MCQ 11 Mark
If $\left|\begin{array}{ccc}5 & 3 & -1 \\ -7 & x & -3 \\ 9 & 6 & -2\end{array}\right|=0$, then the value of $x$ is
Answer$\begin{array}{l}\text {:We have, }\left|\begin{array}{ccc}5 & 3 & -1 \\ -7 & x & -3 \\ 9 & 6 & -2\end{array}\right|=0 \\ \Rightarrow \quad 5(-2 x+18)-3(14+27)-1(-42-9 x)=0 \\ \Rightarrow \quad-x+9=0 \Rightarrow x=9\end{array}$
View full question & answer→MCQ 21 Mark
If $\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right]^{-1}=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$, then
- A
$a=1=b$
- B
$a=\cos 2 \theta, b=\sin 2 \theta$
- C
$a=\sin 2 \theta, b=\cos \theta$
- D
$a=\cos \theta, b=\sin \theta$
AnswerWe have,
$
\begin{array}{l}
{\left[\begin{array}{cc}
1 & -\tan \theta \\
\tan \theta & 1
\end{array}\right]\left[\begin{array}{cc}
1 & \tan \theta \\
-\tan \theta & 1
\end{array}\right]^{-1}=\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right]} \\
\text { Now, }\left[\begin{array}{cc}
1 & \tan \theta \\
-\tan \theta & 1
\end{array}\right]^{-1}=\frac{1}{\sec ^2 \theta}\left[\begin{array}{cc}
1 & -\tan \theta \\
\tan \theta & 1
\end{array}\right] \\
\quad=\left[\begin{array}{cc}
\cos ^2 \theta & -\tan \theta \cos ^2 \theta \\
\cos ^2 \theta \tan \theta & \cos ^2 \theta
\end{array}\right] \\
\Rightarrow\left[\begin{array}{cc}
1 & -\tan \theta \\
\tan \theta & 1
\end{array}\right]\left[\begin{array}{cc}
\cos ^2 \theta & -\tan \theta \cos ^2 \theta \\
\cos ^2 \theta \tan \theta & \cos ^2 \theta
\end{array}\right]=\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right] \\
\Rightarrow\left[\begin{array}{cc}
\cos ^2 \theta-\cos ^2 \theta \tan ^2 \theta & -2 \tan \theta \cos ^2 \theta \\
2 \tan \theta \cos ^2 \theta & \cos ^2 \theta-\cos ^2 \theta \tan ^2 \theta
\end{array}\right] \\
\quad=\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right] \\
\therefore a=\cos ^2 \theta-\cos ^2 \theta \tan ^2 \theta \text { and } b=2 \tan ^2 \cos ^2 \theta \\
\Rightarrow a=\cos ^2 \theta\left(1-\frac{\sin ^2 \theta}{\cos ^2 \theta}\right) \text { and } b=\frac{2 \sin \theta}{\cos \theta} \cdot \cos ^2 \theta \\
\Rightarrow a=\cos ^2 \theta-\sin ^2 \theta=\cos 2 \theta \text { and } b=2 \sin \theta \cos \theta=\sin 2 \theta
\end{array}
$
View full question & answer→MCQ 31 Mark
The inverse of $\left[\begin{array}{cc}-4 & 3 \\ 7 & -5\end{array}\right]$ is
- A
$\left[\begin{array}{cc}-5 & 3 \\ 7 & -4\end{array}\right]$
- B
$\left[\begin{array}{ll}5 & 3 \\ 7 & 4\end{array}\right]$
- C
$\left[\begin{array}{cc}-5 & 7 \\ 3 & -4\end{array}\right]$
- D
$\left[\begin{array}{ll}-5 & -3 \\ -7 & -4\end{array}\right]$
AnswerGiven, $A=\left[\begin{array}{cc}-4 & 3 \\ 7 & -5\end{array}\right] \therefore|A|=20-21=-1$
And adj $A=\left[\begin{array}{ll}-5 & -7 \\ -3 & -4\end{array}\right]^{\top}=\left[\begin{array}{ll}-5 & -3 \\ -7 & -4\end{array}\right]$
$
\therefore \quad A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\left[\begin{array}{ll}
5 & 3 \\
7 & 4
\end{array}\right]
$
View full question & answer→MCQ 41 Mark
If $A=\left[\begin{array}{ccc}1 & -2 & 4 \\ 2 & -1 & 3 \\ 4 & 2 & 0\end{array}\right]$ is the adjoint of a square matrix $B$, then $B^{-1}$ is equal to
