2$
(c) $-3$
$(d)-1$
Answer:
(d) $-1$
Hint. We know $\mathrm{AA}^{-1}=1$
(i.e) $\left(\begin{array}{ccc}3 & 1 & -1 \\ 2 & -2 & 0 \\ 1 & 2 & -1\end{array}\right)\left(\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
Now comparing the element of $\mathrm{C}_3$ on both sides we get
$
\begin{aligned}
3 a_{13}+a_{23}-a_{33} & =0 \\
2 a_{13}-2 a_{23} & =0 \\
a_{13}+2 a_{23}-a_{33} & =1 \\
2 a_{13} & =2 a_{23}
\end{aligned}
$

Question 9.
If $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are invertible matrices of some order, then which one of the following is not true?
(a) $\operatorname{adj} \mathrm{A}=|\mathrm{A}| \mathrm{A}^{-1}$
(b) $\operatorname{adj}(\mathrm{AB})=(\operatorname{adj} \mathrm{A})(\operatorname{adj} \mathrm{B})$
(c) $\operatorname{det} \mathrm{A}^{-1}=(\operatorname{det} \mathrm{A})^{-1}$
(d) $(\mathrm{ABC})^{-1}=\mathrm{C}^{-1} \mathrm{~B}^{-1} \mathrm{~A}^{-1}$
Answer:
(b) $\operatorname{adj}(\mathrm{AB})=(\operatorname{adj} \mathrm{A})(\operatorname{adj} \mathrm{B})$
Question 10.
If $(A B)^{-1}=\left[\begin{array}{cc}12 & -17 \\ -19 & 27\end{array}\right]$ and $A^{-1}=\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]$ then $B^{-1}=$
(a) $\left[\begin{array}{cc}2 & -5 \\ -3 & 8\end{array}\right]$
(b) $\left[\begin{array}{ll}8 & 5 \\ 3 & 2\end{array}\right]$
(c) $\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]$
$(d)\left[\begin{array}{cc}8 & -5 \\ -3 & 2\end{array}\right]$
Answer:
(a) $\left[\begin{array}{cc}2 & -5 \\ -3 & 8\end{array}\right]$
Hint. We know $(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}$
Here $\quad\left(\mathrm{AB}^{-1}\right)=\left(\begin{array}{cc}12 & -17 \\ -19 & 27\end{array}\right) \quad$ and $\mathrm{A}^{-1}=\left(\begin{array}{cc}1 & 1 \\ 2 & 3\end{array}\right)$
Let
$
\mathrm{B}^{-1}=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)
$

$
\text { Now } \quad \begin{aligned}
\left(\begin{array}{cc}
12 & -17 \\
-19 & 27
\end{array}\right) & =\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)\left(\begin{array}{cc}
1 & -1 \\
-2 & 3
\end{array}\right) \\
a-2 b & =12 \\
-a+3 b & =-17 \\
c-2 d & =-19 \\
-c+3 d & =27
\end{aligned}
$
Solving (1) and (2) we get $a=2, b=-5$
Solving (3) and (4) we get $c=-3$ and $d=8$
$
\therefore \quad \mathrm{B}^{-1}=\left(\begin{array}{cc}
2 & -5 \\
-3 & 8
\end{array}\right)
$

Question 11.
If $\mathrm{A}^{\mathrm{T}} \mathrm{A}^{-1}$ is symmetric, then $\mathrm{A}^2=$
(a) $\mathrm{A}^{-1}$
(b) $\left(A^T\right)^2$
(c) $\mathrm{A}^{\mathrm{T}}$
(d) $\left(\mathrm{A}^{-1}\right)^2$
Answer:
(b) $\left(\mathrm{A}^{\mathrm{T}}\right)^2$
Hint. Given $\mathrm{A}^{\mathrm{T}} \mathrm{A}^{-1}$ is symmetric

Question 12 .
If $A$ is a non-singular matrix such that $A^{-1}=\left[\begin{array}{rr}5 & 3 \\ -2 & -1\end{array}\right]$, then $\left(A^{\mathrm{T}}\right)^{-1}=$
(a) $\left[\begin{array}{rr}-5 & 3 \\ 2 & 1\end{array}\right]$
(b) $\left[\begin{array}{rr}5 & 3 \\ -2 & -1\end{array}\right]$
(c) $\left[\begin{array}{rr}-1 & -3 \\ 2 & 5\end{array}\right]$
(d) $\left[\begin{array}{ll}5 & -2 \\ 3 & -1\end{array}\right]$
Answer:
(d) $\left[\begin{array}{ll}5 & -2 \\ 3 & -1\end{array}\right]$
Hint. Given
Now
$
\left(A^{\mathrm{T}}\right)^{-1}=\left(\mathrm{A}^{-1}\right)^{\mathrm{T}}=\left(\begin{array}{rr}
5 & 3 \\
-2 & -1
\end{array}\right)^{\mathrm{T}}=\left[\begin{array}{ll}
5 & -2 \\
3 & -1
\end{array}\right]
$
Question 13.
13. If $\mathrm{A}=\left[\begin{array}{cc}\frac{3}{5} & \frac{4}{5} \\ x & \frac{3}{5}\end{array}\right]$ and $\mathrm{A}^{\mathrm{T}}=\mathrm{A}^{-1}$, then the value of $x$ is
(a) $\frac{-4}{5}$
(b) $\frac{-3}{5}$
(c) $\frac{3}{5}$
(d) $\frac{4}{5}$

