MCQ 511 Mark
For an invertible matrix $A$ if $A(\operatorname{adj} A)$ $=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$, then $|A|$ is
Answer(c) : We have, $A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]=10\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=10 I$
We know that $A(\operatorname{adj} A)=|A| I \Rightarrow|A|=10$
View full question & answer→MCQ 521 Mark
Find the cofactors of elements $a_{12}, a_{22}, a_{32}$ respectively of the matrix $\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$.
- ✓
$0,2,-2 \sin \theta$
- B
$2,0,2 \sin \theta$
- C
$2,0,-2 \sin \theta$
- D
$-2 \sin \theta, 2,0$
AnswerCorrect option: A. $0,2,-2 \sin \theta$
View full question & answer→MCQ 531 Mark
Find the minor of $a_{22}$ of the matrix
$
\left[\begin{array}{lll}
1 & 6 & 1 \\
5 & 3 & 0 \\
2 & 2 & 9
\end{array}\right] \text {. }
$
Answer(c) : $M_{22}=\left|\begin{array}{ll}1 & 1 \\ 2 & 9\end{array}\right|=9-2=7$
View full question & answer→MCQ 541 Mark
$\left|\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right|=$
Answer(b) : We have, $\left|\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right|=\cos ^2 \theta+\sin ^2 \theta=1$
View full question & answer→MCQ 551 Mark
The area of a triangle with vertices $(-3,0)$, $(3,0)$ and $(0, k)$ is 9 sq. units. The value of $k$ will be
Answer(b) : Area of triangle $=\frac{1}{2}\left|\begin{array}{ccc}-3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1\end{array}\right|= \pm 9$
$
\Rightarrow \quad-k(-3-3)= \pm 18 \Rightarrow 6 k= \pm 18 \Rightarrow k= \pm 3
$
View full question & answer→MCQ 561 Mark
If $\Delta=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$, then write the minor of the element $a_{22}$.
Answer(a) : $M_{22}=\left|\begin{array}{ll}5 & 8 \\ 1 & 3\end{array}\right|=15-8=7$
View full question & answer→MCQ 571 Mark
If $A$ is a square matrix of order 3 and $|2 A|=k|A|$, then find the value of $k$.
Answer(c) : Given, $A$ is a square matrix of order 3
$\therefore \quad|2 A|=2^3|A|=8|A|=k|A| \quad$ (given)
$\Rightarrow \quad k=8$
View full question & answer→MCQ 581 Mark
If the equations $a x+4 y+z=0, b x+3 y+z=0$, $c x+2 y+z=0$ have non-trivial solution, then find the value of $a-2 b+c$.
Answer(d) : For non-trivial solution, $\left|\begin{array}{lll}a & 4 & 1 \\ b & 3 & 1 \\ c & 2 & 1\end{array}\right|=0$
$
\begin{array}{l}
\Rightarrow \quad a(3-2)-4(b-c)+1(2 b-3 c)=0 \\
\Rightarrow \quad a-2 b+c=0
\end{array}
$
View full question & answer→MCQ 591 Mark
If $c_{i j}$ is the cofactor of the element $a_{i j}$ of the determinant $\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$, then write the value of $a_{32} \cdot c_{32}$.
Answer(a) : Here, $a_{32}=5$, then,
$
\begin{array}{l}
c_{32}=(-1)^{3+2}\left|\begin{array}{ll}
2 & 5 \\
6 & 4
\end{array}\right|=(-1)^5(8-30)=-(-22)=22 \\
\therefore \quad a_{32} \cdot c_{32}=5 \times 22=110
\end{array}
$
View full question & answer→MCQ 601 Mark
For what value of $x$, matrix $A=\left[\begin{array}{ll}6-x & 4 \\ 3-x & 1\end{array}\right]$ is a singular matrix?
