MCQ Questions for Class 12 Commerce Maths Determinants Quiz 4

If the points

$(a, 0), (0, b)$ and

$(1, 1)$ are collinear, then

$\displaystyle \frac{1}{a} + \frac{1}{b}$ equal to -

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

Explanation

Given points are

$(a,0),(0,b)$ and

$(1,1)$

$x_1 = a, y_1 = 0, x_2 =0, y_2 = b$ and

$x-3=1, y_3=1$
Condition for collinearity

$x_1y_2+x_2y_3+x_3y_1=x_2y_1+x_3y_2+x_1y_3$ gives

$ab+0+0=0 +1.b+a.1 \Rightarrow ab = a+b$

$\Rightarrow \displaystyle 1 = \frac{1}{b}+ \frac{1}{a}$

$\Rightarrow \dfrac{1}{a} + \dfrac{1}{b}=1$

If A is a square matrix so that

$AadjA=diag\left ( k,k,k \right )$ then

$\left | adj\: A \right |=$

0%
$k$
0%
$k^{2}$
0%
$k^{3}$
0%
$k^{4}$

Explanation

Given ,

$AadjA=diag\left ( k,k,k \right )$

$\Rightarrow |A|I_{3}=\begin{bmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{bmatrix}$

$\Rightarrow |A|=k$

Now,

$\left | adj\: A \right |=|A|^{2} (\because |adj A|=|A|^{n-1})$

$\Rightarrow |adj A|=k^{2}$

If A is a square matrix such that

$\displaystyle \left | \begin{matrix}4 &0 &0 \\0 &4 &0 \\0 &0 &4\end{matrix} \right |$ =

0%
$4$
0%
$16$
0%
$64$
0%
$256$

Explanation

We know that the determinant of a matrix

$A=\left[ \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} \right]$ is:

$\left| { A } \right| =a(ei-fh)-b(di-fg)+c(dh-eg)$

Here, the given matrix is

$A=\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right]$ , so, let us find the determinant of matrix

$A$ as shown below:

$\left| { A } \right| =4[(4\times 4)-(0\times 0)]-0[(0\times 4)-(0\times 0)]+0[(0\times 0)-(0\times 4)]=4(16-0)-0(0-0)+0(0-0)$

$=4\times 16=64$

Hence, the determinant of the matrix

$\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right]$ is

$64$ .

$\displaystyle A=\:\left [ \begin{matrix}1 &x \\x^{2} &4y \end{matrix} \right ], B = \left [ \begin{matrix}-3 &1 \\1 &0 \end{matrix} \right ]$ and

$adj \: \left( A \right) +B=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] ,$ then the values of

$x$ and

$y$ are respectively

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$\displaystyle \:\left ( 1,1 \right )$
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$\displaystyle \:\left ( -1,1 \right )$
0%
$\displaystyle \:\left ( 1,0 \right )$
0%
none of these

Explanation

Given,

$\displaystyle A= \:\left [ \begin{matrix}1 &x \\x^{2} &4y \end{matrix} \right ], B = \left [ \begin{matrix}-3 &1 \\1 &0 \end{matrix} \right ]$

$adj(A)=\left[ \begin{matrix} 4y & -x^{ 2 } \\ -x & 1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 4y & -x \\ -x^{ 2 } & 1 \end{matrix} \right]$

But given,

$adj \: \left( A \right) +B=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$

$\Rightarrow\left[ \begin{matrix} 4y-3 & -x+1 \\ -x^{ 2 }+1 & 1 \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$

$\Rightarrow -x+1=0 ,-x^2+1=0,4y-3=1$

$\Rightarrow x=1,y=1$

$A$ is a non-singular matrix of order

$\displaystyle 3\times 3$ , then adj

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

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$\displaystyle \left | A \right |A$
0%
$\displaystyle \left | A \right |^{2}A$
0%
$\displaystyle \left | A \right |^{-1}A$
0%
none of these

Explanation

Given A is a matrix of order 3.
Now,

$adj (adj A)=|A|^{n-2}A$
So,

$adj(adj A)=|A|^{3-2}A =|A|A$

$A=\left[\begin{matrix}1&0&0\\2&1&0\\3&2&1\end{matrix}\right], U_1, U_2$ and

$U_3$ are columns matrices satisfying

$AU_1=\left[\begin{matrix}1\\0\\0\end{matrix}\right], AU_2 = \left[\begin{matrix}2\\3\\0\end{matrix}\right], AU_3 = \left[\begin{matrix}2\\3\\1\end{matrix}\right]$ and

$U$ is

$3\times3$ matrix whose columns are

$U_1, U_2, U_3$ then answer the following question

The value of

$|U|$ is

0%
$3$
0%
$-3$
0%
$\dfrac32$
0%
$2$

Explanation

Given,

$AU_1=\left[\begin{matrix}1\\0\\0\end{matrix}\right]$

$\Rightarrow U_{1}=A^{-1}\left[\begin{matrix}1\\0\\0\end{matrix}\right]$ ...(1)

$AU_2 = \left[\begin{matrix}2\\3\\0\end{matrix}\right]$

$\Rightarrow U_{2}=A^{-1} \left[\begin{matrix}2\\3\\0\end{matrix}\right]$ ...(2)

$AU_3 = \left[\begin{matrix}2\\3\\1\end{matrix}\right]$

$\Rightarrow U_{3}=A^{-1} \left[\begin{matrix}2\\3\\1\end{matrix}\right]$ ...(3)

Also, given

$A=\left[\begin{matrix}1&0&0\\2&1&0\\3&2&1\end{matrix}\right]$

$|A|=1$
Now,

$adj\ A=C^{T}=\begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}^{T}$

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

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

Put the value of

$A^{-1}$ in (1), (2) and (3)

$U_{1}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\left[\begin{matrix}1\\0\\0\end{matrix}\right]$

$\Rightarrow U_{1}=\begin{bmatrix}1\\-2\\1\end{bmatrix}$

$U_{2}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix} \left[\begin{matrix}2\\3\\0\end{matrix}\right]$

$\Rightarrow U_{2}=\begin{bmatrix}2\\-1\\-4\end{bmatrix}$

$U_{3}=A^{-1} \left[\begin{matrix}2\\3\\1\end{matrix}\right]$

$\Rightarrow U_{3}=\begin{bmatrix}2\\-1\\-3\end{bmatrix}$

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

$\Rightarrow |U|= -1-14+18=3$

Hence, option A.

