MCQ Questions for Class 12 Commerce Maths Determinants Quiz 1

The number of distinct real roots of the equation,

$\begin{vmatrix} \cos x& \sin x & \sin x\\ \sin x & \cos x & \sin x\\ \sin x & \sin x & \cos x\end{vmatrix} = 0$

in the interval

$\left [-\dfrac {\pi}{4}, \dfrac {\pi}{4}\right ]$ is/are :

0%
$3$
0%
$2$
0%
$1$
0%
$4$

Explanation

The given determinant is :

$\begin{vmatrix} \cos x& \sin x & \sin x\\ \sin x & \cos x & \sin x\\ \sin x & \sin x & \cos x\end{vmatrix} = 0$

Taking

$\cos^3 x$ common in the above determinant, we get,

$\cos^3 x\begin{vmatrix} \ 1& tanx & tanx\\ tanx & \ 1 & tanx\\ tanx & tanx & \ 1\end{vmatrix} = 0$

$\cos^3x[1(1-\tan^2 x)-tanx(tanx-tan^2 x) + tanx(tan^2x-tanx)]=0$

$\cos^3x[1-3tan^2 x+ 2tan^3x]=0$

We notice that if

$\cos^3x = 0 \rightarrow x=\dfrac{\pi}{2}$

This is not satisfied by the given interval

Thus,

$x=\dfrac{\pi}{2}$ is not a solution.

Now,

$1-3tan^2 x+ 2tan^3x=0$

It is obvious from the above equation that

$tanx=1$ is a solution.

Thus,

$x=\dfrac{\pi}{4}$ is a solution.

We just found out that

$(tanx -1)$ is a factor of the polynomial

$1-3tan^2 x+ 2tan^3x=0$

Thus, dividing the polynomial by

$(tanx-1)$ , we get the quotient as

$2tanx^2 - tanx -1$ and remainder

$0$

Thus, this can be written as

$(2tanx+1)(tanx-1)=0$

Or,

$tanx=\dfrac{-1}{2}$ or

$tanx=1$

We already know that

$tanx=1$ is a solution to the above equation.

Thus,

$tanx=\dfrac{-1}{2}$

Or,

$x= tan^{-1}\left(\dfrac{-1}{2}\right)$

This also lies in the interval

$\left[\dfrac{-\pi}{4}, \dfrac{\pi}{4}\right]$
Thus there are two distinct real roots to the above equation.

The points

$\displaystyle \left( 0, \frac{8}{3} \right), (1, 3)$ and

$(82, 30)$ :

0%
form an obtuse angled triangle.
0%
form a right angled triangle.
0%
lie on a straight line.
0%
form an acute angled triangle.

Explanation

Given

$\displaystyle \left( 0, \frac{8}{3} \right), (1, 3)$ and

$(82, 30)$

$\begin{vmatrix} 0 & \dfrac { 8 }{ 3 } & 1 \\ 1 & 3 & 1 \\ 82 & 30 & 1 \end{vmatrix}$

$=216-216=0$

Hence, the given points are collinear.

$A=\begin {bmatrix} 2 & -3\\ -4 & 1\end{bmatrix}$ , then adj

$(3A^2+12A)$ is equal to.

0%
$\begin{bmatrix} 72 & -84\\ -63 & 51\end{bmatrix}$
0%
$\begin{bmatrix} 51 & 63\\ 84 & 72\end{bmatrix}$
0%
$\begin{bmatrix} 51 & 84 \\ 63 & 72\end{bmatrix}$
0%
$\begin {bmatrix} 72 & -63\\ -84 & 51\end{bmatrix}$

Explanation

It is given that

$A =\begin{bmatrix} 2 & -3\\ -4 & 1 \end{bmatrix}$

Hence,

$A^2 = \begin {bmatrix} 2 & -3 \\ -4 &1 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} =$

$\begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix}$

Now,

$3A^2 = \begin{bmatrix} 48 & -27 \\ -36 & 29\end{bmatrix}$

$12A=\begin{bmatrix}24 & -36 \\ -48 & 12\end{bmatrix}$

Consider,

$3A^2+12A$

$=\begin{bmatrix}48 & -27 \\ -36 & 29\end{bmatrix}+\begin{bmatrix}24 & -36 \\ -48 & 12\end{bmatrix}$

$=\begin{bmatrix}72 & -63 \\ -84 & 51\end{bmatrix}$

$\therefore 3A^2+12A=\begin{bmatrix}72 & -63 \\ -84 & 51\end{bmatrix}$

Hence,

$\text{adj}(3A^2+12A)=\begin{bmatrix} 51 & 84 \\ 63 & 72\end{bmatrix}^T=\begin{bmatrix}51&63\\84&72\end{bmatrix}$

Let

$P = [a_{ij}]$ be a 3

$\times$ 3 matrix and let

$Q = [b_{ij}]$ , where

$b_{ij} = 2^{i + j} a_{ij}$ for

$1 \leq i, j \leq 3$ . If the determinant of P is 2, then the determinant of the matrix Q is

0%
$2^{10}$
0%
$2^{11}$
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$2^{12}$
0%
$2^{13}$

Explanation

Given,

$[P]=a_{ij}$ is a order 3 matrix.