Answer(b) : Matrix $A$ is singular, when $|A|=0$
$\Rightarrow\left|\begin{array}{ll}6-x & 4 \\ 3-x & 1\end{array}\right|=0$
$\Rightarrow 6-x-12+4 x=0 \Rightarrow 3 x=6 \Rightarrow x=2$
View full question & answer→MCQ 611 Mark
Matrix $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right]$, then the value of $a_{31} A_{31}+a_{32} A_{32}+a_{33} A_{33}$ is
Answer(c) : $a_{31} A_{31}+a_{32} A_{32}+a_{33} A_{33}=|A|$
Now, $|A|=1(7-20)-2(7-10)+3(4-2)=-1$
View full question & answer→MCQ 621 Mark
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
$\frac{1}{5}\left[\begin{array}{rr}4 & 2 \\ 4 & -1\end{array}\right]$
- ✓
$\frac{1}{5}\left[\begin{array}{lr}-3 & 2 \\ 4 & -1\end{array}\right]$
- D
$\frac{1}{5}\left[\begin{array}{ll}-1 & 2 \\ -1 & 4\end{array}\right]$
AnswerCorrect option: C. $\frac{1}{5}\left[\begin{array}{lr}-3 & 2 \\ 4 & -1\end{array}\right]$
(c) : We have, $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$ such that $A X=I$
$
\begin{array}{l}
\Rightarrow \quad X=A^{-1} I \\
\because \quad|A|=3-8=-5 \neq 0 \Rightarrow A^{-1} \text { 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}{cc}
1 & 0 \\
0 & 1
\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}
-3 & 2 \\
4 & -1
\end{array}\right]
\end{array}
$
View full question & answer→MCQ 631 Mark
If $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|=8$, then find the value of $x$.
Answer(b) : Expanding the given determinant, we get
$
\begin{array}{l}
x\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+ \\
x \cos \theta)=8 \\
\Rightarrow-x^3-x+x=8 \Rightarrow x^3+8=0 \\
\Rightarrow(x+2)\left(x^2-2 x+4\right)=0 \\
\Rightarrow x+2=0 \\
\Rightarrow x=-2
\end{array} \quad\left[\because x^2-2 x+4>0 \forall x\right]
$ $\left[\because x^2-2 x+4>0 \forall x\right]$
View full question & answer→MCQ 641 Mark
If $\left|\begin{array}{cc}2 x & 5 \\ 8 & x\end{array}\right|=\left|\begin{array}{cc}6 & -2 \\ 7 & 3\end{array}\right|$, then find the value of $x$.
- A
$\pm 2$
- B
$\pm 3$
- C
$\pm 6$
- D
$\pm 4$
Answer$
\begin{array}{l}
\text { (c) : }\left|\begin{array}{cc}
2 x & 5 \\
8 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & -2 \\
7 & 3
\end{array}\right| \Rightarrow 2 x^2-40=18+14 \\
\Rightarrow x^2=36 \Rightarrow x= \pm 6
\end{array}
$
View full question & answer→MCQ 651 Mark
Evaluate: $\left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\ \sin 75^{\circ} & \cos 75^{\circ}\end{array}\right|$
Answer(a) : We have, $\left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\ \sin 75^{\circ} & \cos 75^{\circ}\end{array}\right|$
$
\begin{array}{l}
=\cos 15^{\circ} \times \cos 75^{\circ}-\sin 15^{\circ} \times \sin 75^{\circ}=\cos \left(15^{\circ}+75^{\circ}\right) \\
=\cos 90^{\circ}=0
\end{array}
$
View full question & answer→MCQ 661 Mark
If $\Delta=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$, write the minor of the element $a_{23}$.
Answer(c) : Minor of $a_{23}=\left|\begin{array}{ll}5 & 3 \\ 1 & 2\end{array}\right|=10-3=7$
View full question & answer→MCQ 671 Mark
Write the cofactor of the element $a_{31}$ in
$
A=\left(\begin{array}{lll}
3 & 2 & 6 \\
5 & 0 & 7 \\
3 & 8 & 5
\end{array}\right).
