MCQ 2011 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?
- A
$\pm 3$
- B
$-3$
- ✓
$\pm 2$
- D
$2$
AnswerCorrect option: C. $\pm 2$
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 2 x^2+15=20+3 $
$\Rightarrow 2 x^2=23-15=8$
$\Rightarrow x^2=4 $
$\Rightarrow x= \pm 2$
View full question & answer→MCQ 2021 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 2031 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 2041 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 2051 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 2061 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 2071 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 2081 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 2091 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 2101 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 2111 Mark
If $A$ is a non-singular square matrix of order $3$ such that $A^2=3 A$, then value of $|A|$ is
AnswerGiven $A^2=3 A$
$\Rightarrow\left|A^2\right|=|3 A|$
$\Rightarrow|A|^2=3^3|A|$
$\Rightarrow|A|^2-27|A|=0$
$\Rightarrow|A|[|A|-27]=0$
As, $A$ is a non$-$singular matrix
$\therefore |A| \neq 0$
$\Rightarrow|A|-27=0$
$\Rightarrow|A|=27$
View full question & answer→MCQ 2121 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. $\frac{1}{5}\left[\begin{array}{lr}-3 & 2 \\ 4 & -1\end{array}\right]$
We have, $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$ such that $A X=I$
$ \Rightarrow X=A^{-1} I$
$\because |A|=3-8=-5 \neq 0$
$ \Rightarrow A^{-1} \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] $
$\therefore X=\frac{1}{5}\left[\begin{array}{cc} -3 & 2 \\ 4 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right]$
$=\frac{1}{5}\left[\begin{array}{cc} -3 & 2 \\ 4 & -1 \end{array}\right]$
View full question & answer→MCQ 2131 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. $\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]$
$A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$
$|A|=4+6=10 \neq 0 $ and $|B|=0+1=1 \neq 0$
$\therefore 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]$
$=\left[\begin{array}{cc}0+2 & -2+0 \\ 0+2 & 3+0\end{array}\right]$
$=\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]$
View full question & answer→MCQ 2141 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 2151 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 2161 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 2171 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 2181 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 2191 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 2201 Mark
Find the area of the triangle whose vertices are $(-2,6),(3,-6)$ and $(1,5)$.
- A
$30$ sq. units
- B
$35$ sq. units
- C
$40$ sq. units
- ✓
$15.5$ sq. units
AnswerCorrect option: D. $15.5$ sq. units
Let $\Delta$ be the area of the triangle then,
$\Delta=\frac{1}{2}\left|\begin{array}{ccc}-2 & 6 & 1 \\ 3 & -6 & 1 \\ 1 & 5 & 1\end{array}\right|$
$=\frac{1}{2}|-2(-6-5)-6(3-1)+1(15+6)|$
$=15.5 \text { sq. units }$
View full question & answer→MCQ 2211 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 2221 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 2231 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 2241 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 2251 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 2261 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. $\Delta=0$
We have, $\Delta=(l+m)[2 l-2 m]$
$-(m+n)[2 n-2 m]+(n+l)[2 n-2 l]$
$=2\left[l^2-m^2-n^2+m^2+n^2-l^2\right]=0$
View full question & answer→MCQ 2271 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 2281 Mark
Find the value of $\Delta=\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$.
AnswerWe 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| $
$\Rightarrow \Delta=(a-b)\left[(b-c)(c-a)-(a-b)^2\right]-(b-c)\left[(b-c)^2\right.$
$-(a-b)(c-a)]+(c-a)\left[(b-c)(a-b)-(c-a)^2\right]$
$=0$
View full question & answer→MCQ 2291 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
$\pm 3$
- ✓
$\pm 6$
- D
$\pm 4$
AnswerCorrect option: C. $\pm 6$
$\left|\begin{array}{cc} 2 x & 5 \\ 8 & x \end{array}\right|=\left|\begin{array}{cc} 6 & -2 \\ 7 & 3
\end{array}\right|$
$ \Rightarrow 2 x^2-40=18+14$
$\Rightarrow x^2=36$
$ \Rightarrow x= \pm 6$
View full question & answer→MCQ 2301 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 2311 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 2321 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
AnswerThe given system has non $-$ zero solution, if
$\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$
View full question & answer→MCQ 2331 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 2341 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 2351 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 2361 Mark
Evaluate: $\left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\ \sin 75^{\circ} & \cos 75^{\circ}\end{array}\right|$
AnswerWe have, $\left|\cos 15^{\circ} \sin 15^{\circ} \sin 75^{\circ} \cos 75^{\circ}\right|$
$=\cos 15^{\circ} \times \cos 75^{\circ}-\sin 15^{\circ} \times \sin 75^{\circ}=\cos \left(15^{\circ}+75^{\circ}\right)$
$=\cos 90^{\circ}$
$=0$
View full question & answer→MCQ 2371 Mark
Find the cofactor of each element of the first column of matrix $A=\left[\begin{array}{ccc}2 & 5 & -1 \\ -3 & 0 & 1 \\ 1 & 1 & -1\end{array}\right]$.
- ✓
$-1,4,5$
- B
$-4,5,-1$
- C
$4,5,1$
- D
$-4,-5,1$
AnswerCorrect option: A. $-1,4,5$
$M_{11}=\left|\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right|$
$=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 $
$\Rightarrow C_{31}=M_{31}=5$
View full question & answer→MCQ 2381 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 2391 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 2401 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. $\lambda \neq-8 / 5$
$A^{-1}$ exists if $|A| \neq 0$
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 $
$\Rightarrow 5 \lambda \neq-8 \text { i.e., } \lambda \neq \frac{-8}{5}$
View full question & answer→MCQ 2411 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 2421 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 2431 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 2441 Mark
Using determinants, find the area $($in sq. units$)$ of triangle with vertices $(-3,5),(3,-6)$ and $(7,2)$.
AnswerArea of triangle
$ =\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)] $
$=\frac{1}{2}[24+20+48]=\frac{1}{2} \times 92=46 \text { sq. units }$
View full question & answer→MCQ 2451 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 2461 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 2471 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 2481 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 2491 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 $a_{32} \cdot c_{32}$.
AnswerHere, $a_{32}=5$, then,
$ c_{32}=(-1)^{3+2}\left|\begin{array}{ll} 2 & 5 \\ 6 & 4 \end{array}\right|$
$=(-1)^5(8-30)=-(-22)=22$
$\therefore a_{32} \cdot c_{32}=5 \times 22=110$
View full question & answer→MCQ 2501 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 $a-2 b+c$.
AnswerFor non$-$trivial solution, $\left|\begin{array}{lll}a & 4 & 1 \\ b & 3 & 1 \\ c & 2 & 1\end{array}\right|=0$
$\Rightarrow a(3-2)-4(b-c)+1(2 b-3 c)=0$
$\Rightarrow a-2 b+c=0$
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