MCQ Questions for Class 12 Commerce Maths Determinants Quiz 7

$A = \begin{vmatrix} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3}\end{vmatrix}$ and

$B = \begin{vmatrix} c_{1}& c_{2} & c_{3}\\ a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\end{vmatrix}$ then.

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$A = -B$
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$A = B$
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$B = 0$
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$B = A^{2}$

Explanation

Let determinant of matrix,

$\begin{vmatrix} c_{ 1 } & c_{ 2 } & c_{ 3 } \\ { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \end{vmatrix}$ be

$y$

$\begin{vmatrix} c_{ 1 } & c_{ 2 } & c_{ 3 } \\ { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \end{vmatrix}$

exchanging row

$1$ and row

$2$ , the determinant becomes

$-y$

$\begin{vmatrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ c_{ 1 } & c_{ 2 } & c_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \end{vmatrix}$

exchanging row

$2$ and row

$3$ the determinant becomes

$y$

$\begin{vmatrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ c_{ 1 } & c_{ 2 } & c_{ 3 } \end{vmatrix}$

Now taking transpose the determinant will remain

$y$

$\begin{vmatrix} { a }_{ 1 } & b_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & b_{ 3 } & c_{ 3 } \end{vmatrix}\\ \\$

So,

$A=B$

(b)

$\begin{vmatrix}1 & sin \theta &1 \\ -sin \theta & 1 & sin \theta\\ -1 & -sin \theta & 1\end{vmatrix}$ then,

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$\Delta =0$
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$\Delta \in (0, \infty )$
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$\Delta \in [-1, 2]$
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$\Delta \in [2, 4]$

Explanation

Let

$\Delta = \begin{vmatrix}1 & sin \theta &1 \\ -sin \theta & 1 & sin \theta\\ -1 & -sin \theta & 1\end{vmatrix}$

$=1(1+sin^2 \theta)-sin \theta (0)+1(sin^2 \theta +1)$

$= 2 (1+sin^2 \theta)$

$\because 0 \leq sin^2 \theta \leq 1$

$\Rightarrow 1\leq 1+sin^2 \theta\leq 2$

$\Rightarrow 2 \leq 2 (1+sin^2 \theta)\leq 4$

$\Rightarrow 2 \leq \Delta \leq 4$

$\Rightarrow \Delta \in [2,4]$

$A=\begin{bmatrix} x & 1 & -x \\ 0 & 1 & -1 \\ x & 0 & 7 \end{bmatrix}$ and

$det(A)=\begin{vmatrix} 3 & 0 & 1 \\ 2 & -1 & 0 \\ 0 & 6 & 7 \end{vmatrix}$ then the value of

$x$ is

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$-3$
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$3$
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$2$
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$-8$
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$-2$

Explanation

Given,

$A=\begin{bmatrix} x & 1 & -x \\ 0 & 1 & -1 \\ x & 0 & 7 \end{bmatrix}$ . Then

det

$(A) = x7-1(0+x)-x(0-x)$

$=7x-x+x^2=x^2+6x$

$...(i)$
Also

$det(A)=\begin{vmatrix} 3 & 0 & 1 \\ 2 & -1 & 0 \\ 0 & 6 & 7 \end{vmatrix}$

$=3(-7-0)-0+1(12)=-9$

$...(ii)$
From Eqs(i) and (ii), we get

$\quad { x }^{ 2 }+6x=-9$

$\Rightarrow { x }^{ 2 }+6x+9=0\Rightarrow { (x+3) }^{ 2 }=0$

$x+3=0$

$x=-3$

$A=\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix}$ , then the value of

$\left| A \right| \left| adj\left( A \right) \right|$ is

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${ a }^{ 3 }$
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${ a }^{ 6 }$
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${ a }^{ 9 }$
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${ a }^{ 27 }$

Explanation

$A=\left| \begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix} \right| =\\ \Rightarrow |A|={ a }^{ 3 }\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right| ={ a }^{ 3 }\\ \left| adj(A) \right| ={ \left| A \right| }^{ n-1 }$

Here

$n=3$

$\Rightarrow \left| adj(A) \right| ={ \left| A \right| }^{ 3-1 }={ \left| A \right| }^{ 2 }={ a }^{ 6 }\\ \Rightarrow |A|.\left| adj(A) \right| ={ a }^{ 3 }.{ a }^{ 6 }={ a }^{ 9 }$

So option

$C$ is correct

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

$\text{det} (\text{adj}(\text{adj} A))$ is equal to.

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$14^4$
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$14^3$
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$14^2$
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$14$

Explanation

We have,

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

$\Rightarrow |A|=1(1+2)-2(-1-4)-1(1-2)$

$=3+10+1=14$
We know that, for a square matrix of order

$n$ ,

$\text{adj}(\text{adj} A)=|A|^{n-2}A,$ if

$|A|\neq 0$

$\Rightarrow \text{det}(\text{adj} (\text{adj} A))=||A|^{n-2}A|$

$\Rightarrow \text{det}(\text{adj} (\text{adj} A))=(|A|^{n-2})^n|A|$

$\Rightarrow \text{det}(\text{adj} (\text{adj} A))=|A|^{n^2-2n+1}$
Here,

$n=3$ and

$|A|=14$
Therefore,

$\text{det}(\text{adj}(\text{adj} A))=(14)^{3^2-2\times 3+1}=14^4$ .