$\displaystyle \omega$ is an imaginary cube root of unity,then the value of

$\left | \begin{matrix} a &b\omega ^{2} & a\omega \\ b\omega & c &b\omega ^{2} \\ c\omega ^{2}&a\omega &c \end{matrix} \right |$ ,is ?

0%
$\displaystyle a^{3}+b^{3}+c^{3}$
0%
$\displaystyle a^{2}b-b^{2}c$
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0
0%
$\displaystyle a^{3}b+b^{3}+3abc$

Explanation

$\triangle =\begin{vmatrix} a & b{ \omega }^{ 2 } & a\omega \\ b\omega & c & b{ \omega }^{ 2 } \\ c{ \omega }^{ 2 } & a{ \omega } & c \end{vmatrix}\quad \quad 1+\omega +{ \omega }^{ 2 }=0;{ \omega }^{ 3 }=1;{ \omega }^{ 4 }=\omega$

$\Rightarrow a\left( { c }^{ 2 }-ab{ \omega }^{ 3 } \right) -b{ \omega }^{ 2 }\left( bc{ \omega }-bc\omega^4 \right) +a\omega \left( ab{ \omega }^{ 2 }-{ c }^{ 2 }{ \omega }^{ 2 } \right)$

$\Rightarrow a\left( { c }^{ 2 }-ab{ \omega }^{ 3 } \right) -b{ \omega }^{ 2 }\left( bc{ \omega }-bc\omega \right) +a\left( ab-{ c }^{ 2 } \right)$

$=a\left( { c }^{ 2 }-ab \right) -a\left( c^2-ab \right) =0$

$\therefore \triangle =0$

Option C

$A\displaystyle= \left | \begin{matrix}a &b &c \\ x &y &z \\ p &q &r \end{matrix} \right |$ and

$B=\left | \begin{matrix}q &-b &y \\ -p&a &-x \\ r&-c &z \end{matrix} \right |$ , then

0%
$A= 2B$
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$A= B$
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$A= -B$
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none of these

Explanation

$B=\begin{vmatrix}q &-b &y \\ -p&a &-x \\ r&-c &z \end{vmatrix}$

$=-\begin{vmatrix}q &b &y \\ -p&-a &-x \\ r&c &z \end{vmatrix}$

$=\begin{vmatrix}q &b &y \\ p&a &x \\ r&c &z \end{vmatrix}$

$=-\begin{vmatrix}p &a &x \\ q&b &y \\ r&c &z \end{vmatrix}$

$=-\begin{vmatrix}a &x &p \\ b&y &q \\ c&z &r \end{vmatrix}$

$=-\begin{vmatrix}a &b &c \\ x &y &z \\ p &q &r \end{vmatrix}=-A$ (

$\because |A|=|A^{T}|$ )

Hence, option C.

The value of the determinant

$\displaystyle \left | \begin{matrix} 1 &\omega ^{3} &\omega ^{5} \\ \omega ^{3}&1 &\omega ^{4} \\ \omega ^{5}&\omega ^{4} &1 \end{matrix} \right |$ , where

$\omega$ is an imaginary cube root of unity,is

0%
$\displaystyle (1-\omega )^{2}$
0%
3
0%
-3
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none of these

Explanation

$\left| \begin{matrix} 1 & \omega ^{ 3 } & \omega ^{ 5 } \\ \omega ^{ 3 } & 1 & \omega ^{ 4 } \\ \omega ^{ 5 } & \omega ^{ 4 } & 1 \end{matrix} \right| =\left| \begin{matrix} 1 & 1 & \omega ^{ 2 } \\ 1 & 1 & \omega \\ \omega ^{ 2 } & \omega & 1 \end{matrix} \right| \\ \Rightarrow \omega ^{ 2 }-2\omega +1$

$(\because \omega^{3}=1\quad and\quad 1+\omega+\omega^{2}=0)$

$\Rightarrow (\omega-1)^2$

Let

$\displaystyle \omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ ,then the value of the determinant

$\left | \begin{matrix} 1 & 1 &1 \\ 1& -1-\omega ^{2} &\omega ^{2} \\ 1& \omega ^{2} & \omega ^{4} \end{matrix} \right |,$ is

0%
$\displaystyle 3\omega$
0%
$\displaystyle 3\omega(\omega -1)$
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$\displaystyle 3\omega ^{2}$
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$\displaystyle 3 (-2\omega-1 )$

Explanation

$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1-{ \omega }^{ 2 } & { \omega }^{ 2 } \\ 1 & { \omega }^{ 2 } & { \omega }^{ 4 } \end{vmatrix}$

$\omega =$ cube root of unity

$\Rightarrow { \omega }^{ 3 }=1$

$1+\omega +{ \omega }^{ 2 }=0$

$\Rightarrow -1-{ \omega }^{ 2 }=\omega$

${ \omega }^{ 4 }=\left( { \omega }^{ 3 } \right) \omega =\omega$

$\triangle =\begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & { \omega }^{ 2 } \\ 1 & { \omega }^{ 2 } & \omega \end{vmatrix}$

$=1\left( { \omega }^{ 2 }-{ \omega }^{ 4 } \right) -1\left( \omega -{ \omega }^{ 2 } \right) +1\left( { \omega }^{ 2 }-\omega \right)$

$={ \omega }^{ 2 }-\omega -+{ \omega }^{ 2 }+{ \omega }^{ 2 }-\omega$

$=3({ \omega }^{ 2 }-\omega )$

$=3\omega \left( \omega -1 \right) =3\left( -2\omega -1 \right)$

Option B and D

Let

$\displaystyle A=\left [ \begin{matrix}1 &0 &0 \\ 2 &1 &0 \\ 3 &2 &1 \end{matrix} \right ]$ and

$\displaystyle U_{1}, U_{2}, U_{3}$ be column
matrices satisfying

$\displaystyle AU_{1}=\left [ \begin{matrix}1\\ 0\\ 0\end{matrix} \right ], AU_{2}=\left [ \begin{matrix}2\\ 3\\ 0\end{matrix} \right ], AU_{3}=\left [ \begin{matrix}2\\ 3\\ 1\end{matrix} \right ]$ . If U is

