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CBSE Questions for Class 12 Commerce Maths Determinants Quiz 4 - MCQExams.com
CBSE
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 -
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$$1$$
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$$2$$
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$$3$$
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$$4$$
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 $$\left | adj\: A \right |=$$
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$$k$$
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$$k^{2}$$
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$$k^{3}$$
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$$k^{4}$$
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 $$\displaystyle \left | \begin{matrix}4 &0 &0 \\0 &4 &0 \\0 &0 &4\end{matrix} \right |$$=
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$$4$$
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$$16$$
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$$64$$
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$$256$$
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
$$64$$.
If $$\displaystyle A=\:\left [ \begin{matrix}1 &x \\x^{2} &4y \end{matrix} \right ], B = \left [ \begin{matrix}-3 &1 \\1 &0 \end{matrix} \right ]$$ 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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$$\displaystyle \:\left ( -1,1 \right )$$
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$$\displaystyle \:\left ( 1,0 \right )$$
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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, $$adj \: \left( A \right) +B=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] $$
$$\Rightarrow\left[ \begin{matrix} 4y-3 & -x+1 \\ -x^{ 2 }+1 & 1 \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] $$
$$\Rightarrow -x+1=0 ,-x^2+1=0,4y-3=1$$
$$\Rightarrow x=1,y=1$$
If $$A$$ 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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$$\displaystyle \left | A \right |A$$
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$$\displaystyle \left | A \right |^{2}A$$
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$$\displaystyle \left | A \right |^{-1}A$$
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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
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$$3$$
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$$-3$$
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$$\dfrac32$$
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$$2$$
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.
If $$\displaystyle \omega$$ 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 ?
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$$\displaystyle a^{3}+b^{3}+c^{3}$$
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$$\displaystyle a^{2}b-b^{2}c$$
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0
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$$\displaystyle a^{3}b+b^{3}+3abc $$
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
If $$A\displaystyle= \left | \begin{matrix}a &b &c \\ x &y &z \\ p &q &r \end{matrix} \right |$$ and $$B=\left | \begin{matrix}q &-b &y \\ -p&a &-x \\ r&-c &z \end{matrix} \right |$$, then
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$$A= 2B$$
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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
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$$\displaystyle (1-\omega )^{2} $$
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3
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-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
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$$\displaystyle 3\omega$$
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$$\displaystyle 3\omega(\omega -1)$$
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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 $$\displaystyle \left | U \right |=$$
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$$\displaystyle 3$$
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$$\displaystyle -3$$
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$$\displaystyle\dfrac{3}{2}$$
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$$\displaystyle 2$$
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
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1
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0
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-1
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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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$$\sqrt{2}$$
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$$2$$
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$$0$$
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
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1
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0
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-1
