MCQ 11 Mark
If $A^2=-\frac{1}{2}\left[\begin{array}{cc}1 & -4 \\ -1 & 2\end{array}\right]$ then $A=$
- A
$\left[\begin{array}{rr}2 & 4 \\ -1 & 1\end{array}\right]$
- B
$\left[\begin{array}{rr}2 & 4 \\ 1 & -1\end{array}\right]$
- C
$\left[\begin{array}{rr}2 & -4 \\ 1 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{rr}2 & 4 \\ 1 & 1\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{rr}2 & 4 \\ 1 & 1\end{array}\right]$
$-\frac{1}{2}\left[\begin{array}{ll}2 & 4 \\ 1 & 1\end{array}\right]$
View full question & answer→MCQ 21 Mark
For a $2 \times 2$ matrix $A$, if $A(\operatorname{adj} A)=\left(\begin{array}{ll}10 & 0 \\ 0 & 10\end{array}\right)$ then determinant $A$ equals
View full question & answer→MCQ 31 Mark
The inverse of a symmetric matrix is
MCQ 41 Mark
The inverse of $A=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$
View full question & answer→MCQ 51 Mark
If $F(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$ where $\alpha \in R$ then $[F(\alpha)]^{-1}$ is =
View full question & answer→MCQ 61 Mark
If $\mathrm{A}=\left[\begin{array}{ll}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$ then $\mathrm{A}^{-1}=\ldots$
- A
$\left[\begin{array}{rr}1 / \cos \alpha & -1 / \sin \alpha \\ 1 / \sin \alpha & 1 / \cos \alpha\end{array}\right]$
- ✓
$\left[\begin{array}{rr}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
- C
$\left[\begin{array}{ll}-\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
- D
$\left[\begin{array}{rr}-\cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{rr}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
$\left[\begin{array}{rr}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
View full question & answer→MCQ 71 Mark
If $A=\left(\begin{array}{rr}\lambda & 1 \\ -1 & -\lambda\end{array}\right)$, then $A^{-1}$ does not exist if $\lambda=$
View full question & answer→MCQ 81 Mark
If $A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)$ and $A(\operatorname{adj} A)=k l$, then the value of $k$ is
View full question & answer→MCQ 91 Mark
If $A=\left(\begin{array}{ll}2 & -4 \\ 3 & 1\end{array}\right)$, then the adjoint of matrix $A$ is
- A
$\left[\begin{array}{ll}-1 & 3 \\ -4 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{rr}1 & 4 \\ -3 & 2\end{array}\right]$
- C
$\left[\begin{array}{rr}1 & 3 \\ 4 & -2\end{array}\right]$
- D
$\left[\begin{array}{rr}-1 & -3 \\ -4 & 2\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{rr}1 & 4 \\ -3 & 2\end{array}\right]$
$\left[\begin{array}{rr}1 & 4 \\ -3 & 2\end{array}\right]$
View full question & answer→MCQ 101 Mark
If $A=\left(\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right)$ and $A(\operatorname{adj} A)=k 1$, then the value of $k$ is
Answer-3 [Hint : A(adj A) = |A| ∙ I]
View full question & answer→MCQ 111 Mark
The inverse of $\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)$ is
- A
$\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
- C
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- D
AnswerCorrect option: B. $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
View full question & answer→MCQ 121 Mark
If $\mathrm{A}=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right), \operatorname{adj}=\left(\begin{array}{ll}4 & a \\ -3 & b\end{array}\right)$ then the values of $\mathrm{a}$ and $\mathrm{b}$ are,
View full question & answer→MCQ 132 Marks
If $A=\left[\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right], B=\left[\begin{array}{cc}1 & 1 \\ 4 & -1\end{array}\right]$, then $(A+B)^{-1}$ is
- A
$\left[\begin{array}{cc}\frac{-1}{2} & 0 \\ \frac{-3}{2} & \frac{1}{2}\end{array}\right]$
- ✓
$\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{-1}{2}\end{array}\right]$
- C
$\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{-3}{2} & \frac{1}{2}\end{array}\right]$
- D
$\left[\begin{array}{ll}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{1}{2}\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{-1}{2}\end{array}\right]$
(b) : $A=\left[\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 1 \\ 4 & -1\end{array}\right]$
$
\begin{aligned}
& A+B=\left[\begin{array}{cc}
1+1 & -1+1 \\
2+4 & -1-1
\end{array}\right]=\left[\begin{array}{cc}
2 & 0 \\
6 & -2
\end{array}\right] \\
& \Rightarrow|A+B|=-4
\end{aligned}
$
Now, $\operatorname{adj}(A+B)=\left[\begin{array}{ll}-2 & 0 \\ -6 & 2\end{array}\right]$
$\therefore(A+B)^{-1}=\frac{\operatorname{adj}(A+B)}{|A+B|}=\frac{-1}{4}=\left[\begin{array}{ll}-2 & 0 \\ -6 & 2\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right]$
View full question & answer→MCQ 142 Marks