$\begin{vmatrix} x & 2 & 8 \\ 2 & 8 & x \\ 8 & x & 2 \end{vmatrix}=\begin{vmatrix} 3 & x & 7 \\ x & 7 & 3 \\ 7 & 3 & x \end{vmatrix}=\begin{vmatrix} 5 & 5 & x \\ 5 & x & 5 \\ x & 5 & 5 \end{vmatrix}=0$ then

$x$ is equal to

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$0$
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$-10$
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$3$
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None of these

Explanation

On applying

${ R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$ in each determinant, we can take out

$\left( x+10 \right)$ common.
Then, we get

$x+10=0$

$\Rightarrow x=-10$

$A = \begin{bmatrix} \dfrac {-1 + i\sqrt {3}}{2i}& \dfrac {-1 - i\sqrt {3}}{2i}\\ \dfrac {1 + i\sqrt {3}}{2i} & \dfrac {1 - i\sqrt {3}}{2i}\end{bmatrix}, i = \sqrt {-i}$ and

$f(x) = x^{2} + 2$ , then

$f(A)$ is equal to

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$\left (\dfrac {5 - i\sqrt {3}}{2}\right ) \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$
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$\left (\dfrac {3 - i\sqrt {3}}{2}\right ) \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$
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$\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$
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$(2 + i\sqrt {3})\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$

Explanation

From the given matrix, we have

$A = \begin{bmatrix}\dfrac {\omega}{i} & \dfrac {\omega^{2}}{i}\\ -\dfrac {\omega^{2}}{i} & -\dfrac {\omega}{i}\end{bmatrix} = \dfrac {\omega}{i} \begin{bmatrix}1 & \omega\\ -\omega & -1\end{bmatrix}$
Thus

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

$= -\begin{bmatrix}-\omega^{2} + \omega^{4}& 0\\ 0 & -\omega^{2} + \omega\end{bmatrix}$

$= \begin{bmatrix}-\omega^{2} + \omega & 0\\ 0 & -\omega^{2} + \omega\end{bmatrix}$ ....

$\text{Since} \ f(x) = x^{2} + 2$
Therefore,

$f(A) = A^{2} + 2 = \begin{bmatrix}-\omega^{2} + \omega & 0\\ 0 & -\omega^{2} + \omega\end{bmatrix} + \begin{bmatrix}2 & 0\\ 0 & 2\end{bmatrix}$

$- [\omega^{2} + \omega + 2] \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$

$= (3 + 2\omega) \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$

$= (2 + i\sqrt {3})\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$

$3x+2y=I$ and

$2x-y=O$ , where

$I$ and

$O$ are unit and null matrices of order

$3$ respectively, then

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$x=\dfrac { 1 }{ 7 } , y=\dfrac { 2 }{ 7 }$
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$x=\dfrac { 2 }{ 7 } , y=\dfrac { 1 }{ 7 }$
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$x=\left( \dfrac { 1 }{ 7 } \right) I, y=\left( \dfrac { 2 }{ 7 } \right) I$
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$x=\left( \dfrac { 2 }{ 7 } \right) I, y=\left( \dfrac { 1 }{ 7 } \right) I$

Explanation

We have,

$3x+2y=I$ ...(i)

$2x-y=O$ ....(ii)
On multiplying equation (ii) by

$2$ , we get

$4x-2y=2\cdot O=O$ ...(iii)
On adding equations (i) and (iii), we get

$3x+2y=I$

$4x-2y=O$
____________

$7x=I$

$\Rightarrow x=\dfrac { I }{ 7 }$ or

$\dfrac { 1 }{ 7 } I$
From equation (i), we get

$2y=I-\dfrac { 3 }{ 7 } I=\dfrac { 4 }{ 7 } I$

$\Rightarrow y=\dfrac { 2 }{ 7 } I$

$\begin{vmatrix} { \left( { a }^{ x }+{ a }^{ -x } \right) }^{ 2 } & { \left( { a }^{ x }-{ a }^{ -x } \right) }^{ 2 } & 1 \\ { \left( b^{ x }+{ b }^{ -x } \right) }^{ 2 } & { \left( { b }^{ x }-{ b }^{ -x } \right) }^{ 2 } & 1 \\ { \left( { c }^{ x }+{ c }^{ -x } \right) }^{ 2 } & { \left( { c }^{ x }-{ c }^{ -x } \right) }^{ 2 } & 1 \end{vmatrix}$ is equal to

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$0$
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$2abc$
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${ a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }$
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None of these

Explanation

Let

$\Delta=\begin{vmatrix} { \left( { a }^{ x }+{ a }^{ -x } \right) }^{ 2 } & { \left( { a }^{ x }-{ a }^{ -x } \right) }^{ 2 } & 1 \\ { \left( b^{ x }+{ b }^{ -x } \right) }^{ 2 } & { \left( { b }^{ x }-{ b }^{ -x } \right) }^{ 2 } & 1 \\ { \left( { c }^{ x }+{ c }^{ -x } \right) }^{ 2 } & { \left( { c }^{ x }-{ c }^{ -x } \right) }^{ 2 } & 1 \end{vmatrix}$

Put

$x=0$ in the given determinant, we get

$\Delta =\begin{vmatrix} { \left( 1+1 \right) }^{ 2 } & { \left( 1-1 \right) }^{ 2 } & 1 \\ { \left( 1+1 \right) }^{ 2 } & { \left( 1-1 \right) }^{ 2 } & 1 \\ { \left( 1+1 \right) }^{ 2 } & { \left( 1-1 \right) }^{ 2 } & 1 \end{vmatrix}$

$=\begin{vmatrix} 4 & 0 & 1 \\ 4 & 0 & 1 \\ 4 & 0 & 1 \end{vmatrix}$

$=0$

If the determinant

$\Delta =\begin{vmatrix} 3 & -2 & \sin { 3\theta } \\ -7 & 8 & \cos { 2\theta } \\ -11 & 14 & 2 \end{vmatrix}=0$ , then the value of

$\sin { \theta }$ is

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$\dfrac { 1 }{ 3 }$ or $1$
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$\dfrac { 1 }{ \sqrt { 2 } }$ or $\dfrac { \sqrt { 3 } }{ 2 }$
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$0$ or $\dfrac { 1 }{ 2 }$
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None of these

Explanation

Applying

${ R }_{ 2 }\rightarrow { R }_{ 2 }+4{ R }_{ 1 }$ and

${ R }_{ 3 }\rightarrow { R }_{ 3 }+7{ R }_{ 1 }$ we get

$\begin{vmatrix} 3 & -2 & \sin { 3\theta } \\ 5 & 0 & \cos { 2\theta } +4\sin { 3\theta } \\ 10 & 0 & 2+7\sin { 3\theta } \end{vmatrix}=0$