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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 $$x=0$$
$$\Rightarrow \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}={ a }_{ 0 }$$
$$\Rightarrow a_{0}=-1$$
If $$p+q+r=0=a+b+c$$, then the value of the determinant $$\begin{vmatrix} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{vmatrix}$$ is
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$$0$$
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$$pq+qb+ra$$
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$$1$$
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none of these
Explanation
Given , $$p+q+r=0=a+b+c$$
$$\begin{vmatrix} pa & qb & rc \\ qc & ra & pb \\ rb & pc & qa \end{vmatrix}$$
$$=pa(qr{ a }^{ 2 }-{ p }^{ 2 }bc)-qb({ q }^{ 2 }ac-pr{ b }^{ 2 })+rc(pq{ c }^{ 2 }-{ r }^{ 2 }ab)$$
$$=pqr{ a }^{ 3 }-{ p }^{ 3 }abc-{ q }^{ 3 }abc+pqr{ b }^{ 3 }+pqr{ c }^{ 3 }-{ r }^{ 3 }abc$$
$$=pqr({ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 })-abc({ p }^{ 3 }+{ q }^{ 3 }+{ r }^{ 3 })$$
$$=pqr[{ (a+b+c) }^{ 3 }+3abc(a+b+c)]-abc[{ (p+q+r) }^{ 3 }+3pqr(p+q+r)] $$
$$=0$$
If $$\displaystyle a\neq b\neq c,$$ 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
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$$x=0$$
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$$x=c$$
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$$x=b$$
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$$x=a$$
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
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zero
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$$-16\sqrt { 2 }$$
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$$-8\sqrt { 2 }$$
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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
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0
0%
1
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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
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zero
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-16
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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
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equilateral
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isosceles
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obtuse angled
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none of these
Explanation
$$\begin{vmatrix} 1 & 1 & 1 \\ \cot { \cfrac { A }{ 2 } } & \cot { \cfrac { B }{ 2 } } & \cot { \cfrac { C }{ 2 } } \\ \tan { \cfrac { B }{ 2 } } +\tan { \cfrac { C }{ 2 } } & \tan { \cfrac { C }{ 2 } } +\tan { \cfrac { A }{ 2 } } & \tan { \cfrac { A }{ 2 } } +\tan { \cfrac { B }{ 2 } } \end{vmatrix}=0$$
$$\Rightarrow \tan { \cfrac { A }{ 2 } } \cot { \cfrac { B }{ 2 } } -\tan { \cfrac { A }{ 2 } } \cot { \cfrac { C }{ 2 } } +\tan { \cfrac { B }{ 2 } } \cot { \cfrac { C }{ 2 } } -\tan { \cfrac { B }{ 2 } } \cot { \cfrac { A }{ 2 } } +\tan { \cfrac { C }{ 2 } } \cot { \cfrac { A }{ 2 } } -\tan { \cfrac { C }{ 2 } } \cot { \cfrac { B }{ 2 } } =0$$
$$\tan { \cfrac { A }{ 2 } } (\cot { \cfrac { B }{ 2 } } -\cot { \cfrac { C }{ 2 } } )+\tan { \cfrac { B }{ 2 } } (\cot { \cfrac { C }{ 2 } } -\cot { \cfrac { A }{ 2 } } )+\tan { \cfrac { C }{ 2 } } (\cot { \cfrac { A }{ 2 } } -\cot { \cfrac { B }{ 2 } } )=0$$
$$(\cot { \cfrac { B }{ 2 } } -\cot { \cfrac { C }{ 2 } } )=0 , (\cot { \cfrac { C }{ 2 } } -\cot { \cfrac { A }{ 2 } } )=0, (\cot { \cfrac { A }{ 2 } } -\cot { \cfrac { B }{ 2 } } )=0$$
$$\Rightarrow A=B=C$$
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
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$$1$$
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$$-1$$
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$$0$$
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none of these
Explanation
$$\begin{vmatrix} 1 & 1 & 1 \\ { ^{ m }{ C } }_{ 1 } & { ^{ m+1 }{ C } }_{ 1 } & { ^{ m+2 }{ C } }_{ 1 } \\ { ^{ m }{ C } }_{ 2 } & { ^{ m+1 }{ C } }_{ 2 } & { ^{ m+2 }{ C } }_{ 2 } \end{vmatrix}$$
$$=\begin{vmatrix} 1 & 1 & 1 \\ m & m+1 & m+2 \\ \dfrac { m(m-1) }{ 2 } & \dfrac { m(m+1) }{ 2 } & \dfrac { (m+2)(m+1) }{ 2 } \end{vmatrix}$$
$${ C }_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 2 },{ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 }\quad $$
$$\begin{vmatrix} 0 & 0 & 1 \\ -1 & -1 & m+2 \\ -m & -(m+1) & \dfrac { (m+2)(m+1) }{ 2 } \end{vmatrix}$$
$$ 1(-1(-(m+1))-(-1)(-m))$$
$$=1$$
If $$a,b,c$$ 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
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$$a$$
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$$c$$
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$$b$$
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$$0$$
$$\Delta =\begin{vmatrix} a & { a }^{ 2 } & 0 \\ 1 & 2a+b & (a+b) \\ 0 & 1 & 2a+3b \end{vmatrix}$$ is divisible by
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$$a+b$$
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$$a+2b$$
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$$2a+3b$$
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$${ a }^{ 2 }$$
Explanation
Expanding by $$R_{3}$$,