$
Answer(b) : We have, $A=\left(\begin{array}{lll}3 & 2 & 6 \\ 5 & 0 & 7 \\ 3 & 8 & 5\end{array}\right)$
Cofactor of $a_{31}=(-1)^{3+1}\left|\begin{array}{ll}2 & 6 \\ 0 & 7\end{array}\right|=14$
View full question & answer→MCQ 681 Mark
If $A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$ then $\left(B^{-1} A^{-1}\right)^{-1}=$
- ✓
$\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]$
- B
$\left[\begin{array}{cc}2 & 2 \\ -2 & 3\end{array}\right]$
- C
$\left[\begin{array}{cc}2 & -3 \\ 2 & 2\end{array}\right]$
- D
$\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]$
(a) : $A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$
$\begin{array}{l}|A|=4+6=10 \neq 0 \text { and }|B|=0+1=1 \neq 0 \\ \therefore \quad A^{-1}, B^{-1} \text { exists } \\ \text { So, }\left(B^{-1} A^{-1}\right)^{-1}=\left(A^{-1}\right)^{-1}\left(B^{-1}\right)^{-1}=A B \\ =\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}0+2 & -2+0 \\ 0+2 & 3+0\end{array}\right]=\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]\end{array}$
View full question & answer→MCQ 691 Mark
If $A$ and $B$ are invertible matrices, then which of the following is not correct?
- A
$\operatorname{adj} A=|A| \cdot A^{-1}$
- B
$\operatorname{det}(A)^{-1}=[\operatorname{det}(A)]^{-1}$
- C
$(A B)^{-1}=B^{-1} A^{-1}$
- ✓
$(A+B)^{-1}=B^{-1}+A^{-1}$
AnswerCorrect option: D. $(A+B)^{-1}=B^{-1}+A^{-1}$
View full question & answer→MCQ 701 Mark
Using determinants, find the area (in sq. units) of triangle with vertices $(-3,5),(3,-6)$ and $(7,2)$.
Answer(d) : Area of triangle
$
\begin{array}{l}
=\frac{1}{2}\left|\begin{array}{rrr}
-3 & 5 & 1 \\
3 & -6 & 1 \\
7 & 2 & 1
\end{array}\right|=\frac{1}{2}[-3(-6-2)-5(3-7)+1(6+42)] \\
=\frac{1}{2}[24+20+48]=\frac{1}{2} \times 92=46 \text { sq. units }
\end{array}
$
View full question & answer→MCQ 711 Mark
Find the cofactor of the element of third row and second column of the following determinant
$
\left|\begin{array}{lll}
1 & x & y+z \\
1 & y & z+x \\
1 & z & x+y
\end{array}\right| \text {. }
$
Answer(b) : $M_{32}=\left|\begin{array}{ll}1 & y+z \\ 1 & z+x\end{array}\right|=z+x-y-z=x-y$
$
\Rightarrow c_{32}=-M_{32}=y-x
$
View full question & answer→MCQ 721 Mark
If $A=\left[\begin{array}{cc}1+2 i & i \\ -i & 1-2 i\end{array}\right]$, where $i=\sqrt{-1}$, then $A(\operatorname{adj} A)=$
- ✓
$-2 I$
- B
$2 I$
- C
$5 I$
- D
$4 I$
AnswerCorrect option: A. $-2 I$
(a) : $|A|=6+1=7 \neq 0, \therefore A^{-1}$ exists.
$
\operatorname{adj} A=\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right] \therefore A^{-1}=\frac{1}{7}\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right]
$
View full question & answer→MCQ 731 Mark
Evaluate the following determinant:
$
\left|\begin{array}{cc}
x & -5 x \\
1 & x+10
\end{array}\right|
$
- A
$5 x^2+4$
- ✓
$x(x+15)$
- C
$x(x-15)$
- D
$x(15-x)$
AnswerCorrect option: B. $x(x+15)$
(b): We have, $\left|\begin{array}{cc}x & -5 x \\ 1 & x+10\end{array}\right|=x(x+10)+5 x$ $=x^2+10 x+5 x=x^2+15 x=x(x+15)$
View full question & answer→MCQ 741 Mark
If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, then find $A^{-1}$.
- ✓
$\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$
- B
$\frac{1}{7}\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$
- C
$\frac{1}{7}\left[\begin{array}{cc}-2 & 1 \\ -1 & -3\end{array}\right]$
- D
$\frac{1}{7}\left[\begin{array}{cc}2 & 1 \\ 1 & -3\end{array}\right]$
AnswerCorrect option: A. $\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$
(a) : $|A|=6+1=7 \neq 0, \therefore A^{-1}$ exists.
$\operatorname{adj} A=\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right] \quad \therefore A^{-1}=\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$
View full question & answer→MCQ 751 Mark
If the points $(2,-3),(k,-1)$ and $(0,4)$ are collinear, then find the value of $4 k$.
