MCQ

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
$\left[\begin{array}{rr}2 & -4 \\ 1 & 1\end{array}\right]$
✓
$\left[\begin{array}{rr}2 & 4 \\ 1 & 1\end{array}\right]$

Answer

Correct option: D.

$\left[\begin{array}{rr}2 & 4 \\ 1 & 1\end{array}\right]$

$-\frac{1}{2}\left[\begin{array}{ll}2 & 4 \\ 1 & 1\end{array}\right]$

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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

A
20
✓
10
C
30
D
40

Answer

Correct option: B.

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MCQ 31 Mark

The inverse of a symmetric matrix is

✓
Symmetric
B
Non-symmetric
C
Null matrix
D
Diagonal matrix

Answer

Correct option: A.

Symmetric

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MCQ 41 Mark

The inverse of $A=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$

✓
I
B
A
C
A’
D
-I

Answer

Correct option: A.

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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 =

✓
F(-∝)
B
$F\left(\alpha^{-1}\right)$
C
F(2∝)
D
None of these

Answer

Correct option: A.

F(-∝)

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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
$\left[\begin{array}{rr}-\cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha\end{array}\right]$

Answer

Correct option: B.

$\left[\begin{array}{rr}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$

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MCQ 71 Mark

If $A=\left(\begin{array}{rr}\lambda & 1 \\ -1 & -\lambda\end{array}\right)$, then $A^{-1}$ does not exist if $\lambda=$

A
0
✓
± 1
C
2
D
3

Answer

Correct option: B.

± 1

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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

A
2
✓
-2
C
10
D
-10

Answer

Correct option: B.

-2

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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
$\left[\begin{array}{rr}-1 & -3 \\ -4 & 2\end{array}\right]$

Answer

Correct option: B.

$\left[\begin{array}{rr}1 & 4 \\ -3 & 2\end{array}\right]$

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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

A
1
B
-1
C
3
✓
-3

Answer

Correct option: D.

-3

-3 [Hint : A(adj A) = |A| ∙ I]

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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
None of these

Answer

Correct option: B.

$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

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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,

✓
a = – 2, b = 1
B
a = 2, b = 4
C
a = 2, b = –1
D
a = 1, b = –2

Answer

Correct option: A.

a = – 2, b = 1

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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
$\left[\begin{array}{ll}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{1}{2}\end{array}\right]$

Answer

Correct 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, $\operatorname{adj}(A+B)=\left[\begin{array}{ll}-2 & 0 \\ -6 & 2\end{array}\right]$
$\therefore(A+B)^{-1}=\frac{\operatorname{adj}(A+B)}{|A+B|}=\frac{-1}{4}=\left[\begin{array}{ll}-2 & 0 \\ -6 & 2\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right]$

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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

A
-1
B
-3
C
-4
✓
-5

Answer

Correct option: D.

-5

(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 $B=-1-3-1=-5$

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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 $\frac{|\operatorname{adj} B|}{|C|}=$

A
1
B
5
✓
25
D
-1

Answer

Correct option: C.

(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, $\frac{|\operatorname{adj} B|}{|C|}=\frac{625}{25}=25$

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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
$\frac{1}{3}\left[\begin{array}{cc}-2 & 1 \\ 1 & -2\end{array}\right]$
✓
$\frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$

Answer

Correct 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}
$

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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
$5\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$

Answer

Correct 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]
$

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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 )

A
-25
B
-85
C
85
✓
25

Answer

Correct option: D.

(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
$

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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 $i=\sqrt{-1}$ )

A
$1+2 i$
✓
-7
C
7
D
$1-2 i$

Answer

Correct option: B.

-7

(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}
$

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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
$\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]$
✓
$-\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]$

Answer

Correct 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 $A=\frac{-1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]$

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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
$x=-2, y=0, z=2$

Answer

Correct option: A.

$x=2, y=0, z=2$

(a) $x=2, y=0, z=2$

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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

Correct option: D.

$4 I$

(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, $A(\operatorname{adj} A)=|A| I=4 I$

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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
$\left[\begin{array}{ccc}-1 & -2 & 1 \\ 4 & -2 & -3 \\ 1 & 4 & -2\end{array}\right]$

Answer

Correct 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]
$

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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

A
100
B
-100
✓
10
D
-10

Answer

Correct option: C.

(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 $A(\operatorname{adj} A)=|A| I$
$\Rightarrow|A|=10$

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MCQ 252 Marks

If the inverse of the matrix $\left[\begin{array}{ccc}\alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3\end{array}\right]$ does not exist, then the value of $\alpha$ is

A
1
B
-1
C
0
✓
-2

Answer

Correct option: D.

-2

(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 $|A|=0$
$\Rightarrow 7 \alpha+14=0 \Rightarrow \alpha=-2$.

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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
$-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ -3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$

Answer

Correct 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 $A=-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$

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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
$\frac{1}{5}\left[\begin{array}{ll}-1 & 2 \\ -1 & 4\end{array}\right]$

Answer

Correct 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.
$A^{-1}=\frac{1}{-5}\left[\begin{array}{cc}3 & -2 \\ -4 & 1\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]$
$\therefore \quad X=\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]$

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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
$\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]$

Answer

Correct 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, $\left(B^{-1} A^{-1}\right)^{-1}=\left(A^{-1}\right)^{-1}\left(B^{-1}\right)^{-1}=A B$
$=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}0+2 & -2+0 \\ 0+2 & 3+0\end{array}\right]=\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]$

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MCQ 292 Marks

If $A=\left[\begin{array}{lll}1 & 1 & 0 \\ 2 & 1 & 5 \\ 1 & 2 & 1\end{array}\right]$, then $a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23}=$

A
1
✓
0
C
-1
D
2

Answer

Correct option: B.

(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}
$

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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
$\left(\begin{array}{c}-1 \\ 1 \\ -1\end{array}\right)$

Answer

Correct 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] \\ $

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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
MN

Answer

Correct 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 $)]$
$=- M \cdot I \cdot M =- M ^2$

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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

A
$B ^{-1}$
B
$\left( B ^{-1}\right)^{\prime}$
C
$I+B$
✓
I

Answer

Correct option: D.

(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) $]$
$\begin{array}{l}=\left( A ^{-1} A\right)\left( A ^{\prime}\left( A ^{-1}\right)^{\prime}\right) \\ = I \left( A ^{-1} A\right)^{\prime}= I . I = I ^2= I \end{array}$

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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
$-\frac{1}{92}\left[\begin{array}{ccc}90 & -94 & 8 \\ -138 & 46 & 0 \\ 2 & 2 & -8\end{array}\right]$

Answer

Correct 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 $A =-\frac{1}{92}\left[\begin{array}{ccc}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{array}\right]$

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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
$B (-2 \alpha)$

Answer

Correct 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) $]$

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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
$\left[\begin{array}{cc}0 & i \\ 2 i & 0\end{array}\right]$

Answer

Correct 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]

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MCQ 362 Marks

Let $F(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$, where $\alpha \in R$.
Then $[ F (\alpha)]^{-1}$ is equal to

✓
$F (-\alpha)$
B
$F \left(\alpha^{-1}\right)$
C
$F (2 \alpha)$
D
None of these

Answer

Correct option: A.

$F (-\alpha)$

(A) $F(\alpha) \cdot F(-\alpha)$
$=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]= I$
$\therefore[ F (\alpha)]^{-1}= F (-\alpha)$

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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
None of these

Answer

Correct 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]
$AB = I \Rightarrow B = IA ^{-1} \Rightarrow B= A ^{-1}$
$\therefore B=\frac{1}{\sec ^2 \frac{\theta}{2}}\left[\begin{array}{cc}1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1\end{array}\right]=\cos ^2 \frac{\theta}{2} \cdot A^{ T }$

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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

A
$0$
✓
1
C
$\sin \alpha \cos \alpha$
D
$\cos 2 \alpha$

Answer

Correct option: B.

(B) Using Shortcut 4(i),
$A (\operatorname{adj} A )=| A | \cdot I$
$\therefore \quad\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]=\left(\cos ^2 \alpha+\sin ^2 \alpha\right)\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow k=1$

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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
$\left[\begin{array}{cc}-2 & 0 \\ 0 & -2\end{array}\right]$

Answer

Correct 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),
$\begin{aligned} & A (\operatorname{adj} A )=| A | . I \\ \therefore \quad & A (\operatorname{adj} A )=1\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\end{aligned}$

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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
$\cos ^2 \alpha\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]$

Answer

Correct 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]
$\begin{array}{l}=\frac{1}{\cos ^2 \alpha}\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right] \\ =\sec ^2 \alpha\left[\begin{array}{cc}1 & \sin \alpha \\ -\sin \alpha & -1\end{array}\right]\end{array}$

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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
$\left[\begin{array}{lll}\omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1\end{array}\right]$

Answer

Correct option: C.

$\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$

(C) $A=\left[\begin{array}{ccc}\omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1\end{array}\right]$
$|A|=\omega^3=1 \neq 0$
$\operatorname{adj} A =\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & \omega^3\end{array}\right]^{ T }$
$=\left[\begin{array}{ccr}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & \omega^3\end{array}\right]$
$=\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$
$A^{-1}=\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$

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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

A
6
✓
0 (zero)
C
1
D
2/3

Answer

Correct option: B.