Answer$\begin{array}{l}|A|=\left|\begin{array}{ccc}1 & -2 & 4 \\ 2 & -1 & 3 \\ 4 & 2 & 0\end{array}\right| \\ =1(0-6)+2(0-12)+4(4+4)=2 \\ \text { Given, } A=\operatorname{adj} B \\ \Rightarrow|A|=|\operatorname{adj} B| \Rightarrow|\operatorname{adj} B|=2 \\ \Rightarrow|B|^2=2 \quad\left[\because|\operatorname{adj} B|=|B|^{3-1} \text {, where } B \text { is } 3 \times 3 \text { matrix }\right] \\ \Rightarrow|B|= \pm \sqrt{2} \\ \therefore B^{-1}= \pm \frac{1}{\sqrt{2}} A\left[\because B^{-1}=\frac{1}{|B|}(\operatorname{adj} B)\right]\end{array}$
View full question & answer→MCQ 51 Mark
If $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 59 & 69 & -1\end{array}\right]$, then $A^{-1}$
AnswerGiven, $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 59 & 69 & -1\end{array}\right]$
Here, $|A|=-1$
$\begin{array}{l}\text { And adj } A=\left[\begin{array}{ccc}-1 & 0 & -59 \\ 0 & -1 & -69 \\ 0 & 0 & 1\end{array}\right]^{\top}=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -1 & 0 \\ -59 & -69 & 1\end{array}\right] \\ \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 59 & 69 & -1\end{array}\right]=A\end{array}$
View full question & answer→MCQ 61 Mark
The value of $\left|\begin{array}{ccc}\lfloor 1 & \lfloor 2 & \lfloor 3 \\ 2\lfloor 2 & 3 \underline{3} & 4\lfloor 4 \\ \lfloor 3 & \lfloor 4 & \lfloor 5\end{array}\right|$ is
AnswerGiven, $\left|\begin{array}{ccc}\lfloor 1 & \lfloor 2 & \lfloor 3 \\ 2\lfloor\underline{2} & 3 \underline{3} & 4\lfloor 4 \\ \underline{3} & \lfloor 4 & \lfloor 5\end{array}\right|=\left|\begin{array}{ccc}1 & 2 & 6 \\ 4 & 18 & 96 \\ 6 & 24 & 120\end{array}\right|$
\[\begin{array}{l}
=1(2160-2304)-2(480-576)+6(96-108) \\
=-144-2(-96)+6(-12)=-144+192-72=-24
\end{array}\]
View full question & answer→MCQ 71 Mark
If $A=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right]$ and $\left|A^3\right|=27$, then the value of $\alpha$ is
- A
$\pm 1$
- B
$\pm 2$
- C
$\pm \sqrt{5}$
- D
$\pm \sqrt{7}$
Answer$\begin{array}{l}
\text { Given, } A=\left[\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right] \\
\left|A^3\right|=27 \\
\Rightarrow|A|^3=27\left[\because\left|A^n\right|=|A|^n\right] \Rightarrow|A|=3
\end{array}
$
From (i) and (ii), we get
$
\Rightarrow \alpha^2-4=3 \Rightarrow \alpha^2=7 \Rightarrow \alpha= \pm \sqrt{7}
$
View full question & answer→MCQ 81 Mark
For $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, then $14 A^{-1}$ is given by
- A
$14\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$
- B
$\left[\begin{array}{cc}4 & -2 \\ 2 & 6\end{array}\right]$
- C
$2\left[\begin{array}{ll}2 & -1 \\ 1 & -3\end{array}\right]$
- D
$2\left[\begin{array}{cc}-3 & -1 \\ 1 & -2\end{array}\right]$
AnswerWe have, $|A|=6+1=7$
Also, $\operatorname{adj} A=\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$
Now, $A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)=\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$
$
\therefore 14 A^{-1}=\left[\begin{array}{cc}
4 & -2 \\
2 & 6
\end{array}\right]
$
View full question & answer→MCQ 91 Mark