Answer:
(a) $\frac{-4}{5}$
Hint.
$
\begin{aligned}
& \mathrm{A}=\left(\begin{array}{ll}
\frac{3}{5} & \frac{4}{5} \\
x & \frac{3}{5}
\end{array}\right) \\
& \therefore A^T=\left(\begin{array}{ll}
\frac{3}{5} & x \\
\frac{4}{5} & \frac{3}{5}
\end{array}\right) \\
& |A|=1 \\
& \operatorname{adj} \mathrm{A}=\left(\begin{array}{cc}
\frac{3}{5} & -\frac{4}{5} \\
-x & \frac{3}{5}
\end{array}\right) \\
&
\end{aligned}
$
$
\text { Given } \quad A^T=A^{-1}
$
$\Rightarrow \quad-x=\frac{4}{5} \quad \Rightarrow x=-\frac{4}{5}$
Question 14.
If $\mathrm{A}=\left[\begin{array}{cc}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right]$ and $\mathrm{AB}=\mathrm{I}_2$, then $\mathrm{B}=$
(a) $\left(\cos ^2 \frac{\theta}{2}\right) \mathrm{A}$
(b) $\left(\cos ^2 \frac{\theta}{2}\right) \mathrm{A}^{\top}$
(c) $\left(\cos ^2 \theta\right) \mathrm{I}$
(d) $\left(\sin ^2 \frac{\theta}{2}\right) \mathrm{A}$
Answer:

(b) $\left(\cos ^2 \frac{\theta}{2}\right) \mathrm{A}^{\mathrm{T}}$
Hint. Given $A=\left(\begin{array}{cc}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right)$ and $A B=I \quad \Rightarrow B=A^{-1}$
[(i.e) $\mathrm{A}$ and $\mathrm{B}$ are inverses] So $\mathrm{B}=\mathrm{A}^{-1}$
Now
$
A=\left(\begin{array}{cc}
1 & \tan \frac{\theta}{2} \\
-\tan \frac{\theta}{2} & 1
\end{array}\right)
$
$|A|=\left|\begin{array}{cc}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right|=1+\tan ^2 \frac{\theta}{2}=\sec ^2 \frac{\theta}{2}$
$\operatorname{adj} A=\left(\begin{array}{cc}1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1\end{array}\right)$
$
\begin{aligned}
& \mathrm{A}^{-1}=\frac{1}{\sec ^2 \frac{\theta}{2}}\left(\begin{array}{cc}
1 & -\tan \frac{\theta}{2} \\
\tan \frac{\theta}{2} & 1
\end{array}\right)=\cos ^2 \frac{\theta}{2}\left(\begin{array}{cc}
1 & -\tan \frac{\theta}{2} \\
\tan \frac{\theta}{2} & 1
\end{array}\right)=\cos ^2 \frac{\theta}{2} A^T \\
& \mathrm{~B}=\left(\cos ^2 \frac{\theta}{2}\right) \mathrm{A}^{\mathrm{T}} \\
&
\end{aligned}
$
Question 15.
If $\mathrm{A}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$ and $\mathrm{A}(\operatorname{adj} \mathrm{A})\left[\begin{array}{cc}k & 0 \\ 0 & k\end{array}\right]$, then $k=$.
(a) 0
(b) $\sin \theta$
(c) $\cos \theta$
(d) 1