0 (zero)

(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 $U ^{-1}=0$

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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

A
7
B
8
✓
9
D
6

Answer

Correct option: C.

(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 $M = a +y+ n$
$=0+2+7=9$

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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
$\left[\begin{array}{c}\frac{8}{3} \\ \frac{1}{3} \\ -1\end{array}\right]$

Answer

Correct 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 $R _3 \rightarrow R _3- R _1$,
$\left[\begin{array}{ccc}3 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 0 & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}8 \\ 5 \\ 8\end{array}\right]$
$\begin{aligned} \therefore \quad & 3 x=8 \Rightarrow x=\frac{8}{3} \\ & 3 x-z=8 \Rightarrow z=0\end{aligned}$
$2 x+y+z=5 \Rightarrow y=\frac{-1}{3}$
$\therefore \quad X=\left[\begin{array}{c}\frac{8}{3} \\ \frac{-1}{3} \\ 0\end{array}\right]$

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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
$\left[\begin{array}{l}-1 \\ -2 \\ -3\end{array}\right]$

Answer

Correct 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 $R _1 \rightarrow R _1-5 R _2$,
$\left[\begin{array}{ccc}-5 & 0 & 0 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]-\left[\begin{array}{l}5 \\ 1 \\ 4\end{array}\right]$
$\begin{array}{ll}\therefore & -5 x_1=5 \Rightarrow x_1=-1 \\ & 2 x_1+x_3=1 \Rightarrow x_3=3\end{array}$
$3 x_1+2 x_2+x_3=4 \Rightarrow x_2=2$
$\therefore \quad X =\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$

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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

A
$A^{-n} B^n A^n$
B
$A^n B^n A^{-n}$
✓
$A ^{-1} B^n A$
D
$n\left(A^{-1} B A\right)$

Answer

Correct 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,
$\left(A^{-1} B A\right)^n=A^{-1} B^n A$

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MCQ 472 Marks

If a matrix $A$ is such that $4 A^3+2 A^2+7 A+I=0$, then $A^{-1}$ equals

A
$4 A^2+2 A+71$
✓
$-\left(4 A^2+2 A+71\right)$
C
$-\left(4 A^2-2 A+71\right)$
D
$4 A^2+2 A-71$

Answer

Correct option: B.

$-\left(4 A^2+2 A+71\right)$

(B) $4 A^3+2 A^2+7 A+I=0$
$\Rightarrow 4 A^{-1} A^3+2 A^{-1} A^2+7 A^{-1} A+ A ^{-1} I =0$
$\Rightarrow 4 A^2+2 A+7 I + A ^{-1}=0$
$\Rightarrow A^{-1}=-\left(4 A^3+2 A+7 I\right)$

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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 $A ^{-1}=$

A
$-\frac{1}{m}( A + nl )$
✓
$-\frac{1}{n}(A+m I)$
C
$-\frac{1}{n}(I+m A)$
D
$\frac{1}{ n }( I + mA )$

Answer

Correct option: B.

$-\frac{1}{n}(A+m I)$

(B) $A^2+m A+n I=0$
$\Rightarrow A \cdot A + mA + nI =0$
$\Rightarrow A ^{-1} \cdot A \cdot A + mA ^{-1} \cdot A+ n A ^{-1} \cdot I =0$
$\Rightarrow A + mI + nA ^{-1}=0$
$\Rightarrow nA ^{-1}=- A - mI$
$\Rightarrow A ^{-1}=\frac{-1}{ n }( A + ml )$

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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
$I + A$

Answer

Correct option: B.

$\frac{1}{2}( I - A )$

(B) $A^2-A+2 I=0$
$\Rightarrow A \cdot A - A +2 I =0$
$\Rightarrow A ^{-1} \cdot A \cdot A - A ^{-1} \cdot A+2 A^{-1} \cdot I =0$
$\Rightarrow A - I +2 A^{-1}=0$
$\Rightarrow 2 A^{-1}= I - A$
$\Rightarrow A ^{-1}=\frac{1}{2}( I - A )$

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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
3B
✓
$\frac{1}{3} B$
C
$3 B^{-1}$
D
$\frac{1}{3} B^{-1}$

Answer

Correct 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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MCQ 512 Marks

If $A =\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]$ and $A ^2= I$, then $A ^{-1}$ is equal to

✓
$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
B
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
C
$\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
D
$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

Answer

Correct option: A.

$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

(A) $A ^2=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}x^2+1 & x \\ x & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow x=0$
$\therefore A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$A ^2= I$
$\therefore \quad A^{-1} A \cdot A=A^{-1} I$
$\Rightarrow I \cdot A = A ^{-1} I$
$\Rightarrow A ^{-1}= A$

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MCQ 522 Marks

If A is solution of $x^2-3 x-7=0$ and $A=\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]$, then $A^{-1}$ equals

✓
$\left[\begin{array}{cc}\frac{2}{7} & \frac{3}{7} \\ \frac{-1}{7} & \frac{-5}{7}\end{array}\right]$
B
$\left[\begin{array}{cc}2 & 3 \\ -1 & -5\end{array}\right]$
C
$\left[\begin{array}{cc}\frac{1}{7} & \frac{1}{7} \\ \frac{-1}{7} & \frac{-5}{7}\end{array}\right]$
D
$\left[\begin{array}{cc}3 & -1 \\ 1 & 2\end{array}\right]$

Answer

Correct option: A.

$\left[\begin{array}{cc}\frac{2}{7} & \frac{3}{7} \\ \frac{-1}{7} & \frac{-5}{7}\end{array}\right]$

(A) $A^2-3 A-7 I=0$
$\Rightarrow A -3 I -7 A^{-1}=0 \Rightarrow A^{-1}=\frac{1}{7}(A-3 I )$
$\therefore \quad A ^{-1}=\frac{1}{7}\left\{\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]\right\}$
$=\left[\begin{array}{cc}\frac{2}{7} & \frac{3}{7} \\ -\frac{1}{7} & -\frac{5}{7}\end{array}\right]$

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MCQ 532 Marks

If A and B are two square matrices such that $B=-A^{-1} B A$, then $(A+B)^2=$

A
$0$
✓
$A ^2+ B ^2$
C
$A^2+2 A B+B^2$
D
$A+B$

Answer

Correct option: B.

$A ^2+ B ^2$

(B) Given, $B =- A ^{-1} BA$
$\therefore \quad AB =- AA ^{-1} BA$
$\Rightarrow AB =- I ( BA ) \Rightarrow AB =- BA$
Now $(A+B)^2=(A+B)(A+B)$
$= A ^2+ AB + BA + B ^2$
$= A ^2+ B ^2 \quad[\because BA =- AB ]$

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MCQ 542 Marks

If $A ^2- A + I =0$, then $A ^{-1}=$

A
$A^{-2}$
B
$A + I$
✓
$I - A$
D
$A - I$

Answer

Correct option: C.

$I - A$

(C) $A^2-A+I=0$
$\Rightarrow A \cdot A - A + I =0$
$\Rightarrow A ^{-1} \cdot A \cdot A - A ^{-1} \cdot A+ A ^{-1} \cdot I =0$
$\Rightarrow A - I + A ^{-1}=0$
$\Rightarrow A ^{-1}= I - A$

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MCQ 552 Marks

If for the matrix $A, A^3=I$, then $A^{-1}=$

✓
$A ^2$
B
$A ^3$
C
A
D
none of these

Answer

Correct option: A.

$A ^2$

(A) $A ^3= I$
$\Rightarrow A ^{-1} A^3= A ^{-1} \cdot I$
$\Rightarrow\left( A ^{-1} A\right) A ^2= A ^{-1}$
$\Rightarrow IA ^2= A ^{-1} \Rightarrow A^2= A ^{-1}$

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MCQ 562 Marks

Let for any matrix $M , M ^{-1}$ exist, then which of the following is not true?

A
$\left( M ^{-1}\right)=( M )^{-1}$
B
$\left(M^2\right)^{-1}=\left(M^{-1}\right)^2$
✓
$\left( M ^{-1}\right)^{-1}=\left( M ^{-1}\right)^1$
D
$\left( M ^{-1}\right)^{-1}= M$

Answer

Correct option: C.

$\left( M ^{-1}\right)^{-1}=\left( M ^{-1}\right)^1$

(C) $\left(M^{-1}\right)^{-1} \neq\left(M^{-1}\right)^1$
$\therefore \quad\left( M ^{-1}\right)^{-1}=\left( M ^{-1}\right)^1$ is not true

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MCQ 572 Marks

If $A=\left[\begin{array}{cc}x & -2 \\ 3 & 7\end{array}\right]$ and $A^{-1}=\left[\begin{array}{cc}\frac{7}{34} & \frac{1}{17} \\ \frac{-3}{34} & \frac{2}{17}\end{array}\right]$, then the value of $x$ is

A
2
B
3
C
-4
✓
4

Answer

Correct option: D.