For matrix $A=\left[\begin{array}{cc}2 & 5 \\ -11 & 7\end{array}\right],(\operatorname{adj} A)^{\prime}$ is equal to
- A
$\left[\begin{array}{ll}-2 & -5 \\ 11 & -7\end{array}\right]$
- B
$\left[\begin{array}{cc}7 & 5 \\ 11 & 2\end{array}\right]$
- C
$\left[\begin{array}{cc}7 & 11 \\ -5 & 2\end{array}\right]$
- D
$\left[\begin{array}{cc}7 & -5 \\ 11 & 2\end{array}\right]$
AnswerWe know that, $(\operatorname{adj} A)^{\prime}=$ cofactor matrix of $A$
Here, cofactor matrix of $A=\left[\begin{array}{cc}7 & 11 \\ -5 & 2\end{array}\right]=(\operatorname{adj} A)^{\prime}$
View full question & answer→MCQ 101 Mark
Given that A = [$a_{i j}$]is a square matrix of order $3 \times 3$ and $|A|=-7$, then the value of $\sum_{i=1}^3 a_{i 2} A_{i 2}$, where$A_{i j}$denotes the cofactor of element $a_{i j}$ is
AnswerWe have, $|A|=-7$
\[\therefore \sum_{i=1}^3 a_{i 2} A _{ i 2}=a_{12} A _{12}+a_{22} A _{22}+a_{32} A _{32}=| A |=-7\]
View full question & answer→MCQ 111 Mark
Given that $A$ is a square matrix of order 3 and $|A|=-4$, then $|\operatorname{adj} A|$ is equal to
AnswerWe know that, $|\operatorname{adj} A|=\left.|A|\right|^{n-1}$, where $n$ is the order of $A$.
Here, $|\operatorname{adj} A|=|A|^2=(-4)^2=16$
View full question & answer→MCQ 121 Mark
Value of $k$, for which $A=\left[\begin{array}{cc}k & 8 \\ 4 & 2 k\end{array}\right]$ is a singular matrix is
Answer$\because$ A is a singular matrix.
$
\begin{array}{l}
\therefore|A|=0 \\
\Rightarrow\left|\begin{array}{cc}
k & 8 \\
4 & 2 k
\end{array}\right|=0 \Rightarrow 2 k^2-32=0 \Rightarrow k^2=16 \Rightarrow k= \pm 4
\end{array}$
View full question & answer→MCQ 131 Mark
The determinant $\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|$ is equal to
- A
$k\left(3 y+k^2\right)$
- B
$3 y+k^3$
- C
$3 y+k^2$
- D
$k^2(3 y+k)$
Answer$\begin{array}{l}\text {Let } D=\left|\begin{array}{ccc} y + k & y & y \\ y & y + k & y \\ y & y & y + k \end{array}\right| \\ =(y+k)\left((y+k)^2-y^2\right)-y\left(y^2+k y-y^2\right)+y\left(y^2-y^2-y k\right) \\ =k^3+3 k^2 y=k^2(k+3 y)\end{array}$
View full question & answer→MCQ 141 Mark
Given that $A$ is a non-singular matrix of order 3 such that $A^2=2 A$, then value of $|2 A|$ is
Answer$\begin{array}{l}\text{}\quad \text{: We have, } A^2=2 A \Rightarrow\left|A^2\right|=|2 A| \\ \Rightarrow \quad|A|^2=2^3|A| \quad[A s|k A|=k n|A| \text { for a matrix of order } n] \\ \Rightarrow \quad|A|[|A|-8]=0 \\ \Rightarrow \quad \text { either }|A|=0 \text { or }|A|=8\end{array}$
But $A$ is non-singular matrix $\Rightarrow|A| \neq 0$
$
\therefore \quad|2 A|=2^3 \cdot|A|=64
$
View full question & answer→MCQ 151 Mark
For the matrix $A=\left[\begin{array}{ccc}2 & -1 & 1 \\ \lambda & 2 & 0 \\ 1 & -2 & 3\end{array}\right]$ to be invertible, the value of $\lambda$ is
- A
$0$
- B
- C
$R -\{10\}$
- D
$R -\{-10\}$
AnswerGiven, $A=\left[\begin{array}{ccc}2 & -1 & 1 \\ \lambda & 2 & 0 \\ 1 & -2 & 3\end{array}\right]$
For invertible matrix, $|A| \neq 0$
So, $2(6-0)+1(3 \lambda-0)+1(-2 \lambda-2) \neq 0$
$\Rightarrow \quad 12+3 \lambda-2 \lambda-2 \neq 0$
$\Rightarrow \lambda+10 \neq 0 \Rightarrow \lambda \neq-10$
$\therefore \quad$ Required value of $\lambda$ is $R-\{-10\}$.