$$=-1(a(a+b))+(2a+3b)(2a^{2}+ab-a^{2})$$
$$=-1(a(a+b))+(2a+3b)a(a+b)$$
$$=a(a+b)(2a+3b-1)$$
If $$\Delta =\begin{vmatrix} \sin { \theta } \cos { \phi } & \sin { \theta } \sin { \phi } & \cos { \theta } \\ \cos { \theta } \cos { \phi } & \cos { \theta } \sin { \phi } & -\sin { \theta } \\ -\sin { \theta } \sin { \phi } & \sin { \theta } \cos { \phi } & 0 \end{vmatrix}$$, then
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$$\Delta$$ is independent of $$\theta$$
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$$\Delta$$ is independent of $$\phi$$
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$$\Delta$$ is a constant
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$$ \displaystyle \cfrac { d\Delta }{ d\theta }| {_{ \theta =\pi / 2 }{ =0 }_{ } } $$
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 $$\phi$$
$$\frac{\mathrm{d\Delta } }{\mathrm{d} \theta }=cos\theta |_{\frac{\pi }{2}}=0$$
If $$\alpha$$ is a characteristic root of a nonsingular matrix, then the corresponding characteristic root of adj A is
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$$\displaystyle \frac{|A|}{\alpha} $$
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$$\displaystyle |\frac{A}{\alpha}| $$
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$$\displaystyle \frac{|adj A|}{\alpha} $$
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$$\displaystyle |\frac{adj A}{\alpha}| $$
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
$$AX = \alpha X$$
$$(adj A) (AX) = (adj A) (\alpha X)$$
$$[(adj A)A] X = \alpha (adj A)X$$
$$|A| IX = \alpha (adj A)X$$ $$[\because (adj A) A = |A| I]$$
$$|A| X = \alpha (adj A) X$$
$$|A| X = \alpha (adj A) X$$
$$\displaystyle \frac{|A|}{\alpha} X = (adj A)X$$
$$(adj A) X = \displaystyle \frac{|A|}{\alpha} X$$
Since X is a nonzero vector, $$|A / \alpha|$$ is a characteristic root of the matrix adj A.
If $$A$$ is a square matrix of order $$m\times n$$, then $$adj(adj\space A)$$ is equal to
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$$|A|^n A$$
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$$|A|^{n-1} A$$
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$$|A|^{n-2} A$$
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$$|A|^{n-3} A$$
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.
If $$A = \begin{bmatrix}-1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1\end{bmatrix}$$, Then $$adj(A)$$ equals
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$$A$$
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$$A^T$$
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$$3A$$
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$$3A^T$$
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 $$\begin{vmatrix}2 & 3 & 4\\ 7 & 2 & -5\\ 8 & -1 & 3\end{vmatrix}$$
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$$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$$ and $$\begin{vmatrix}1 & 61 & -23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$$
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$$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$$ and $$\begin{vmatrix}1 & -61 & 23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$$
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$$\begin{vmatrix}1 & 61 & -23\\ 13 & -26 & -26\\ -23 & -38 & -17\end{vmatrix}$$ and $$\begin{vmatrix}1 & 61 & 23\\ -13 & -26 & 26\\ -23 & 38 & -17\end{vmatrix}$$
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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
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$$\frac{1}{11} \begin{bmatrix}9 & -1 & -4\\ -3 & 4 & 5\\ 5 & -3 & -1\end{bmatrix}$$
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$$\begin{bmatrix}9 & 1 & -4\\ 3 & 4 & -5\\ 5 & 3 & -1\end{bmatrix}$$
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$$\begin{bmatrix}9 & -3 & 5\\ -1 & 4 & -3\\ -4 & 5 & -1\end{bmatrix}$$
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$$\begin{bmatrix}9 & -1 & -4\\ -3 & 4 & 5\\ 5 & -3 & -1\end{bmatrix}$$
Explanation
Let $$C_{ij}$$ be a cofactor of $$a_{ij}$$ in A. Then, the cofactors of elements of A are given by
$$C_{11} = \begin{bmatrix} 1& -3\\ 2 & 3\end{bmatrix} = 9$$
$$C_{12} = - \begin{bmatrix} 2& -3\\ -1 & 3\end{bmatrix} = - 3$$
$$C_{13} = \begin{bmatrix}2 & 1\\ -1 & 2\end{bmatrix} = 5$$
$$C_{21} = - \begin{bmatrix}1 & 1\\ 2 & 3\end{bmatrix} = - 1$$
$$C_{22} = \begin{bmatrix}1 & 1\\ -1 & 3\end{bmatrix} = 4$$
$$C_{23} = - \begin{bmatrix} 1& 1\\ -1 & 2\end{bmatrix} = - 3$$
$$C_{31} = \begin{bmatrix}1 & 1\\ 1 & -3\end{bmatrix} = - 4$$
$$C_{32} = - \begin{bmatrix} 1& 1\\ 2 & -3\end{bmatrix} = 5$$
$$C_{33} = \begin{bmatrix} 1& 1\\ 2 &1 \end{bmatrix} = -1$$
$$\therefore
adj A = \begin{bmatrix}9 & -3 & 5\\ -1 & 4 &-3 \\ -4
& 5 & -1\end{bmatrix} '= \begin{bmatrix}9 & -1 & -4\\ -3
& 4 & 5\\ 5 & -3 & -1\end{bmatrix}$$
If $$A^2 = I$$, then the value of $$det(A - I)$$ is (where $$A$$ has order $$3$$)
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$$1$$
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$$-1$$
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$$0$$
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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.
