Question 511 Mark
Evaluate : $\left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\ \sin 75^{\circ} & \cos75^{\circ}\end{array}\right|$
AnswerWe have, $\left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\ \sin 75^{\circ} & \cos 75^{\circ}\end{array}\right|$
$=\cos 15^{\circ} \times \cos 75^{\circ}-\sin 15^{\circ} \times \sin 75^{\circ}=\cos \left(15^{\circ}+75^{\circ}\right)$
$=\cos 90^{\circ}=0$
View full question & answer→Question 521 Mark
Evaluate the following determinant : $\left|\begin{array}{cc}x & -7 \\ x & 5 x+1\end{array}\right|$
Answer(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→Question 531 Mark
Find the area of the triangle whose vertices are $(-2,6),(3,-6)$ and $(1,5)$.
AnswerLet $\Delta$ 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→Question 541 Mark
$\left|\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right|=$
Answer(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→Question 551 Mark
If $\Delta=\left|\begin{array}{ccc}l+m & m+n & n+l \\ n & l & m \\ 2 & 2 & 2\end{array}\right|,$ then
AnswerWe 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→Question 561 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 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| $
$\Rightarrow \Delta=(a-b)\left[(b-c)(c-a)-(a-b)^2\right]-(b-c)\left[(b-c)^2\right.$
$-(a-b)(c-a)]+(c-a)\left[(b-c)(a-b)-(c-a)^2\right]$
$=0$
View full question & answer→Question 571 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→Question 581 Mark
Find the cofactor of each element of the first column of matrix $A=\left[\begin{array}{ccc}2 & 5 & -1 \\ -3 & 0 & 1 \\ 1 & 1 & -1\end{array}\right].$
Answer$M_{11}=\left|\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right|=0-1=-1 \Rightarrow C_{11}=M_{11}=-1$
$\begin{aligned} & M_{21}=\left|\begin{array}{cc}5 & -1 \\ 1 & -1\end{array}\right|=-5+1=-4 \Rightarrow C_{21}=-M_{21}=4 \end{aligned} $
$ 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→Question 591 Mark
The inverse of the matrix $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{array}\right]$
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→Question 601 Mark
If $A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]$, then adj $A$ is equal to
Answer(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→Question 611 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
AnswerExpanding along $R_1$, we get
$\Delta=0-\sin \alpha(0-\sin \beta \cos \alpha)-\cos \alpha(\sin \alpha \sin \beta-0)$
$=\sin \alpha \sin \beta \cos \alpha-\cos \alpha \sin \alpha \sin \beta=0$
View full question & answer→Question 621 Mark
If $A$ and $B$ are invertible matrices, then which of the following is not correct?
View full question & answer→Question 631 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→Question 641 Mark
If $A=\left[\begin{array}{cc}1+2 i & i \\ -i & 1-2 i\end{array}\right]$, where $i=\sqrt{-1}$, then $A(\operatorname{adj} A)=$
Answer(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→Question 651 Mark
Evaluate the following determinant:
$
\left|\begin{array}{cc}
x & -5 x \\
1 & x+10
\end{array}\right|
$
Answer(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→Question 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→Question 671 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→Question 681 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 $x.$
AnswerExpanding the given determinant, we get
$x\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+ x \cos \theta)=8$
$\Rightarrow-x^3-x+x=8 $
$\Rightarrow x^3+8=0$
$\Rightarrow(x+2)\left(x^2-2 x+4\right)=0$
$\Rightarrow x+2=0$
$\Rightarrow x=-2 \left[\because x^2-2 x+4>0 \forall x\right]$
$\left[\because x^2-2 x+4>0 \forall x\right]$
View full question & answer→Question 691 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→Question 701 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→Question 711 Mark
Find the cofactors of elements $a_{12}, a_{22}, a_{32}$ respectively of the matrix $\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$.
View full question & answer→Question 721 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→Question 731 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→Question 741 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→Question 751 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→Question 761 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→Question 771 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, $\left|\begin{array}{lll}a & 4 & 1 \\ b & 3 & 1 \\ c & 2 & 1\end{array}\right|=0$
$\Rightarrow a(3-2)-4(b-c)+1(2 b-3 c)=0$
$\Rightarrow a-2 b+c=0$
View full question & answer→Question 781 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→Question 791 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→Question 801 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
$=\frac{1}{2}\left|\begin{array}{rrr} -3 & 5 & 1 \\3 & -6 & 1 \\ 7 & 2 & 1 \end{array}\right|$
$=\frac{1}{2}[-3(-6-2)-5(3-7)+1(6+42)] $
$=\frac{1}{2}[24+20+48]$
$=\frac{1}{2} \times 92=46^2 \text { units }$
View full question & answer→Question 811 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→Question 821 Mark
If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, then find $A^{-1}$.
Answer(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→Question 831 Mark
If the points $(2,-3),(k,-1)$ and $(0,4)$ are collinear, then find the value of $4 k$.
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→Question 841 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→Question 851 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→Question 861 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→Question 871 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
Answer(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→Question 881 Mark
Find the values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$.
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→Question 891 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
Answer$A^{-1}$ exists if $|A| \neq 0$
$\text { i.e., }\left|\begin{array}{ccc} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3\end{array}\right| \neq 0$
$ \Rightarrow 2(6-5)+1(5 \lambda+6) \neq 0$
$\Rightarrow 2+5 \lambda+6 \neq 0 $
$\Rightarrow 5 \lambda \neq-8 $
$\text { i.e., } \lambda \neq \frac{-8}{5}$
View full question & answer→MCQ 901 Mark
If $\left|\begin{array}{ll}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{ll}6 & 2 \\ 18 & 6\end{array}\right|$ then the value of $x$ is-
AnswerCorrect option: B. $\pm 6$
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