(D) Since $AA ^{-1}= I$,
$\left[\begin{array}{cc}x & -2 \\ 3 & 7\end{array}\right]\left[\begin{array}{cc}\frac{7}{34} & \frac{1}{17} \\ \frac{-3}{34} & \frac{2}{17}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}\frac{7 x+6}{34} & \frac{x-4}{17} \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
By equality of matrices,
$\frac{x-4}{17}=0 \Rightarrow x-4=0$
$\Rightarrow x=4$

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MCQ 582 Marks

Let $A =\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$ and $10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$. If B is the inverse of matrix A , then $\alpha$ is

A
$0$
B
1
C
2
✓
5

Answer

Correct option: D.

(D) $10 A^{-1}=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right] \quad \ldots\left[\because B = A ^{-1}\right]$
$\Rightarrow 10 A^{-1} A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right] A$
$\Rightarrow 10 I =\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10\end{array}\right]=\left[\begin{array}{ccc}10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10\end{array}\right]$
$\therefore \quad-5+\alpha=0 \Rightarrow \alpha=5$

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MCQ 592 Marks

If the inverse of the matrix $A =\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ is $\frac{1}{5}\left[\begin{array}{ccc}-3 & 2 & 2 \\ 2 & -3 & \alpha \\ 2 & 2 & -3\end{array}\right]$, then $\alpha=$ __________

A
3
B
4
✓
2
D
-2

Answer

Correct option: C.

(C) $AA ^{-1}= I$
$\therefore \quad \frac{1}{5}\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{ccc}-3 & 2 & 2 \\ 2 & -3 & \alpha \\ 2 & 2 & -3\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}1 & 0 & \frac{2 \alpha-4}{5} \\ 0 & 1 & \frac{\alpha-2}{5} \\ 0 & 0 & \frac{1+2 \alpha}{5}\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
∴ By equality of matrices,
$\frac{\alpha-2}{5}=0 \Rightarrow \alpha=2$

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MCQ 602 Marks

From the matrix equation $AB = AC$ we can conclude that $B = C$, provided

A
A is singular.
✓
A is non-singular
C
A is symmetric.
D
A is square

Answer

Correct option: B.

A is non-singular

(B) $AB = AC$
$\Rightarrow A ^{-1} AB = A ^{-1} AC$
$\Rightarrow IB = IC$
$\Rightarrow B = C$
$\therefore \quad$ For $B = C , A ^{-1}$ must exist
$\Rightarrow A$ is non-singular

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MCQ 612 Marks

$A =\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right] ; I =\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$A^{-1}=\frac{1}{6}\left(A^2+c A+d I\right)$ where $c, d \in R$, then pair of values ( $c, d$ ) is

A
$(6,11)$
B
$(6,-11)$
✓
$(-6,11)$
D
$(-6,-11)$

Answer

Correct option: C.

$(-6,11)$

(C) $|A|=\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right|=6 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1\end{array}\right]$
$\therefore \quad A ^{-1}=\frac{1}{6}\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1\end{array}\right]$
$A^2=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right]$
$=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14\end{array}\right]$
$\therefore \quad A ^2+ cA + dI$
$=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14\end{array}\right]+\left[\begin{array}{ccc} c & 0 & 0 \\ 0 & c & c \\ 0 & -2 c & 4 c \end{array}\right]+\left[\begin{array}{ccc} d & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & d\end{array}\right]$
$=\left[\begin{array}{ccc}1+ c + d & 0 & 0 \\ 0 & -1+ c + d & 5+ c \\ 0 & -10-2 c & 14+4 c + d \end{array}\right]$
Since $6 A^{-1}=A^2+c A+d I$
$\therefore\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1\end{array}\right]=\left[\begin{array}{ccc}1+ c + d & 0 & 0 \\ 0 & -1+ c + d & 5+ c \\ 0 & -10-2 c & 14+4 c + d \end{array}\right]$
∴ by equality of matrices,
$1+c+d=6$ and $5+c=-1$
$\therefore \quad c=-6$ and $d=11$

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MCQ 622 Marks

If the inverse of product of the matrix
$B=\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$ with a matrix $A$ is
$C=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]$, then $A^{-1}$ equals

A
$\left[\begin{array}{ccc}-3 & -5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 6\end{array}\right]$
B
$\left[\begin{array}{ccc}-3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16\end{array}\right]$
✓
$\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$
D
$\left[\begin{array}{ccc}-3 & -3 & 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$

Answer

Correct option: C.

$\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$

(C) $( BA )^{-1}= C$
$\Rightarrow A ^{-1} B^{-1}= C \Rightarrow A ^{-1}= CB$
$\therefore \quad A^{-1}=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$
$=\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$

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MCQ 632 Marks

If $\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right] A \left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then matrix A equals

✓
$\left[\begin{array}{cc}7 & 5 \\ -11 & -8\end{array}\right]$
B
$\left[\begin{array}{ll}2 & 1 \\ 5 & 3\end{array}\right]$
C
$\left[\begin{array}{cc}7 & 1 \\ 34 & 5\end{array}\right]$
D
$\left[\begin{array}{cc}5 & 3 \\ 13 & 8\end{array}\right]$

Answer

Correct option: A.

$\left[\begin{array}{cc}7 & 5 \\ -11 & -8\end{array}\right]$

(A) If $XAY = I$, then $A = X ^{-1} Y ^{-1}=( YX )^{-1}$
Here, $YX =\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$
$=\left[\begin{array}{cc}8 & 5 \\ -11 & -7\end{array}\right]$
$\therefore \quad A=\left[\begin{array}{cc}8 & 5 \\ -11 & -7\end{array}\right]^{-1}$
$=\left[\begin{array}{cc}7 & 5 \\ -11 & -8\end{array}\right]$

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MCQ 642 Marks

The matrix A satisfying $A \left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}3 & -1 \\ 6 & 0\end{array}\right]$ is

A
$\left[\begin{array}{cc}3 & 2 \\ 6 & -3\end{array}\right]$
✓
$\left[\begin{array}{ll}3 & -16 \\ 6 & -30\end{array}\right]$
C
$\left[\begin{array}{cc}3 & -16 \\ 6 & 30\end{array}\right]$
D
$\left[\begin{array}{cc}3 & -3 \\ 6 & 2\end{array}\right]$

Answer

Correct option: B.

$\left[\begin{array}{ll}3 & -16 \\ 6 & -30\end{array}\right]$

(B) If $AC = B$, then $A = BC ^{-1}$
$\therefore \quad A=\left[\begin{array}{cc}3 & -1 \\ 6 & 0\end{array}\right]\left[\begin{array}{cc}1 & 5 \\ 0 & 1\end{array}\right]^{-1}$
$=\left[\begin{array}{cc}3 & -1 \\ 6 & 0\end{array}\right]\left[\begin{array}{cc}1 & -5 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{ll}3 & -16 \\ 6 & -30\end{array}\right]$

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MCQ 652 Marks

The element of second row and third column in the inverse of $\left[\begin{array}{ccc}1 & 2 & 1 \\ 2 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$ is

A
$-2$
✓
$-1$
C
$1$
D
$2$

Answer

Correct option: B.

$-1$

(B) Let $A=\left[\begin{array}{ccc}1 & 2 & 1 \\ 2 & 1 & 0 \\ -1 & 0 & 1\end{array}\right] \Rightarrow|A|=-2 \neq 0$
Now, co-factor of element $a _{32}$ of $A = A _{32}$
$\therefore \quad A _{32}=(-1)^{3+2}\left|\begin{array}{ll}1 & 1 \\ 2 & 0\end{array}\right|=2$
$\therefore \quad$ Element $a _{23}$ of $A ^{-1}=\frac{ A _{32}}{|A|}=\frac{2}{-2}=-1$
Alternate method:
$|A|=-2 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}1 & -2 & -1 \\ -2 & 2 & 2 \\ 1 & -2 & -3\end{array}\right]$
$\therefore \quad A^{-1}=\left[\begin{array}{ccc}-\frac{1}{2} & 1 & \frac{1}{2} \\ 1 & -1 & -1 \\ -\frac{1}{2} & 1 & \frac{3}{2}\end{array}\right]$
$\therefore \quad$ Element $a _{23}$ of $A ^{-1}=-1$.

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MCQ 662 Marks

The inverse of matrix $A=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$ is

✓
A
B
$A ^{ T }$
C
$\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
D
$\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]$

Answer

Correct option: A.

(A) $|A|=\left|\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right|=-1 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1\end{array}\right]$
$\therefore \quad A ^{-1}=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]= A$

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MCQ 672 Marks

If $P=\left[\begin{array}{lll}1 & 2 & 3 \\ 3 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], Q=\left[\begin{array}{ccc}1 & -2 & -3 \\ -3 & 1 & 9 \\ 0 & 0 & -5\end{array}\right]$ then $( PQ )^{-1}$ equals to

A
zero matrix
B
$I _3$
C
$\operatorname{diag}[-5,-5,-5]$
✓
$-\frac{1}{5} I _3$

Answer

Correct option: D.