$\Rightarrow 2\left[ 5\left( 2+7\sin { 3\theta } \right) -10\left( \cos { 2\theta } +4\sin { 3\theta } \right) \right] =0$

$\Rightarrow 2+7\sin { 3\theta } -2\cos { 2\theta } -8\sin { 3\theta } =0$

$\Rightarrow 2-2\cos { 2\theta } -\sin { 3\theta } =0$

$\Rightarrow \sin { \theta } \left( 4\sin ^{ 2 }{ \theta } +4\sin { \theta } -3 \right) =0$

$\Rightarrow \sin { \theta } =0$ or

$\left( 2\sin { \theta } -1 \right) =0$ or

$\left( 2\sin { \theta } +3 \right) =0$

$\Rightarrow \sin { \theta } =0$ or

$\sin { \theta } =\dfrac { 1 }{ 2 }$

$A, B, C$ are collinear points such that

$A(3, 4), C(11, 10)$ and

$AB = 2.5$ then point

$B$ is

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$\left(5,\dfrac{11}{2}\right)$
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$\left(\dfrac{5}{2},11\right)$
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$(5, 5)$
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$(5, 6)$

Explanation

Given that the points

$A,B,C$ are collinear such that

$A(3,4),C(11,10)$ and

$AB=2.5$ ,

Consider the line joining

$A$ and

$C$ ,

Slope of that line is

$\tan { \alpha } =\dfrac { 10-4 }{ 11-3 } =\dfrac { 3 }{ 4 }$ ,

$\Longrightarrow \cos { \alpha } =\dfrac { 4 }{ 5 }$ and

$\sin { \alpha } =\dfrac { 3 }{ 5 }$ ,

We know that a point in a line at a distance

$r$ from that point is

$y={ y }_{ 1 }+r\sin { \alpha }$ and

$x={ x }_{ 1 }+r\cos { \alpha }$ ,

Where

$\tan{\alpha}$ is the slope of that line

Let the coordinates of the point

$B$ be

$(x,y)$ ,

$\Longrightarrow x=3+2.5(0.2)=5$ and

$y=3+2.5(0.6)=\dfrac { 11 }{ 2 }$ ,

$\therefore$ The coordinates of

$B$ are

$\left(5,\dfrac{11}{2}\right)$

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

$A$ is _______.

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

Explanation

Let

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

Let

$C$ be the cofactor matrix of matrix

$A$ .

${C}_{11}={\left(-1\right)}^{1+1}{M}_{11}=1$

${C}_{12}={\left(-1\right)}^{1+2}{M}_{12}=-4$

${C}_{21}={\left(-1\right)}^{2+1}{M}_{21}=3$

${C}_{22}={\left(-1\right)}^{2+2}{M}_{22}=2$

$\therefore {C}_{ij}=\begin{bmatrix} {C}_{11} & {C}_{12} \\ {C}_{21} & {C}_{22}\end{bmatrix}$

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

Adj

$\left({C}_{ij}\right)={\begin{bmatrix} 1 & -4 \\ 3 & 2\end{bmatrix}}^{T}$

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

If maximum and minimum values of

$D = \begin{vmatrix}1 & -\cos \theta & 1\\ \cos \theta & 1 & -\cos \theta\\ 1 & \cos \theta & 1\end{vmatrix}$ are

$p$ and

$q$ respectively, then the value of

$2p + 3q$ is _________.

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$16$
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$6$
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$14$
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$8$

Explanation

$D = \begin{vmatrix}1 & -\cos \theta & 1\\ \cos \theta & 1 & -\cos \theta\\ 1 & \cos \theta & 1\end{vmatrix}$

$\Rightarrow D = 1(1 + \cos^{2}\theta) + \cos \theta (\cos \theta + \cos \theta) - 1 (\cos^{2}\theta - 1)$

$= 1 + \cos^{2}\theta + 2\cos^{2}\theta -\cos^{2} \theta + 1$

$= 2 (1 + \cos^{2} \theta)$

Now,

$-1\le \cos \theta\le 1\Rightarrow 0\le \cos^{2}\theta\le 1$

$\therefore p = 4, q = 2$

$\therefore 2p + 3q = 14$ .

The value of the determinant

$\begin{vmatrix}b^2-ab & b-c & bc-ac\\ ab-a^2 & a-b & b^2-ab\\ bc-ac & c-a & ab-a^2\end{vmatrix}$

$=$ ____________.

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$abc$
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$a+b+c$
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$0$
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$ab+bc+ca$

If the vectors

$\vec {a}, \vec {b}, \vec {c}$ are coplanar, then the value of

$\begin{vmatrix}\vec {a}& \vec {b} & \vec {c}\\ \vec {a}.\vec {a} & \vec {a}.\vec {b} & \vec {a}.\vec {c}\\ \vec {b}.\vec {a} & \vec {b}.\vec {b} & \vec {b} . \vec {c}\end{vmatrix} =$

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$1$
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$0$
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$-1$
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$\vec {a} + \vec {b} + \vec {c}$

Explanation

$\bar { a } \cdot \bar { b } =\bar { a } \bar { b } \cos { \theta }$

Co planar

$\Rightarrow \theta =0$

$\Rightarrow \bar { a } \cdot \bar { b } =0$

Parallelly

$\bar { a } \cdot \bar { c } =\bar { a } \cdot \bar { c } =0$

$\begin{vmatrix} \bar { a } & \bar { b } & \bar { c } \\ \bar { a } \cdot \bar { a } & \bar { a } \cdot \bar { b } & \bar { a } \cdot \bar { c } \\ \bar { b } \cdot \bar { a } & \bar { b } \cdot \bar { b } & \bar { b } \cdot \bar { c } \end{vmatrix}$

$=\begin{vmatrix} \bar { a } & \bar { b } & \bar { c } \\ \bar { a } \cdot \bar { a } & 0 & 0 \\ 0 & \bar { b } \cdot \bar { b } & 0 \end{vmatrix}$

$=\bar { a } \left( 0 \right) -\bar { b } \left( 0 \right) +\bar { c } \left( \bar { a } \cdot \bar { a } -\bar { b } \cdot \bar { b } \right)$

$=0$

Let

$A^{-1}\begin{bmatrix} 1 & 2017 & 2\\ 1 & 2017 & 4 \\ 1 & 2018 & 8\end {bmatrix}$ . Then

$|2A|-|2A^{-1}|$ is equal to.