Let $A=\left[\begin{array}{cc}2 & -1 \\ 0 & 2\end{array}\right]$. If $B=I-{ }^3 C_1(\operatorname{adj} A +C_2(\operatorname{adj} A)^2$ $-{ }^3 C_3(\operatorname{adj} A)^3$, then the sum of all elements of the matrix $B$ is
Answer(d) : $A=\left[\begin{array}{cc}2 & -1 \\ 0 & 2\end{array}\right]$, then $\operatorname{adj} A=\left[\begin{array}{ll}2 & 1 \\ 0 & 2\end{array}\right]$
Given, $B=I-{ }^3 C _1(\operatorname{adj} A)+{ }^3 C _2(\operatorname{adj} A)^2-{ }^3 C _3(\operatorname{adj} A)^3$
Then, $B=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]-3\left[\begin{array}{ll}2 & 1 \\ 0 & 2\end{array}\right]+3\left[\begin{array}{ll}2 & 1 \\ 0 & 2\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 0 & 2\end{array}\right]$
$
-\left[\begin{array}{ll}
2 & 1 \\
0 & 2
\end{array}\right]\left[\begin{array}{ll}
2 & 1 \\
0 & 2
\end{array}\right]\left[\begin{array}{ll}
2 & 1 \\
0 & 2
\end{array}\right]
$
$
\begin{aligned}
& =\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{ll}
6 & 3 \\
0 & 6
\end{array}\right]+3\left[\begin{array}{ll}
4 & 4 \\
0 & 4
\end{array}\right]-\left[\begin{array}{ll}
4 & 4 \\
0 & 4
\end{array}\right]\left[\begin{array}{ll}
2 & 1 \\
0 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
-5 & -3 \\
0 & -5
\end{array}\right]+\left[\begin{array}{cc}
12 & 12 \\
0 & 12
\end{array}\right]-\left[\begin{array}{cc}
8 & 12 \\
0 & 8
\end{array}\right]=\left[\begin{array}{cc}
-1 & -3 \\
0 & -1
\end{array}\right]
\end{aligned}
$
Sum of elements of $B=-1-3-1=-5$
View full question & answer→MCQ 152 Marks
If $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0\end{array}\right]$ and $B=\operatorname{adj} A, C=5 A$, then $\frac{|\operatorname{adj} B|}{|C|}=$
Answer(c) : Given, $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0\end{array}\right]$
$
|A|=5, B=\operatorname{adj} A
$
$
|\operatorname{adj} B|=|\operatorname{adj} \operatorname{adj}(A)|=|A|^{(3-1) 2}=(5)^{(3-1) 2}=5^4=625
$
$
|C|=5|A|=5 \times 5=25
$
So, $\frac{|\operatorname{adj} B|}{|C|}=\frac{625}{25}=25$
View full question & answer→MCQ 162 Marks
If $A=\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]$ such that $A^2-4 A+3 I=0$ where $I$ is a unit matrix of order, 2 , then $A^{-1}$ is
- A
$\frac{1}{3}\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]$
- B
$\frac{1}{3}\left[\begin{array}{cc}-1 & 2 \\ 2 & -1\end{array}\right]$
- C
$\frac{1}{3}\left[\begin{array}{cc}-2 & 1 \\ 1 & -2\end{array}\right]$
- ✓
$\frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
AnswerCorrect option: D. $\frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
(d) : Given, $A^2-4 A+3 I=0$
Postmultiply by $A^{-1}$ both sides,
$
\begin{aligned}
& A-4 I+3 A^{-1}=0 \\
& \text { or } 3 A^{-1}=4 I-A \\
& A^{-1}=\frac{1}{3}(4 I-A)=\frac{1}{3}\left[\left[\begin{array}{ll}
4 & 0 \\
0 & 4
\end{array}\right]-\left[\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right]\right]=\frac{1}{3}\left[\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right]
\end{aligned}
$
View full question & answer→MCQ 172 Marks
If $A=\left[\begin{array}{ll}2 & -3 \\ 5 & -7\end{array}\right]$, then $A-A^{-1}=$
- A
$\left[\begin{array}{cc}-5 & 0 \\ 0 & -5\end{array}\right]$
- B
$3\left[\begin{array}{cc}3 & 2 \\ 10 & 3\end{array}\right]$
- ✓
$3\left[\begin{array}{cc}3 & -2 \\ \frac{10}{3} & -3\end{array}\right]$
- D
$5\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
AnswerCorrect option: C. $3\left[\begin{array}{cc}3 & -2 \\ \frac{10}{3} & -3\end{array}\right]$
(c) : Given, $A=\left[\begin{array}{ll}2 & -3 \\ 5 & -7\end{array}\right]$, then $A^{-1}=\left[\begin{array}{ll}-7 & 3 \\ -5 & 2\end{array}\right]$
Now, $A-A^{-1}=\left[\begin{array}{ll}2 & -3 \\ 5 & -7\end{array}\right]-\left[\begin{array}{ll}-7 & 3 \\ -5 & 2\end{array}\right]$
$
=\left[\begin{array}{cc}
9 & -6 \\
10 & -9
\end{array}\right]=3\left[\begin{array}{cc}
3 & -2 \\
10 / 3 & -3
\end{array}\right]
$
View full question & answer→MCQ 182 Marks
If $A=\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]$ and $A(\operatorname{adj} A)=K I$, then the value of $K$ is (where $I$ is unit matrix of order 3 )
Answer(d) : $A=\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]$ and $A(\operatorname{adj} A)=K I$
We know that, $A(\operatorname{adj} A)=|A| \cdot I \Rightarrow K=|A|$
Now, $|A|=1(0+25)-3(0+10)-2(-15)$ (Expanding along $R_1$ )
$
=25-30+30=25 \quad \therefore K=25
$
View full question & answer→MCQ 192 Marks
If $A=\left[\begin{array}{ccc}0 & 1+2 i & i-2 \\ -1-2 i & 0 & K \\ 2-i & 7 & 0\end{array}\right]$ and $A^{-1}$ does not exists, then $K=$ (where $i=\sqrt{-1}$ )
Answer(b) : $A=\left[\begin{array}{ccc}0 & 1+2 i & i-2 \\ -1-2 i & 0 & K \\ 2-i & 7 & 0\end{array}\right]$
Given that $A^{-1}$ does not exist.