View full question & answer→MCQ 161 Mark
If $A$ is a square matrix of order 3 such that the value of $|\operatorname{adj} A|=8$, then the value of $\left|A^{\top}\right|$ is
- A
$\sqrt{2}$
- B
$-\sqrt{2}$
- C
- D
$2 \sqrt{2}$
AnswerGiven, $|\operatorname{adj} A|=8$
We know that $|\operatorname{adj} A|=|A|^{n-1}, n$ is the order of matrix.
$
\begin{array}{l}
\Rightarrow|A|^{3-1}=8 \Rightarrow|A|^2=8 \Rightarrow|A|=2 \sqrt{2} \\
\therefore\left|A^T\right|=2 \sqrt{2}\left[\because|A|=\left|A^T\right|\right]
\end{array}
$
View full question & answer→MCQ 171 Mark
If $A=\left[\begin{array}{ccc}-2 & 0 & 0 \\ 1 & 2 & 3 \\ 5 & 1 & -1\end{array}\right]$, then the value of $|A(\operatorname{adj} . A)|$ is:
AnswerWe have, $A=\left[\begin{array}{ccc}-2 & 0 & 0 \\ 1 & 2 & 3 \\ 5 & 1 & -1\end{array}\right] \Rightarrow|A|=(-2)(-2-3)=10$
As, $|A(\operatorname{adj} \cdot A)|=|A||\operatorname{adj} A|=|A||A|^{n-1}=|A|^n$
$\Rightarrow|A(\operatorname{adj} \cdot A)|=(10)^3=1000$
View full question & answer→MCQ 181 Mark
If inverse of matrix $\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$ is the matrix $\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & \lambda & 3 \\ 1 & 3 & 4\end{array}\right]$, then value of $\lambda$ is
AnswerLet $A=\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$ and $A^{-1}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & \lambda & 3 \\ 1 & 3 & 4\end{array}\right]$ $|A|=7(1-0)+3(-1)-3(0+1)=7-3-3=1$ The cofactor matrix is $C=\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 4 & 3 \\ 3 & 3 & 4\end{array}\right]$
Thus, the adjoint of $A$ is $C^{\top}$.
So, $A^{-1}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & \lambda & 3 \\ 1 & 3 & 4\end{array}\right]$ (Given)
\[\Rightarrow \lambda=4\]
View full question & answer→MCQ 191 Mark
Given that $A^{-1}=\frac{1}{7}\left[\begin{array}{cc}2 & 1 \\ -3 & 2\end{array}\right]$, matrix $A$ is :
- A
$7\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
- B
$\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
- C
$\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
- D
$\frac{1}{49}\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
AnswerWe have, $A^{-1}=\frac{1}{7}\left[\begin{array}{cc}2 & 1 \\ -3 & 2\end{array}\right]$
$\because\left(A^{-1}\right)^{-1}=A$
Now, $\left(A^{-1}\right)^{-1}=\frac{\operatorname{adj}\left(A^{-1}\right)}{\left|A^{-1}\right|}=\frac{\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]}{\frac{1}{7}}=\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
$
\Rightarrow \quad A=\left[\begin{array}{cc}
2 & -1 \\
3 & 2\end{array}\right]
$
View full question & answer→MCQ 201 Mark
$\left|\begin{array}{ccc}-a & b & c \\ a & -b & c \\ a & b & -c\end{array}\right|=k a b c$, then the value of $k$ is:
AnswerWe have,
\[\begin{aligned}
\left|\begin{array}{ccc}
-a & b & c \\
a & -b & c \\
a & b & -c
\end{array}\right| & =-a(b c-b c)-b(-a c-a c)+c(a b+a b) \\