AnswerCorrect option: B. $7 / 140$
(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 761 Mark
If the system of equations $x+k y-z=0$, $3 x-k y-z=0$ and $x-3 y+z=0$ has non-zero solution, then $k$ is equal to
Answer(c) : The given system has non-zero solution, if
$
\begin{array}{l}
\left|\begin{array}{ccc}
1 & k & -1 \\
3 & -k & -1 \\
1 & -3 & 1
\end{array}\right|=0 \Rightarrow 1(-k-3)-k(3+1)-1(-9+k)=0 \\
\Rightarrow-6 k+6=0 \Rightarrow k=1
\end{array}
$
View full question & answer→MCQ 771 Mark
The value of $\Delta=\left|\begin{array}{ccc}3 & 7 & 13 \\ -5 & 0 & 0 \\ 0 & 11 & -2\end{array}\right|$ is
Answer(d) : On expanding along $R_2$, we get
$
\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
3 & 7 & 13 \\
-5 & 0 & 0 \\
0 & 11 & -2
\end{array}\right| \\
& =5(-14-143)+0-0=-785 .
\end{aligned}
$
View full question & answer→MCQ 781 Mark
The value of $\left|\begin{array}{rrr}\cos (\alpha+\beta) & -\sin (\alpha+\beta) & \cos 2 \beta \\ \sin \alpha & \cos \alpha & \sin \beta \\ -\cos \alpha & \sin \alpha & \cos \beta\end{array}\right|$ is independent of
- ✓
$\alpha$
- B
$\beta$
- C
$\alpha, \beta$
- D
AnswerCorrect option: A. $\alpha$
(a) : Expanding given determinant along $R_1$, we get $\cos ^2(\alpha+\beta)+\sin ^2(\alpha+\beta)+\cos 2 \beta=1+\cos 2 \beta$,
Which is independent of $\alpha$.
View full question & answer→MCQ 791 Mark
Find the values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$.
- A
$2 \sqrt{2}$
- B
$-2 \sqrt{2}$
- C
- ✓
Answer(d) : We have, $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$
$
\Rightarrow \quad 3-x^2=3-8 \quad \Rightarrow \quad x^2=8 \text {. Hence, } x= \pm 2 \sqrt{2}
$
View full question & answer→MCQ 801 Mark
If $A=\left[\begin{array}{ccc}2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3\end{array}\right]$, then $A^{-1}$ exists only if
- A
$\lambda=2$
- B
$\lambda \neq 2$
- C
$\lambda \neq-2$
- ✓
$\lambda \neq-8 / 5$
AnswerCorrect option: D. $\lambda \neq-8 / 5$
(d) : $A^{-1}$ exists if $|A| \neq 0$
$
\begin{array}{l}
\text { i.e., }\left|\begin{array}{ccc}
2 & \lambda & -3 \\
0 & 2 & 5 \\
1 & 1 & 3
\end{array}\right| \neq 0 \Rightarrow 2(6-5)+1(5 \lambda+6) \neq 0 \\
\Rightarrow 2+5 \lambda+6 \neq 0 \Rightarrow 5 \lambda \neq-8 \text { i.e., } \lambda \neq \frac{-8}{5}
\end{array}
$
View full question & answer→MCQ 811 Mark
Find the area of the triangle whose vertices are $(-2,6),(3,-6)$ and $(1,5)$.
AnswerCorrect option: D. $15.5 sq$. units
(d) : Let $\Delta$ be the area of the triangle then,
$\begin{array}{l}\Delta=\frac{1}{2}\left|\begin{array}{ccc}-2 & 6 & 1 \\ 3 & -6 & 1 \\ 1 & 5 & 1\end{array}\right|=\frac{1}{2}|-2(-6-5)-6(3-1)+1(15+6)| \\ =15.5 \text { sq. units }\end{array}$
View full question & answer→MCQ 821 Mark
$\left|\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right|=$
AnswerCorrect option: D. $a^2+b^2+c^2+d^2$
(d): We have, $\left|\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right|$
View full question & answer→MCQ 831 Mark
If $\Delta=\left|\begin{array}{ccc}l+m & m+n & n+l \\ n & l & m \\ 2 & 2 & 2\end{array}\right|$, then
- A
$-l+m+n=0$
- B
$l+m-n=0$
- C
$l-m-n=0$
- ✓
$\Delta=0$
AnswerCorrect option: D. $\Delta=0$
(d) : We have, $\Delta=(l+m)[2 l-2 m]$
$
\begin{array}{l}
-(m+n)[2 n-2 m]+(n+l)[2 n-2 l] \\
=2\left[l^2-m^2-n^2+m^2+n^2-l^2\right]=0
\end{array}
$
View full question & answer→MCQ 841 Mark
Find the value of $\Delta=\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$.