Let

$Q=\begin{bmatrix} { 2 }^{ 2 }{ a }_{ 11 } & { 2 }^{ 3 }{ a }_{ 12 } & { 2 }^{ 4 }{ a }_{ 13 } \\ { 2 }^{ 3 }{ a }_{ 21 } & { 2 }^{ 4 }{ a }_{ 22 } & { 2 }^{ 5 }{ a }_{ 23 } \\ { 2 }^{ 4 }{ a }_{ 31 } & { 2 }^{ 5 }{ a }_{ 32 } & { 2 }^{ 6 }{ a }_{ 33 } \end{bmatrix}$

$|Q|=\begin{vmatrix} { 2 }^{ 2 }{ a }_{ 11 } & { 2 }^{ 3 }{ a }_{ 12 } & { 2 }^{ 4 }{ a }_{ 13 } \\ { 2 }^{ 3 }{ a }_{ 21 } & { 2 }^{ 4 }{ a }_{ 22 } & { 2 }^{ 5 }{ a }_{ 23 } \\ { 2 }^{ 4 }{ a }_{ 31 } & { 2 }^{ 5 }{ a }_{ 32 } & { 2 }^{ 6 }{ a }_{ 33 } \end{vmatrix}$

$\Rightarrow Q={ 2 }^{ 2 }{ 2 }^{ 3 }{ 2 }^{ 4 }\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { 2 }{ a }_{ 21 } & { 2 }{ a }_{ 22 } & { 2 }{ a }_{ 23 } \\ { 2 }^{ 2 }{ a }_{ 31 } & { 2 }^{ 2 }{ a }_{ 32 } & { 2 }^{ 2 }{ a }_{ 33 } \end{bmatrix}$

$\Rightarrow |Q|={ 2 }^{ 9 }{ \times 2 }\times { 2 }^{ 2 }\begin{vmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{vmatrix}$

$\Rightarrow |Q|=2^{12}|P|$

$\Rightarrow |Q|=2^{13}$

Consider three points

${P}=(-\sin(\beta-\alpha), -\cos\beta) , {Q}=(\cos(\beta-\alpha), \sin\beta)$ and

${R}=(\cos(\beta-\alpha +\theta), \sin(\beta-\theta))$ , where

$0< \alpha,\ \beta,\ \theta <\displaystyle \frac{\pi}{4}$ . Then

0%
$P$ lies on the line segment $RQ$
0%
$Q$ lies on the line segment $PR$
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$R$ lies on the line segment $QP$
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$P, Q, R$ are non-collinear

Explanation

$\mathrm{P}\equiv(-\sin(\beta-\alpha), -\cos\beta)\equiv(\mathrm{x}_{1}, \mathrm{y}_{1})$

$\mathrm{Q}\equiv(\cos(\beta-\alpha), \sin\beta)\equiv(\mathrm{x}_{2}, \mathrm{y}_{2})$
and

$\mathrm{R}\equiv(\mathrm{x}_{2}\cos\theta+\mathrm{x}_{1}\sin\theta, \mathrm{y}_{2}\cos\theta+\mathrm{y}_{1}\sin\theta)$
We see that

$\displaystyle \mathrm{T}\equiv(\frac{\mathrm{x}_{2}\cos\theta+\mathrm{x}_{1}\sin\theta}{\cos\theta+\sin\theta}, \frac{\mathrm{y}_{2}\cos\theta+\mathrm{y}_{1}\sin\theta}{\cos\theta+\sin\theta})$ and

$\mathrm{P},\ \mathrm{Q},\ \mathrm{T}$ are collinear

$\Rightarrow \mathrm{P},\ \mathrm{Q},\ \mathrm{R}$ are non-collinear.

The number of

$A$ in

$T_p$ such that

$A$ is either symmetric or skew-symmetric or both, and det

$(A)$ divisible by

$p$ , is

0%
$(p - 1)^2$
0%
$2(p - 1)$
0%
$(p - 1)^2 + 1$
0%
$2 p - 1$

Explanation

Given

$A=\begin{bmatrix} a\quad & b \\ c & a \end{bmatrix},a,b,c\in \left\{ 0,1,2,...,p-1 \right\}$
If A is skew -symmetric matrix, then a =0, b=-c

$\therefore \left| A \right| =-{ \left| b \right| }^{ 2 }$
Thus, p divides

$\left| A \right|$ only when

$b=0$ ...(1)
Again, if A is symmetric matrix, then

$v=c$ and

$\left| A \right| ={ a }^{ 2 }-{ b }^{ 2 }$
Thus p divides

$\left| A \right|$ is either p divides

$(a-b)$ or o divides

$(a+b)$
p divides

$(a-b)$ only when

$a=b$
i.e

$a=b\in \left\{ 0,1,2,...,\left( p-1 \right) \right\}$
i.e p choice
p divides

$(a+b)$

$\Rightarrow$ p choice including

$a=b=0$ in (1)
Therefore total number of choices are

$\left( p+p+1 \right) =2p-1$

Let k be a positive real number and let

$A = \begin{bmatrix}2k-1 & 2\sqrt{k} & 2\sqrt{k}\\ 2\sqrt{k} & 1 & -2k\\ -2\sqrt{k} & 2k & -1\end{bmatrix}$ and

$B=\begin{bmatrix}0 & 2k-1 & \sqrt{k}\\ 1-2k & 0 & 2\sqrt{k}\\ -\sqrt{k} & -2\sqrt{k} & 0\end{bmatrix}$ .