$-\frac{1}{5} I _3$

(D) Since $PQ =-5 I _3$,
$(P Q)^{-1}=-\frac{1}{5} I_3$

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MCQ 682 Marks

If $A =\left[\begin{array}{ccc}\frac{ k }{2} & 0 & 0 \\ 0 & \frac{l}{3} & 0 \\ 0 & 0 & \frac{m}{4}\end{array}\right]$ and $A ^{-1}=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{array}\right]$ then $k +l+ m =$

A
1
B
9
C
14
✓
29

Answer

Correct option: D.

(D) Using Shortcut 1,
$A ^{-1}=\left[\begin{array}{ccc}\frac{2}{ k } & 0 & 0 \\ 0 & \frac{3}{l} & 0 \\ 0 & 0 & \frac{4}{m}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{array}\right]$
$\Rightarrow \frac{2}{k}=\frac{1}{2} \Rightarrow k=4$,
$\frac{3}{l}=\frac{1}{3} \Rightarrow l=9$ and
$\frac{4}{m}=\frac{1}{4} \Rightarrow m=16$
$\therefore \quad k +l+ m =4+9+16=29$

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MCQ 692 Marks

The inverse of the matrix $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$ is

A
$\frac{1}{24}\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$
B
$\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$
C
$\frac{1}{24}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
✓
$\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{array}\right]$

Answer

Correct option: D.

$\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{array}\right]$

(D) The inverse of the given diagonal matrix is
$A ^{-1}=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{array}\right]$
...[Using Shortcut 1]

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MCQ 702 Marks

If $A=\left[\begin{array}{ccc}1 & -4 & -1 \\ 6 & 3 & 0 \\ 2 & 0 & 0\end{array}\right]$, then $6\left|A^{-1}\right|$ is equal to

✓
1
B
2
C
3
D
4

Answer

Correct option: A.

(A) $A=\left[\begin{array}{ccc}1 & -4 & -1 \\ 6 & 3 & 0 \\ 2 & 0 & 0\end{array}\right]$
$|A|=6 \neq 0$
$\therefore \quad\left| A ^{-1}\right|=\frac{1}{|A|}$
$=\frac{1}{6}$
$\therefore \quad 6\left|A^{-1}\right|=1$

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MCQ 712 Marks

If $A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$, then the sum of all the diagonal entries of $A ^{-1}$ is

A
2
B
3
C
-3
✓
4

Answer

Correct option: D.

(D) $A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$
$A^{-1}=\left[\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right]$
$\therefore \quad$ sum of all the diagonal entries $=\frac{1}{2}+3+\frac{1}{2}=4$

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MCQ 722 Marks

If $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$, then $A^{-1}=$

A
A
B
$A ^2$
✓
$A ^3$
D
$A^4$

Answer

Correct option: C.

$A ^3$

(C) $|A|=\left|\begin{array}{ccc}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right|=1 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
$\therefore \quad A^{-1}=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
$A^2=\left[\begin{array}{ccc}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3\end{array}\right]$
$A^3=A^2 \cdot A=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
$= A ^{-1}$

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MCQ 732 Marks

If $A=\left[\begin{array}{ccc}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1\end{array}\right]$, then $A^{-1}$ is

✓
$A ^{ T }$
B
$A ^2$
C
A
D
I

Answer

Correct option: A.

$A ^{ T }$

(A) $|A|=\left|\begin{array}{ccc}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1\end{array}\right|=-1 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\therefore \quad A ^{-1}=\frac{1}{|A|}(\operatorname{adj} A )$
$=\left[\begin{array}{ccc}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1\end{array}\right]= A ^{ T }$

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MCQ 742 Marks

The inverse of the matrix $A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$ is

A
$\left[\begin{array}{ccc}\frac{3}{2} & \frac{6}{2} & \frac{-5}{2} \\ \frac{-15}{2} & \frac{-1}{2} & \frac{1}{2} \\ 5 & -1 & 1\end{array}\right]$
B
$\left[\begin{array}{ccc}3 & 6 & 2 \\ -15 & -1 & 1 \\ 5 & -2 & -5\end{array}\right]$
✓
$\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
D
$\left[\begin{array}{ccc}\frac{3}{2} & \frac{-1}{2} & \frac{1}{2} \\ \frac{-15}{2} & \frac{6}{2} & \frac{-5}{2} \\ 5 & -1 & 1\end{array}\right]$

Answer

Correct option: C.

$\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$

(C) $|A|=\left|\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right|=1 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
$\therefore \quad A ^{-1}=\frac{1}{|A|}(\operatorname{adj} A )$
$=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$

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MCQ 752 Marks

If $A=\left[a_{i j}\right]_{2 \times 2}$, where $a_{i j}=\left\{\begin{array}{cc}i+j, & \text { if } i \neq j \\ i^2-2 j, & \text { if } i=j\end{array}\right.$ then $A ^{-1}$ is equal to

A
$\frac{1}{9}\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right]$
B
$\frac{1}{9}\left[\begin{array}{cc}0 & -3 \\ -3 & -1\end{array}\right]$
✓
$\frac{1}{9}\left[\begin{array}{ll}0 & 3 \\ 3 & 1\end{array}\right]$
D
None of these

Answer

Correct option: C.

$\frac{1}{9}\left[\begin{array}{ll}0 & 3 \\ 3 & 1\end{array}\right]$

(C) $A=\left[a_{i j}\right]_{2 \times 2} \Rightarrow A=\left[\begin{array}{cc}-1 & 3 \\ 3 & 0\end{array}\right]$
$|A|=-9 \neq 0$
$\therefore \quad A^{-1}=\frac{-1}{9}\left[\begin{array}{cc}0 & -3 \\ -3 & -1\end{array}\right]=\frac{1}{9}\left[\begin{array}{ll}0 & 3 \\ 3 & 1\end{array}\right]$
...[Using Shortcut 2]

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MCQ 762 Marks

Matrix $A=\left[\begin{array}{ccc}1 & 0 & -k \\ 2 & 1 & 3 \\ k & 0 & 1\end{array}\right]$ is invertible for

A
$k=1$
B
$k=-1$
C
$k =0$
✓
All real k

Answer

Correct option: D.

All real k

(D) $| A |= k ^2+1$, which can be never zero.
Hence matrix A is invertible for all real k .

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MCQ 772 Marks

Let A be a square matrix of order 3 whose all entries are 1 and let $I_3$ be the identity matrix of order 3 . Then the matrix $A -3 I _3$ is

A
invertible
B
orthogonal
✓
non-invertible
D
real skew symmetric matrix

Answer

Correct option: C.

non-invertible

(C) $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right], I_3=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$A-3 I_3=\left[\begin{array}{ccc}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{array}\right]$
$\left|A-3 I_3\right|=\left|\begin{array}{ccc}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{array}\right|$
$=0$
$\therefore \quad$ the matrix $A -3 I _3$ is non-invertible.

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MCQ 782 Marks

The matrix $\left[\begin{array}{ccc}\lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2\end{array}\right]$ is invertible, if

A
$\lambda \neq-15$
✓
$\lambda \neq-17$
C
$\lambda \neq-16$
D
$\lambda \neq-18$

Answer

Correct option: B.

$\lambda \neq-17$

(B) The given matrix will be invertible, if
$\left|\begin{array}{ccc}\lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2\end{array}\right| \neq 0$
$\begin{array}{l}\Rightarrow \lambda(0-1)+1(-6+1)+4(-3) \neq 0 \\ \Rightarrow-\lambda-5-12 \neq 0 \\ \Rightarrow \lambda \neq-17\end{array}$

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MCQ 792 Marks

The matrix $\left[\begin{array}{lll}1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1\end{array}\right]$ is not invertible, if ' $a$ ' has the value

A
2
✓
1
C
$0$
D
$-1$

Answer

Correct option: B.

(B) The matrix is not invertible if $\left|\begin{array}{lll}1 & \text { a } & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1\end{array}\right|=0$
$\Rightarrow 1(2-5)-a(1-10)+2(1-4)=0$
$\Rightarrow-3+9 a-6=0$
$\Rightarrow a=1$

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MCQ 802 Marks

If A is a square matrix such that $A(\operatorname{adj} A)=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$, then $\frac{|\operatorname{adj}(\operatorname{adj} A)|}{|\operatorname{adj} A|}$ is equal to

A
256
✓
16
C
32
D
64

Answer

Correct option: B.

(B) A. $(\operatorname{adj} A)=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$ . . .(i)
$=4\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$=4 . I$
Using Shortcut 4(i),
$A (\operatorname{adj} A )=| A | . I$,
$|A|=4$
From (i), $| A | \cdot|\operatorname{adj} A |=64$
$\Rightarrow|\operatorname{adj} A|=\frac{64}{4}=16$
Using Shortcut 4(xiv),
$|\operatorname{adj}(\operatorname{adj} A )|=| A |^{( n -1)^2}$
$=| A |^{(3-1)^2}$
$=(4)^4=256$
$\therefore \frac{|\operatorname{adj}(\operatorname{adj} A)|}{|\operatorname{adj} A|}=\frac{256}{16}=16$

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MCQ 812 Marks

Let A be a $2 \times 2$ matrix
Statement-1: $\operatorname{adj}(\operatorname{adj} A )= A$
Statement-2 : $|\operatorname{adj} A |=| A |$

A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
✓
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
C
Statement- 1 is true, Statement- 2 is false
D
Statement-1 is false, Statement-2 is true

Answer

Correct option: B.