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$3$
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$-3$
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$12$
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$-12$

Explanation

$\left | A^{-1} \right | = \dfrac{1}{\left | A \right |}$

given matrix

$A^{-1} =$

$\begin{bmatrix} 1&2017 &2 \\ 1& 2017 &4 \\ 1&2018 &8 \end{bmatrix}$

solving the determinant

$\left | A^{-1} \right | = \dfrac{1}{\left | A \right |} = -2$

$\left | kA \right | = k^{n}\left | A \right |$

where n is the order of the matrix

$-\left | 2A^{-1} \right | + \left | 2A \right |$

$=-(-2)\times 2^{3}+\dfrac{-1}{2}\times 2^{3} =12$

The value of the determinant

$\begin{vmatrix} \cos^2 \dfrac{\theta}{2}&\sin^2\dfrac{\theta}{2}\\ \sin^2\dfrac{\theta}{2} &\cos^2\dfrac{\theta}{2} \end{vmatrix}$
for all values of

$\theta$ , is

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$1$
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$\cos\theta$
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$\sin\theta$
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$\cos2\theta$

Explanation

$\begin{vmatrix}\cos ^2\dfrac{\theta}{2}&\sin ^2\dfrac{\theta}{2}\\ \sin ^2\dfrac{\theta}{2} & {\cos }^2 \dfrac{\theta}{2}\end{vmatrix}=\cos ^4\dfrac{\theta}{2}-\sin ^4\dfrac{\theta}{2}$

$=\left({\cos }^2 \dfrac{\theta}2+{\sin }^2 \dfrac{\theta}2\right)\left({\cos }^2 \dfrac{\theta}2-{\sin }^2 \dfrac{\theta}2\right)$

$=1\times \cos \theta=\cos \theta$

Hence, the answer is

$\cos \theta$ .

Let

$z = \begin{vmatrix} 1& 1 + 2i & -5i\\ 1 - 2i & -3 & 5 + 3i\\ 5i & 5 - 3i & 7\end{vmatrix}$ , then

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$z$ is purely real
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$z$ is purely imaginary
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$(z - \overline {z})i = 0$
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$(z + \overline {z})i = 0$

Explanation

$z=\left| \begin{matrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{matrix} \right| \\ \quad =1(-21-36)\quad -(1+2i)(7-14i-25i+15)\quad -5i(-13i-1)\\ \quad =-57-(1+2i)(22-39i)-65+5i\\ \quad =-57-(22-39i+44i+78)-65+5i\\ \quad =-57-100-5i-65+5i\\ \quad =-222$

$\therefore z$ is purely real.

The adjoint of the matric

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

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$\begin{bmatrix}-1 & 6 & 2\\ -2 & 1 & -4\\ 6 & 3 & 1\end{bmatrix}$
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$\begin{bmatrix}1 & 6 & -2\\ -2 & 1 & 4\\ 6 & -3 & 1\end{bmatrix}$
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$\begin{bmatrix}6 & 1 & 2\\ 4 & -1 & 2\\ 6 & 3 & -1\end{bmatrix}$
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$\begin{bmatrix}-6 & 2 & 1\\ 4 & -2 & 1\\ 3 & 1 & -6\end{bmatrix}$

Explanation

$adj(A)=cofactor(A)^T=\begin{bmatrix} A_{11} &A_{21}&A_{31}\\A_{12} &A_{22} &A_{32}\\A_{13} &A_{23} &A_{33}\end{bmatrix}=\begin{bmatrix} 1 &6&-2\\-2 &1 &4\\6 &-3 &1\end{bmatrix}$

Three distinct points A, B and C are given in the 2-dimensional coordinate plane such that the ratio of the distance of any one of them from the point

$(1, 0)$ to the distance from the point

$(-1, 0)$ is equal to

$\displaystyle \frac{1}{3}$ . Then the circumcentre of the triangle ABC is at the point:

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$\displaystyle \left( \frac{5}{2}, 0 \right)$
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$\displaystyle \left( \frac{5}{3}, 0 \right)$
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$\displaystyle \left( 0, 0 \right)$
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$\displaystyle \left( \frac{5}{8}, 0 \right)$

Explanation

Let (h,k) be the locus of points A, B & C

using given condition

$\dfrac{\sqrt{(h-1)^{2}+K^{2}}}{\sqrt{(h+1)^{2}+K^{2}}}=\dfrac{1}{3}$

$9(h-1)^{2}+9K^{2}=(h+1)^{2}+K^{2}$

or,

$8h^{2}+8K^{2}-20h+8=0$

$h^{2}+K^{2}-\dfrac{5h}{4}+1=0$

this is a circle and it will be same circle which circumscribe

$\Delta ABC$ as it contains all three pts on it.

$\therefore$ center of circumcircle

$\equiv \left(\dfrac{5}{8}, 0\right)$

$A + B + C = \pi$ , then

$\begin{vmatrix} \sin (A + B + C)& \sin B & \cos C\\ -\sin B & 0 & \tan A\\ \cos (A + B) & -\tan A & 0\end{vmatrix}$ is equal to

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$0$
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$2\sin B \tan A \cos C$
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$1$
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None of these

Explanation

Given

$A+B+C= \pi$

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

$=\begin{vmatrix} \sin \pi & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ \cos (\pi-C) & -\tan A & 0 \end{vmatrix}$

$=\sin\pi [\tan^2A]-\sin B[-\tan A \cos (\pi-C)]+\cos C[\sin B\tan A]$

$=0+\sin B\, \tan A\, \cos(\pi-C)+\cos C \sin B \tan A$

$=\sin B\, \tan A[\cos\pi \cos C+ \sin\pi \sin C]+\cos C\sin B\tan A$

$=-\sin B \tan A\cos C +\cos C \sin B \tan A$

$=0$

$\left[\therefore \sin \pi=0 ,\cos \pi =-1\right]$

$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ satisfies the equation

$x^2 - (a + d) x + k = 0$ , then ?