Answer:
(d) 1
Hint. Given
$
\begin{aligned}
\mathrm{A} & =\left(\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right) \Rightarrow|\mathrm{A}|=\left|\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right|=\cos ^2 \theta+\sin ^2 \theta=1 \\
\mathrm{~A}(\operatorname{adj} \mathrm{A}) & =|\mathrm{A}| \mathrm{I}=(1) \mathrm{I} \\
& =\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=\left(\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right) \Rightarrow k=1
\end{aligned}
$
Question 16.
If $A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]$ be such that $\lambda A^{-1}=A$, then $\lambda$ is
(a) 17
(b) 14
(c) 19
(d) 21
Answer:
(c) 19
Hint.
$
A=\left(\begin{array}{cc}
2 & 3 \\
5 & -2
\end{array}\right) \quad \Rightarrow|A|=\left|\begin{array}{cc}
2 & 3 \\
5 & -2
\end{array}\right|=-4-15=-19
$
Now
$
\operatorname{adj} A=\left(\begin{array}{cc}
-2 & -3 \\
-5 & 2
\end{array}\right)
$
$
\Rightarrow \quad \mathrm{A}^{-1}=\frac{1}{19} \mathrm{~A} \quad \Rightarrow 19 \mathrm{~A}^{-1}=\mathrm{A}
$
But we are given $\lambda \mathrm{A}^{-1}=\mathrm{A} \quad \Rightarrow \lambda=19$
Question 17.
If adj $\mathrm{A}=\left[\begin{array}{cc}2 & 3 \\ 4 & -1\end{array}\right]$ and adj $\mathrm{B}=\left[\begin{array}{cc}1 & -2 \\ -3 & 1\end{array}\right]$ then adj $(\mathrm{AB})$ is
(a) $\left[\begin{array}{rr}-7 & -1 \\ 7 & -9\end{array}\right]$
(b) $\left[\begin{array}{rr}-6 & 5 \\ -2 & -10\end{array}\right]$
(c) $\left[\begin{array}{rr}-7 & 7 \\ -1 & -9\end{array}\right]$
$(d)\left[\begin{array}{rr}-6 & -2 \\ 5 & -10\end{array}\right]$
Answer:
(b) $\left[\begin{array}{rr}-6 & 5 \\ -2 & -10\end{array}\right]$
Hint.
$
\operatorname{adj}(A B)=(\operatorname{adj} B)(\operatorname{adj} A)=\left(\begin{array}{rr}
1 & -2 \\
-3 & 1
\end{array}\right)\left(\begin{array}{rr}
2 & 3 \\
4 & -1
\end{array}\right)=\left(\begin{array}{rr}
-6 & 5 \\
-2 & -10
\end{array}\right)
$

Question 18.
The rank of the matrix $\left[\begin{array}{rrrr}1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ -1 & -2 & -3 & -4\end{array}\right]$ is .
(a) 1
(b) 2
(c) 4
(d) 3
Answer:
(a) 1
Hint. Let
$
A=\left(\begin{array}{rrrr}
1 & 2 & 3 & 4 \\
2 & 4 & 6 & 8 \\
-1 & -2 & -3 & -4
\end{array}\right) \sim\left(\begin{array}{rrrr}
1 & 2 & 3 & 4 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right) \begin{aligned}
& R_2 \rightarrow R_2-2 R_1 \\
& R_3 \rightarrow R_3+R_1
\end{aligned}
$
The matrix is in echelon form and the number of non zero rows $=1$.
$
\therefore \quad \rho(\mathrm{A})=1
$
Question 19.
If $x^a y^b=e^m, x^c y^d=e^n, \Delta_1=\left|\begin{array}{ll}m & b \\ n & d\end{array}\right|, \Delta_2=\left|\begin{array}{cc}a & m \\ c & n\end{array}\right|, \Delta_3=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$ then the values of $x$ and $y$ are respectively,
(a) $e^{\left(\Delta_2 / \Delta_1\right)}, e^{\left(\Delta_3 / \Delta_1\right)}$
(b) $\log \left(\Delta_1 / \Delta_3\right), \log \left(\Delta_2 / \Delta_3\right)$
(c) $\log \left(\Delta_2 / \Delta_1\right), \log \left(\Delta_3 / \Delta_1\right)$
(d) $e^{\left(\Delta_1 / \Delta_3\right)}, e^{\left(\Delta_2 / \Delta_3\right)}$

Answer:
(d) $e^{\left(\Delta_1 / \Delta_3\right)}, e^{\left(\Delta_2 / \Delta_3\right)}$
Hint.
$
\begin{aligned}
& x^a y^b=e^m \\
& x^c y^d=e^n
\end{aligned}
$
Taking $\log$ on both sides we get
$
\begin{aligned}
& a \log x+b \log y=m \\
& c \log x+d \log y=n \quad \text { (i.e) }\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)\left(\begin{array}{l}
\log x \\
\log y
\end{array}\right)=\left(\begin{array}{l}
m \\
n
\end{array}\right)
\end{aligned}
$
Here
So
$
\begin{aligned}
& \left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|=\Delta_3,\left|\begin{array}{ll}
m & b \\
n & d
\end{array}\right|=\Delta_1, \text { and }\left|\begin{array}{cc}
a & m \\
c & n
\end{array}\right|=\Delta_2 \\
& \log x=\frac{\Delta_1}{\Delta_3} \quad \Rightarrow x=e^{\frac{\Delta_1}{\Delta_3}} \\
& \log y=\frac{\Delta_2}{\Delta_3} \quad \Rightarrow y=e^{\frac{\Delta_2}{\Delta_3}} \\
&
\end{aligned}
$
Question 20.
Which of the following is/are correct?
(i) Adjoint of a symmetric matrix is also a symmetric matrix
(ii) Adjoint of a diagonal matrix is also a diagonal matrix.
(iii) If $\mathrm{A}$ is a square matrix of order $\mathrm{n}$ and $\lambda$, is a scalar, then $\operatorname{adj}(\lambda \mathrm{A})=\lambda^{\mathrm{n}} \operatorname{adj}(\mathrm{A})$.
(iv) $\mathrm{A}(\operatorname{adj} \mathrm{A})=(\operatorname{adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}$
(a) Only (i)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i), (ii) and (iv)
Answer:
(d) (i), (ii) and (iv)
Question 21.
If $\rho(A)=\rho([A \mid B])$, then the system $A X=B$ of linear equations is
(a) consistent and has a unique solution
(b) consistent
(c) consistent and has infinitely many solution
(d) inconsistent
Answer:
(b) consistent