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

(B) $|\operatorname{adj} A |=| A |^{ n -1}=| A |^{2-1}=| A |$
…[Using Shortcut 4(viii)]
$\operatorname{adj}(\operatorname{adj} A )=| A |^{ n -2} A=| A |^0 A= A$
...[Using Shortcut 4(xiii)]
$\therefore$ option (B) is the correct answer.

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MCQ 822 Marks

If A is a matrix of order 3, such that $A(\operatorname{adj} A)=10 I$, then $|\operatorname{adj} A |=$

A
1
B
10
✓
100
D
10 I

Answer

Correct option: C.

100

(C) Using Shortcut 4(i),
$A (\operatorname{Adj} A )=| A | I$
$\therefore|A|=10$
Using Shortcut 4(viii),
$\begin{aligned}|\operatorname{Adj} A | & =| A |^{ n -1} \\ \therefore \quad|\operatorname{Adj} A| & =| A |^{3-1}=| A |^2=10^2=100\end{aligned}$

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MCQ 832 Marks

If $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, then $\operatorname{adj}(\operatorname{adj} A)$ is equal to

A
$\operatorname{adj} A$
✓
A
C
$A ^{ T }$
D
$-A$

Answer

Correct option: B.

(B) $\operatorname{adj} A=\left[\begin{array}{cc} d & - b \\ - c & a \end{array}\right]$
$\therefore \quad \operatorname{adj}(\operatorname{adj} A )=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]= A$

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MCQ 842 Marks

If $A$ is a singular matrix of order $n$, then $A \cdot(\operatorname{adj} A )$ is

✓
zero matrix
B
row matrix
C
unit matrix
D
column matrix

Answer

Correct option: A.

zero matrix

(A) A is a singular matrix.
$\therefore|A|=0$ and $A \cdot(\operatorname{adj} A)=|A| . I=0 . I=0$
$\Rightarrow A (\operatorname{adj} A )$ is a zero matrix.

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MCQ 852 Marks

If $A=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$, then $A(\operatorname{adj} A)=$

✓
$\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$
B
$\left[\begin{array}{cc}0 & 10 \\ 10 & 0\end{array}\right]$
C
$\left[\begin{array}{cc}10 & 1 \\ 1 & 10\end{array}\right]$
D
None of these

Answer

Correct option: A.

$\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$

(A) Using Shortcut 4(i),
$A (\operatorname{adj} A )=| A | \cdot I _{ n }$
$\therefore \quad A (\operatorname{adj} A )=\left|\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right|\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$

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MCQ 862 Marks

If the adjoint of a $3 \times 3$ matrix $P$ is $\left[\begin{array}{lll}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{array}\right]$, then the possible values of the determinant of $P$ are

✓
$\pm 2$
B
$\pm 1$
C
$\pm 3$
D
$\pm 4$

Answer

Correct option: A.

$\pm 2$

(A) $\operatorname{adj} P=\left[\begin{array}{lll}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{array}\right]$
Using Shortcut 4(viii),
$|\operatorname{adj} P |=| P |^2$
$\Rightarrow\left|\begin{array}{lll}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{array}\right|=| P |^2$
$\begin{array}{l}\Rightarrow|P|^2=1(-4)-4(-1)+4(1) \\ \Rightarrow|P|^2=4 \Rightarrow|P|= \pm 2\end{array}$

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MCQ 872 Marks

If $A$ is a singular matrix, then $\operatorname{adj} A$ is

✓
singular
B
non-singular
C
symmetric
D
not defined

Answer

Correct option: A.

singular

(A) Given, A is a singular matrix.
$\therefore \quad| A |=0$
Using Shortcut 4(viii),
$|\operatorname{adj} A |=| A |^{ n -1}$
$\Rightarrow|\operatorname{adj} A|=0$
$\Rightarrow \operatorname{adj} A$ is also singular.

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MCQ 882 Marks

If X is a square matrix of order $3 \times 3$ and $\lambda$ is a scalar, then adj $(\lambda X)$ is equal to

A
$\lambda \operatorname{adj} X$
B
$\lambda^3 \operatorname{adj} X$
✓
$\lambda^2 \operatorname{adj} X$
D
$\lambda^4 \operatorname{adj} X$

Answer

Correct option: C.

$\lambda^2 \operatorname{adj} X$

(C) $\operatorname{adj}(\lambda X )=\lambda^{3-1}(\operatorname{adj} X )$
…[Using Shortcut 4(vi)]
$=\lambda^2 \operatorname{adj} X$

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MCQ 892 Marks

If the value of a third order determinant is 16, then the value of the determinant formed by replacing each of its elements by its cofactor is

A
48
B
16
C
96
✓
256

Answer

Correct option: D.

256

(D) If A is a square matrix of order 3 , then
$|\operatorname{adj} A |=| A |^2$
...[Using Shortcut 4(x)]
$\begin{array}{l}=(16)^2 \\ =256\end{array}$

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MCQ 902 Marks

If $M$ is any square matrix of order 3 over $R$ and if $M^{\prime}$ be the transpose of $M$, then $\operatorname{adj}\left( M ^{\prime}\right)-(\operatorname{adj} M )^{\prime}$ is equal to

A
M
B
$M ^{\prime}$
✓
null matrix
D
identity matrix

Answer

Correct option: C.

null matrix

(C) $\operatorname{adj}\left(M^{\prime}\right)-(\operatorname{adj} M)^{\prime}$ is a null matrix as $\operatorname{adj}\left( M ^{\prime}\right)=\left(\operatorname{adj} M )^{\prime}\right.$

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MCQ 912 Marks

If $A =\left[\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right], B =\left[\begin{array}{ll}6 & 7 \\ 8 & 9\end{array}\right]$, then $\operatorname{adj}( AB )$ is equal to

✓
$\left[\begin{array}{cc}94 & -39 \\ -82 & 34\end{array}\right]$
B
$\left[\begin{array}{ll}94 & -39 \\ 82 & -34\end{array}\right]$
C
$\left[\begin{array}{cc}94 & -82 \\ -39 & 34\end{array}\right]$
D
$\left[\begin{array}{cc}-94 & -39 \\ 82 & 34\end{array}\right]$

Answer

Correct option: A.

$\left[\begin{array}{cc}94 & -39 \\ -82 & 34\end{array}\right]$

(A) $AB =\left[\begin{array}{ll}34 & 39 \\ 82 & 94\end{array}\right] \Rightarrow \operatorname{adj}( AB )=\left[\begin{array}{cc}94 & -39 \\ -82 & 34\end{array}\right]$

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MCQ 922 Marks

Adjoint of the matrix $N=\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]$ is

✓
N
B
2N
C
-N
D
$-2 N$

Answer

Correct option: A.

(A) Matrix of co-factors,
$\left[ A _{ ij }\right]_{3 \times 3}=\left[\begin{array}{ccc}-4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & 3\end{array}\right]$
$\therefore \quad$ adj $N=\left[A_{i j}\right]_{3 \times 3}^T=\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]=N$

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MCQ 932 Marks

If $A=\left[\begin{array}{lll}1 & 2 & 1 \\ 3 & 2 & 3 \\ 2 & 1 & 2\end{array}\right]$, then $a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}=$

A
1
B
2
✓
$|A|$
D
3

Answer

Correct option: C.

$|A|$

(C) $a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}$
$=1(4-3)+3[-(4-1)]+2(6-2)=0$
and $|A|=1(4-3)-2(6-6)+1(3-4)=0$
$\therefore \quad a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}=|A|$

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MCQ 942 Marks

If $A=\left[\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$, then $C_2+2 C_1$ and then $R_1+R_3$ gives

A
$\left[\begin{array}{lll}2 & 1 & 3 \\ 2 & 6 & 3 \\ 3 & 4 & 2\end{array}\right]$
B
$\left[\begin{array}{ccc}2 & -1 & 3 \\ 2 & 4 & 3 \\ 3 & 1 & 2\end{array}\right]$
C
$\left[\begin{array}{lll}5 & 5 & 5 \\ 3 & 4 & 2 \\ 2 & 6 & 3\end{array}\right]$
✓
$\left[\begin{array}{lll}5 & 5 & 5 \\ 2 & 6 & 3 \\ 3 & 4 & 2\end{array}\right]$

Answer

Correct option: D.