$\displaystyle 3\times 3$ matrix whose columns are

$\displaystyle U_{1}, U_{2}, U_{3}$ , then

$\displaystyle \left | U \right |=$

0%
$\displaystyle 3$
0%
$\displaystyle -3$
0%
$\displaystyle\dfrac{3}{2}$
0%
$\displaystyle 2$

Explanation

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

$U_1=\begin{bmatrix} a_{ 1 } \\ b_{ 1 } \\ c_{ 1 } \end{bmatrix}$ ,

$U_2=\begin{bmatrix} a_{ 2 } \\ b_{ 2 } \\ c_{ 2 } \end{bmatrix}$ and

$U_3=\begin{bmatrix} a_{ 3 } \\ b_{ 3 } \\ c_{ 3 } \end{bmatrix}$
Given

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

$\Rightarrow a_1=1; 2a_1+b_1=0; 3a_1+2b_1+c_1=0$
simplifying gives

$a_1=1; b_1=-2; c_1=1$

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

$\Rightarrow a_2=2; 2a_2+b_2=3; 3a_2+2b_2+c_2=0$
simplifying gives

$a_2=2; b_2=-1; c_2=-4$

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

$\Rightarrow a_3=2; 2a_3+b_3=3; 3a_3+2b_3+c_3=1$
simplifying gives

$a_3=2; b_3=-1; c_3=-3$

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

$|U|=\begin{vmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{vmatrix}=3$
Hence, option A.

The value of

$\displaystyle \left | \begin{matrix} 11 & 12 &13 \\ 12&13 &14 \\ 13&14 &15 \end{matrix} \right |$ ,is

0%
1
0%
0
0%
-1
0%
67

Explanation

$\left| \begin{matrix} 11 & 12 & 13 \\ 12 & 13 & 14 \\ 13 & 14 & 15 \end{matrix} \right|$

${ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }$

$=\left| \begin{matrix} 11 & 12 & 13 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{matrix} \right|$

$=2\left| \begin{matrix} 11 & 12 & 13 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right|$

$=0$

If the lines

$L_{1}:\lambda ^{2}x-y-1=0$

$L_{2}:x-\lambda ^{2}y+1=0$

$L_{3}:x+y-\lambda ^{2}=0$ pass through the same point the value(s) of

$\lambda$ equals

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$1$
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$\sqrt{2}$
0%
$2$
0%
$0$

Explanation

The given equations passes through same point.So, they are concurrent lines

$\Rightarrow \left| \begin{matrix} { \lambda }^{ 2 } & -1 & -1 \\ 1 & -{ \lambda }^{ 2 } & 1 \\ 1 & 1 & -{ \lambda }^{ 2 } \end{matrix} \right| =0$

$\Rightarrow { \lambda }^{ 6 }-3{ \lambda }^{ 2 }-2=0$

By doing synthetic division we get,

$(\lambda^4-2\lambda^2+1)(\lambda^2-2)=0$

$(\lambda^2-1)^2(\lambda^2-2)=0$

$\lambda=\pm \sqrt 2 or \lambda=\pm1$

But here

$\lambda=\pm1\ does\ not\ satisfies\ ,hence\ \lambda=\sqrt2$
Option B satisfies above equation

When the determinant

$\begin{vmatrix} \cos { 2x } & \sin ^{ 2 }{ x } & \cos { 4x } \\ \sin ^{ 2 }{ x } & \cos { 2x } & \cos ^{ 2 }{ x } \\ \cos { 4x } & \cos ^{ 2 }{ x } & \cos { 2x } \end{vmatrix}$ is expanded in powers of

$\sin { x }$ , then the constant term in that expression is

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

Explanation

$\begin{vmatrix} \cos { 2x } & \sin ^{ 2 }{ x } & \cos { 4x } \\ \sin ^{ 2 }{ x } & \cos { 2x } & \cos ^{ 2 }{ x } \\ \cos { 4x } & \cos ^{ 2 }{ x } & \cos { 2x } \end{vmatrix}={ a }_{ 0 }+{ a }_{ 1 }\sin { x } +{ a }_{ 2 }\sin ^{ 2 }{ x } +.....$

Put

$x=0$

$\Rightarrow \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}={ a }_{ 0 }$

$\Rightarrow a_{0}=-1$

$p+q+r=0=a+b+c$ , then the value of the determinant

$\begin{vmatrix} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{vmatrix}$ is

0%
$0$
0%
$pq+qb+ra$
0%
$1$
0%
none of these

Explanation

Given ,

$p+q+r=0=a+b+c$

$\begin{vmatrix} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{vmatrix}$

$=pa(qr{ a }^{ 2 }-{ p }^{ 2 }bc)-qb({ q }^{ 2 }ac-pr{ b }^{ 2 })+rc(pq{ c }^{ 2 }-{ r }^{ 2 }ab)$

$=pqr{ a }^{ 3 }-{ p }^{ 3 }abc-{ q }^{ 3 }abc+pqr{ b }^{ 3 }+pqr{ c }^{ 3 }-{ r }^{ 3 }abc$

$=pqr({ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 })-abc({ p }^{ 3 }+{ q }^{ 3 }+{ r }^{ 3 })$

$=pqr[{ (a+b+c) }^{ 3 }+3abc(a+b+c)]-abc[{ (p+q+r) }^{ 3 }+3pqr(p+q+r)]$

$=0$

$\displaystyle a\neq b\neq c,$ are value of x which satisfies the equation

$\displaystyle \left | \begin{matrix}0 &x-a &x-b \\ x+a &0 &x-c \\ x+b &x+c &0 \end{matrix} \right |=0$ is given by

0%
$x=0$
0%
$x=c$
0%
$x=b$
0%
$x=a$

Explanation

$\displaystyle \left | \begin{matrix}0 &x-a &x-b \\ x+a &0 &x-c \\ x+b &x+c &0 \end{matrix} \right |=0$

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

We can now check by options

$for \,\,option\,\, A,$ put

$x=0$ in above equation we get,

$\Rightarrow (0-a)(0+b)(0-c)+(0-b)(0+a)(0+c)=0$

$\Rightarrow abc-abc=0$

So,

$x=0$ satisfies the equation.