$
\begin{aligned}
& \Rightarrow|A|=0 \\
& \Rightarrow 0-(1+2 i)(-K(2-i))+(i-2)(-7-14 i)=0
\end{aligned}
$
(Expanding along $R_1$ )
$
\begin{aligned}
& \Rightarrow(1+2 i)(2 K-K i)-7 i+14+14+28 i=0 \\
& \Rightarrow 2 K-K i+4 K i+2 K-7 i+28+28 i=0 \\
& \Rightarrow(28+4 K)+(3 K+21) i=0 \\
& \Rightarrow 28+4 K=0 \text { and } 3 K+21=0 \Rightarrow k=-7
\end{aligned}
$
View full question & answer→MCQ 202 Marks
If $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 3 & 4 \\ 3 & 4 & 3\end{array}\right]$, then $A^{-1}=$
- A
$-\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & 1 \\ 9 & -1 & 1 \\ -5 & 2 & 1\end{array}\right]$
- B
$-\frac{1}{4}\left[\begin{array}{ccc}-7 & -6 & -1 \\ 9 & 6 & -1 \\ -5 & -2 & 1\end{array}\right]$
- C
$\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]$
- ✓
$-\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]$
AnswerCorrect option: D. $-\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]$
(d) : $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 3 & 4 \\ 3 & 4 & 3\end{array}\right]$ $|A|=1(9-16)-2(3-12)+3(4-9)$
[Expanding along $R_1$ ]
$
=-7+18-15=-4 \neq 0
$
Let us find the cofactor of matrix $A$
$
\begin{aligned}
& A_{11}=-7, A_{12}=9, A_{13}=-5, \\
& A_{21}=6, A_{22}=-6, A_{23}=2, \\
& A_{31}=-1, A_{32}=-1, A_{33}=1
\end{aligned}
$
Now, $\operatorname{adj} A=\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]$
$\therefore \quad A^{-1}=\frac{1}{|A|}$ adj $A=\frac{-1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]$
View full question & answer→MCQ 212 Marks
Solve the system of equations $x+2 y+z=4$, $-x+y+z=0$ and $x-3 y+z=4$.
- ✓
$x=2, y=0, z=2$
- B
$x=2, y=0, z=-2$
- C
$x=-2, y=2, z=0$
- D
$x=-2, y=0, z=2$
AnswerCorrect option: A. $x=2, y=0, z=2$
(a) $x=2, y=0, z=2$
View full question & answer→MCQ 222 Marks
If $A=\left[\begin{array}{cc}1+2 i & i \\ -i & 1-2 i\end{array}\right]$, where $i=\sqrt{-1}$, then $A(\operatorname{adj} A)$ $=$
- A
$-2 I$
- B
$2 I$
- C
$5 I$
- ✓
$4 I$
Answer(d) : We have, $A=\left[\begin{array}{cc}1+2 i & i \\ -i & 1-2 i\end{array}\right]$
$
\therefore|A|=(1+2 i)(1-2 i)+i^2=1-4 i^2-1=4
$
Now, $A(\operatorname{adj} A)=|A| I=4 I$
View full question & answer→MCQ 232 Marks
If $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$, then $\left(A^2-5 A\right) A^{-1}=$
- A
$\left[\begin{array}{ccc}4 & 2 & 3 \\ -1 & 4 & 2 \\ 1 & 2 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{ccc}-4 & 2 & 3 \\ -1 & -4 & 2 \\ 1 & 2 & -1\end{array}\right]$
- C
$\left[\begin{array}{ccc}-4 & -1 & 1 \\ 2 & -4 & 2 \\ 3 & 2 & -1\end{array}\right]$
- D
$\left[\begin{array}{ccc}-1 & -2 & 1 \\ 4 & -2 & -3 \\ 1 & 4 & -2\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{ccc}-4 & 2 & 3 \\ -1 & -4 & 2 \\ 1 & 2 & -1\end{array}\right]$
(b) : $|A|=1(4-4)-2(-4-2)+3(-2-1)$
$
=12-9=3 \neq 0
$
$\Rightarrow A^{-1}$ exists.
Now, $\left(A^2-5 A\right) A^{-1}=A^2 A^{-1}-5 A A^{-1}=A-5 I$
$
=\left[\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]-\left[\begin{array}{lll}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5
\end{array}\right]=\left[\begin{array}{ccc}
-4 & 2 & 3 \\
-1 & -4 & 2 \\
1 & 2 & -1
\end{array}\right]
$
View full question & answer→MCQ 242 Marks
For an invertible matrix $A$ if $A($ adj $A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$, then $|A|$ is
Answer(c) : We have, $A($ adj $A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]=10\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=10 I$
We know that $A(\operatorname{adj} A)=|A| I$
$\Rightarrow|A|=10$
View full question & answer→MCQ 252 Marks
Answer(d) : Let $A=\left[\begin{array}{ccc}\alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3\end{array}\right]$
$\Rightarrow|A|=\left|\begin{array}{ccc}\alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3\end{array}\right|=\alpha(9-2)-14(6-6)-1(4-18)$
$\Rightarrow|A|=7 \alpha+14$
Now, $A^{-1}$ does not exist if $|A|=0$
$\Rightarrow 7 \alpha+14=0 \Rightarrow \alpha=-2$.