$\left[\begin{array}{lll}5 & 5 & 5 \\ 2 & 6 & 3 \\ 3 & 4 & 2\end{array}\right]$

(D) $A=\left[\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
Applying $C _2 \rightarrow C _2+2 C _1$,
$A \sim\left[\begin{array}{lll}2 & 1 & 3 \\ 2 & 6 & 3 \\ 3 & 4 & 2\end{array}\right]$
Applying $R _1 \rightarrow R _1+ R _3$,
$A \sim\left[\begin{array}{lll}5 & 5 & 5 \\ 2 & 6 & 3 \\ 3 & 4 & 2\end{array}\right]$

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MCQ 952 Marks

If $\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]= A \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
then $C _2 \rightarrow C _2-3 C _1$ and $C _3 \rightarrow C _3+2 C _1$ gives

A
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & -9 & 11 \\ -2 & 1 & 4\end{array}\right]=A\left[\begin{array}{ccc}-1 & 3 & 2 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$
B
$\left[\begin{array}{ccc}1 & 0 & 0 \\ -3 & -1 & -4 \\ 2 & -9 & 11\end{array}\right]-A\left[\begin{array}{ccc}1 & -3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
C
$\left[\begin{array}{ccc}1 & 0 & 0 \\ -3 & 1 & 4 \\ -2 & 9 & -11\end{array}\right]=A\left[\begin{array}{ccc}-1 & 3 & 2 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$
✓
$\left[\begin{array}{ccc}1 & 0 & 0 \\ -3 & 9 & -11 \\ 2 & -1 & 4\end{array}\right]=A\left[\begin{array}{ccc}1 & -3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

Answer

Correct option: D.

$\left[\begin{array}{ccc}1 & 0 & 0 \\ -3 & 9 & -11 \\ 2 & -1 & 4\end{array}\right]=A\left[\begin{array}{ccc}1 & -3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

(D) $\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]= A \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Applying $C _2 \rightarrow C _2-3 C _1$ and $C _3 \rightarrow C _3+2 C _1$,
$\left[\begin{array}{ccc}1 & 3-3 & -2+2 \\ -3 & 0+9 & -5-6 \\ 2 & 5-6 & 0+4\end{array}\right]= A \left[\begin{array}{lll}1 & 0-3 & 0+2 \\ 0 & 1-0 & 0+0 \\ 0 & 0-0 & 1+0\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ -3 & 9 & -11 \\ 2 & -1 & 4\end{array}\right]= A \left[\begin{array}{ccc}1 & -3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

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MCQ 962 Marks

If $3 x-4 y+2 z=-1,2 x+3 y+5 z=7$, $x+z=2$, then $x=$

✓
3
B
2
C
1
D
-1

Answer

Correct option: A.

(A) $A X=B$
$\therefore \quad\left[\begin{array}{ccc}3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-1 \\ 7 \\ 2\end{array}\right]$
$R _2-5 R _3 \Rightarrow\left[\begin{array}{ccc}3 & -4 & 2 \\ -3 & 3 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-1 \\ -3 \\ 2\end{array}\right]$
$R _1-2 R _3 \Rightarrow\left[\begin{array}{ccc}1 & -4 & 0 \\ -3 & 3 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-5 \\ -3 \\ 2\end{array}\right]$
$\Rightarrow x-4 y=-5$, and …(i)
$-3 x+3 y=-3$ …(ii)
Solving (i) and (ii), we get $x=3$

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MCQ 972 Marks

If $x+2 y=3$ and $2 x+3 y=4$, then the values of $x$ and $y$ respectively are

A
$1,-2$
B
$-2,1$
✓
$-1,2$
D
$2,-1$

Answer

Correct option: C.

$-1,2$

(C) The given system of equations can be written in matrix form AX = B, where
$A =\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right], X =\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B =\left[\begin{array}{l}3 \\ 4\end{array}\right]$
$\therefore \quad$ Now, $\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}3 \\ 4\end{array}\right]$
Applying $R _2 \rightarrow R _2-2 R _{ 1 }$,
$\left[\begin{array}{cc}1 & 2 \\ 0 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}3 \\ -2\end{array}\right]$
$\Rightarrow x+2 y=3$, and $\ldots$(i)
$- y = - 2$ $\ldots$(ii)
$\Rightarrow y=2$
putting y = 2 in (i), we get
x + 2(2) = 3
$\Rightarrow x=-1$
Alternate method:
$\begin{array}{l}A X=B \Rightarrow X=A^{-1} B \\ |A|=-1 \neq 0 \\ A^{-1}=\frac{1}{-1}\left[\begin{array}{cc}3 & -2 \\ -2 & 1\end{array}\right]\end{array}$
$\begin{aligned} A^{-1} & =\frac{1}{-1}\left[\begin{array}{cc}3 & -2 \\ -2 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}-3 & 2 \\ 2 & -1\end{array}\right] \end{aligned}$
$\begin{aligned} X & =\left[\begin{array}{cc}-3 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{l}3 \\ 4\end{array}\right] \\ & =\left[\begin{array}{c}-1 \\ 2\end{array}\right]\end{aligned}$
$\therefore x=-1, y=2$

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MCQ 982 Marks

If $\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}2 \\ 4\end{array}\right]$, then the values of $x$ and $y$ respectively are

A
$-3,-1$
B
1,3
C
3,1
✓
$-1,3$

Answer

Correct option: D.

$-1,3$

(D) $\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}2 \\ 4\end{array}\right]$
$\begin{array}{l}\Rightarrow x+y=2 \text { and }-x+y=4 \\ \Rightarrow x=-1, y=3\end{array}$

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MCQ 992 Marks

If A and B are non-singular matrices, then

A
$( AB )^{-1}= A ^{-1} B^{-1}$
B
$AB = BA$
C
$( AB )^{\prime}= A ^{\prime} B ^{\prime}$
✓
$( AB )^{-1}= B ^{-1} A^{-1}$

Answer

Correct option: D.

$( AB )^{-1}= B ^{-1} A^{-1}$

(D)

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MCQ 1002 Marks

Let A be an invertible matrix. Which of the following is not true?

A
$\left(A^T\right)^{-1}=\left(A^{-1}\right)^T$
✓
$A ^{-1}=| A |^{-1}$
C
$\left(A^2\right)^{-1}=\left(A^{-1}\right)^2$
D
$\left| A ^{-1}\right|=| A |^{-1}$

Answer

Correct option: B.

$A ^{-1}=| A |^{-1}$

(B) Consider option (B),
$A ^{-1}$ is a matrix and $| A |^{-1}$ is a number.
∴ option (B) is not true.

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MCQ 1012 Marks

If $I _3$ is the identity matrix of order 3 , then $I _3^{-1}$ is

A
$0$
B
$3 I _3$
✓
$I_3$
D
does not exist

Answer

Correct option: C.

$I_3$

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MCQ 1022 Marks

If matrix $A=\left[\begin{array}{ccc}3 & 2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right]$ and $A^{-1}=\frac{1}{k} \operatorname{adj} A$, then k is

A
7
B
$-7$
C
$\frac{1}{7}$
✓
11

Answer

Correct option: D.

(D) Given, $A ^{-1}=\frac{1}{ k } \operatorname{adj} A$
$\therefore \quad k =| A |$
$\therefore \quad| A |=\left|\begin{array}{ccc}3 & 2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1\end{array}\right|$
$=3(2+1)-2(1-0)+4(1-0)$
$=9-2+4$
$=11$
$\Rightarrow k =11$

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MCQ 1032 Marks

$A =\left[\begin{array}{ll}0 & 3 \\ 2 & 0\end{array}\right]$ and $A ^{-1}=\lambda(\operatorname{adj}( A ))$, then $\lambda=$

✓
$\frac{-1}{6}$
B
$\frac{1}{3}$
C
$\frac{-1}{3}$
D
$\frac{1}{6}$

Answer

Correct option: A.

$\frac{-1}{6}$

(A) $| A |=\left|\begin{array}{ll}0 & 3 \\ 2 & 0\end{array}\right|=-6 \neq 0$
$A ^{-1}=\frac{1}{|A|} \operatorname{adjA}$
$=\lambda(\operatorname{adj} A )$ ....[Given]
$\therefore \quad \lambda=\frac{1}{|A|}=-\frac{1}{6}$

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MCQ 1042 Marks

If $A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]$ and $a b c \neq 0$, then $A^{-1}=$

A
$\left[\begin{array}{ccc}a & 0 & 0 \\ 0 & \frac{1}{b} & 0 \\ 0 & 0 & c\end{array}\right]$
B
$\left[\begin{array}{ccc}\frac{1}{0} & 0 & 0 \\ 0 & \frac{1}{b} & 0 \\ 0 & 0 & c\end{array}\right]$
C
$\left[\begin{array}{ccc} a & 0 & 0 \\ 0 & \frac{1}{b} & 0 \\ 0 & 0 & \frac{1}{ c }\end{array}\right]$
✓
$\left[\begin{array}{ccc}\frac{1}{a} & 0 & 0 \\ 0 & \frac{1}{b} & 0 \\ 0 & 0 & \frac{1}{c}\end{array}\right]$

Answer

Correct option: D.