If det

$(adj A) + det (adj B) = 10^{6}, then [k]$ is equal to

[Note: adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k].

0%
$3$
0%
$4$
0%
$5$
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$6$

Explanation

$|\mathrm{A}|=(2\mathrm{k}+1)^{3},\ |\mathrm{B}|=0$ (Since

$\mathrm{B}$ is a skew-symmetric matrix of order 3)

$\Rightarrow \det(adj \mathrm{A}) =|\mathrm{A}|^{\mathrm{n}-1}=((2\mathrm{k}+1)^{3})^{2}=10^6\Rightarrow 2\mathrm{k}+1=10\Rightarrow 2\mathrm{k}=9$

$[\mathrm{k}]=4$ .

The value of

$|\mathrm{U}|$ is

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3
0%
$-3$
0%
3/2
0%
2

Explanation

$A=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$

$|A|=1$

$Adj\: A=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$

$\Rightarrow A^{-1}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$

$AU_1=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$

$\Rightarrow U_1=A^{-1}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}=\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$

$AU_2=\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$

$\Rightarrow U_2=A^{-1}\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}=\begin{bmatrix} 2 \\ -1 \\ -4 \end{bmatrix}$

$AU_3=\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$

$\Rightarrow U_3=A^{-1}\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}=\begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix}$

$\therefore U=\begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix}$

$U^{-1}=\displaystyle\frac{1}{3}\begin{bmatrix} -1 & -2 & 0 \\ -7 & -5 & -3 \\ 9 & 6 & 3 \end{bmatrix}$

$\therefore$ Sum of elements of

$U^{-1}$ is

$0$ .

Hence, option B.

Hence

$U=\begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix}\Rightarrow \left| U \right| =3$

$A = \begin{bmatrix}1&2\\3&4\end{bmatrix}, B = \begin{bmatrix}2&1\\3&4\end{bmatrix}$ then

$\left|(B^TA^T)^{-1}\right|$ is equal to

0%
$10$
0%
$\dfrac{1}{10}$
0%
$1$
0%
$- 1$

Explanation

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

$B = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$

$B^{T}A^{T} = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 2 + 6 & 6 + 12 \\ 1 + 8 & 3 + 16 \end{bmatrix} = \begin{bmatrix} 8 & 18 \\ 9 & 19 \end{bmatrix}$

$det (B^{T}A^{T}) = 8 \times 19 - 9 \times 18 = 152 - 162 = -10$

$(B^TA^T)^{-1} = \dfrac1{det(B^TA^T)} adj(B^TA^T) = \dfrac{1}{-10}\begin{bmatrix}19&-18\\-9&8\end{bmatrix}$

$det( (B^{T}A^{T})^{-1}) =\dfrac{19\times8 - 18\times9}{-10} = 1$

If three points

$(k, 2k), (2k, 3k), (3, 1)$ are collinear, then

$k$ is equal to:

0%
$-2$
0%
$1$
0%
$\displaystyle\frac{1}{2}$
0%
$-\displaystyle\frac{1}{2}$

Explanation

If given points are collinear, then

$\begin{vmatrix} k&2k&1\\2k&3k&1\\3&1&1\end{vmatrix}=0$

$\Rightarrow \begin{vmatrix} k&2k&1\\k&k&0\\3&1&1\end{vmatrix}=0, R_2\rightarrow R_2-R_1$

$\Rightarrow 1(k-3k)+1(k^2-2k^2)=0$

$\Rightarrow k^2+2k=0$

$\Rightarrow k=0,-2$

$\begin{vmatrix} 4\sin^{2}\theta & \cos^{2}\theta \\ 3\sec^{2}\theta & \text{cosec}^{2}\theta \end{vmatrix}=$

0%
$8\sin^{2}\theta \cos^{2}\theta$
0%
$4\sin 2\theta \cos 2\theta$
0%
$1$
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$4\cos^{3}\theta -3\cos \theta$

Explanation

The value of

$\begin{vmatrix} 4\sin^{2}\theta & \cos^{2}\theta \\ 3\sec^{2}\theta & \text{cosec}^{2}\theta \end{vmatrix}$ is

$=4\sin^{2}\theta .\text{cosec}^{2}\theta -3\sec^{2}\theta .\cos^{2}\theta$

$=4-3=1$

$\left|\begin{array}{lllll} 0 & & \mathrm{c}\mathrm{o}\mathrm{s}\alpha & \mathrm{c}\mathrm{o}\mathrm{s} & \beta\\ \mathrm{c}\mathrm{o}\mathrm{s} & \alpha & 0 & \mathrm{c}\mathrm{o}\mathrm{s} & \gamma\\ \mathrm{c}\mathrm{o}\mathrm{s} & \beta & \mathrm{c}\mathrm{o}\mathrm{s}\gamma & 0 & \end{array}\right|=$

0%
$\cos\alpha+\cos\beta+\cos\gamma$
0%
$\cos\alpha\cos\beta\cos\gamma$
0%
$2\cos\alpha\cos\beta\cos\gamma$
0%
$2 \displaystyle \sum \cos\alpha\cos\beta$

Explanation

Let

$A=\begin{vmatrix} 0 & cos \alpha & \cos \beta\\ \cos \alpha & 0 & cos \gamma\\ cos \beta & \cos \gamma & 0 \end{vmatrix}$