View full question & answer→MCQ 262 Marks
The inverse of the matrix $\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$ is
- A
$-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 2 & -3\end{array}\right]$
- ✓
$-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$
- C
$-\frac{1}{3}\left[\begin{array}{ccc}3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$
- D
$-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ -3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$
AnswerCorrect option: B. $-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$
(b) : Let $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$
$
\therefore|A|=\left|\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right|=1(-3-0)=-3 \neq 0
$
Thus $A$ is invertible.
$A^{-1}=\frac{1}{|A|}$ adj $A=-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$
View full question & answer→MCQ 272 Marks
If matrix $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$ such that $A X=I$, then $X=$
- A
$\frac{1}{5}\left[\begin{array}{rr}1 & 3 \\ 2 & -1\end{array}\right]$
- B
$\frac{1}{5}\left[\begin{array}{rr}4 & 2 \\ 4 & -1\end{array}\right]$
- ✓
$\frac{1}{5}\left[\begin{array}{cr}-3 & 2 \\ 4 & -1\end{array}\right]$
- D
$\frac{1}{5}\left[\begin{array}{ll}-1 & 2 \\ -1 & 4\end{array}\right]$
AnswerCorrect option: C. $\frac{1}{5}\left[\begin{array}{cr}-3 & 2 \\ 4 & -1\end{array}\right]$
(c) : $\because A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$, such that $A X=I$
$\Rightarrow X=A^{-1} I$
$\because|A|=3-8=-5 \neq 0 \Rightarrow A^{-1}$ exists.
$A^{-1}=\frac{1}{-5}\left[\begin{array}{cc}3 & -2 \\ -4 & 1\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]$
$\therefore \quad X=\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]$
View full question & answer→MCQ 282 Marks
If $A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$ then $\left(B^{-1} A^{-1}\right)^{-1}=$
- ✓
$\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]$
- B
$\left[\begin{array}{cc}2 & 2 \\ -2 & 3\end{array}\right]$
- C
$\left[\begin{array}{cc}2 & -3 \\ 2 & 2\end{array}\right]$
- D
$\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]$
(a) : $A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$
$|A|=4+6=10 \neq 0$ and $|B|=0+1=1 \neq 0$
$\therefore \quad A^{-1}, B^{-1}$ exists
So, $\left(B^{-1} A^{-1}\right)^{-1}=\left(A^{-1}\right)^{-1}\left(B^{-1}\right)^{-1}=A B$
$=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}0+2 & -2+0 \\ 0+2 & 3+0\end{array}\right]=\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]$
View full question & answer→MCQ 292 Marks
If $A=\left[\begin{array}{lll}1 & 1 & 0 \\ 2 & 1 & 5 \\ 1 & 2 & 1\end{array}\right]$, then $a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23}=$
Answer(b) : Given : $A=\left[\begin{array}{lll}1 & 1 & 0 \\ 2 & 1 & 5 \\ 1 & 2 & 1\end{array}\right]$
Here, $a_{11}=1, a_{12}=1$, and $a_{13}=0$
$
\begin{aligned}
& A_{21}=(-1)^{2+1} M_{21}=(-1)\left|\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right|=-1[1-0]=-1 \\
& A_{22}=(-1)^{2+2} M_{22}=(1)\left|\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right|=1-0=1 \\
& A_{23}=(-1)^{2+3} M_{23}=(-1)\left|\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right|=(-1)|2-1|=-1 \\
& \therefore a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23} \\
& =(1)(-1)+(1)(1)+(0)(-1)=-1+1+0=0 \\
&
\end{aligned}
$
View full question & answer→MCQ 302 Marks
Let $A=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)$. If $u_1$ and $u_2$ are column matrices such that $A u_1=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)$ and $A u_2=\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)$, then $u_1+u_2$ is equal to
- A
$\left(\begin{array}{c}-1 \\ -1 \\ 0\end{array}\right)$
- ✓
$\left(\begin{array}{c}1 \\ -1 \\ -1\end{array}\right)$
- C
$\left(\begin{array}{c}-1 \\ 1 \\ 0\end{array}\right)$
- D
$\left(\begin{array}{c}-1 \\ 1 \\ -1\end{array}\right)$
AnswerCorrect option: B. $\left(\begin{array}{c}1 \\ -1 \\ -1\end{array}\right)$
$A=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right], A u_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right], A u_2=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$
Let $u_1=\left[\begin{array}{l}a \\ b \\ c\end{array}\right]$
Let $u_2=\left[\begin{array}{l}p \\ q \\ r\end{array}\right]$
$ \begin{array}{l} A u_2=\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right] \Rightarrow \begin{array}{c} p=0,2 p+q=1 \Rightarrow q=1, \\ 3 p+2 q+r=0 \Rightarrow r=-2 \end{array}\end{array} $
$ u_1+u_2=\left[\begin{array}{r} 1 \\ -2 \\ 1 \end{array}\right]+\left[\begin{array}{r} 0 \\ 1 \\ -2 \end{array}\right]=\left[\begin{array}{r} 1 \\ -1 \\ -1 \end{array}\right] \\ $
View full question & answer→MCQ 312 Marks
Let M and N be two $3 \times 3$ non-singular skew-symmetric matrices such that $MN = NM$. If $P ^{ T }$ denotes the transpose of. P , then $M ^2 N^2\left( M ^{ T } N \right)^{-1}\left( MN ^{-1}\right)^{ T }$ is equal to
- A
$M ^2$
- B
$- N ^2$
- ✓
$- M ^2$
- D
AnswerCorrect option: C. $- M ^2$