The value of

$\begin{vmatrix} -1 & 2 & 1 \\ 3+2\sqrt { 2 } & 2+2\sqrt { 2 } & 1 \\ 3-2\sqrt { 2 } & 2-2\sqrt { 2 } & 1 \end{vmatrix}$ is equal to

0%
zero
0%
$-16\sqrt { 2 }$
0%
$-8\sqrt { 2 }$
0%
one of these

Explanation

$\begin{vmatrix} -1 & 2 & 1 \\ 3+2\sqrt { 2 } & 2+2\sqrt { 2 } & 1 \\ 3-2\sqrt { 2 } & 2-2\sqrt { 2 } & 1 \end{vmatrix}$

$-1(2+2\sqrt2-2+2\sqrt2)-2(3+2\sqrt2-3+2\sqrt2)+1[(3+2\sqrt2)(2-2\sqrt2)-(2+2\sqrt2)(3-2\sqrt2)]$

$=-1(4\sqrt { 2 } )-2(4\sqrt { 2 } )+1(-4\sqrt { 2 } )=-16\sqrt { 2 }$

Number of values of

$a$ for which the lines

$2x+y-1=0, ax+3y-3=0, 3x+2y-2=0$ are concurrent is

0%
0
0%
1
0%
2
0%
$\infty$

Explanation

Here coefficient matrix,

$\Delta = \begin{vmatrix} 2 & 1 & -1 \\ a & 3 & -3 \\ 3 & 2 & -2 \end{vmatrix}$
Using

$C_2 \to C_2+C_3$

$\Delta = \begin{vmatrix} 2 & 0 & -1 \\ a & 0 & 3 \\ 3 & 0 & -2 \end{vmatrix}=0$
Clearly

$\Delta=0$ , Hence given lines are concurrent for all values of

$a$

Let

$\begin{vmatrix} x & 2 & x \\ { x }^{ 2 } & x & 6 \\ x & x & 6 \end{vmatrix}=A{ x }^{ 4 }+B{ x }^{ 3 }+C{ x }^{ 2 }+Dx+E$ . Then the value of

$5A+4B+3C+2D+E$ is equal to

0%
zero
0%
-16
0%
16
0%
-11

Explanation

$\begin{vmatrix} x & 2 & x \\ { x }^{ 2 } & x & 6 \\ x & x & 6 \end{vmatrix}=A{ x }^{ 4 }+B{ x }^{ 3 }+C{ x }^{ 2 }+Dx+E$

Consider

$LHS=\begin{vmatrix} x & 2 & x \\ { x }^{ 2 } & x & 6 \\ x & x & 6 \end{vmatrix}$

$=x(6x-6x)-2(6x^2-6x)+x(x^3-x^2)$

$={ x }^{ 4 }-{ x }^{ 3 }-12{ x }^{ 2 }+12x$

Comparing with RHS, we get

$A=1,B=-1, C=-12, D=12, E=0$

So,

$5A+4B+3C+2D+E=-11$

In triangle

$ABC$ , if

$\begin{vmatrix} 1 & 1 & 1 \\ \cot { \cfrac { A }{ 2 } } & \cot { \cfrac { B }{ 2 } } & \cot { \cfrac { C }{ 2 } } \\ \tan { \cfrac { B }{ 2 } } +\tan { \cfrac { C }{ 2 } } & \tan { \cfrac { C }{ 2 } } +\tan { \cfrac { A }{ 2 } } & \tan { \cfrac { A }{ 2 } } +\tan { \cfrac { B }{ 2 } } \end{vmatrix}=0$ , then the triangle must be

0%
equilateral
0%
isosceles
0%
obtuse angled
0%
none of these

Explanation

$\Rightarrow \tan { \cfrac { A }{ 2 } } \cot { \cfrac { B }{ 2 } } -\tan { \cfrac { A }{ 2 } } \cot { \cfrac { C }{ 2 } } +\tan { \cfrac { B }{ 2 } } \cot { \cfrac { C }{ 2 } } -\tan { \cfrac { B }{ 2 } } \cot { \cfrac { A }{ 2 } } +\tan { \cfrac { C }{ 2 } } \cot { \cfrac { A }{ 2 } } -\tan { \cfrac { C }{ 2 } } \cot { \cfrac { B }{ 2 } } =0$

$\tan { \cfrac { A }{ 2 } } (\cot { \cfrac { B }{ 2 } } -\cot { \cfrac { C }{ 2 } } )+\tan { \cfrac { B }{ 2 } } (\cot { \cfrac { C }{ 2 } } -\cot { \cfrac { A }{ 2 } } )+\tan { \cfrac { C }{ 2 } } (\cot { \cfrac { A }{ 2 } } -\cot { \cfrac { B }{ 2 } } )=0$

$(\cot { \cfrac { B }{ 2 } } -\cot { \cfrac { C }{ 2 } } )=0 , (\cot { \cfrac { C }{ 2 } } -\cot { \cfrac { A }{ 2 } } )=0, (\cot { \cfrac { A }{ 2 } } -\cot { \cfrac { B }{ 2 } } )=0$

$\Rightarrow A=B=C$

The value of the determinant

$\begin{vmatrix} 1 & 1 & 1 \\ { _{ }^{ m }{ C } }_{ 1 } & { _{ }^{ m+1 }{ C } }_{ 1 } & { _{ }^{ m+2 }{ C } }_{ 1 } \\ { _{ }^{ m }{ C } }_{ 2 } & { _{ }^{ m+1 }{ C } }_{ 2 } & { _{ }^{ m+2 }{ C } }_{ 2 } \end{vmatrix}$ is equal to

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

Explanation

$\begin{vmatrix} 1 & 1 & 1 \\ { ^{ m }{ C } }_{ 1 } & { ^{ m+1 }{ C } }_{ 1 } & { ^{ m+2 }{ C } }_{ 1 } \\ { ^{ m }{ C } }_{ 2 } & { ^{ m+1 }{ C } }_{ 2 } & { ^{ m+2 }{ C } }_{ 2 } \end{vmatrix}$

$=\begin{vmatrix} 1 & 1 & 1 \\ m & m+1 & m+2 \\ \dfrac { m(m-1) }{ 2 } & \dfrac { m(m+1) }{ 2 } & \dfrac { (m+2)(m+1) }{ 2 } \end{vmatrix}$

${ C }_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 2 },{ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 }\quad$

$\begin{vmatrix} 0 & 0 & 1 \\ -1 & -1 & m+2 \\ -m & -(m+1) & \dfrac { (m+2)(m+1) }{ 2 } \end{vmatrix}$