Expanding the determinant,

$A=-cos \alpha[-cos \beta cos \gamma]+cos \beta[cos \alpha \cos \gamma]$

$=2 \cos \alpha \cos \beta \cos \gamma$

$A=\left\{\begin{array}{ll} 8 & 9\\ 10 & 11 \end{array}\right\}$ , then cofactor of

$\mathrm{a}_{12}$ is:

0%
11
0%
10
0%
-11
0%
-10

Explanation

By property of cofactor (2),

$A=\begin{bmatrix} 8 & 9\\ 10 & 11 \end{bmatrix}$
minor of

$a_{12}=M_{12}=10$
So, cofactor of

$a_{12}=M_{12}(-1)^{1+2}=-M_{12}=-10$

If P =

$\begin{bmatrix} 1 & 4\\ 2 & 6 \end{bmatrix}$ ,then adj (P)

0%
$\begin{bmatrix}1 & 4\\ 2 & 6\end{bmatrix}$
0%
$\begin{bmatrix}6& -4\\ -2 & 1\end{bmatrix}$
0%
$\begin{bmatrix}6& -2\\ -4 & 1\end{bmatrix}$
0%
$\begin{bmatrix}2& 1\\ 6 & 4\end{bmatrix}$

Explanation

Cofactor of the martix is

$\begin{bmatrix} M11 & -M12 \\ -M21 & M22 \end{bmatrix}\\$

Where

$M11 = 6$

$M12 = 2$

$M21 = 4$

$M22 = 1$

Substituting all value we get

Cofactor of the matrix as

$\begin{bmatrix} 6 & -2 \\ -4 & 1 \end{bmatrix}\\$

The adjoint of the matrix is transpose of cofactor of the given matrix

$\therefore$ Adjoint of the matrix is

$\begin{bmatrix} 6 & -4 \\ -2 & 1 \end{bmatrix}$

Hence the answer is option B.

$\begin{vmatrix} x^{2}+3 &x-1 &x+3 \\ x+3 & -2x &x-4 \\ x-3& x+4 & 3x \end{vmatrix}$

$=px^{4}+qx^3+rx^{2}+sx+t,$ then

$t =$

0%
$72$
0%
$0$
0%
$24$
0%
$-48$

Explanation

On expanding the determinant we get,

$(x^2+3)(-6x^2-x^2+16)-(x-1)(3x^2+9x-x^2+7x-12)+(x+3)(x^2+7x+12-2x^2+6x)$

Let us consider only the constant terms,

since

$t$ is constant term

$\therefore (16)(3)-(-1)(-12)+3(12) =72$

Hence,

$t = 72$

Maximum value of a second order determinant whose every element is either 0,1 or 2 only is:

0%
0
0%
1
0%
2
0%
4

Explanation

So,

$A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ Given a,b,c & D can only be 0,1,2
det A= ad-bc
So for max. value of A,
a=2 and d=2 and

$b,c \epsilon {0,0}$
So, Max value of det

$A=\begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix}=4$

$\begin{vmatrix} cos(A+B) & -sin(A+B) &cos2B \\ sin A& cos A &sin B \\ -cos A& sin A & cos B \end{vmatrix}$ =0 then B=

0%
(2n+1) $\frac{\pi }{2}$
0%
n $\pi$
0%
(2n+1) $\pi$
0%
2n $\pi$

Explanation

$\Rightarrow cos(A+B)(cosAcos B-sin A sin B)+sin^2(A+B)+cos2B+0$

$\Rightarrow cos^2(A+B)+sin^2(A+B)+cos2B=0$

$\Rightarrow cos (2B)=-1$

$2B=(2n+1)\pi$

$B=(2n+1)\pi/2$

so

$(2n+1)\pi/2$

If A

$=\begin{bmatrix} 0 & c &-b \\ -c& 0& a\\ b & -a & 0 \end{bmatrix}$ then

$\left ( a^{2}+b^{2}-c^{2} \right )\left | A \right |=$

0%
$abc$
0%
$a + b + c$
0%
$\left ( a^{3}+b^{3}+c^{3} \right )$
0%
$0$

Explanation

$|A|=0\times(a^2)-c(-ab)-b(ac)$

$=0+abc-abc$

$=0$

Find x if it is given that:

$\det \left[\begin{array}{lll} 2 & 0 & 0\\ 4 & 3 & 0\\ 4 & 6 & x \end{array}\right]=42$

0%
$8$
0%
$7$
0%
$6$
0%
$21/4$

Explanation

Given

$det\begin{pmatrix} 2 & 0 & 0\\ 4 & 3 & 0\\ 4 & 6 & x \end{pmatrix}=42$
By operation of matrix (5)

$x(6)=42$

$\Rightarrow x=7$

$\mathrm{If}$

$\left|\begin{array}{lll} 1 & 0 & 0\\ 2 & 3 & 4\\ 5 & -6 & x \end{array}\right|$

$= 45$

$\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}$

$\mathrm{x}=$

0%
$4$
0%
$7$
0%
$- 5$
0%
$-7$

Explanation

Given

$\begin{vmatrix} 1 & 0 & 0\\ 2 & 3 & 4\\ 5 & -6 & x \end{vmatrix}=45$
By operation of matrix (5),

$1(3x+24)=45$

$3x=21$

$\Rightarrow x=7$

If A is a singular matrix, then A (adj A) is a

0%
scalar matrix
0%
zero matrix
0%
identity matrix
0%
orthogonal matrix

Explanation

Given

$A$ is a singular matrix.

