Question 11 Mark
If $\left|\begin{array}{ll}3 & 3 \\ x & 1\end{array}\right|=\left|\begin{array}{cc}-3 & x \\ 1 & 1\end{array}\right|$ then value of $x$ is :
Answer(B)$\quad\left|\begin{array}{ll}3 & 3 \\ x & 1\end{array}\right|=\left|\begin{array}{cc}-3 & x \\ 1 & 1\end{array}\right|$
$
\begin{aligned}
\Rightarrow & & 3-3 x & =-3-x \\
\Rightarrow & & 3+3 & =3 x-x \\
\Rightarrow & & 6 & =2 x \Rightarrow x=3
\end{aligned}
$
View full question & answer→Question 21 Mark
If $x=-1$ is a root of $\left|\begin{array}{lll}x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x\end{array}\right|=0$ then find other two roots of this equation :
Answer(A)
Suppose $P ( X )=(x+4)(x-a)(x-b)$ is a trinomial expression.
$\therefore \alpha=-4, \beta=a, \gamma=b$, is three roots :Here$
\begin{aligned}
P(X) & =\left|\begin{array}{lll}
x & 2 & 3 \\
1 & x & 1 \\
3 & 2 & x
\end{array}\right| \\
& =x\left(x^2-2\right)-2(x-3)+3(2-3 x) \\
& =x^3-2 x-2 x+6+6-9 x \\
& =x^2-13 x+12
\end{aligned}
$$\therefore$ Sum of roots $=\alpha+\beta+\gamma=\frac{-b}{a}$
$\therefore-4+a+b=0 \Rightarrow a+b=4$
View full question & answer→Question 31 Mark
For any matrix $A , A =\left[\begin{array}{cc}\alpha & -2 \\ -2 & \alpha\end{array}\right],\left| A ^3\right|=125$ then value of $\alpha$ is :
Answer(A) $\pm 3$
Here $A =\left[\begin{array}{cc}\alpha & -2 \\ -2 & \alpha\end{array}\right]$ and $\left| A ^3\right|=125$$
\begin{aligned}
\left|A^n\right|=|A|^n & \Rightarrow\left|A^3\right|=|A|^3 \\
& \Rightarrow 125=|A|^3 \\
& \Rightarrow 125=\left(\alpha^2-4\right)^3 \\
& \Rightarrow \alpha^2-4=5 \\
& \Rightarrow \alpha^2=9 \\
& \Rightarrow \alpha= \pm 3
\end{aligned}
$
View full question & answer→Question 41 Mark
If A has $3 \times 3$ order square matrix and $| A |=-5$, then $|\operatorname{adj} A |$ is :
Answer(C)25
Given : $| A |=-5$ and $n=3$
$\therefore|\operatorname{adj} A |=| A |^{n-1}=(-5)^{3-1}=(-5)^2=25$
View full question & answer→Question 51 Mark
Inverse of matrix $X=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$ is $:$
AnswerHere $|X|=\left|\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right|=2(3 \times 4)=24 \neq 0$
and $C_{11}=12, C_{12}=0, C_{13}=0$
$ C_{21}=0, C_{22}=8, C_{23}=0$
$ C_{31}=0, C_{32}=0, C_{33}=6$
$\therefore $ adj $X=\left[\begin{array}{ccc}12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6\end{array}\right]$
$\therefore X^{-1}=\frac{1}{24}\left[\begin{array}{ccc}12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6\end{array}\right]=\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]$
View full question & answer→Question 61 Mark
System of linear equation 5x + ky = 5, 3x + 3y = 5 is consistent, if :
Answer(D) System of linear equation is consistent if
$
\begin{aligned}
\left|\begin{array}{cc}
5 & k \\
3 & 3
\end{array}\right| & \neq 0 \\
15-3 k & \neq 0 \\
3 k & \neq 15 \\
k & \neq 5
\end{aligned}
$
View full question & answer→Question 71 Mark
Three points $P (2 x, x+3), Q (0, x)$ and $R (x+3$, $x+6$ ) are collinear, then value of $x:$
Answer$P, Q$ and $R$ are collinear.