$1(-1(-(m+1))-(-1)(-m))$

$=1$

$a,b,c$ are different, then the value of

$x$ satisfying

$\begin{vmatrix} 0 & { x }^{ 2 }-a & { x }^{ 3 }-b \\ { x }^{ 2 }+a & 0 & { x }^{ 2 }+c \\ { x }^{ 4 }+b & x-c & 0 \end{vmatrix}=0$ is

0%
$a$
0%
$c$
0%
$b$
0%
$0$

$\Delta =\begin{vmatrix} a & { a }^{ 2 } & 0 \\ 1 & 2a+b & (a+b) \\ 0 & 1 & 2a+3b \end{vmatrix}$ is divisible by

0%
$a+b$
0%
$a+2b$
0%
$2a+3b$
0%
${ a }^{ 2 }$

Explanation

Expanding by

$R_{3}$ ,

$=-1(a(a+b))+(2a+3b)(2a^{2}+ab-a^{2})$

$=-1(a(a+b))+(2a+3b)a(a+b)$

$=a(a+b)(2a+3b-1)$

$\Delta =\begin{vmatrix} \sin { \theta } \cos { \phi } & \sin { \theta } \sin { \phi } & \cos { \theta } \\ \cos { \theta } \cos { \phi } & \cos { \theta } \sin { \phi } & -\sin { \theta } \\ -\sin { \theta } \sin { \phi } & \sin { \theta } \cos { \phi } & 0 \end{vmatrix}$ , then

0%
$\Delta$ is independent of $\theta$
0%
$\Delta$ is independent of $\phi$
0%
$\Delta$ is a constant
0%
$\displaystyle \cfrac { d\Delta }{ d\theta }| {_{ \theta =\pi / 2 }{ =0 }_{ } }$

Explanation

$\Delta =\begin{vmatrix}sin\theta cos\phi & sin\theta sin\phi & cos\theta\\ cos\theta cos\phi & cos\theta sin\phi & -sin\theta\\ -sin\theta sin\phi & sin\theta cos\phi & 0 \end{vmatrix}$
Expanding by

$C_{3}$ ,

$\Delta =cos\theta(cos\theta sin\theta (cos^{2}\phi+sin^{2}\phi))+sin\theta (sin^{2}\theta (cos^{2}\phi+sin^{2}\phi))$

$=cos\theta(cos\theta sin\theta )+sin\theta (sin^{2}\theta)=sin\theta$
So,

$\Delta$ is independent of

$\phi$

$\frac{\mathrm{d\Delta } }{\mathrm{d} \theta }=cos\theta |_{\frac{\pi }{2}}=0$

$\alpha$ is a characteristic root of a nonsingular matrix, then the corresponding characteristic root of adj A is

0%
$\displaystyle \frac{|A|}{\alpha}$
0%
$\displaystyle |\frac{A}{\alpha}|$
0%
$\displaystyle \frac{|adj A|}{\alpha}$
0%
$\displaystyle |\frac{adj A}{\alpha}|$

Explanation

Since

$\alpha$ is a characteristic root of a nonsingular matrix,
therefore

$\alpha \neq 0$ . Also

$\alpha$ is a characteristic root of
A implies that there exists a nonzero vector X such that

$AX = \alpha X$

$(adj A) (AX) = (adj A) (\alpha X)$

$[(adj A)A] X = \alpha (adj A)X$

$|A| IX = \alpha (adj A)X$

$[\because (adj A) A = |A| I]$

$|A| X = \alpha (adj A) X$

$\displaystyle \frac{|A|}{\alpha} X = (adj A)X$

$(adj A) X = \displaystyle \frac{|A|}{\alpha} X$
Since X is a nonzero vector,

$|A / \alpha|$ is a characteristic root of the matrix adj A.

$A$ is a square matrix of order

$m\times n$ , then

$adj(adj\space A)$ is equal to

0%
$|A|^n A$
0%
$|A|^{n-1} A$
0%
$|A|^{n-2} A$
0%
$|A|^{n-3} A$

Explanation

$(adj\: A)^{-1}=\displaystyle\frac{adj(adj\: A)}{|adj\: A|}$

$\Rightarrow (|A|A^{-})^{-1}=\displaystyle\frac{adj(adj\: A)}{|adj\: A|}$

$\Rightarrow \displaystyle\frac{A}{|A|}=\displaystyle\frac{adj(adj\: A)}{|adj\: A|}$

$\Rightarrow adj(adj\: A)= \displaystyle\frac{|adj\: A|A}{|A|}$

$\therefore adj(adj\: A)=|A|^{n-2}A$
Hence, option C.

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

$adj(A)$ equals

0%
$A$
0%
$A^T$
0%
$3A$
0%
$3A^T$

Explanation

Given

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

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

Now,

$adj A=C^{T}={\begin{bmatrix} -3 & -6 & -6 \\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{bmatrix}}^{T}$

$\Rightarrow adj A=\begin{bmatrix} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}$

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

$\Rightarrow adj A=3A^{T}$

Find the determinants of minors and cofactors of the determinant

$\begin{vmatrix}2 & 3 & 4\\ 7 & 2 & -5\\ 8 & -1 & 3\end{vmatrix}$

0%
$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$ and $\begin{vmatrix}1 & 61 & -23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$
0%
$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$ and $\begin{vmatrix}1 & -61 & 23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$
0%
$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$ and $\begin{vmatrix}1 & 61 & 23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$
0%
None of these.