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$k = bc$
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$k = ad$
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$k = a^2 + b^2 + c^2 + d^2$
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$ad - bc$

Explanation

If a matrix

$A$ satisfies a polynomial equation, then the product of the roots of that equation is the determinant of

$A$ .

Here, product of roots of the equation =

$k$ , also

$det(A)=ad-bc$ .
Therefore,

$k=ad-bc$

For a positive numbers

$x, y$ and

$z$ the numerical value of the determinant

$\begin{bmatrix}1 & \log_{x} y & \log_{x} z \\ \log_{y} x & 1 & \log_{y} z\\ \log_{z} x & \log_{z} y & 1\end{bmatrix}$ is:

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$0$
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$1$
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$\log_{e} xyz$
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$-\log_{e} xyz$

Explanation

$D=\left| \begin{matrix} 1 & \log _{ x }{ y } & \log _{ x }{ z } \\ \log _{ y }{ x } & 1 & \log _{ y }{ z } \\ \log _{ z }{ x } & \log _{ z }{ y } & 1 \end{matrix} \right| =(1-\log _{ z }{ y } .\log _{ y }{ z } )-\log _{ x }{ y } \left( \log _{ y }{ x } -\log _{ z }{ x } .\log _{ y }{ z } \right) +\log _{ x }{ z } \left( \log _{ y }{ x } .\log _{ z }{ y } -\log _{ z }{ x } \right) \\ =\left( 1-1 \right) -\log _{ x }{ y } \left( \log _{ y }{ x } -\log _{ y }{ x } \right) +\log _{ x }{ z } \left( \log _{ z }{ x } -\log _{ z }{ x } \right) =0-0+0=0$

The value of the determinant

$\begin{vmatrix}b^{2}-ab\,\, b-c\,\, bc-ac & & \\ ab-a^{2}\,\, a-b\,\, b^{2}-ab& & \\ bc-ac\,\, c-a\,\, ab-a^{2}& & \end{vmatrix}$ =

0%
$abc$
0%
$a+b+c$
0%
$0$
0%
$ab+bc+ca$

Explanation

$\begin{vmatrix}b^{2}-ab\,\, b-c\,\, bc-ac & & \\ ab-a^{2}\,\, a-b\,\, b^{2}-ab& & \\ bc-ac\,\, c-a\,\, ab-a^{2}& & \end{vmatrix}=0$

$a=b=c=1$

If the points

$A(-2,1),B(a,b)$ and

$C(4,-1)$ are collinear and

$a-b=1$ , find the values of

$a$ and

$b$ .

0%
$a=1,b=5$
0%
$a=1,b=0$
0%
$a=2,b=0$
0%
None of these

Explanation

Three points

$A(-2,1),\ B(a,b)$ and

$C(4,-1)$ are collinear so the area of the triangle is equal to zero

Here the equation

$(1)$ is

$\to a-b=1$

Here

$x_1=-2\ y_1=1$

$x_1 =a\ y_1=b$

$x_3=4\ y_3=-1$

Area of the triangle

$\triangle ABC=\dfrac {1}{2} [x_1 (y_2 -y_3)+x_2 (y_3 -y_1)+x_3 (y_1 -y_2)]=0$

$\Rightarrow \ \dfrac {1}{2} [-2(b+1)+a(-1 -1)+4(1-b)]=0$

$\Rightarrow \ \dfrac {1}{2}[-2b-2-2b+4-4b]=0$

$\Rightarrow \ \dfrac {1}{2} [-6b+2-2b]=0$

$\Rightarrow \ -3b+1-a=0$

$\Rightarrow \ -a-3b=-1$ ......Equation

$2$

Equation

$(1)\times 1\longrightarrow a-b=1$

Equation

$(2)\times 1\to \dfrac {-a-2b=-1}{-4b=0\\ \Rightarrow b=0}$

Putting the value of

$b=0$ in Equation

$1:-$

$a-b=1$

$\Rightarrow \ a=1$

So the value of

$a=1, b=0$

State true or false.

The determinants

$\left | \begin{array}{111} 1 & a & bc \\ 1 & b & cd \\ 1 & c & ab \\ \end {array} \right |$ and

$\left | \begin{array}{111} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end {array} \right |$ are not identically equal.

0%
True
0%
False

$\triangle$ =

$\left | \begin{array}{111} x_1+y_1\omega & x_1\omega^2+y_1 & x_1+y_1\omega+z_1\omega^2 \\ x_2+y_2\omega & x_2\omega^2+y_2 & x_2+y_2\omega+z_2\omega^2 \\ x_3+y_3\omega & x_3\omega^2+y_3 & x_3+y_3\omega+z_3\omega^2 \\ \end {array} \right |$
where

$1,\omega,\omega^2$ are cube roots of unity then

$\triangle$ is equal to

0%
$0$
0%
$1$
0%
$-1$
0%
None of these

$\left| \begin{matrix} 1+a & 1 & 1 & 1 \\ 1 & 1+b & 1 & 1 \\ 1 & 1 & 1+c & 1 \\ 1 & 1 & 1 & 1+d \end{matrix} \right|$

$=abcd\left( 1+\dfrac { 1 }{ a } +\dfrac { 1 }{ b } +\dfrac { 1 }{ c } +\dfrac { 1 }{ d } \right)\\ =s-r$
if

$a,b,c,d$ are the roots of

${ x }^{ 4 }+p{ x }^{ 3 }+q{ x }^{ 2 }+rx+s=0.$

0%
True
0%
False

$\left |\begin{array}{111}6i & -3i & 1 \\4 & 3i & -1 \\20 & 3 & i \\\end {array}\right | =x+iy$ , then

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

Explanation

$\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i \end{vmatrix}$

$\Rightarrow 6i((3i. i) - (-3)) + 3i (4 . i + 20) + 1 (12 - 60 i)$

$\Rightarrow 6i (-3 + 3) + 12 (-1) + 60 i + 12 - 60 i$

$\Rightarrow 0$

$[\because i^2 = -1]$

$x + i y = 0 + i(0)$

$x = 0 , y = 0$

The value of determinant

$\begin{vmatrix} x+1 & x+2 & x+4\\ x+3 & x+5 & x+8\\ x+7 & x+10 & x+14\end{vmatrix}$ is?