Question 22.
If $0 \leq \theta \leq \pi$, the system of equations $x+(\sin \theta) y-(\cos \theta) z=0,(\cos \theta) x-y+z=0,(\sin \theta) x+y-z=0$ has a non-trivial solution then $\theta$ is
(a) $\frac{2 \pi}{3}$
(b) $\frac{3 \pi}{4}$
(c) $\frac{5 \pi}{4}$
(d) $\frac{\pi}{4}$
Answer:
(d) $\frac{\pi}{4}$
Since the system has non trivial solutions $\left|\begin{array}{rrr}1 & \sin \theta & -\cos \theta \\ \cos \theta & -1 & 1 \\ \sin \theta & 1 & -1\end{array}\right|=0$
(i.e) $1(1-1)-\sin \theta(-\cos \theta-\sin \theta)-\cos \theta(\cos \theta+\sin \theta)=0$
(i.e) $\sin \theta(\sin \theta+\cos \theta)-\cos \theta(\sin \theta+\cos \theta)=0$
$\Rightarrow(\sin \theta+\cos \theta)(\sin \theta-\cos \theta)=0$
$\begin{aligned} \sin ^2 \theta-\cos ^2 \theta & =0 \\ \Rightarrow \quad \theta & =\frac{\pi}{4}\end{aligned}$
Question 23.
The augmented matrix of a system of linear equations is $\left[\begin{array}{cccc}1 & 2 & 7 & 3 \\ 0 & 1 & 4 & 6 \\ 0 & 0 & \lambda-7 & \mu+5\end{array}\right]$. The system has infinitely many solutions if
(a) $\lambda=7, \mu \neq 5$
(b) $\lambda=-7, \mu=5$
(c) $\lambda \neq 7, \mu \neq-5$
(d) $\lambda=7, \mu=-5$
Answer:
(d) $\lambda=7, \mu=-5$
Question 24.
24. Let $\mathrm{A}=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$ and $4 \mathrm{~B}=\left[\begin{array}{ccc}3 & 1 & -1 \\ 1 & 3 & x \\ -1 & 1 & 3\end{array}\right]$. If $\mathrm{B}$ is the inverse of $\mathrm{A}$, then the value of $x$ is .
(a) 2
(b) 4
(c) 3
(d) 1

Answer:
(d) 1
Hint. Given $B$ is the inverse of $A \quad \Rightarrow A B=I$
(i.e) $\left(\begin{array}{rrr}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right) \frac{1}{4}\left(\begin{array}{rrr}3 & 1 & -1 \\ 1 & 3 & x \\ -1 & 1 & 3\end{array}\right)=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
Comparing the element $\mathrm{A}_3$ on both side
$
\begin{aligned}
\frac{1}{4}(-2-x+3) & =0 \quad \Rightarrow 1-x=0 \\
\Rightarrow \quad x & =1
\end{aligned}
$
Question 25.
If $\mathrm{A}=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$, then $\operatorname{adj}(\operatorname{adj} \mathrm{A})$ is
(a) $\left[\begin{array}{rrr}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$
(b) $\left[\begin{array}{rrr}6 & -6 & 8 \\ 4 & -6 & 8 \\ 0 & -2 & 2\end{array}\right]$
(c) $\left[\begin{array}{ccc}-3 & 3 & -4 \\ -2 & 3 & -4 \\ 0 & 1 & -1\end{array}\right]$
(d) $\left[\begin{array}{lll}3 & -3 & 4 \\ 0 & -1 & 1 \\ 2 & -3 & 4\end{array}\right]$
Answer:
(a) $\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$
Hint. $\quad \operatorname{adj}(\operatorname{adj} A)=|A|^{3-2}(\mathrm{~A})=|\mathrm{A}|(\mathrm{A})$
Here
$|A|=1$
$\therefore \operatorname{adj}(\operatorname{adj} A)=\mathrm{A}$