(C) $M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T$
$\begin{array}{l}=M^2 N^2 N^{-1}\left(M^T\right)^{-1}\left(N^{-1}\right)^T M^T \\ =M^2 N\left(M^T\right)^{-1}\left(N^T\right)^{-1} M^T\end{array}$
$= M ^2 N(- M )^{-1}(- N )^{-1}(- M )$ $\ldots[\because$ For skew-symmetric matrices M and $\left.N , M ^{ T }=- M , N ^{ T }=- N \right]$
$=- M ^2 NM ^{-1} N^{-1} M \quad \ldots\left[\because( kA )^{-1}=\frac{1}{ k } A ^{-1}\right]$
$\begin{array}{l}=- M ( MN )( NM )^{-1} M \\ =- M ( NM )( NM )^{-1} M \end{array}$
$\ldots[\because MN = NM ($ given $)]$
$=- M \cdot I \cdot M =- M ^2$
View full question & answer→MCQ 322 Marks
If A is an $3 \times 3$ non-singular matrix such that $A A^{\prime}=A^{\prime} A$ and $B=A^{-1} A^{\prime}$, then $B B^{\prime}$ equals
Answer(D) $BB ^{\prime}=\left( A ^{-1} A^{\prime}\right)\left( A ^{-1} A^{\prime}\right)^{\prime}$
$=\left( A ^{-1} A^{\prime}\right)\left( A \left( A ^{-1}\right)^{\prime}\right)$
$= A ^{-1} AA ^{\prime}\left( A ^{-1}\right)^{\prime}$ $\ldots\left[\because AA ^{\prime}= A ^{\prime} A\right.$ (given) $]$
$\begin{array}{l}=\left( A ^{-1} A\right)\left( A ^{\prime}\left( A ^{-1}\right)^{\prime}\right) \\ = I \left( A ^{-1} A\right)^{\prime}= I . I = I ^2= I \end{array}$
View full question & answer→MCQ 332 Marks
If $k$ is one of the roots of the equation $x^2-25 x+24=0$ such that $A =\left[\begin{array}{lll}1 & 2 & 1 \\ 3 & 2 & 3 \\ 1 & 1 & k \end{array}\right]$ is a non-singular matrix, then $A ^{-1}=$
- A
$-\frac{1}{46}\left[\begin{array}{ccc}90 & -94 & 8 \\ -138 & 46 & 0 \\ 2 & 2 & -8\end{array}\right]$
- ✓
$-\frac{1}{92}\left[\begin{array}{ccc}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{array}\right]$
- C
$-\frac{1}{46}\left[\begin{array}{ccc}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{array}\right]$
- D
$-\frac{1}{92}\left[\begin{array}{ccc}90 & -94 & 8 \\ -138 & 46 & 0 \\ 2 & 2 & -8\end{array}\right]$
AnswerCorrect option: B. $-\frac{1}{92}\left[\begin{array}{ccc}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{array}\right]$
(B) $x^2-25 x+24=0$
$\begin{array}{l}\Rightarrow(x-1)(x-24)=0 \\ \Rightarrow x=1 \text { or } x=24 \\ \Rightarrow k =1 \text { or } k =24\end{array}$
For $k =1$,
$| A |=\left|\begin{array}{lll}1 & 2 & 1 \\ 3 & 2 & 3 \\ 1 & 1 & 1\end{array}\right|=0$
For $k =24$,
$| A |=\left|\begin{array}{ccc}1 & 2 & 1 \\ 3 & 2 & 3 \\ 1 & 1 & 24\end{array}\right|=-92 \neq 0$
$A=\left[\begin{array}{ccc}1 & 2 & 1 \\ 3 & 2 & 3 \\ 1 & 1 & 24\end{array}\right]$
$\operatorname{adj} A=\left[\begin{array}{ccc}45 & -69 & 1 \\ -47 & 23 & 1 \\ 4 & 0 & -4\end{array}\right]^T=\left[\begin{array}{ccc}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{array}\right]$
$\therefore \quad A ^{-1}=\frac{1}{|A|}$ adj $A =-\frac{1}{92}\left[\begin{array}{ccc}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{array}\right]$
View full question & answer→MCQ 342 Marks
Given $A =\left[\begin{array}{cc}0 & -\tan \alpha \\ \tan \alpha & 0\end{array}\right]$ and $B (\alpha)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$, then $( I + A )( I - A )^{-1}$ is equal to
- A
$B (\alpha)$
- B
$B (-\alpha)$
- ✓
$B (2 \alpha)$
- D
$B (-2 \alpha)$
AnswerCorrect option: C. $B (2 \alpha)$
(C) $I+A=\left[\begin{array}{cc}1 & -\tan \alpha \\ \tan \alpha & 1\end{array}\right]$
$I-A=\left[\begin{array}{cc}1 & \tan \alpha \\ -\tan \alpha & 1\end{array}\right]$
$\therefore| I - A |=1+\tan ^2 \alpha=\sec ^2 \alpha \neq 0$
$\operatorname{adj}(I-A)=\left[\begin{array}{cc}1 & \tan \alpha \\ -\tan \alpha & 1\end{array}\right]^{\top}=\left[\begin{array}{cc}1 & -\tan \alpha \\ \tan \alpha & 1\end{array}\right]$
$\Rightarrow( I - A )^{-1}=\frac{1}{| I - A |}[\operatorname{adj}( I - A )]$
$=\left[\begin{array}{cc}\cos ^2 \alpha & -\sin \alpha \cdot \cos \alpha \\ \sin \alpha \cdot \cos \alpha & \cos ^2 \alpha\end{array}\right]$
$\therefore(I+A)(I-A)^{-1}$
$\begin{array}{l}=\left[\begin{array}{cc}1 & -\tan \alpha \\ \tan \alpha & 1\end{array}\right]\left[\begin{array}{cc}\cos ^2 \alpha & -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha & \cos ^2 \alpha\end{array}\right] \\ =\left[\begin{array}{cc}\cos ^2 \alpha-\sin ^2 \alpha & -2 \sin \alpha \cos \alpha \\ 2 \sin \alpha \cos \alpha & \cos ^2 \alpha-\sin ^2 \alpha\end{array}\right]\end{array}$
$\therefore \quad(I+A)(I-A)^{-1}=\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right]$ ...(i)
$B (\alpha)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$\therefore \quad B (2 \alpha)=\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right]$ ...(ii)
$\therefore \quad(I+A)(I-A)^{-1}=B(2 \alpha) \ldots[$ From (i) and (ii) $]$
View full question & answer→MCQ 352 Marks
If $A =\left[\begin{array}{cc} i & 0 \\ 0 & \frac{ i }{2}\end{array}\right]$ where $( i =\sqrt{-1})$, then $A ^{-1}=$
- A
$\left[\begin{array}{ll} i & 0 \\ 0 & \frac{ i }{2}\end{array}\right]$
- ✓
$\left[\begin{array}{cc}- i & 0 \\ 0 & -2 i \end{array}\right]$
- C
$\left[\begin{array}{cc} i & 0 \\ 0 & 2 i \end{array}\right]$
- D