0%
$-2$
0%
$x^2+2$
0%
$2$
0%
None of these

Explanation

Operating

$C_3-C_2$ and

$C_2-C_1$ , we get

$\Delta =\begin{vmatrix} x+1 & 1 & 2\\ x+3 & 2 & 3\\ x+7 & 3 & 4\end{vmatrix}$
Apply

$R_3-R_2, R_2-R_1$

$\Delta =\begin{vmatrix} x+1 & 1 & 2\\ 2 & 1 & 1\\ 4 & 1 & 1\end{vmatrix}$ apply

$C_3-C_2$

$\begin{vmatrix} x+1 & 1 & 1\\ 2 & 1 & 0 \\ 4 & 1 & 0\end{vmatrix}=1.(2-4)=-2$

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

$\Delta_2=\begin{vmatrix} x & b\\ a & x\end{vmatrix}$ are the given determinants, then.

0%
$\Delta_1=3(\Delta_2)^2$
0%
$(d/dx)\Delta_1=3\Delta_2$
0%
$(d/dx)\Delta_1=3(\Delta_2)^2$
0%
$\Delta_1=3\Delta_2^{3/2}$

If A =

$\begin{bmatrix} a & 0 & 0 \\[0.3em] 0 & a & 0 \\[0.3em] 0 & 0 & a \end{bmatrix}$ , then the value of |A| |Adj. A|

0%
$a^3$
0%
$a^6$
0%
$a^9$
0%
$a^{27}$

Explanation

$|A|=(a)^3$

$|A|(Adj. A) = |A|$

$\begin{bmatrix} a & 0 & 0 \\[0.3em] 0 & a & 0 \\[0.3em] 0 & 0 & a \end{bmatrix}$

$\begin{bmatrix} |A| & 0 & 0 \\[0.3em] 0 & |A| & 0 \\[0.3em] 0 & 0 & |A| \end{bmatrix}$ where

$|A|$ =

$a^3$

Take determinant of both sides

|A|. |Adj. A| =

$\begin{vmatrix} |A| & 0 & 0 \\[0.3em] 0 & |A| & 0 \\[0.3em] 0 & 0 & |A| \end{vmatrix} |A|^3 = (a^3)^3 = a^9$

$f(x) =$

$\left | \begin{array}{111} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \\ \end {array} \right |$
then f(100) is equal to

0%
0
0%
1
0%
100
0%
-100

Explanation

$f(x) = \begin{vmatrix} 1 & x & x + 1 \\ 2x & x(x - 1) & (x + 1)x \\ 3x(x -1) & x(x - 1) (x - 2) & (x + 1) x(x - 1) \end{vmatrix}$

$x \times x \times (x - 1) \, \begin{vmatrix} 1 & x & x + 1 \\ 2 & x - 1 & x + 1 \\ 3 & x - 2 & x + 1 \end{vmatrix}$

$x^2 (x - 1)(x + 1) \, \begin{vmatrix} 1 & x & 1 \\ 2 & x - 1 & 1 \\ 3 & x - 2 & 1 \end{vmatrix}$

operating :

$R_1 \rightarrow R_2 - R_1$ &

$R_2 \rightarrow R_3 - R_2$

$x^2 (x^2 - 1) \, \begin{vmatrix} 1 & -1 & 0 \\ 1 & -1 & 0 \\ 3 & x - 2 & 1 \end{vmatrix}$

As two rows of the determinant is zero

$\therefore$ Determinant

$= 0$

Hence,

$\boxed {f(x) = 0}$

$\therefore$ for any value of

$x ;\ f(x) = 0$

Hence,

$\boxed {f(100) = 0}$

If A =

$\begin{bmatrix} \alpha & 2 \\ 2 & \alpha\end{bmatrix}$ and | A

$^3$ | = 125 then

$\alpha$ is

0%
$\pm1$
0%
=2
0%
$\pm3$
0%
$\pm5$

Explanation

$A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ .

$|A^3| = 125$

$|A| = \alpha^2 - 4$

we know by properties of determinants that

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

$\Rightarrow \ (\alpha^2 - 4)^3 = 125$

$\Rightarrow \ \alpha^2 - 4 = 5$

$\Rightarrow \ a^2 = 9$

$\Rightarrow \ \boxed {\alpha = \pm 3}$

Hence, option

$(C)$ is correct

If a, b, c are three non-zero distinct numbers in A.P., then

$\triangle =$

$\left | \begin{array}{111} (b-c)(c-a) & (a-b)(c-a) & (a-b)(b-c) \\ (c-a)(a-b) & (b-c)(a-b) & (b-c)(c-a) \\ (a-b)(b-c) & (c-a)(b-c) & (c-a)(a-b) \\ \end {array} \right |$ is always +ve.

0%
True
0%
False

If the points

$(-2, -5), (2, -2), (8, a)$ are collinear, then the value of

$a$ is ________.

0%
$\dfrac 52$
0%
$\dfrac 32$
0%
$\dfrac 72$
0%
None of these

Explanation

If three points are collinear, then the slope of the line joining the first two points is equal to that of the last two.

Apply the slope formula

$\dfrac{(y2-y1)}{(x2-x1).}$

If collinear then

$\dfrac { -2+5 }{ 2+2 } =\dfrac { a+2 }{ 6 } \\ \dfrac { 9 }{ 2 } =a+2\\ a=\dfrac { 5 }{ 2 }$

State true or false

Following points are collinear.

$(-2, 1) , (0, 5) , (-1, 2)$ .

0%
True
0%
False

The points

$(-a, -b), (0, 0), (a, b)$

$(a^2,ab)$ are

0%
collinear
0%
vertices of rectangle
0%
vertices of parallelgram
0%
none of these

$2x - 3y + z = 0$

$x + 2y - 3z =0$

$4x - y - 2z = 0$

The system of equations have a non trivial solution

0%
True
0%
False

If the lines

$p_{ 1 }x+q_{ 1 }y=1, p_{ 2 }x+q_{ 2 }y=1$ and

$p_{3}x+q_{3}y=1$ be concurrent, show that the points

$(p_{1},q_{1}), (p_{2}, q_{2})$ and

$(p_{3}, q_{3})$ are collinear.