$\Rightarrow |A|=0$

$A(adj\:A)=|A|I=0I=O$

$\therefore A(adj\:A)$ is a zero matrix.
Hence, option B.

If A is a singular matrix, then adj A is

0%
$\displaystyle non-singular$
0%
singular
0%
symmetric
0%
not defined

Explanation

Given

$|A|=0$

We know

$|adj A|=|A|^{n-1}$

$\therefore |adj A|=0$
Hence, adj A is singular

If the value of the determinant

$\begin{vmatrix}m & 2\\ -5 & 7\end{vmatrix}$ is

$31$ , find

$m$ .

0%
$3$
0%
$8$
0%
$6$
0%
$2$

Explanation

Given,

$\begin{vmatrix}m & 2\\ -5 & 7\end{vmatrix}$

$\Rightarrow 7m- (-10)=31$

$\Rightarrow 7m+ 10= 31$

$\Rightarrow 7m= 21$

$\Rightarrow m= 3$

$A$ is a

$3\times 3$ matrix and

$\text{det} (3A)=k(\text{det} A)$ , then

$k=$

0%
$9$
0%
$6$
0%
$1$
0%
$27$

Explanation

The non zero determinant of a scalar multiple of a n×n matrix is given by the following property.

$|kA|=kn|A|$

⟹

$|3A|=3^3|A|=27|A|$

$k=27$

If A and B are similar matrices such that

$det (AB)=0,$ then

0%
$det (A)=0$ and $det (B)=0$
0%
$det (A)=0$ and $det (B) \neq 0$
0%
$A=0$ and $B=0$
0%
$A=0$ or $B=0$

Explanation

Given. two n

$\times$ n square matrices A and B

Given,

$B=P^{-1}AP$

$det\left( B \right) =det\left( { P }^{ -1 }AP \right)$

$det\left( B \right) =\left( det\left( { P } \right) \right) ^{ -1 }\left( det\left( A \right) \right) \left( det\left( P \right) \right)$

$det\left( B \right) =det\left( A \right)$

$det\left( AB \right) =\left( det\left( A \right) \right) \left( det\left( B \right) \right)$

$\left( det\left( A \right) \right) =0$ then

$\left( det\left( B \right) \right)$ also zero.

Let a, b, c be three complex numbers, and let

$z=\begin{vmatrix} 0 & -b & -c\\ b & 0 & -a\\ c & a & 0 \end{vmatrix}$
then z equal

0%
$0$
0%
purely imaginary
0%
$abc$
0%
none of these

Explanation

$z=\begin{vmatrix} 0 & -b & -c \\ b & 0 & -a \\ c & a & 0 \end{vmatrix}=0\left( 0+{ a }^{ 2 } \right) +b\left( 0+ac \right) -c\left( ab-0 \right) =0$

Two

$n\times n$ square matrices A and B are said to be similar if there exists a non-singular matrix P such that

$P^{-1}A P=B$ .

If A and B are similar matrices such that

$det (A)=1$ , then

0%
$det (B)=1$
0%
$det (A)=-det (B)$
0%
$det (B)=-1$
0%
none of these

Explanation

Given, two n

$\times$ n square matrices A and B

Given,

$B=P^{-1}AP$

$det\left( B \right) =det\left( { P }^{ -1 }AP \right)$

$det\left( B \right) =\left( det\left( { P }^{ -1 } \right) \right) \left( det\left( A \right) \right) \left( det\left( P \right) \right)$

$det\left( B \right) =\left( det\left( { P } \right) \right) ^{ -1 }\left( det\left( A \right) \right) \left( det\left( P \right) \right)$

$[\because \quad|A||A^{-1}|=|AA^{-1}|=|I|=1]$

$det\left( B \right) =det\left( A \right) =1$

If A is a square matrix of order

$n$ then adj

$\left ( adj A \right )$ is equal to

0%
$\displaystyle \left | A \right |^{n}A$
0%
$\displaystyle \left | A \right |^{n-1}A$
0%
$\displaystyle \left | A \right |^{n-2}A$
0%
$\displaystyle \left | A \right |^{n-3}A$

Explanation

${ A }^{ -1 }=\cfrac { AdjA }{ detA } =\cfrac { I }{ A } =\cfrac { AdjA }{ \left| A \right| }$

$A(adjA)=\left| A \right|$

Replacing

$A=adjA$

$adjA(AdjAdjA)=\left| AdjA \right|$

$\Rightarrow Adj\left( AdjA \right) =\cfrac { \left| Adj(A) \right| }{ \left| A \right| } \times A$

$=\cfrac { { \left| A \right| }^{ n-1 } }{ { \left| A \right| } } \times A$

$={ \left| A \right| }^{ n-2 }\times A$

$Adj(AdjA)={ \left| A \right| }^{ n-2 }\times 2$

$n\rightarrow$ order

Option C is correct

$\displaystyle \left | \begin{matrix}-12 &0 &\lambda \\ 0& 2& -1\\ 2& 1 &15 \end{matrix} \right |=-360$ , then the value of