$\left[\begin{array}{cc}0 & i \\ 2 i & 0\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{cc}- i & 0 \\ 0 & -2 i \end{array}\right]$
(B) $|A|=-\frac{1}{2} \neq 0$
$\therefore \quad A ^{-1}=\frac{1}{\frac{-1}{2}}\left[\begin{array}{ll}\frac{ i }{2} & 0 \\ 0 & i \end{array}\right]=\left[\begin{array}{cc}- i & 0 \\ 0 & -2 i \end{array}\right]$ ...[Using Shortcut 2]
View full question & answer→MCQ 362 Marks
AnswerCorrect option: A. $F (-\alpha)$
(A) $F(\alpha) \cdot F(-\alpha)$
$=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]= I$
$\therefore[ F (\alpha)]^{-1}= F (-\alpha)$
View full question & answer→MCQ 372 Marks
If $A=\left[\begin{array}{cc}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right]$ and $A B=I$, then $B=$
- A
$\cos ^2 \frac{\theta}{2} \cdot A$
- ✓
$\cos ^2 \frac{\theta}{2} \cdot A^{ T }$
- C
$\cos ^2 \frac{\theta}{2} \cdot I$
- D
AnswerCorrect option: B. $\cos ^2 \frac{\theta}{2} \cdot A^{ T }$
(B) $| A |=1+\tan ^2 \frac{\theta}{2}=\sec ^2 \frac{\theta}{2} \neq 0$
$\therefore \quad A ^{-1}=\frac{1}{\sec ^2 \frac{\theta}{2}}\left[\begin{array}{cc}1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1\end{array}\right]$ ...[Using Shortcut 2]
$AB = I \Rightarrow B = IA ^{-1} \Rightarrow B= A ^{-1}$
$\therefore B=\frac{1}{\sec ^2 \frac{\theta}{2}}\left[\begin{array}{cc}1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1\end{array}\right]=\cos ^2 \frac{\theta}{2} \cdot A^{ T }$
View full question & answer→MCQ 382 Marks
If $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$ and $A \cdot \operatorname{adj} A=\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]$, then $k$ is equal to
Answer(B) Using Shortcut 4(i),
$A (\operatorname{adj} A )=| A | \cdot I$
$\therefore \quad\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]=\left(\cos ^2 \alpha+\sin ^2 \alpha\right)\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow k=1$
View full question & answer→MCQ 392 Marks
If $A=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$, then $A(\operatorname{adj} A)=$
- ✓
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- B
$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
- C
$\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$
- D
$\left[\begin{array}{cc}-2 & 0 \\ 0 & -2\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
(A) $|A|=\left|\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right|=\cos ^2 x+\sin ^2 x=1$
Using Shortcut 4(i),
$\begin{aligned} & A (\operatorname{adj} A )=| A | . I \\ \therefore \quad & A (\operatorname{adj} A )=1\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\end{aligned}$
View full question & answer→MCQ 402 Marks
The inverse of $\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]$ is
- A
$\left[\begin{array}{cc}1 & -\sin \alpha \\ -\sin \alpha & -1\end{array}\right]$
- B
$-\sec ^2 \alpha\left[\begin{array}{cc}1 & -\sin \alpha \\ \sin \alpha & -1\end{array}\right]$
- ✓
$\sec ^2 \alpha\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]$
- D
$\cos ^2 \alpha\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]$
AnswerCorrect option: C. $\sec ^2 \alpha\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]$
(C) $|A|=\left|\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right|=-1+\sin ^2 \alpha \neq 0$
$\therefore \quad A ^{-1}=\frac{1}{-1+\sin ^2 \alpha}\left[\begin{array}{cc}-1 & -\sin \alpha \\ \sin \alpha & 1\end{array}\right]$ ...[Using Shortcut 2]
$\begin{array}{l}=\frac{1}{\cos ^2 \alpha}\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right] \\ =\sec ^2 \alpha\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]\end{array}$
View full question & answer→MCQ 412 Marks
If $\omega$ is a complex cube root of unity and $A=\left[\begin{array}{lll}\omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1\end{array}\right]$, then $A^{-1}$ is equal to
- A
$\left[\begin{array}{llr}1 & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & \omega\end{array}\right]$
- B
$\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$
- D
$\left[\begin{array}{lll}\omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1\end{array}\right]$
AnswerCorrect option: C. $\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$
(C) $A=\left[\begin{array}{ccc}\omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1\end{array}\right]$
$|A|=\omega^3=1 \neq 0$
$\operatorname{adj} A =\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & \omega^3\end{array}\right]^{ T }$
$=\left[\begin{array}{ccr}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & \omega^3\end{array}\right]$
$=\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$
$A^{-1}=\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$
View full question & answer→MCQ 422 Marks
For a matrix
$A=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right]$ if $U_1, U_2$ and $U_3$ are $3 \times 1$
column matrices satisfying
$AU _1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right], AU _2=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right], AU _3=\left[\begin{array}{l}2 \\ 3 \\ 1\end{array}\right]$ and U is
$3 \times 3$ matrix whose columns are $U _1, U _2$ and $U _3$.