$\left[\begin{array}{ccc}\frac{1}{a} & 0 & 0 \\ 0 & \frac{1}{b} & 0 \\ 0 & 0 & \frac{1}{c}\end{array}\right]$

(D)

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MCQ 1052 Marks

Inverse of the matrix $\left[\begin{array}{ccc}3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1\end{array}\right]$ is

A
$\left[\begin{array}{ccc}1 & 2 & 3 \\ 3 & 3 & 7 \\ -2 & -4 & -5\end{array}\right]$
B
$\left[\begin{array}{ccc}1 & -3 & 5 \\ 7 & 4 & 6 \\ 4 & 2 & 7\end{array}\right]$
✓
$\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\right]$
D
$\left[\begin{array}{ccc}1 & -3 & 5 \\ 7 & 4 & 6 \\ 4 & 2 & -7\end{array}\right]$

Answer

Correct option: C.

$\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\right]$

(C) Let $A=\left[\begin{array}{ccc}3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1\end{array}\right] \Rightarrow|A|=1 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\right]$
$\therefore \quad A^{-1}=\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\right]$

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MCQ 1062 Marks

If $U =\left[\begin{array}{ll}\frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right]$, then $U ^{-1}$ is

✓
$U ^{ T }$
B
U
C
I
D
$0$

Answer

Correct option: A.

$U ^{ T }$

(A) $|U|=\left[\begin{array}{ll}\frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right]=1 \neq 0$
$\therefore \quad U ^{-1}=\left[\begin{array}{ll}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right] \quad \ldots[$ Using Shortcut 2]
$= U ^{ T }$

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MCQ 1072 Marks

The inverse of the matrix $\left[\begin{array}{cc}3 & -2 \\ 1 & 4\end{array}\right]$ is

✓
$\left[\begin{array}{cc}\frac{4}{14} & \frac{2}{14} \\ \frac{-1}{14} & \frac{3}{14}\end{array}\right]$
B
$\left[\begin{array}{cc}\frac{3}{14} & \frac{-2}{14} \\ \frac{1}{14} & \frac{4}{14}\end{array}\right]$
C
$\left[\begin{array}{cc}\frac{4}{14} & \frac{-2}{14} \\ \frac{1}{14} & \frac{3}{14}\end{array}\right]$
D
$\left[\begin{array}{cc}\frac{3}{14} & \frac{2}{14} \\ \frac{1}{14} & \frac{4}{14}\end{array}\right]$

Answer

Correct option: A.

$\left[\begin{array}{cc}\frac{4}{14} & \frac{2}{14} \\ \frac{-1}{14} & \frac{3}{14}\end{array}\right]$

(A) Let $A=\left[\begin{array}{cc}3 & -2 \\ 1 & 4\end{array}\right] \Rightarrow|A|=14 \neq 0$
$\therefore \quad A ^{-1}=\frac{1}{14}\left[\begin{array}{cc}4 & 2 \\ -1 & 3\end{array}\right]=\left[\begin{array}{ll}\frac{4}{14} & \frac{2}{14} \\ \frac{-1}{14} & \frac{3}{14}\end{array}\right]$
...[Using Shortcut 2]

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MCQ 1082 Marks

The multiplicative inverse of matrix $A=\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$ is

A
$\left[\begin{array}{cc}4 & -1 \\ -7 & -2\end{array}\right]$
B
$\left[\begin{array}{cc}-4 & -1 \\ 7 & -2\end{array}\right]$
C
$\left[\begin{array}{cc}4 & -7 \\ 7 & 2\end{array}\right]$
✓
$\left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right]$

Answer

Correct option: D.

$\left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right]$

(D) The multiplicative inverse of $A = A ^{-1}$
$|A|=\left|\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right|=1 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right]$
$\therefore \quad A ^{-1}=\frac{1}{|A|} \cdot \operatorname{adj} A =\left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right]$
Alternate Method:
Using Shortcut 2,
$\begin{aligned} A ^{-1} & =\frac{1}{1}\left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right]\end{aligned}$

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MCQ 1092 Marks

If $A=\left[\begin{array}{ll}2 & 3 \\ 4 & 6\end{array}\right]$, then $A^{-1}=$

A
$\left[\begin{array}{cc}1 & 2 \\ -\frac{3}{2} & 3\end{array}\right]$
B
$\left[\begin{array}{cc}2 & -3 \\ 4 & 6\end{array}\right]$
C
$\left[\begin{array}{ll}-2 & 4 \\ -3 & 6\end{array}\right]$
✓
Does not exist

Answer

Correct option: D.

Does not exist

(D) $|A|=\left|\begin{array}{ll}2 & 3 \\ 4 & 6\end{array}\right|=12-12=0$
$\therefore \quad A ^{-1}$ does not exist.

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MCQ 1102 Marks

A square matrix has inverse, if $|A|$ is

A
$0$
B
1
C
-1
✓
non zero

Answer

Correct option: D.

non zero

(D) If $|A| \neq 0$, then $A^{-1}$ exists
$\therefore|A|$ is non zero

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MCQ 1112 Marks

If $A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]$, then the value of $|A||\operatorname{adj} A|$ is

A
$a ^3$
B
$a^6$
✓
$a^9$
D
$a ^{27}$

Answer

Correct option: C.

$a^9$

(C) $| A |= a ^3$
Using Shortcut 4(viii),
$| A ||\operatorname{adj} A |=| A | \cdot| A |^{3-1}$
$=| A |^3=\left( a ^3\right)^3= a ^9$

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MCQ 1122 Marks

If $A=\left[\begin{array}{ll}4 & 2 \\ 5 & 3\end{array}\right]$, then $|\operatorname{adj}(\operatorname{adj} A)|=$

A
1
✓
2
C
3
D
4

Answer

Correct option: B.

(B) Using Shortcut 4(xiv),
$|\operatorname{adj}(\operatorname{adj} A )|=| A |^{(2-1)^2}$
$=| A |=12-10=2$

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MCQ 1132 Marks

If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is

A
$d ^{ n }$
✓
$d ^{ n -1}$
C
$d^{n+1}$
D
d

Answer

Correct option: B.

$d ^{ n -1}$

(B) $\mid$ Adj $A \left|=| A |^{ n -1}= d ^{ n -1}\right.$
...[Using Shortcut 4(viii)]

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MCQ 1142 Marks

If $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & 4 & 9 \\ 1 & 8 & 27\end{array}\right]$, then $|\operatorname{adj} A|$ is equal to

A
12
✓
144
C
72
D
64

Answer

Correct option: B.

144

(B) A is a $3 \times 3$ matrix.
$\therefore \quad|\operatorname{adj} A |=| A |^2=(12)^2=144$
...[Using Shortcut 4(x)]

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MCQ 1152 Marks

If $A=\left[\begin{array}{ll}4 & 2 \\ 3 & 4\end{array}\right]$, then $|\operatorname{adj} A|$ is equal to

A
16
✓
10
C
6
D
8

Answer

Correct option: B.

(B) A is a $2 \times 2$ matrix.
$\therefore|\operatorname{adj} A |=| A |=10$
...[Using Shortcut 4(ix)]

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MCQ 1162 Marks

Which of the following statements is false?

A
If $|A|=0$, then $|\operatorname{adj} A|=0$
B
Adjoint of a diagonal matrix of order $3 \times 3$ is a diagonal matrix
C
Product of two upper triangular matrices is a upper triangular matrix
✓
$\operatorname{adj}( AB )=\operatorname{adj}( A ) \operatorname{adj}( B )$

Answer

Correct option: D.

$\operatorname{adj}( AB )=\operatorname{adj}( A ) \operatorname{adj}( B )$

(D) $\operatorname{adj}( AB )=\operatorname{adj}( B ) \operatorname{adj}( A )$

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MCQ 1172 Marks

$\operatorname{adj} A B-(\operatorname{adj} B)(\operatorname{adj} A)=$

A
$\operatorname{adj} A-\operatorname{adj} B$
B
I
✓
O
D
none of these

Answer

Correct option: C.

(C) $\operatorname{adj} A B-(\operatorname{adj} B)(\operatorname{adj} A)$
$=(\operatorname{adj} B)(\operatorname{adj} A)-(\operatorname{adj} B)(\operatorname{adj} A)$
...[Using Shortcut 4(xvi)]
$= O$

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MCQ 1182 Marks

If k is a scalar and I is a unit matrix of order 3 , then $\operatorname{adj}( kI )=$

A
$k ^3 I$
✓
$k ^2 I$
C
$-k^3 I$
D
$- k ^2 I$

Answer

Correct option: B.

$k ^2 I$

(B) Let $I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$, then $k I=\left[\begin{array}{lll}k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k\end{array}\right]$
$\Rightarrow \operatorname{adj}( kI )=\left[\begin{array}{ccc} k ^2 & 0 & 0 \\ 0 & k ^2 & 0 \\ 0 & 0 & k ^2\end{array}\right]= k ^2 I$

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MCQ 1192 Marks

If $X =\left[\begin{array}{cc}-x & -y \\ z & t \end{array}\right]$, then transpose of $\operatorname{adj} X$ is

A
$\left[\begin{array}{cc} t & z \\ -y & -x\end{array}\right]$
B
$\left[\begin{array}{cc}t & y \\ -z & -x\end{array}\right]$
✓
$\left[\begin{array}{cc} t & - z \\ y & -x\end{array}\right]$
D
None of these

Answer

Correct option: C.