$\lambda$ ,is

0%
$-1$
0%
$-2$
0%
$-3$
0%
$4$

Explanation

$\left| \begin{matrix} -12 & 0 & \lambda \\ 0 & 2 & -1 \\ 2 & 1 & 15 \end{matrix} \right| =-360$

Expanding along first column

$\Rightarrow (-12)(31)-4\lambda =-360$

$\Rightarrow \lambda =-3$

$A$ is any skew-symmetric matrix of odd order then

$\left| A \right|$ equals

0%
$-1$
0%
$0$
0%
$1$
0%
none of these

Explanation

$A$ is skew symmetric matrix
then

$A=-A^T$
Therefore,

$|A|=-|A^T|=-|A|$

$\Rightarrow 2|A|=0$

$\Rightarrow|A|=0$

Ans: B

$\begin{vmatrix} x & y\\ 4 & 2 \end{vmatrix}=7$ and

$\begin{vmatrix} 2 & 3\\ y & x \end{vmatrix}=4$ then

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$x=-3, y=-\dfrac {5}{2}$
0%
$x=-\dfrac {5}{2}, y=-3$
0%
$x=-3, y=\dfrac {5}{2}$
0%
$x=-\dfrac {5}{2}, y=3$

Explanation

$\begin{vmatrix} x & y\\ 4 & 2 \end{vmatrix}=7$

$\Rightarrow 2x-4y=7$

$\begin{vmatrix} 2 & 3\\ y & x \end{vmatrix}=4$

$\Rightarrow 2x-3y=4$
Therefore,

$x=-5/2,y=-3$

Ans: B

$\begin{vmatrix}a & -b & -c\\-a & b & -c \\ -a & -b & -c\end{vmatrix}+\lambda abc=0$ , then

$\lambda$ is equal to

0%
$2$
0%
$4$
0%
$-2$
0%
$-4$

Explanation

$\begin{vmatrix} a & -b & -c \\ -a & b & -c \\ -a & -b & -c \end{vmatrix}+\lambda abc=0$

Substitute

$a=b=c=1$

$\begin{vmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & -1 \end{vmatrix}+\lambda =0$

$(-2-2)+\lambda=0$

$\Rightarrow \lambda =4$

The cofactors of elements in second row of the determinant

$\begin{vmatrix} 2 & -1 & 4 \\ 4 & 2 & -3 \\1 & 1 & 2 \end{vmatrix}$ are

0%
$5, 6, 4$
0%
$6, 0, -3$
0%
$5, 1, 8$
0%
$6, 0, 3$

Explanation

$\begin{vmatrix} 2 & -1 & 4 \\ 4 & 2 & -3 \\ 1 & 1 & 2 \end{vmatrix}\\ { C }_{ 21 }=-\begin{vmatrix} -1 & 4 \\ 1 & 2 \end{vmatrix}=6\\ { C }_{ 22 }=\begin{vmatrix} 2 & 4 \\ 1 & 2 \end{vmatrix}=0\\ { C }_{ 23 }=-\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}=-3$

A set of points which do not lie on the same line are called as

0%
collinear
0%
non-collinear
0%
concurrent
0%
square

$A$ and

$B$ are two points and

$C$ is any point collinear with

$A$ and

$B$ . IF

$AB=10$ ,

$BC=5$ , then

$AC$ is equal to:

0%
either $15$ or $5$
0%
necessarily $5$
0%
necessarily $16$
0%
none of these

Explanation

Since

$C$ is collinear with

$A$ and

$B, C$ lies either
(i) to the left of point B or
(ii) to the right of point B

$\therefore$ In case (i)

$AC=AB-BC=10-5=5$
In case (ii)

$AC=AB+BC=10+5=15$

$A=\begin{bmatrix} 3 & -5 \\ -1 & 0 \end{bmatrix}$ , then adj.

$A$ is equal to

0%
$\begin{bmatrix} 3 & 5 \\ 1 & 0 \end{bmatrix}$
0%
$\begin{bmatrix} 0 & -5 \\ -1 & 3 \end{bmatrix}$
0%
$\begin{bmatrix} 0 & 5 \\ 1 & 3 \end{bmatrix}$
0%
$\begin{bmatrix} 0 & -1 \\ -5 & 3 \end{bmatrix}$

Explanation

co-factor of

$A_{11}=0$ ; co-factor of

$A_{12}=5$ ; co-factor of

$A_{21}=1$ ; co-factor of

$A_{22}=3$

$adjA=\begin{bmatrix} 0 & 5 \\ 1 & 3 \end{bmatrix}$

Ans: C

The value of the determinant

$\begin{vmatrix}a & b & 0\\ 0 & a & b\\ b & 0 &a \end{vmatrix}$ is equal to

0%
$a^3-b^3$
0%
$a^3+b^3$
0%
0
0%
none of these

Explanation

$\begin{vmatrix} a & b & 0 \\ 0 & a & b \\ b & 0 & a \end{vmatrix}=a\left( { a }^{ 2 }-0 \right) -b\left( 0-{ b }^{ 2 } \right) +0={ a }^{ 3 }+{ b }^{ 3 }$