Then sum of the elements of $U ^{-1}$ is
Answer(B) Let $U _1=\left[\begin{array}{l} a _1 \\ b_1 \\ c _1\end{array}\right], U _2=\left[\begin{array}{l} a _2 \\ b_2 \\ c _2\end{array}\right]$ and $U _3=\left[\begin{array}{l} a _3 \\ b_3 \\ c _3\end{array}\right]$
$AU _1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$
$\therefore\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}a_1 \\ b_1 \\ c_1\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$
$\therefore\left[\begin{array}{l}a_1 \\ 2 a_1+b_1 \\ 3 a_1+2 b_1+c_1\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$
∴ by the equality of matrices,
$a_1=1, b_1=-2$ and $c_1=1$
Similarly $a _2=2, b_2=-1$ and $c _2=-4$
$a_3=2, b_3=-1$ and $c_3=-3$
$\therefore \quad U=\left[\begin{array}{ccc}1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3\end{array}\right]$
$|U|=\left|\begin{array}{ccc}1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3\end{array}\right|=3$
$\therefore U ^{-1}$ exists
$\therefore \quad U^{-1}=\left[\begin{array}{ccc}\frac{-1}{3} & \frac{-2}{3} & 0 \\ \frac{-7}{3} & \frac{-5}{3} & -1 \\ 3 & 2 & 1\end{array}\right]$
$\therefore \quad$ sum of elements of $U ^{-1}=0$
View full question & answer→MCQ 432 Marks
Let M be a $3 \times 3$ matrix satisfying $M \left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right], M \left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}1 \\ 1 \\ -1\end{array}\right]$ and $M \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{c}0 \\ 0 \\ 12\end{array}\right]$ Then the sum of the diagonal entries of $M$ is
Answer(C) Let $M =\left[\begin{array}{lll} a & b & c \\ x & y & z \\ l & m & n \end{array}\right]$, then
$M \left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right] \Rightarrow\left[\begin{array}{l} b \\ y \\ m\end{array}\right]=\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$
∴ by the equality of matrices,
$\begin{array}{l} b =-1, y=2, m=3 \\ M \left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}1 \\ 1 \\ -1\end{array}\right] \Rightarrow\left[\begin{array}{l} a - b \\ x-y \\ l- m \end{array}\right]=\left[\begin{array}{c}1 \\ 1 \\ -1\end{array}\right]\end{array}$
∴ by the equality of matrices,
$\begin{array}{l} a - b =1, x-y=1, l- m =-1 \\ \Rightarrow a =0, x=3, l=2\end{array}$
$M \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{c}0 \\ 0 \\ 12\end{array}\right] \Rightarrow\left[\begin{array}{l} a + b + c \\ x+y+ z \\ l+ m + n \end{array}\right]=\left[\begin{array}{c}0 \\ 0 \\ 12\end{array}\right]$
∴ by the equality of matrices,
$a + b + c =0, x+y+ z =0, l+ m + n =12$
$\Rightarrow c=1, z=-5, n=7$
∴ sum of diagonal elements of $M = a +y+ n$
$=0+2+7=9$
View full question & answer→MCQ 442 Marks
Let $X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right], D =\left[\begin{array}{c}3 \\ 5 \\ 11\end{array}\right]$ and $A =\left[\begin{array}{rrr}1 & -1 & -2 \\ 2 & 1 & 1 \\ 4 & -1 & -2\end{array}\right]$, if $X=A^{-1} D$, then $X$ is equal to
- A
$\left[\begin{array}{l}1 \\ 0 \\ 2\end{array}\right]$
- ✓
$\left[\begin{array}{c}\frac{8}{3} \\ \frac{-1}{3} \\ 0\end{array}\right]$
- C
$\left[\begin{array}{c}\frac{-8}{3} \\ 1 \\ 0\end{array}\right]$
- D
$\left[\begin{array}{c}\frac{8}{3} \\ \frac{1}{3} \\ -1\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{c}\frac{8}{3} \\ \frac{-1}{3} \\ 0\end{array}\right]$
(B) $X = A ^{-1} D$
$\Rightarrow A X=D$
$\left[\begin{array}{ccc}1 & -1 & -2 \\ 2 & 1 & 1 \\ 4 & -1 & -2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}3 \\ 5 \\ 11\end{array}\right]$
Applying $R _1 \rightarrow R _1+ R _2, R _3 \rightarrow R _3+ R _2$,
$\left[\begin{array}{ccc}3 & 0 & -1 \\ 2 & 1 & 1 \\ 6 & 0 & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}8 \\ 5 \\ 16\end{array}\right]$
Applying $R _3 \rightarrow R _3- R _1$,