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}2 x & x+3 & 1 \\ 0 & x & 1 \\ x+3 & x+6 & 1\end{array}\right|=0$
$ \Rightarrow \frac{1}{2}[2 x(x-x-6)-(x+3)(0-x-3) \left.+(1)\left(0-x^2-3 x\right)\right]=0$
$\Rightarrow \frac{1}{2}\left[-12 x+(x+3)^2-x^2-3 x\right]=0$
$\Rightarrow-12 x+x^2+6 x+9-x^2-3 x=0$
$\Rightarrow-9 x+9=0$
$\Rightarrow x=1$
View full question & answer→Question 81 Mark
If cofactor $C _{i j}$ represent for element $p_{i j},$ of matrix $P =\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & 2 & 4\end{array}\right]$ then value of $C_{31} . C_{23}$ is $:$
Answer$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-1 & 2 \\ 2 & -3\end{array}\right|=3-4=-1 $
$ C_{23}=(-1)^{2+3}\left|\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right|=-(2+3)=-5$
Hence $C_{31} \cdot C_{23}$
$=(-1)(-5)$
$=5$
View full question & answer→Question 91 Mark
Suppose $X =\left[x_{i j}\right]$ a matrix, where
$
X=\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]
$
Then matrix $Y =\left[m_{i j}\right]$, where $m_{i j}=$ minor of $x_{i j}$ :
Answer(D)
Here $X=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right]$$
\begin{aligned}
\therefore \quad m_{11} & =\left|\begin{array}{cc}
4 & -5 \\
-1 & 3
\end{array}\right|=12-5=7 \\
m_{12} & =\left|\begin{array}{cc}
3 & -5 \\
2 & 3
\end{array}\right|=9+10=19 \\
m_{13} & =\left|\begin{array}{cc}
3 & 4 \\
2 & -1
\end{array}\right|=-3-8=-11
\end{aligned}
$
Thus $\quad m_{21}=-1, m_{22}=-1, m_{23}=1$$
\begin{aligned}
m_{31} & =-3, m_{32}=-11, m_{33}=7 \\
Y=\left[m_{i j}\right] & =\left[\begin{array}{lll}
m_{11} & m_{12} & m_{13} \\
m_{21} & m_{22} & m_{23} \\
m_{31} & m_{32} & m_{33}
\end{array}\right]=\left[\begin{array}{ccc}
7 & 19 & -11 \\
-1 & -1 & 1 \\
-3 & -11 & 7
\end{array}\right]
\end{aligned}
$
View full question & answer→Question 101 Mark
If $\left[\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right]=\left[\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right]$ then $x$ is equal to :
Answer(B)$\pm 6$
$x^2-36=36-36 \Rightarrow x^2-36=0$
or $x^2=36 \Rightarrow x=\sqrt{36}= \pm 6$
View full question & answer→Question 111 Mark
Determinant $\left|\begin{array}{cc}1 & \log _b a \\ \log _a b & 1\end{array}\right|=$
Answer(B)
$
\begin{aligned}
1-\log _b a \times \log _a b & =1-\log _a a=1-1=0 \\
\log _a a & =1
\end{aligned}
$
Hence correct option is (B).
View full question & answer→Question 121 Mark
If $A=\left[\begin{array}{ccc}3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & -3 & K\end{array}\right]$ is non-invertible matrix, then value of K :
Answer(B)
For a non-invertible matrix
$|A|=0 \therefore\left|\begin{array}{rrr}3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & -3 & K\end{array}\right|=0$
$3(K+9)+1(2 K-3)+2(-6-1)=0$
$\Rightarrow \quad 3 K+27+2 K-3-14=0$
$\Rightarrow \quad 5 K+10=0 \Rightarrow K=-2$
Hence correct option is (B).
View full question & answer→Question 131 Mark
For same order square matrix $A , B , C , AB = AC$ $\Rightarrow B=C$, then $A$ will be :
Answer(D)
Because in invertible matrix $|A| \neq 0$.
View full question & answer→Question 141 Mark
If $A =\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$, then value of $A ^{-1}$ :
Answer(B)
$A_{11}=4, A_{12}=-3, A_{21}=-2, A_{22}=1$$
\operatorname{adj} A=\left[\begin{array}{cc}
4 & -3 \\
-2 & 1
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]
$
$
\begin{aligned}
|A| & =4-6=-2 \neq 0 \\
A^{-1}=\frac{\operatorname{adj} A}{|A|} & =\frac{-1}{2}\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]=\left[\begin{array}{cc}
-2 & 1 \\
\frac{3}{2} & -1 / 2
\end{array}\right]
\end{aligned}
$
Hence correct option is (B).
View full question & answer→Question 151 Mark
If $A =\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$ and $A (\operatorname{adj} A )= KI$ then value of $K$ will be :
Answer$\operatorname{adj} A=\left[\begin{array}{ll}A_{11} & A_{12} \\A_{21} & A_{22}\end{array}\right]^{\prime}=\left[\begin{array}{ll} A_{11} & A_{21} \\A_{12} & A_{22} \end{array}\right] $
$\therefore \quad \operatorname{adj} A=\left[\begin{array}{cc} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right] $
$\therefore A(\operatorname{adj} A)=\left[\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right]\left[\begin{array}{cc} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right]=KI $
$ \Rightarrow\left[\begin{array}{cc} \cos ^2 x+\sin ^2 x & -\cos x \sin x+\sin x \cos x \\ -\sin x \cos x+\cos x \sin x & \sin ^2 x+\cos ^2 x \end{array}\right] $
$ =K\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $
$\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=K\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $
$\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=KI$
$\Rightarrow I=KI $
$\therefore K=1 $
View full question & answer→Question 161 Mark
The value of determinant $\left|\begin{array}{ll}\cos 50^{\circ} & \sin 10^{\circ} \\ \sin 50^{\circ} & \cos 10^{\circ}\end{array}\right|$ :
Answer(C)
$\cos 50^{\circ} \cos 10^{\circ}-\sin 10^{\circ} \sin 50^{\circ}$
$
=\cos \left(50^{\circ}+10^{\circ}\right)=\cos 60^{\circ}=\frac{1}{2}
$
Hence correct option is (C).
View full question & answer→Question 171 Mark
If $A =\left[\begin{array}{ll}0 & 1 \\ 1 & 2\end{array}\right]$, then adj A will be :
Answer(A)
$A _{11}=2, A_{12}=-1, A_{21}=-1, A_{22}=0$$
\operatorname{adj} A=\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
2 & -1 \\
-1 & 0
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
2 & -1 \\
-1 & 0
\end{array}\right]
$
Hence correct option is (A).
View full question & answer→