& =2 a b c+2 a b c=4 a b c
\end{aligned}\]
$\Rightarrow k a b c=4 a b c \Rightarrow k=4$
View full question & answer→MCQ 211 Mark
Let $f(x)=\left|\begin{array}{cc}x^2 & \sin x \\ p & -1\end{array}\right|$, where $p$ is a constant. The value of $p$ for which $f^{\prime}(0)=1$ is
AnswerGiven, $f(x)=\left|\begin{array}{cc}x^2 & \sin x \\ p & -1\end{array}\right|=-x^2-p \sin x$
$
\begin{array}{l}
\Rightarrow f^{\prime}(x)=-2 x-p \cos x \\
\text { We have, } f^{\prime}(0)=1 \Rightarrow-2(0)-p \cos (0)=1 \Rightarrow-p=1 \Rightarrow p=-1
\end{array}
$
View full question & answer→MCQ 221 Mark
If the area of the triangle with vertices $(-3,0),(3,0)$ and $(0, k)$ is 9 sq units, then the value/s of $k$ is will be
AnswerArea $\left.=\left|\frac{1}{2}\right| \begin{array}{ccc}-3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1\end{array} \right\rvert\,$, given that the area
$
\begin{array}{l}
=9 \text { sq. unit. } \\
\Rightarrow \quad \pm 9=\frac{1}{2}\left|\begin{array}{ccc}
-3 & 0 & 1 \\
3 & 0 & 1 \\
0 & k & 1
\end{array}\right| \text { expanding along } C_2 \text {, we get } \\
\Rightarrow \pm 18=-k(-3-3) \Rightarrow \pm 18=6 k \Rightarrow k= \pm 3 \\
\end{array}
$
View full question & answer→MCQ 231 Mark
$V$ is a matrix of order 3 such that $\mid$ adj $V \mid=7$.
Which of these could be $|V|$ ?
- A
$7^2$
- B
- ✓
$\sqrt{7}$
- D
$\sqrt[3]{7} \quad$
AnswerCorrect option: C. $\sqrt{7}$
$\sqrt{7}$
View full question & answer→MCQ 241 Mark
If $A$ and $B$ are invertible square matrices of the same order, then which of the following is not correct?
- A
$\left|A B^{-1}\right|=\frac{|A|}{|B|}$
- B
$\left|(A B)^{-1}\right|=\frac{1}{|A||B|}$
- C
$(A B)^{-1}=B^{-1} A^{-1}$
- D
$(A+B)^{-1}=B^{-1}+A^{-1}$
Answer$(A+B)^{-1} \neq B^{-1}+A^{-1}$
View full question & answer→MCQ 251 Mark
Given that $A$ is a square matrix of order 3 and $|A|=-2$, then $|\operatorname{adj}(2 A)|$ is equal to
Answer$|\operatorname{adj}(2 A)|=|(2 A)|^2=\left(2^3|A|\right)^2=2^6|A|^2$ $=2^6 \times(-2)^2=2^8$
View full question & answer→MCQ 261 Mark
The points $D, E$ and $F$ are the mid-points of $A B, B C$ and $C A$ respectively.
What is the area of the shaded region? AnswerCorrect option: B. $\frac{3}{2}$ squnits
$\frac{3}{2}$ squnits
View full question & answer→MCQ 271 Mark
The value of $|A|$, if $A=\left[\begin{array}{ccc}0 & 2 x-1 & \sqrt{x} \\ 1-2 x & 0 & 2 \sqrt{x} \\ -\sqrt{x} & -2 \sqrt{x} & 0\end{array}\right]$, where $x \in R ^{+}$, is
- A
$(2 x+1)^2$
- B
- C
$(2 x+1)^3$
- D
$(2 x-1)^2$
AnswerGiven,
$
A=\left[\begin{array}{ccc}
0 & 2 x-1 & \sqrt{x} \\
1-2 x & 0 & 2 \sqrt{x} \\
-\sqrt{x} & -2 \sqrt{x} & 0
\end{array}\right], A^{\prime}=\left[\begin{array}{ccc}
0 & 1-2 x & -\sqrt{x} \\
2 x-1 & 0 & -2 \sqrt{x} \\
\sqrt{x} & 2 \sqrt{x} & 0
\end{array}\right]
$
Since, $A^{\prime}=-A$,
Matrix $A$ is a skew symmetric matrix of odd order.