$\left[\begin{array}{ccc}3 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 0 & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}8 \\ 5 \\ 8\end{array}\right]$
$\begin{aligned} \therefore \quad & 3 x=8 \Rightarrow x=\frac{8}{3} \\ & 3 x-z=8 \Rightarrow z=0\end{aligned}$
$2 x+y+z=5 \Rightarrow y=\frac{-1}{3}$
$\therefore \quad X=\left[\begin{array}{c}\frac{8}{3} \\ \frac{-1}{3} \\ 0\end{array}\right]$
View full question & answer→MCQ 452 Marks
Let $X =\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right], A =\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]$ and $B =\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$. If $AX = B$, then $X =$
- A
$\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
- ✓
$\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$
- C
$\left[\begin{array}{c}-1 \\ -2 \\ 3\end{array}\right]$
- D
$\left[\begin{array}{l}-1 \\ -2 \\ -3\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$
(B) $A X=B$
$\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$
Applying $R _1 \rightarrow 2 R _1+ R _3$,
$\left[\begin{array}{lll}5 & 0 & 5 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{c}10 \\ 1 \\ 4\end{array}\right]$
Applying $R _1 \rightarrow R _1-5 R _2$,
$\left[\begin{array}{ccc}-5 & 0 & 0 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]-\left[\begin{array}{l}5 \\ 1 \\ 4\end{array}\right]$
$\begin{array}{ll}\therefore & -5 x_1=5 \Rightarrow x_1=-1 \\ & 2 x_1+x_3=1 \Rightarrow x_3=3\end{array}$
$3 x_1+2 x_2+x_3=4 \Rightarrow x_2=2$
$\therefore \quad X =\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$
View full question & answer→MCQ 462 Marks
If $A$ and $B$ are square matrices of the same order and $A$ is non-singular, then for a positive integer $n ,\left( A ^{-1} BA \right)^{ n }$ is equal to
AnswerCorrect option: C. $A ^{-1} B^n A$
(C) $\left( A ^{-1} BA \right)^2=\left( A ^{-1} BA \right)\left( A ^{-1} BA \right)$
$= A ^{-1} B\left( AA ^{-1}\right) BA$
$= A ^{-1} BIBA = A ^{-1} B^2 A$
$\left( A ^{-1} BA \right)^3=\left( A ^{-1} B^2 A\right)\left( A ^{-1} BA \right)$
$= A ^{-1} B^2\left( AA ^{-1}\right) BA$
$= A ^{-1} B^2 IBA = A ^{-1} B^3 A$
In general,
$\left(A^{-1} B A\right)^n=A^{-1} B^n A$
View full question & answer→MCQ 472 Marks
If a matrix $A$ is such that $4 A^3+2 A^2+7 A+I=0$, then $A^{-1}$ equals
AnswerCorrect option: B. $-\left(4 A^2+2 A+71\right)$
(B) $4 A^3+2 A^2+7 A+I=0$
$\Rightarrow 4 A^{-1} A^3+2 A^{-1} A^2+7 A^{-1} A+ A ^{-1} I =0$
$\Rightarrow 4 A^2+2 A+7 I + A ^{-1}=0$
$\Rightarrow A^{-1}=-\left(4 A^3+2 A+7 I\right)$
View full question & answer→MCQ 482 Marks
If $A$ is a square matrix such that $|A| \neq 0$ and $m$, n $(\neq 0)$ are scalars such that $A ^2+ mA + nI =0$, then $A ^{-1}=$
AnswerCorrect option: B. $-\frac{1}{n}(A+m I)$
(B) $A^2+m A+n I=0$
$\Rightarrow A \cdot A + mA + nI =0$
$\Rightarrow A ^{-1} \cdot A \cdot A + mA ^{-1} \cdot A+ n A ^{-1} \cdot I =0$
$\Rightarrow A + mI + nA ^{-1}=0$
$\Rightarrow nA ^{-1}=- A - mI$
$\Rightarrow A ^{-1}=\frac{-1}{ n }( A + ml )$
View full question & answer→MCQ 492 Marks
A square, non-singular matrix A satisfies $A ^2- A +2 I =0$, then $A ^{-1}$ is equal to
- A
$I - A$
- ✓
$\frac{1}{2}( I - A )$
- C
$\frac{1}{2}( I + A )$
- D
$I + A$
AnswerCorrect option: B. $\frac{1}{2}( I - A )$
(B) $A^2-A+2 I=0$
$\Rightarrow A \cdot A - A +2 I =0$
$\Rightarrow A ^{-1} \cdot A \cdot A - A ^{-1} \cdot A+2 A^{-1} \cdot I =0$
$\Rightarrow A - I +2 A^{-1}=0$
$\Rightarrow 2 A^{-1}= I - A$
$\Rightarrow A ^{-1}=\frac{1}{2}( I - A )$
View full question & answer→MCQ 502 Marks
If A and B are square matrices of the same order and $AB =3 I$, then $A ^{-1}$ is equal to
- A
- ✓
$\frac{1}{3} B$
- C
$3 B^{-1}$
- D
$\frac{1}{3} B^{-1}$
AnswerCorrect option: B. $\frac{1}{3} B$
(B) $A B=3 I$
$\Rightarrow A ^{-1} AB =3 A^{-1} I$
$\Rightarrow B =3 A^{-1}$
$\therefore \quad A ^{-1}=\frac{1}{3} B$
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