Explanation

Here

$M_{11} = \begin{vmatrix}2 & -5 \\ -1 & 3\end{vmatrix}$ (Delete 1st row and first column)

$= 6-5$

$M_{11} = 1$

$\therefore C_{11} = 1$

$(\because (-1)^{1 + 1} = 1)$

$M_2 = \begin{vmatrix}7 & -5\\ 8 & 3 \end{vmatrix}$ (Delete 1st row and 2nd column)

$= 21 - (-40)$

$M_2 = 61$

$\therefore C_{12} = - 61$ ,

$(\because (-1)^{1+2} = - 1)$

$M_{13} = \begin{vmatrix}7 & 2 \\ 8 & -1\end{vmatrix}$ (Delete 1st row and 3rd column)

$= - 7 - 16$

$M_{13} = - 23$

$\therefore C_{13} = - 23, (\because (-1)^{1 + 3} = 1)$

$M_{21} = \begin{vmatrix}3 & 4 \\ -1 & 3\end{vmatrix}$ (Delete 2nd row and 1st column)

$= 9 - (-4)$

$M_{21} = 13$

$\therefore C_{21} = - 13, (\because (-1)^{2 + 1} = - 1)$

$M_{22} = \begin{vmatrix}2 & 4 \\ 8 & 3\end{vmatrix}$ (Delete 2nd row and 2nd column)

$= 6 - 32$

$M_{22} = - 26$

$\therefore C_{22} = - 26, (\because (-1)^{2 + 2} = 1)$

$M_{23} = \begin{vmatrix}2 & 3 \\ 8 & -1\end{vmatrix}$ (Delete 2nd row and 3rd column)

$= -2 - 24$

$M_{23} = - 26$

$\therefore C_{23} = 26, (\because (-1)^{2 + 3} = - 1)$

$M_{31} = \begin{vmatrix}3 & 4 \\ 2 & -5\end{vmatrix}$ (Delete 3rd row and 1st column)

$= -15 -8$

$M_{31} = - 23$

$\therefore C_{31} = - 23, (\because (-1)^{3 + 1} = 1)$

$M_{32} = \begin{vmatrix}2 & 4 \\ 7 & -5\end{vmatrix}$ (Delete 3rd row and 2nd column)

$= -10 - 28$

$M_{32} = - 38$

$\therefore C_{32} = 38, (\because (-1)^{3 + 2} = - 1)$

$M_{33} = \begin{vmatrix}2 & 3 \\ 7 & 2\end{vmatrix}$ (Delete 3rd row and 3rd column)

$= 4 - 21$

$M_{33} = - 17$

$\therefore C_{33} = - 17, (\because (-1)^{3 + 3} = 1)$
Hence Determinants of Minors and Cofactors are

$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$ and

$\begin{vmatrix}1 & -61 & -23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$

The adjoint of the matrix

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

0%
$\frac{1}{11} \begin{bmatrix}9 & -1 & -4\\ -3 & 4 & 5\\ 5 & -3 & -1\end{bmatrix}$
0%
$\begin{bmatrix}9 & 1 & -4\\ 3 & 4 & -5\\ 5 & 3 & -1\end{bmatrix}$
0%
$\begin{bmatrix}9 & -3 & 5\\ -1 & 4 & -3\\ -4 & 5 & -1\end{bmatrix}$
0%
$\begin{bmatrix}9 & -1 & -4\\ -3 & 4 & 5\\ 5 & -3 & -1\end{bmatrix}$

Explanation

Let

$C_{ij}$ be a cofactor of

$a_{ij}$ in A. Then, the cofactors of elements of A are given by

$C_{11} = \begin{bmatrix} 1& -3\\ 2 & 3\end{bmatrix} = 9$

$C_{12} = - \begin{bmatrix} 2& -3\\ -1 & 3\end{bmatrix} = - 3$

$C_{13} = \begin{bmatrix}2 & 1\\ -1 & 2\end{bmatrix} = 5$

$C_{21} = - \begin{bmatrix}1 & 1\\ 2 & 3\end{bmatrix} = - 1$

$C_{22} = \begin{bmatrix}1 & 1\\ -1 & 3\end{bmatrix} = 4$

$C_{23} = - \begin{bmatrix} 1& 1\\ -1 & 2\end{bmatrix} = - 3$

$C_{31} = \begin{bmatrix}1 & 1\\ 1 & -3\end{bmatrix} = - 4$

$C_{32} = - \begin{bmatrix} 1& 1\\ 2 & -3\end{bmatrix} = 5$

$C_{33} = \begin{bmatrix} 1& 1\\ 2 &1 \end{bmatrix} = -1$

$\therefore adj A = \begin{bmatrix}9 & -3 & 5\\ -1 & 4 &-3 \\ -4 & 5 & -1\end{bmatrix} '= \begin{bmatrix}9 & -1 & -4\\ -3 & 4 & 5\\ 5 & -3 & -1\end{bmatrix}$

$A^2 = I$ , then the value of

$det(A - I)$ is (where

$A$ has order

$3$ )

0%
$1$
0%
$-1$
0%
$0$
0%
cannot say anything

Explanation

$A^2=I$

$\Rightarrow A^2-I=O$

$\Rightarrow (A-I)(A+I)=O$

$\Rightarrow |A-I||A+I|=0$

$\therefore$ Either

$|A-I|=0$ or

$|A+I|=0$
Hence, cannot say anything about

$|A-I|$ .
Hence, option D.

$A=\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}$ and

$f(x) = \displaystyle \frac{1 + x}{1- x}$ , then

$f(|A|)$ is

0%
$\dfrac{-1}{2}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{-1}{3}$
0%
none of these

Explanation

Here

$|A| =1\times 1-2\times 2 = -3$

$\therefore f(|A|)=\cfrac{1+(-3)}{1+3}=-\cfrac{1}{2}$

$A$ is a square matrix of order

$n\times n$ and

$k$ is a scalar, then

$adj(kA)$ is equal to _____________.

0%
$k^{n-1}adj\space A$
0%
$k^{n}adj\space A$
0%
$k^{n+1}adj\space A$
0%
$kadj\space A$

Explanation

$(kA)^{-1}=\displaystyle\frac{adj(kA)}{|kA|}$

$\Rightarrow \displaystyle\frac{1}{k}A^{-1}=\frac{adj(kA)}{|kA|}$

$\Rightarrow \displaystyle\frac{adj A}{k|A|}=\frac{adj(kA)}{k^n|A|}$

$\therefore adj(kA)=k^{n-1}adj A$
Hence, option A.

Find the adjoint of the matrix

$A = \begin{bmatrix}1 & 2 & 3 \\ 1 & 3 & 5 \\ 1 & 5 & 12\end{bmatrix}$ .
If

$\mbox{Adjoint A: } \begin{bmatrix}a & -9 & 1 \\ b & 9 & -2 \\ 2 & c & 1\end{bmatrix} \\$ , find the value of

$abc$ .