The value of the determinant

$\begin{vmatrix} 1 & 2 & 3\\ 3 & 5 & 7\\ 8 & 14 & 20\end{vmatrix}$ is equal to

0%
$20$
0%
$10$
0%
$0$
0%
$45$

Explanation

The value of the determinant is

$\begin{vmatrix} 1 & 2 & 3 \\ 3 & 5 & 7 \\ 8 & 14 & 20 \end{vmatrix}\\ =1\left( 100-98 \right) -2\left( 60-56 \right) +3\left( 42-40 \right)$

$=2-8+6=0$

$\begin{vmatrix} 6i & -3i & 1\\ 4 & 3i & -1 \\ 20 & 3 & i\end{vmatrix}=x+iy$ , then

0%
$x = 3, y = 1$
0%
$x = 1, y = 3$
0%
$x = 0, y = 3$
0%
$x = 0, y = 0$

Explanation

$\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix}=x+iy$

$\Rightarrow 6i\left( 3{ i }^{ 2 }+3 \right) +3i\left( 4i+20 \right) +1\left( 12-60i \right) =x+iy$

$\Rightarrow 0=x+iy$
Therefore,

$x=y=0$

Ans: D

The points which are not collinear are:

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$(0, 1),\ (8, 3)\ and\ (6, 7)$
0%
$(4, 3),\ (5, 1)\ and\ (1, 9)$
0%
$(2, 5),\ (-1, 2)\ and\ (4, 7)$
0%
$(-3, 2)\, (1, -2)\ and\ (9, -10)$

Explanation

$Let\quad us\quad take\quad three\quad points\quad A,\quad B\quad \& \quad C.\\ Three\quad points\quad A,\quad B\quad \& \quad C\quad will\quad be\quad collinear\\ If\quad AB+BC=AC.\\ Three\quad points\quad A,\quad B\quad \& \quad C\quad will\quad not\quad be\quad collinear\\ If\quad AB+BC=AC.\\ Let\quad us\quad investigate\quad thr\quad given\quad options.\\ Option\quad A\longrightarrow The\quad points\quad are\\ A(0,1)\quad B(8,3)\quad \& \quad C(6,7).\\ \therefore \quad AB=\sqrt { { \left( 0-8 \right) }^{ 2 }+{ \left( 1-3 \right) }^{ 2 } } units=\sqrt { 68 } =8.25units,\\ \quad \quad \quad BC=\sqrt { { \left( 8-6 \right) }^{ 2 }+{ \left( 3-7 \right) }^{ 2 } } units=\sqrt { 28 } =5.29units,\\ and\quad AC=\sqrt { { \left( 0-6 \right) }^{ 2 }+{ \left( 1-7 \right) }^{ 2 } } units=\sqrt { 72 } =8.49units.\\ \therefore \quad AB+BC=\left( 8.25+5.29 \right) units.=13.54units.\\ So\quad AB+BC\neq AC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad not\quad collinear.\\ Option\quad B\longrightarrow The\quad points\quad are\\ A(4,3)\quad B(5,1)\quad \& \quad C(1,9).\\ \therefore \quad AB=\sqrt { { \left( 4-5 \right) }^{ 2 }+{ \left( 3-1 \right) }^{ 2 } } units=\sqrt { 5 } =2.24units,\\ \quad \quad \quad BC=\sqrt { { \left( 5-1 \right) }^{ 2 }+{ \left( 1-9 \right) }^{ 2 } } units=\sqrt { 80 } =8.94units,\\ and\quad AC=\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( 3-9 \right) }^{ 2 } } units=\sqrt { 45 } =6.71units.\\ \therefore \quad AB+AC=\left( 2.24+6.71 \right) units=8.95units.\\ So\quad AB+AC=BC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Option\quad C\longrightarrow The\quad points\quad are\\ A(2,5)\quad B(-1,2)\quad \& \quad C(4,7).\\ \therefore \quad AB=\sqrt { { \left( 2+1 \right) }^{ 2 }+{ \left( 5-2 \right) }^{ 2 } } units=\sqrt { 18 } units=4.24units,\\ \quad \quad \quad BC=\sqrt { { \left( -1-4 \right) }^{ 2 }+{ \left( 2-7 \right) }^{ 2 } } units=\sqrt { 50 } units=7.07units,\\ and\quad AC=\sqrt { { \left( 2-4 \right) }^{ 2 }+{ \left( 5-7 \right) }^{ 2 } } units=\sqrt { 8 } units=2.83units.\\ \therefore \quad AB+AC=\left( 4.24+2.83 \right) units=7.07units.\\ So\quad AB+AC=BC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Option\quad C\longrightarrow The\quad points\quad are\\ A(-3,2)\quad B(1,-2)\quad \& \quad C(9,-10).\\ \therefore \quad AB=\sqrt { { \left( -3-1 \right) }^{ 2 }+{ \left( 2+2 \right) }^{ 2 } } units=\sqrt { 32 } units=5.66units,\\ \quad \quad \quad BC=\sqrt { { \left( 1-9 \right) }^{ 2 }+{ \left( -2+10 \right) }^{ 2 } } units=\sqrt { 128 } units=11.31units,\\ and\quad AC=\sqrt { { \left( -3-9 \right) }^{ 2 }+{ \left( 2+10 \right) }^{ 2 } } units=\sqrt { 288 } units=16.97units.\\ \therefore \quad AB+BC=\left( 5.66+11.31 \right) units=16.97units.\\ So\quad AB+BC=AC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Ans-\quad Option\quad A.\\ \\ \\$