If $$A=\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}$$and $$f(x) = \displaystyle \frac{1 + x}{1- x}$$, then $$f(|A|)$$ is
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$$\dfrac{-1}{2}$$
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$$\dfrac{1}{2}$$
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$$\dfrac{-1}{3}$$
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none of these
Explanation
Here $$|A| =1\times 1-2\times 2 = -3$$
$$\therefore f(|A|)=\cfrac{1+(-3)}{1+3}=-\cfrac{1}{2}$$
If $$A$$ is a square matrix of order $$n\times n$$ and $$k$$ is a scalar, then $$adj(kA)$$ is equal to _____________.
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$$k^{n-1}adj\space A$$
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$$k^{n}adj\space A$$
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$$k^{n+1}adj\space A$$
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$$kadj\space A$$
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 $$abc$$.
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231
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213
0%
321
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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
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$$2^{2006}$$
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$$(-11) 2^{2005}$$
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$$-2^{2005}$$
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$$(-9)2^{2004}$$
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
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$$\displaystyle p+q+r$$
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$$0$$
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$$\displaystyle p-q-r$$
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$$\displaystyle -p+q+r$$
Explanation
$$\left| \begin{matrix} 0 & p-q & p-r \\ q-p & 0 & q-r \\ r-p & r-q & 0 \end{matrix} \right| \\$$
$$ =0\left( 0-\left( r-q \right) \left( q-r \right) \right) -\left( p-q \right) \left( 0-\left( r-p \right) \left( q-r \right) \right) +\left( p-r \right) \left( \left( q-p \right) \left( r-q \right) -0 \right) \\$$
$$ =\left( p-q \right) \left( r-p \right) \left( q-r \right) +\left( p-r \right) \left( q-p \right) \left( r-q \right) =0$$
If $$\displaystyle \left | \begin{matrix}6i &-3i &1 \\ 4 &3i &-1 \\ 20 &3 &i \end{matrix} \right |=x+iy$$ then
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$$\displaystyle x=3, y=1$$
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$$\displaystyle x=1, y=3$$
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$$\displaystyle x=0, y=3$$
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$$\displaystyle x=0, y=0$$
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.
If $$A = \begin{bmatrix}3& -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$$ then find $$Adj(Adj\space A)$$.
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$$\quad \begin{bmatrix}3& -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$$
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$$\quad \begin{bmatrix}3& 3 & 4 \\ 2 & -3 & -4 \\ 0 & -1 & 1\end{bmatrix}$$
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$$\quad \begin{bmatrix}3& 3 & 4 \\ 2 & -3 & 4 \\ 0 & 1 & 1\end{bmatrix}$$
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$$\quad \begin{bmatrix}3& -3 & 4 \\ 2 & -3 & -4 \\ 0 & 1 & 1\end{bmatrix}$$
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
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non-singular
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singular
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skew-symmetric
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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
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$$\displaystyle \left ( 15 \right )^{m}$$
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$$\displaystyle 15^{2m}$$
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$$\displaystyle \left ( 15 \right )^{m-1}$$
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$$\displaystyle \left ( 15 \right )^{2m-1}$$
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
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vertices of an equilateral triangle
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collinear
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concyclic
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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
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0
0%
1
0%
-1
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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