0%
231
0%
213
0%
321
0%
312

If matrix A is given by

$A = \begin{bmatrix}6 & 11\\ 2 & 4\end{bmatrix}$ , then the determinant of

$A^{2005} - 6A^{2004}$ is

0%
$2^{2006}$
0%
$(-11) 2^{2005}$
0%
$-2^{2005}$
0%
$(-9)2^{2004}$

Explanation

Given,

$A = \begin{bmatrix}6 & 11\\ 2 & 4\end{bmatrix}$

$|A|=2$

Now,

$|A^{2005}-6A^{2004}|=|A|^{2004}|A-6I|$

$=2^{2004}\begin{vmatrix} 0 & 11 \\ 2 & -2 \end{vmatrix}$

$=2^{2004}(-22)$

$=2^{2005}(-11)$

$\displaystyle \left | \begin{matrix}0 &p-q &p-r \\ q-p &0 &q-r \\ r-p &r-q &0 \end{matrix} \right |$ is equal to

0%
$\displaystyle p+q+r$
0%
$0$
0%
$\displaystyle p-q-r$
0%
$\displaystyle -p+q+r$

Explanation

$\left| \begin{matrix} 0 & p-q & p-r \\ q-p & 0 & q-r \\ r-p & r-q & 0 \end{matrix} \right| \\$

$=0\left( 0-\left( r-q \right) \left( q-r \right) \right) -\left( p-q \right) \left( 0-\left( r-p \right) \left( q-r \right) \right) +\left( p-r \right) \left( \left( q-p \right) \left( r-q \right) -0 \right) \\$

$=\left( p-q \right) \left( r-p \right) \left( q-r \right) +\left( p-r \right) \left( q-p \right) \left( r-q \right) =0$

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

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

Explanation

$\Delta =\left| \begin{matrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{matrix} \right| \\$

$=6i\left( -3+3 \right) +3i\left( 4i+20 \right) +1\left( 12-60i \right) \\$

$=0-12+60i+12-60i$

$=0\\$

$\Rightarrow x=0,y=0$

Hence, the option 'D' is correct.

$A = \begin{bmatrix}3& -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$ then find

$Adj(Adj\space A)$ .

0%
$\quad \begin{bmatrix}3& -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$
0%
$\quad \begin{bmatrix}3& 3 & 4 \\ 2 & -3 & -4 \\ 0 & -1 & 1\end{bmatrix}$
0%
$\quad \begin{bmatrix}3& 3 & 4 \\ 2 & -3 & 4 \\ 0 & 1 & 1\end{bmatrix}$
0%
$\quad \begin{bmatrix}3& -3 & 4 \\ 2 & -3 & -4 \\ 0 & 1 & 1\end{bmatrix}$

Explanation

We know that

$Adj(Adj\space A) = | A |^{n-2} A$
Since

$\quad A = \begin{bmatrix}3& -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$
Here

$n = 3$ (3 order matrix)

$\therefore \quad |A| = \begin{vmatrix}3& -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{vmatrix}$

$\quad = 3(1) + 3(2-0) + 4(-2-0)$

$\quad = 1 \ne 0, \quad \therefore \quad A$ is non-singular.

$\therefore Adj(Adj A) = | A |^{3-2}.A = | A |.A = 1.A = A = \begin{bmatrix}3& -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$

The matrix

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

0%
non-singular
0%
singular
0%
skew-symmetric
0%
symmetric

Explanation

Given

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

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

$\Rightarrow |A|=1-1=0$

Hence,

$A$ is singular.

The value of the

$\displaystyle m^{th}$ order determinant of a matrix

$\displaystyle A$ is

$\displaystyle 15$ then the value of determinant formed by the cofactors of

$\displaystyle A$ will be

0%
$\displaystyle \left ( 15 \right )^{m}$
0%
$\displaystyle 15^{2m}$
0%
$\displaystyle \left ( 15 \right )^{m-1}$
0%
$\displaystyle \left ( 15 \right )^{2m-1}$

Explanation

If A be the matrix of order m and B is matrix of cofactors of a then determinant of B will be equal to

$\displaystyle \left ( m-1 \right )^{th}$ power of the determinant of A.

$\displaystyle \left | B \right |= \left | A \right |^{m-1}$

$\displaystyle \therefore |B| = \left ( 15 \right )^{m-1}$

The points

$\displaystyle(a, b+c),(b, c+a),(c, a+b)$ are

0%
vertices of an equilateral triangle
0%
collinear
0%
concyclic
0%
none of these

Explanation

To check for collineauty, we can check the slope of the two points. Suppose the points be A, B and C then

$(slope)_{AB} = \dfrac {(c+a)-(b+c)}{b-a} =-1$

$(slope)_{BC}= \dfrac{(a+b)-(c+a)}{c-b} =-1$

$m_{AB}= m_{BC} =$ points are collinear

In a third order determinant

$a_{ij}$ denotes the element in the ith row and the jth column If

$a_{ij} = \left\{\begin{matrix}0, & i = j\\ 1, & i > j\\ -1, & i < j\end{matrix}\right.$ then the value of the determinant

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

Explanation

$a_{ij} = \left\{\begin{matrix}0, & i = j\\ 1, & i > j\\ -1, & i < j\end{matrix}\right.$

Let

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

Now,

$|A|=\begin{vmatrix} 0&-1&-1\\1&0&-1\\1&1&0\end{vmatrix}$

$=1-1=0$

$\Rightarrow |A|=0$

The value of the determinant

$\begin{vmatrix} \sqrt { 6 } & 2i & 3+\sqrt { 6 } \\ \sqrt { 12 } & \sqrt { 3 } +\sqrt { 8 } i & 3\sqrt { 2 } +\sqrt { 6 } i \\ \sqrt { 18 } & \sqrt { 2 } +\sqrt { 12 } i & \sqrt { 27 } +2i \end{vmatrix}$ is

0%
complex
0%
real
0%
irrational
0%
rational

Explanation

$\begin{vmatrix} \sqrt { 6 } & 2i & 3+\sqrt { 6 } \\ \sqrt { 12 } & \sqrt { 3 } +\sqrt { 8 } i & 3\sqrt { 2 } +\sqrt { 6 } i \\ \sqrt { 18 } & \sqrt { 2 } +\sqrt { 12 } i & \sqrt { 27 } +2i \end{vmatrix}$

${ R }_{ 2 }\rightarrow { R }_{ 2 }-\sqrt { 2 } { R }_{ 1 }$

${ R }_{ 3 }\rightarrow { R }_{ 3 }-\sqrt { 3 } { R }_{ 1 }$

$\Rightarrow \triangle =\sqrt { 6 } \begin{vmatrix} 1 & 2i & 3+\sqrt { 6 } \\ 0 & \sqrt { 3 } & \sqrt { 6 } i-2\sqrt { 3 } \\ 0 & \sqrt { 2 } & 2i-3\sqrt { 2 } \end{vmatrix}$

$\Rightarrow \sqrt { 6 } \begin{vmatrix} \sqrt { 3 } & \sqrt { 6 } i-2\sqrt { 3 } \\ \sqrt { 2 } & 2i-3\sqrt { 2 } \end{vmatrix}$

$\Rightarrow \triangle$ is real and complex.