$
\therefore \quad|A|=0
$
View full question & answer→MCQ 281 Mark
The value of the determinant $\left|\begin{array}{ccc}2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1\end{array}\right|$ is
AnswerLet $|A|=\left|\begin{array}{ccc}2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1\end{array}\right|$
Expanding along $R_1$, we get
$
\begin{array}{l}
|A|=2(1-8)-7(1-10)+1(8-10) \\
=2(-7)-7(-9)+1(-2)=-14+63-2=47
\end{array}
$
View full question & answer→MCQ 291 Mark
If $\left|\begin{array}{lll}\alpha & 3 & 4 \\ 1 & 2 & 1 \\ 1 & 4 & 1\end{array}\right|=0$, then the value of $\alpha$ is
Answer$\begin{array}{l}\left|\begin{array}{lll}\alpha & 3 & 4 \\ 1 & 2 & 1 \\ 1 & 4 & 1\end{array}\right|=0 \Rightarrow \alpha(2-4)-3(1-1)+4(4-2)=0 \\ \Rightarrow \quad-2 \alpha+8=0 \Rightarrow 2 \alpha=8 \Rightarrow \alpha=4\end{array}$
View full question & answer→MCQ 301 Mark
For which value of $x$, are the determinants $\left|\begin{array}{ll}2 x & -3 \\ 5 & x\end{array}\right|$ and $\left|\begin{array}{cc}10 & 1 \\ -3 & 2\end{array}\right|$ equal?
Answer$\begin{array}{l}\text {Since, }\left|\begin{array}{cc}2 x & -3 \\ 5 & x\end{array}\right|=\left|\begin{array}{ll}10 & 1 \\ -3 & 2\end{array}\right| \\ \Rightarrow \quad 2 x^2+15=20+3 \Rightarrow 2 x^2=23-15=8 \\ \Rightarrow x^2=4 \Rightarrow x= \pm 2\end{array}$
View full question & answer→MCQ 311 Mark
If $|A|=2$, where $A$ is a $2 \times 2$ matrix, then $\left|4 A^{-1}\right|$ equals:
AnswerGiven, $|A|=2$, Where $A$ is a $2 \times 2$ matrix.
Now, $\left|4 A^{-1}\right|=4|A|^{-1}$
and we know that, $|A|^{-1}=\frac{1}{|A|}$
$
\therefore\left|4 A^{-1}\right|=4 \times \frac{1}{|A|}=4 \times \frac{1}{2}=2
$
View full question & answer→MCQ 321 Mark
Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj} A|=64$. Then $|A|$ is equal to:
AnswerGiven, $|\operatorname{adj} A|=64$, where $A$ is a $3 \times 3$ matrix.
Since, $|\operatorname{adj} A|=|A|^{n-1}$, where $A$ is a $n \times n$ matrix.
$\Rightarrow \quad|\operatorname{adj} A|=|A|^{3-1}=64 \Rightarrow|A|^2=64 \Rightarrow|A|= \pm 8$.
View full question & answer→MCQ 331 Mark
If $A$ is a square matrix of order 3 and $|A|=6$, then the value of $|\operatorname{adj} A |$ is:
AnswerSince, order of a square matrix $A$ is 3 and $|A|=6$ We know that $| Adj A |=| A |^{n-1}=6^{3-1}=6^2=36$
View full question & answer→MCQ 341 Mark
The value of the cofactor of the element of second row and third column in the matrix $\left[\begin{array}{ccc}4 & 3 & 2 \\ 2 & -1 & 0 \\ 1 & 2 & 3\end{array}\right]$ is:
AnswerLet $A=\left[\begin{array}{ccc}4 & 3 & 2 \\ 2 & -1 & 0 \\ 1 & 2 & 3\end{array}\right]$
Minor $M_{23}$ of an element $a_{23}$ of a matrix $A$ is given by
$M_{23}=\left|\begin{array}{ll}4 & 3 \\ 1 & 2\end{array}\right|=8-3=5$
$\because \quad$ Required cofactor $C_{23}=-M_{23}=-5$
View full question & answer→MCQ 351 Mark
If $A$ is a square matrix of order 3 and $|A|=5$, then $|\operatorname{adj} A|=$
AnswerGiven, $|A|=5$, order of matrix, $n=3$.