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complex
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real
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irrational
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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
If $$\displaystyle \left | \begin{matrix}a+x &a &x \\ a-x &a &x \\ a-x &a &-x \end{matrix} \right |=0$$ then $$\displaystyle x$$ is
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$$0$$
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$$\displaystyle a$$
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$$3$$
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$$\displaystyle 2a$$
If $$\Delta =\begin{vmatrix}
\cos \theta /2 & 1 & 1\\
1 & \cos \theta /2 & -\cos \theta /2\\
-\cos \theta /2 & 1 & -1
\end{vmatrix}$$, If the minimun of $$\Delta $$ is $$m_{1}$$ and maximum of $$\Delta $$ is $$m_{2}$$, then $$\left [ m_{1}, m_{2} \right ]$$ are related
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[-4, -2]
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[2, 4]
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[-4, 0]
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[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, $$m_{1}=2,m_{2}=4$$
If $$\displaystyle A= \begin{bmatrix}2 &14 &17 \\0 &\sin 2x &\cos 2x \\0 &\cos 2x &\sin 2x \end{bmatrix}$$ then $$\displaystyle \left | A \right |$$ equals
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$$\displaystyle \cos 2x $$
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$$-2$$
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$$\displaystyle -2 \cos 4x $$
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$$\displaystyle \sin 4x $$
Explanation
$$\displaystyle A= \begin{bmatrix}2 &14 &17 \\0 &\sin 2x &\cos 2x \\0 &\cos 2x &\sin 2x \end{bmatrix}$$
$$\displaystyle \left | A \right |=2(\sin^{2}2x-\cos^{2}2x)=-2\cos 4x$$
If $$x$$ 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
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$$-7$$
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$$-13$$
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$$0$$
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$$-12$$
Explanation
Given, $$x=\sqrt [ 3 ]{ -2 } $$
$$\Rightarrow x^{3}=-2$$
$$\begin{vmatrix} 1 & 2x & 1 \\ x^{ 2 } & 1 & 3x^{ 2 } \\ 2 & 2x & 1 \end{vmatrix}$$
$$=(1-6x^{3})-2x(-5x^{2})+1(2x^{3}-2)$$
$$=13-20-6$$
$$=-13$$
If $$\begin{vmatrix}x^{2}+3x &x+1 &x-2 \\ x-1 &1-2x &x+4 \\ x+3 &x-4 &3x \end{vmatrix}= Ax^{4}+Bx^{3}+Cx^{2}+Dx+\varrho $$
Then value of $$\varrho$$ equals to,
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-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 $$x=0$$
$$\Rightarrow \begin{vmatrix} 0 & 1 & -2 \\ -1 & 1 & 4 \\ 3 & -4 & 0 \end{vmatrix}=\varrho $$
$$=12-2=10$$
If $$A=\begin{pmatrix}
1 & 2 & 1\\
-1 & 0 & 3\\
2 & -1 & 1
\end{pmatrix}$$ then characteristic equation is given by
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$$-\lambda ^{3}+2\lambda ^{2}-4\lambda +18=0$$
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$$\lambda ^{3}+2\lambda ^{2}+4\lambda +18=0$$
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$$2\lambda ^{3}-\lambda ^{2}+6\lambda -2=0$$
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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
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$$A.P.$$
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$$G.P.$$
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$$H.P.$$
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$$k=1/2$$
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.
If $$A=\begin{bmatrix}
-1 & -3 & -3\\
3 & 1 & -3\\
3 & -3 & 1
\end{bmatrix}$$ then adj (A) is
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$$=4\begin{bmatrix} -2 & 3 & 3 \\ -3 & 2 & -3 \\ -3 & 3 & 2 \end{bmatrix}$$
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$$=4\begin{bmatrix} -2 & 3 & 3 \\ 3 & 2 & -3 \\ -3 & -3 & 2 \end{bmatrix}$$
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$$=4\begin{bmatrix} -2 & -3 & 3 \\ -3 & 2 & -3 \\ -3 & -3 & 2 \end{bmatrix}$$
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$$=4\begin{bmatrix} -2 & 3 & 3 \\ -3 & 2 & -3 \\ -3 & -3 & 2 \end{bmatrix}$$
Explanation
Given $$A=\begin{bmatrix} -1 & -3 & -3 \\ 3 & 1 & -3 \\ 3 & -3 & 1 \end{bmatrix}$$
$$ C=\begin{bmatrix} -8 & -12 & -12 \\ 12 & 8 & -12 \\ 12 & -12 & 8 \end{bmatrix}$$
$$adj A=C^{T}=\begin{bmatrix} -8 & 12 & 12 \\ -12 & 8 & -12 \\ -12 & -12 & 8 \end{bmatrix}$$
$$=4\begin{bmatrix} -2 & 3 & 3 \\ -3 & 2 & -3 \\ -3 & -3 & 2 \end{bmatrix}$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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