A and B

$\displaystyle \left | \begin{matrix}a+x &a &x \\ a-x &a &x \\ a-x &a &-x \end{matrix} \right |=0$ then

$\displaystyle x$ is

0%
$0$
0%
$\displaystyle a$
0%
$3$
0%
$\displaystyle 2a$

$\Delta =\begin{vmatrix} \cos \theta /2 & 1 & 1\\ 1 & \cos \theta /2 & -\cos \theta /2\\ -\cos \theta /2 & 1 & -1 \end{vmatrix}$ , If the minimun of

$\Delta$ is

$m_{1}$ and maximum of

$\Delta$ is

$m_{2}$ , then

$\left [ m_{1}, m_{2} \right ]$ are related

0%
[-4, -2]
0%
[2, 4]
0%
[-4, 0]
0%
[0, 2]

Explanation

$\Delta =\begin{vmatrix} \cos \theta /2 & 1 & 1 \\ 1 & \cos \theta /2 & -\cos \theta /2 \\ -\cos \theta /2 & 1 & -1 \end{vmatrix}$

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

$=2+2\cos ^{ 2 }{ \theta /2 }$

$=3+\cos\theta$

Now,

$\because -1\le \cos\theta \le 1$

$\Rightarrow 2\le 3+\cos\theta \le 4$
Hence,

$m_{1}=2,m_{2}=4$

$\displaystyle A= \begin{bmatrix}2 &14 &17 \\0 &\sin 2x &\cos 2x \\0 &\cos 2x &\sin 2x \end{bmatrix}$ then

$\displaystyle \left | A \right |$ equals

0%
$\displaystyle \cos 2x$
0%
$-2$
0%
$\displaystyle -2 \cos 4x$
0%
$\displaystyle \sin 4x$

Explanation

$\displaystyle A= \begin{bmatrix}2 &14 &17 \\0 &\sin 2x &\cos 2x \\0 &\cos 2x &\sin 2x \end{bmatrix}$

$\displaystyle \left | A \right |=2(\sin^{2}2x-\cos^{2}2x)=-2\cos 4x$

$x$ is a non-real cube root of

$-2$ , then the value of

$\begin{vmatrix} 1 & 2x & 1\\ x^{2} & 1 & 3x^{2}\\ 2 & 2x & 1 \end{vmatrix}$ equals to

0%
$-7$
0%
$-13$
0%
$0$
0%
$-12$

Explanation

Given,

$x=\sqrt [ 3 ]{ -2 }$

$\Rightarrow x^{3}=-2$

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

$=(1-6x^{3})-2x(-5x^{2})+1(2x^{3}-2)$

$=13-20-6$

$=-13$

$\begin{vmatrix}x^{2}+3x &x+1 &x-2 \\ x-1 &1-2x &x+4 \\ x+3 &x-4 &3x \end{vmatrix}= Ax^{4}+Bx^{3}+Cx^{2}+Dx+\varrho$
Then value of

$\varrho$ equals to,

0%
-10
0%
10
0%
0
0%
None of these

Explanation

$\begin{vmatrix}x^{2}+3x &x+1 &x-2 \\ x-1 &1-2x &x+4 \\ x+3 &x-4 &3x \end{vmatrix}= Ax^{4}+Bx^{3}+Cx^{2}+Dx+\varrho$

Substitute

$x=0$

$\Rightarrow \begin{vmatrix} 0 & 1 & -2 \\ -1 & 1 & 4 \\ 3 & -4 & 0 \end{vmatrix}=\varrho$

$=12-2=10$

$A=\begin{pmatrix} 1 & 2 & 1\\ -1 & 0 & 3\\ 2 & -1 & 1 \end{pmatrix}$ then characteristic equation is given by

0%
$-\lambda ^{3}+2\lambda ^{2}-4\lambda +18=0$
0%
$\lambda ^{3}+2\lambda ^{2}+4\lambda +18=0$
0%
$2\lambda ^{3}-\lambda ^{2}+6\lambda -2=0$
0%
None of these

Explanation

The characteristic equation of A is given by

$|A-\lambda I|=0$

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

$(1-\lambda)[-\lambda(1-\lambda)+3]-2(-1(1-\lambda)-6))+1(1+2\lambda)=0$

$\Rightarrow -\lambda ^{3}+2\lambda ^{2}-4\lambda +18=0$

If the determinant

$\begin{vmatrix}a & b & at-b\\ b & c & bt-c\\ 2 & 1 & 0\end{vmatrix}=0$ , if

$a, b, c$ are in

0%
$A.P.$
0%
$G.P.$
0%
$H.P.$
0%
$k=1/2$

Explanation

$\begin{vmatrix}a & b & at-b\\ b & c & bt-c\\ 2 & 1 & 0\end{vmatrix}=0$
Expanding it along third row,

$\Rightarrow 2[b(bt-c)-c(at-b)]-1[a(bt-c)-b(at-b)]=0$

$\Rightarrow 2t(b^2-ac)-(b^2-ac)=0$

$\Rightarrow (2t-1)(b^2-ac)=0$

$\Rightarrow t=\dfrac{1}{2}$ or

$b^2=ac$
If

$b^2 =ac$ then

$a,b,c\in$ G.P.

$A=\begin{bmatrix} -1 & -3 & -3\\ 3 & 1 & -3\\ 3 & -3 & 1 \end{bmatrix}$ then adj (A) is

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

Explanation

Given

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

$C=\begin{bmatrix} -8 & -12 & -12 \\ 12 & 8 & -12 \\ 12 & -12 & 8 \end{bmatrix}$

$adj A=C^{T}=\begin{bmatrix} -8 & 12 & 12 \\ -12 & 8 & -12 \\ -12 & -12 & 8 \end{bmatrix}$

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

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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