$|\operatorname{adj} A|=|A|^{n-1} \Rightarrow|\operatorname{adj} A|=25$
View full question & answer→MCQ 361 Mark
If $A$ is a square matrix of order $3,\left|A^{\prime}\right|=-3$, then $\left|A A^{\prime}\right|=$
Answer$\left|A A^{\prime}\right|=|A|\left|A^{\prime}\right|=\left|A^{\prime}\right|^2=(-3)^2=9$
View full question & answer→MCQ 371 Mark
If $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$, then the possible value(s) of ' $x$ ' is/are
- A
- B
$\sqrt{3}$
- C
$-\sqrt{3}$
- D
$\sqrt{3},-\sqrt{3}$
AnswerWe have, $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$
\[\Rightarrow \quad 2-20=2 x^2-24 \Rightarrow 2 x^2=6 \Rightarrow x^2=3 \Rightarrow x= \pm \sqrt{3}\]
View full question & answer→MCQ 381 Mark
If $A=\left[\begin{array}{ccc}-2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right]$, then the value of $|\operatorname{adj} A|$ is
AnswerWe have, $|A|=\left|\begin{array}{ccc}-2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right|$
$\begin{aligned} & |A|=-2(4-0)-0+0=-8 \\ \therefore \quad & |\operatorname{adj} A|=(-8)^2=64\end{aligned}$
View full question & answer→MCQ 391 Mark
If $A$ is a $3 \times 3$ matrix such that $|A|=8$, then $|3 A|$ equals
AnswerWe have, $|3 A|=3^3|A|=3^3-8$ [Given $|A|=8$ ]
$=27 \cdot 8=216$
View full question & answer→MCQ 401 Mark
If $A$ is a skew-symmetric matrix of order 3 , then the value of $|A|$ is
AnswerWe have, $A^T=-A[\because A$ is skew-symmetric matrix $]$
$\begin{array}{ll}
\therefore & \left|A^{\top}\right|=|-A| \Rightarrow|A|=(-1)^3|A| \\
\Rightarrow & |A|=-|A| \Rightarrow 2|A|=0 \Rightarrow|A|=0
\end{array} \quad[\because A \text { is of order } 3]
$
View full question & answer→MCQ 411 Mark
The roots of the equation $\left|\begin{array}{ccc}x & 0 & 8 \\ 4 & 1 & 3 \\ 2 & 0 & x\end{array}\right|$ = 0 are
- A
$-4,4$
- B
$2,-4$
- C
$2.4$
- D
$2.8$
AnswerGiven, $\left|\begin{array}{lll}x & 0 & 8 \\ 4 & 1 & 3 \\ 2 & 0 & x\end{array}\right|=0$
Expanding along $R_1$, we get
\[\begin{aligned}
& x(x-0)-0+8(0-2)=0 \\
\Rightarrow \quad & x^2-16=0 \Rightarrow x^2=16 \Rightarrow x= \pm 4
\end{aligned}\]
View full question & answer→MCQ 421 Mark
If $A$ is a square matrix of order 3 and $|A|=5$, then the value of $\left|2 A^{\prime}\right|$ is
AnswerGiven, $A$ is a $3 \times 3$ matrix and $|A|=5$
Now, $\left|2 A^{\prime}\right|=2^3\left|A^{\prime}\right|=2^3|A|=8 \times 5=40$
View full question & answer→MCQ 431 Mark
If $A$ is a non-singular square matrix of order 3 such that $A^2=3 A$, then value of $|A|$ is
AnswerGiven $A^2=3 A$
$
\begin{array}{l}
\Rightarrow\left|A^2\right|=|3 A| \Rightarrow|A|^2=3^3|A| \Rightarrow|A|^2-27|A|=0 \\
\Rightarrow|A|[|A|-27]=0
\end{array}
$
As, $A$ is a non-singular matrix$
\therefore \quad|A| \neq 0 \Rightarrow|A|-27=0 \Rightarrow|A|=27
$
View full question & answer→MCQ 441 Mark
If $A$ is a square matrix of order 3 , such that $A(\operatorname{adj} A)=$ $10 I$, then $\mid$ adj $A \mid$ is equal to
AnswerGiven, $A(\operatorname{adj} A)=101$
\[\begin{aligned}