Answer$
\begin{array}{l}
\text { (a) : We have, } \Delta=\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right| \\
\Rightarrow \Delta=(a-b)\left[(b-c)(c-a)-(a-b)^2\right]-(b-c)\left[(b-c)^2\right. \\
-(a-b)(c-a)]+(c-a)\left[(b-c)(a-b)-(c-a)^2\right] \\
=0 \\
\end{array}
$
View full question & answer→MCQ 851 Mark
If $A$ is any square matrix of order $3 \times 3 such$ that $|A|=3$, then the value of $|\operatorname{adj} A|$ is
Answer(c) : Given, $A$ is a square matrix of order $3 \times 3$ and $|A|=3$.
Now, $|\operatorname{adj} A|=|A|^{n-1}$, where $n$ is order of $A$.
$\Rightarrow|\operatorname{adj} A|=(3)^2=9$
View full question & answer→MCQ 861 Mark
Find the cofactor of each element of the first column of matrix $A=\left[\begin{array}{ccc}2 & 5 & -1 \\ -3 & 0 & 1 \\ 1 & 1 & -1\end{array}\right]$.
- A
$-1,4,5$
- B
$-4,5,-1$
- C
$4,5,1$
- D
$-4,-5,1$
Answer$
\begin{array}{l}
\text { (a): } M_{11}=\left|\begin{array}{cc}
0 & 1 \\
1 & -1
\end{array}\right|=0-1=-1 \Rightarrow C_{11}=M_{11}=-1 \\
M_{21}=\left|\begin{array}{cc}
5 & -1 \\
1 & -1
\end{array}\right|=-5+1=-4 \Rightarrow C_{21}=-M_{21}=4 \\
M_{31}=\left|\begin{array}{cc}
5 & -1 \\
0 & 1
\end{array}\right|=5-0=5 \Rightarrow C_{31}=M_{31}=5
\end{array}
$
View full question & answer→MCQ 871 Mark
The inverse of the matrix $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{array}\right]$
- A
is $\left[\begin{array}{rrr}1 & -1 & 1 \\ -2 & 3 & 0 \\ 18 & -5 & 10\end{array}\right]$
- B
is $\left[\begin{array}{rrr}-4 & -3 & 2 \\ -8 & -12 & 6 \\ -3 & 4 & \frac{1}{8}\end{array}\right]$
- C
is $\left[\begin{array}{ccc}\frac{1}{4} & \frac{1}{3} & \frac{1}{2} \\ 1 & 0 & 0 \\ 5 & 8 & 9\end{array}\right]$
- ✓
Answer(d) : $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{array}\right]$
$\therefore \quad|A|=1(30-0)+1(20-0)+1(4-54)=30+20-50=0$
So, $A^{-1}$ does not exist.
View full question & answer→MCQ 881 Mark
A matrix $A$ of order $3 \times 3$ has determinant What is the value of $|3 A|$ ?
Answer(a) : Given, $|A|=5$, order of $A$ is $3 \times 3$.
$\therefore \quad|3 A|=3^3|A|=27 \times 5=135$.
View full question & answer→MCQ 891 Mark
What positive value of $x$ makes the following pair of determinants equal?
$
\left|\begin{array}{cc}
2 x & 3 \\
5 & x
\end{array}\right|,\left|\begin{array}{cc}
16 & 3 \\
5 & 2
\end{array}\right|
$
Answer(d) : We have, $\left|\begin{array}{rr}2 x & 3 \\ 5 & x\end{array}\right|=\left|\begin{array}{rr}16 & 3 \\ 5 & 2\end{array}\right|$
$
\begin{array}{l}
\Rightarrow 2 x^2-15=32-15 \Rightarrow 2 x^2=32 \Rightarrow x^2=16 \\
\Rightarrow \quad x=4[\because x>0]
\end{array}
$
View full question & answer→