$\left[\begin{array}{cc} t & - z \\ y & -x\end{array}\right]$

(C) Co-factor matrix of $X =\left[\begin{array}{cc} t & - z \\ y & -x\end{array}\right]$
$\begin{aligned} \therefore \quad \text { Transpose of adj } X & =\text { co-factor matrix of } X \\ & =\left[\begin{array}{ll} t & - z \\ y & -x\end{array}\right]\end{aligned}$

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MCQ 1202 Marks

If $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$, then $\operatorname{adj} A$ is

✓
$\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$
B
$\left[\begin{array}{ccc}0 & 0 & -3 \\ 0 & -1 & 3 \\ 3 & -2 & -9\end{array}\right]$
C
$\left[\begin{array}{ccc}3 & 0 & 0 \\ -3 & 1 & 0 \\ 9 & 2 & -3\end{array}\right]$
D
$\left[\begin{array}{ccc}0 & 0 & 3 \\ 0 & 1 & -3 \\ -3 & 2 & 9\end{array}\right]$

Answer

Correct option: A.

$\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$

(A) Matrix of co-factors is:
$\left[ A _{i j}\right]_{3 \times 3}=\left[\begin{array}{lll} A _{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right]=\left[\begin{array}{ccc}-3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3\end{array}\right]$
$\therefore \quad$ adj $A=\left[A_{i j}\right]_{3 \times 3}^T=\left[\begin{array}{ccc}-3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3\end{array}\right]^T$
$\therefore \quad$ adj $A=\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$

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MCQ 1212 Marks

If $A=\left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]$, then adj $A=$

A
$\left[\begin{array}{cc}2 & 3 \\ -5 & -1\end{array}\right]$
B
$\left[\begin{array}{ll}1 & -5 \\ 3 & -2\end{array}\right]$
✓
$\left[\begin{array}{ll}2 & -5 \\ 3 & -1\end{array}\right]$
D
$\left[\begin{array}{ll}-2 & 5 \\ -3 & 1\end{array}\right]$

Answer

Correct option: C.

$\left[\begin{array}{ll}2 & -5 \\ 3 & -1\end{array}\right]$

(C) Matrix of co-factors $=\left[ A _{ ij }\right]_{2 \times 2}$
$=\left[\begin{array}{ll} A _{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]$
$=\left[\begin{array}{cc}2 & -(-3) \\ -5 & -1\end{array}\right]$
$=\left[\begin{array}{cc}2 & 3 \\ -5 & -1\end{array}\right]$
$\therefore \quad \operatorname{adj} A=\left[A_{i j}\right]_{2 \times 2}^T=\left[\begin{array}{cc}2 & 3 \\ -5 & -1\end{array}\right]^T=\left[\begin{array}{cc}2 & -5 \\ 3 & -1\end{array}\right]$

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MCQ 1222 Marks

If $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 1 & -3 \\ -1 & 2 & 3\end{array}\right]$, then co-factors of $3^{\text {rd }}$ row are

A
$4,-5,1$
✓
$-4,5,-1$
C
$-4,5,1$
D
$4,5,-1$

Answer

Correct option: B.

$-4,5,-1$

(B) $A _{31}=(-1)^{3+1}\left|\begin{array}{cc}1 & 1 \\ 1 & -3\end{array}\right|=-3-1=-4$
$A_{32}=-(-3-2)=-(-5)=5$
$A _{33}-1-2--1$
∴ Co-factors are $-4,5,-1$

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MCQ 1232 Marks

If $A=\left[\begin{array}{ccc}2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1\end{array}\right]$, then co-factor $A_{32}$ is

A
$-2$
B
$-8$
C
4
✓
2

Answer

Correct option: D.

(D) $A_{32}=(-1)^{3+2} \cdot M_{32}=(-1)^5\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=2$

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MCQ 1242 Marks

If $B=\left[\begin{array}{cc}2 & 3 \\ -4 & 3\end{array}\right]$, then co-factor $A_{21}=$

A
2
✓
$-3$
C
$-4$
D
3

Answer

Correct option: B.

$-3$

(B) $A _{21}=(-1)^3 M _{21}=-(3)=-3$

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MCQ 1252 Marks

The co-factor $A_{12}$ for the matrix $A=\left[\begin{array}{ll}-2 & 3 \\ -3 & 5\end{array}\right]$ is

✓
3
B
-3
C
2
D
-5

Answer

Correct option: A.

(A) $A _{12}=(-1)^{1+2} M _{12}=(-1)^3(-3)=3$

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MCQ 1262 Marks

If $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 1 & -3 \\ -1 & 2 & 3\end{array}\right]$, then $M_{23}=$

✓
3
B
4
C
5
D
$-1$

Answer

Correct option: A.

(A) $M _{23}=\left|\begin{array}{cc}1 & 1 \\ -1 & 2\end{array}\right|=3$

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MCQ 1272 Marks

If $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 4 & 2 \\ 0 & 0 & 6\end{array}\right]$, then the minor of the element $a_{31}$ is

A
$0$
✓
$-8$
C
4
D
$-12$

Answer

Correct option: B.

$-8$

(B) $M _{31}=\left|\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right|$$\ldots$ [By leaving $R _3$ and $C _1$ ]
$=-8$

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MCQ 1282 Marks

If $A=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$, then $M_{21}=$

A
1
✓
-1
C
2
D
3

Answer

Correct option: B.

-1

(B) The minor of element $a_{21}=M_{21}=-1$
$\ldots\left[\right.$ By leaving $R _2$ and $\left.C _1\right]$

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MCQ 1292 Marks

The upper triangular matrix of the matrix $\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4\end{array}\right]$ is

✓
$\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 3 & -1 \\ 0 & 0 & \frac{-1}{3}\end{array}\right]$
B
$\left[\begin{array}{ccc}1 & 1 & -2 \\ 0 & 3 & -1 \\ 0 & 0 & \frac{-1}{3}\end{array}\right]$
C
$\left[\begin{array}{ccc}\frac{-1}{3} & 0 & 0 \\ 3 & -1 & 0 \\ -1 & 2 & 0\end{array}\right]$
D
$\left[\begin{array}{ccc}1 & 1 & 2 \\ 0 & -3 & -1 \\ 0 & 0 & \frac{-1}{3}\end{array}\right]$

Answer

Correct option: A.

$\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 3 & -1 \\ 0 & 0 & \frac{-1}{3}\end{array}\right]$

(A) Let $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4\end{array}\right]$
Applying $R _2 \rightarrow R _2-2 R _1$ and $R _3 \rightarrow R _3-3 R _1$,
$A \sim\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 3 & -1 \\ 0 & 5 & -2\end{array}\right]$
Applying $R _3 \rightarrow R _3-\left(\frac{5}{3}\right) R _2$,
$A \sim\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 3 & -1 \\ 0 & 0 & -\frac{1}{3}\end{array}\right]$
which is an upper triangular matrix.

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MCQ 1302 Marks

If $A=\left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & -2 & 5\end{array}\right]$, then $R_1 \leftrightarrow R_2$ and $C_1 \rightarrow C_1+2 C_3$ gives

A
$\left[\begin{array}{ccc}-13 & 2 & -5 \\ -1 & 2 & -1\end{array}\right]$
B
$\left[\begin{array}{ccc}-1 & -2 & -1 \\ 13 & -2 & 5\end{array}\right]$
✓
$\left[\begin{array}{ccc}13 & -2 & 5 \\ -1 & 2 & -1\end{array}\right]$
D
$\left[\begin{array}{ccc}2 & -13 & -5 \\ 2 & -1 & -1\end{array}\right]$

Answer

Correct option: C.

$\left[\begin{array}{ccc}13 & -2 & 5 \\ -1 & 2 & -1\end{array}\right]$

(C) $A=\left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & -2 & 5\end{array}\right]$
Applying $R _1 \leftrightarrow R _2$,
$A \sim\left[\begin{array}{ccc}3 & -2 & 5 \\ 1 & 2 & -1\end{array}\right]$
Applying $C _1 \rightarrow C _1+2 C _3$,
$A \sim\left[\begin{array}{ccc}13 & -2 & 5 \\ -1 & 2 & -1\end{array}\right]$

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MCQ 1312 Marks

$\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ $A, R_2 \rightarrow R_2-2 R_1$ gives

A
$\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right] A$
✓
$\left[\begin{array}{cc}1 & -1 \\ 0 & 5\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right] A$
C
$\left[\begin{array}{cc}1 & -1 \\ 0 & 5\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] A$
D
$\left[\begin{array}{cc}1 & -1 \\ 0 & 5\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -2 & 0\end{array}\right] A$

Answer

Correct option: B.

$\left[\begin{array}{cc}1 & -1 \\ 0 & 5\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right] A$

(B) $\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$
Applying $R _2 \rightarrow R _2-2 R _1$,
$\left[\begin{array}{cc}1 & -1 \\ 2-2(1) & 3-2(-1)\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0-2(1) & 1-2(0)\end{array}\right] A$
$\left[\begin{array}{cc}1 & -1 \\ 0 & 5\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right] A$

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