$A=\begin{bmatrix} a & b \\ b & a \end{bmatrix}$ , then

$\left| A+{ A }^{ T } \right|$ equals

0%
$4({a}^{2}-{b}^{2})$
0%
$2({a}^{2}-{b}^{2}$
0%
${a}^{2}-{b}^{2})$
0%
$4ab$

Explanation

$A+{A}^{T}=\begin{bmatrix} a & b \\ b & a \end{bmatrix}\begin{bmatrix} a & b \\ b & a \end{bmatrix}=\begin{bmatrix} 2a & 2b \\ 2b & 2a \end{bmatrix}$

$\left| A+{ A }^{ T } \right| =4{ a }^{ 2 }-4{ b }^{ 2 }\quad$

Ans: A

$\Delta_1=\begin{vmatrix} 1 & 0\\ a & b\end{vmatrix}$ and

$\Delta_2=\begin{vmatrix} 1 & 0\\ c & d\end{vmatrix}$ then

$\Delta_2 \Delta_1$ is equal to

0%
$ac$
0%
$bd$
0%
$(b-a)(d-c)$
0%
none of these

Explanation

$\Delta_1=\begin{vmatrix} 1 & 0\\ a & b\end{vmatrix}=1.b-0.a=b$

$\Delta_2=\begin{vmatrix} 1 & 0\\ c & d\end{vmatrix}=1.d-0.c=d$
then

$\Delta_2 \Delta_1=bd$

Ans: B

$A$ and

$B$ are two matrices of same order

$3\times 3$ , where

$A=\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 5 & 6 & 8 \end{bmatrix}$ and

$B=\begin{bmatrix} 3 & 2 & 5 \\ 2 & 3 & 8 \\ 7 & 2 & 9 \end{bmatrix}$

The value of

$\text{Adj} (\text{Adj}\, A)$ is equal to

0%
$-A$
0%
$A$
0%
$8A$
0%
$16A$

Explanation

$\left| A \right| =1\left( 3\times8-4\times6 \right) -2\left( 2\times8-5\times4 \right) +3\left( 2\times6-5\times3 \right) =-1$
Use following results

$\text{Adj} (\text{Adj}\, A)={ \left| A \right| }^{ n-2 }A \Rightarrow \text{Adj}(\text{Adj}\,A)={ \left| A \right| }^{ n-2 }A={ \left( -1 \right) }^{ 1 }A=-A$

Ans: $A$

$\omega$ is a cube root of unity and

$\Delta=\begin{vmatrix}1 & 2\omega \\ \omega & \omega^2\end{vmatrix}$ , then

$\Delta^2$ is equal to

0%
$-\omega$
0%
$\omega$
0%
$1$
0%
$\omega^2$

Explanation

$\Delta=\omega^2-2\omega^2=-\omega^2$

$\Delta^2=\omega^4=\omega$

Ans: B

$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 3x & 6 \end{vmatrix}$ , then

$x$ is equal to

0%
$6$
0%
$\pm 6$
0%
$-6$
0%
$0$

Explanation

$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 3x & 6 \end{vmatrix}$

$\Rightarrow x^2-36=36-6x$

$\Rightarrow x^2+6x-72=0$

$\Rightarrow \displaystyle x=\frac{-6\pm\sqrt{6^2+4\times 72}}{2}$

$\quad \displaystyle =\frac{-6\pm \sqrt{36(1+8)}}{2}$

$\quad =-3\pm 9=6\ or\ -12$

Which of the following is correct?

0%
Determinant is a square matrix
0%
Determinant is a number associated to a matrix
0%
Determinant is a number associated to a square matrix
0%
None of these

$|A| \displaystyle \neq 0$ , then

$A$ is

0%
zero matrix
0%
singular matrix
0%
non - singular matrix
0%
diagonal matrix

Explanation

$|A| \neq 0$ then by definition,

$A$ is non-singular matrix

$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and

$A(\text{adj}\,A)=KI$ , then the value of

$K$ is:

0%
$2$
0%
$-2$
0%
$10$
0%
$-10$

Explanation

Given,

$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\quad$
Adjoint of

$A=\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\quad$

$\therefore$

$A(\text{adj}\,A)=KI$

$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 4 & 2 \\ -3 & 1 \end{bmatrix}=K\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}=\begin{bmatrix} K & 0 \\ 0 & K \end{bmatrix}$

$\therefore$

$K=-2$

What is the determinant of the matrix

$\left [\begin{matrix} 3& 6\\ -1 & 2\end {matrix} \right]$ ?

0%
$0$
0%
$12$
0%
$|0|$
0%
$|6|$

Explanation

Given,

$\begin{vmatrix} 3 & 6 \\ -1 & 2 \end{vmatrix}$
Let determinent be

$\left| d \right|$
Value of

$\left| d \right|$ will be

$\left| d \right|$

$=$

$3\times 2-\left( 6\times -1 \right)$

$=6+6=12$

$A = \begin{bmatrix} 1& \log_{b}a\\ \log_{a}b & 1\end{bmatrix}$ then

$|A|$ is equal to

0%
$0$
0%
$\log_{a}b$
0%
$-1$
0%
$\log_{b}a$

Explanation

On solving the given matrix,

$|A| = 1 - \log_{a}b . \log_{b}a = 1 - 1 = 0$

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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