\Rightarrow & |A(\operatorname{adj}) A||=| 10 \mid \\
\Rightarrow & |A||\operatorname{adj} A|=10^3|I| \\
\Rightarrow & |A||A|^2=10^3 \\
& {\left[\because A \text { is matrix of order } n, \text { then }|\operatorname{adj} A|=\left.|A|\right|^{n-1} \text { and } n=3\right] } \\
\Rightarrow & |A|^3=10^3 \Rightarrow|A|=10
\end{aligned}\]
Now, from (i), we get
\[\mid(\text { (adj) }) A||=|A|^2=10^2=100\]
View full question & answer→MCQ 451 Mark
If $A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]$, then adj $A$ is equal to
- A
$\left[\begin{array}{cc}-2 & -5 \\ -3 & 2\end{array}\right]$
- ✓
$\left[\begin{array}{cc}-2 & -3 \\ -5 & 2\end{array}\right]$
- C
$\left[\begin{array}{cc}5 & 3 \\ 2 & -2\end{array}\right]$
- D
$\left[\begin{array}{cc}3 & 5 \\ 2 & -2\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{cc}-2 & -3 \\ -5 & 2\end{array}\right]$
(b) : Given, $A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right] \Rightarrow \operatorname{adj} A=\left[\begin{array}{cc}-2 & -5 \\ -3 & 2\end{array}\right]^{\prime}$
$
\therefore \operatorname{adj} A=\left[\begin{array}{cc}
-2 & -3 \\
-5 & 2
\end{array}\right]
$
View full question & answer→MCQ 461 Mark
The value of $\Delta=\left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right|$ is
Answer(a) : Expanding along $R_1$, we get
$
\begin{array}{l}
\Delta=0-\sin \alpha(0-\sin \beta \cos \alpha)-\cos \alpha(\sin \alpha \sin \beta-0) \\
=\sin \alpha \sin \beta \cos \alpha-\cos \alpha \sin \alpha \sin \beta=0
\end{array}
$
View full question & answer→MCQ 471 Mark
Find the area of the triangle with vertices $P(4,5), Q(4,-2)$ and $R(-6,2)$.
Answer(b) : Let $\Delta$ be the area of triangle $P Q R$. Then,
$\begin{aligned} \Delta & =\frac{1}{2}\left|\begin{array}{ccc}4 & 5 & 1 \\ 4 & -2 & 1 \\ -6 & 2 & 1\end{array}\right|=\frac{1}{2}[4(-2-2)-5(4+6)+1(8-12)] \\ & =\frac{1}{2}|[-16-50-4]|=35 \text { sq. units }\end{aligned}$
View full question & answer→MCQ 481 Mark
If $A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$, then find $(\operatorname{adj} A)^{-1}$.
Answer(a): We have, $|A|=\cos ^2 \alpha+\sin ^2 \alpha=1 \neq 0$
So, $A^{-1}$ exists.
We know adj $A=|A| A^{-1}$
$
\begin{array}{lll}
\Rightarrow & \text { adj } A=A^{-1} & {[\because|A|=1]} \\
\Rightarrow & (\operatorname{adj} A)^{-1}=\left(A^{-1}\right)^{-1}=A &
\end{array}
$
View full question & answer→MCQ 491 Mark
The value of the determinant of a matrix $A$ of order $3 \times 3$ is 4 . Find the value of $|5 A|$.
Answer(c): Given, $A$ is a $3 \times 3$ matrix and $|A|=4$
$
\Rightarrow|5 A|=5^3 \cdot|A|=125 \times 4=500 \text {. }
$
View full question & answer→MCQ 501 Mark
Evaluate the following determinant : $\left|\begin{array}{cc}x & -7 \\ x & 5 x+1\end{array}\right|$
- A
$3 x^2+4$
- ✓
$x(5 x+8)$
- C
$3 x+4 x^2$
- D
$x(3 x+4)$
AnswerCorrect option: B. $x(5 x+8)$
(b) : We have, $\left|\begin{array}{cc}x & -7 \\ x & 5 x+1\end{array}\right|=x(5 x+1)+7(x)$
$
=5 x^2+x+7 x=5 x^2+8 x=x(5 x+8)
$
View full question & answer→