0%
vertices of right angle triangle
0%
vertices of an equilateral triangle
0%
vertices of an isosceles triangle
0%
Collinear

Explanation

Given:

$p_{ 1 }x+q_{ 1 }y=1, p_{ 2 }x+q_{ 2 }y=1$ and

$p_{3}x+q_{3}y=1$

The lines will be concurrent if

$\begin{vmatrix} { p }_{ 1 } & { q }_{ 1 } & -1 \\ { p }_{ 2 } & { q }_{ 2 } & -1 \\ { p }_{ 3 } & { q }_{ 3 } & -1 \end{vmatrix}=0$

$\dfrac { 1 }{ 2 } \begin{vmatrix} { p }_{ 1 } & { q }_{ 1 } & 1 \\ { p }_{ 2 } & { q }_{ 2 } & 1 \\ { p }_{ 3 } & { q }_{ 3 } & 1 \end{vmatrix}=0$

$\triangle=0$

i.e., area of a triangle formed by the points

$({p}_{1}, {q}_{1})$ ,

$({p}_{2}, {q}_{2})$ , and

$({p}_{3}, {q}_{3})$ is zero and as such the points are collinear.

If A =

$\begin{bmatrix} a & 0 & 0 \\[0.3em] 0 & a & 0 \\[0.3em] 0 & 0 & a \end{bmatrix}$ , then the value of |Adj. A| is equal to

0%
$a^3$
0%
$a^6$
0%
$a^9$
0%
$a^{27}$

Explanation

A(Adj. A) = |A|

$I_3$ = |A|

$\begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 & 0 & 1 \end{bmatrix}$
=

$\begin{bmatrix} |A| & 0 & 0 \\[0.3em] 0 & |A| & 0 \\[0.3em] 0 & 0 & |A| \end{bmatrix}$ where |A| =

$a^3$
Take determinant of both sides
|A|. |Adj. A| =

$\begin{vmatrix} |A| & 0 & 0 \\[0.3em] 0 & |A| & 0 \\[0.3em] 0 & 0 & |A| \end{vmatrix} |A|^3 =|Adj. A| = |A|^{n-1} = |A|^2 = a^6$

$\begin{vmatrix}a & a & x \\ m & m & m \\b & x & b\end{vmatrix}=0$ then

$x$ is:

0%
$a$
0%
$b$
0%
$a$ or $b$
0%
$0$

Explanation

Given

$\begin{vmatrix} a & a & x \\ m & m & m \\ b & x & b \end{vmatrix}=0$

Expanding the determinant we get

$a(mb-mx)-a(mb-mb)+x(mx-mb)=0\\ \Rightarrow a(mb-mx)-x(mb-mx)=0\\ \Rightarrow (a-x)(mb-mx)=0\\ \Rightarrow (a-x)(b-x)=0$

The solution for this quadratic equation is either

$x=a$ or

$x=b$

$\begin{vmatrix} 2^{ 3 } & 3^{ 3 } & 3.2^{ 2 }+3.2+1 \\ 3^{ 3 } & 4^{ 3 } & 3.3^{ 2 }+3.3+1 \\ 4^{ 3 } & 5^{ 3 } & 3.4^{ 2 }+3.4+1 \end{vmatrix}$ is equal to

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

Explanation

$\begin{vmatrix} 2^{ 3 } & 3^{ 3 } & 3.2^{ 2 }+3.2+1 \\ 3^{ 3 } & 4^{ 3 } & 3.3^{ 2 }+3.3+1 \\ 4^{ 3 } & 5^{ 3 } & 3.4^{ 2 }+3.4+1 \end{vmatrix}$

$C_{1}\rightarrow C_{1}+C_{2}$

$=\begin{vmatrix} 2^{ 3 } + 3.2^{ 3 }+3.2+1 & 3^{ 3 }& 3.2^{2}+3.2+1 \\ 3^{ 3 } +3.3^{2}+3.3+1& 4^{ 3 } & 3.3^{ 2 }+3.3+1 \\ 4^{ 3 }+3.4^{2}+3.4+1 & 5^{ 3 } & 3.4^{ 2 }+3.4+1 \end{vmatrix}$

$=\begin{vmatrix} 3^{ 3 } & 3^{ 3 } & 3.2^{ 2 }+3.2+1 \\ 4^{ 3 } & 4^{ 3 } & 3.3^{ 2 }+3.3+1 \\ 5^{ 3 } & 5^{ 3 } & 3.4^{ 2 }+3.4+1 \end{vmatrix}$

$= 0$

1153409_1035448_ans_fbfe50cd025b4514b3173a69028b7b23.jpeg

Find

$\begin{vmatrix}\log e & \log e^{2} & \log e^{3} \\ \log e^{2} & \log e^{3} & \log e^{4} \\ \log e^{3} & \log e^{4} & \log e^{5}\end{vmatrix}$ .

0%
$0$
0%
$1$
0%
$4 \log e$
0%
$5 \log e$

Explanation

As we know that,

$loge^n=nloge$

So, We can write given Determinant as

$\begin{vmatrix}log e & 2log e & 3log e \\ 2log e & 3log e & 4log e \\ 3log e &\ 4log e & 5log e \end{vmatrix}=(loge)^3\begin{vmatrix}1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$ .... [Taking

$loge$ common from each row]

Expanding the Given Determinant , we get

$=(loge)^3|(15-16)-2(10-12)+3(8-9)|=(loge)^3\times 0=0$

$\left| {\begin{array}{*{20}{c}}1&3&2\\1&{x - 1}&{2x + 2}\\2&5&9\end{array}} \right| = 0$ , then

$x$ is equal to :-

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

Explanation

Given,

$\left| {\begin{array}{*{20}{c}}1&3&2\\1&{x - 1}&{2x + 2}\\2&5&9\end{array}} \right| = 0$

$\begin{array}{l}\left| {\begin{array}{*{20}{c}}1&3&2\\1&{x - 1}&{2x + 2}\\2&5&9\end{array}} \right|\\{R_2} = {R_2} - {R_1}\\{R_3} = {R_3} - 2{R_1}\\\left| {\begin{array}{*{20}{c}}1&3&4\\0&{x - 4}&{2x - 2}\\0&{ - 1}&1\end{array}} \right|\\\left| {\begin{array}{*{20}{c}}{x - 4}&{2x - 2}\\{ - 1}&1\end{array}} \right| = 0\\x - 4 + 2x - 2 = 0\\3x - 6 = 0\\x = 2\end{array}$

$\Delta = \begin{vmatrix} x-3 & 2x+1 & 2 \\ 3x+2 & x+2 & 1 \\ 5x+1 & 5x+4 & 5 \end{vmatrix}$ , then

$\Delta$ is

0%
multiple of $x^2$
0%
$15$
0%
a multiple of $x$
0%
$-15$

Explanation

$\left| \begin{matrix} x-3 & 2x+1 & 2 \\ 3x+2 & x+2 & 1 \\ 5x+1 & 5x+4 & 5 \end{matrix} \right|$

$=\cfrac { 1 }{ 2 } \left| \begin{matrix} 2(x-3) & 2(2x+1) & 2(2) \\ 3x+2 & x+2 & 1 \\ 5x+1 & 5x+4 & 5 \end{matrix} \right|$

$=\cfrac { 1 }{ 2 } \left| \begin{matrix} 2x-6 & 4x+2 & 4 \\ 3x+2 & x+2 & 1 \\ 5x+1 & 5x+4 & 5 \end{matrix} \right|$

Apply

$R_{1}\rightarrow R_{1}+R_{2}-R_{3}$

$=\cfrac { 1 }{ 2 } \left| \begin{matrix} -5 & 0 & 0 \\ 3x+2 & x+2 & 1 \\ 5x+1 & 5x+4 & 5 \end{matrix} \right|$

$=\cfrac{1}{2}\times (-5)\left \{ 5(x+2)-1(5x+4)\right \}$

$=\cfrac{-5}{2}\times 6$

$=-15$

Find the determinant of given matrix

$\left[ \begin{matrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right]$

0%
$2(a+b+c)^{3}$
0%
$(a-b-c)^{3}$
0%
$2(a-b-c)^{3}$
0%
$(a+b+c)^{3}$

Explanation

$\left| \begin{matrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right|$

$=\left| \begin{matrix} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right| { R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$

Taking out common

$a+b+c$ from

${R_1}$

$=\left( a+b+c \right) \left| \begin{matrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{matrix} \right|$

$=\left( a+b+c \right) \left| \begin{matrix} 1 & 0 & 0 \\ 2b & -\left( a+b+c \right) & 0 \\ 2c & 0 & -\left( a+b+c \right) \end{matrix} \right| { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }$

Expanding along

${R_1}$

$={ \left( a+b+c \right) }^{ 3 }$

$\triangle =\begin{bmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{bmatrix}$ and

${A}_{2},{B}_{2},{C}_{2}$ are respectively cofactors of

${a}_{2},{b}_{2},{c}_{2}$ then

${a}_{1}{A}_{2}+{b}_{1}{B}_{2}+{c}_{1}{C}_{2}$ is equal to ?

0%
$-\triangle$
0%
$0$
0%
$\triangle$
0%
$none\ of\ these$

Explanation

Co-factor of

$\displaystyle a_2 = (-1)^{i+j}\left|\begin{matrix} b_1 & c_1 \\ b_3 & c_3 \end{matrix}\right| = - [b_1 \, c_3 - c_1 \, b_3] = A_2$

i=2 j=1

$\displaystyle b_2 = (-1)^{2+2} \left|\begin{matrix} a_1 & c_1 \\ a_3 & c_3 \end{matrix}\right| = [a_1 \, c_3 - c_1 \, a_3] = B_2$

$\displaystyle c_2 = (-1)^{2+3} \left|\begin{matrix} a_1 & b_1 \\ a_3 & b_3 \end{matrix}\right| = -[a_1 \, b_3 \, - b_1 \, a_3 ] = C_2$

$a_1 \, A_2 + b_1 \, B_2 +c_1 \, C_2$

$\displaystyle -a_1 \, b_1 \, c_3 + a_1 \, c_1 \, b_3 + b_1 \, a_1 \, c_3 - b_1 c_1 \, a_3 -c_1 \, a_1 \, b_3 + c_1 \, b_1 \, a_3$

$\displaystyle = 0$

the below matrix relation is

$\begin{vmatrix} 1 & { a }^{ 2 } & { a }^{ 3 } \\ 1 & { b }^{ 2 } & { b }^{ 3 } \\ 1 & { c }^{ 2 } & { c }^{ 3 } \end{vmatrix}=\begin{vmatrix} 0 & { a }^{ 2 }-{ c }^{ 2 } & { a }^{ 3 }-{ c }^{ 3 } \\ 0 & { b }^{ 2 }-{ c }^{ 2 } & { b }^{ 3 }-{ c }^{ 3 } \\ 1 & { c }^{ 2 } & { c }^{ 3 } \end{vmatrix}$

$=\left( a-b \right) \left( b-c \right) \begin{vmatrix} 0 & { a }-{ c } & { a }^{ 2 }+ac{ +c }^{ 2 } \\ 0 & { b }-{ c } & { b }^{ 2 }+bc+{ c }^{ 2 } \\ 1 & { c }^{ 2 } & { c }^{ 3 } \end{vmatrix}$

0%
True
0%
False

Let the matrix A and B be defined as

$A = \left( {\matrix{ 3 & 2 \cr 2 & 1 \cr } } \right)$ and

$B = \left( {\matrix{ 3 & 1 \cr 7 & 3 \cr } } \right)$ then the value of

$Det.\left( {2{A^9}{b^{ - 1}}} \right),$ is

0%
2
0%
1
0%
-1
0%
-2

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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