MCQ Questions for Class 12 Commerce Maths Determinants Quiz 12

$A=A=\left[ \begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix} \right]$ ,then

$\left| A \right| \left| AdjA \right|$ is equal to

0%
${a}^{9}$
0%
${a}^{27}$
0%
${a}^{61}$
0%
$none \ of \ these$

The adjoining of the matrix

$\begin{bmatrix} 1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2 \end{bmatrix}$ , is

0%
$\begin{bmatrix} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{bmatrix}$
0%
$\begin{bmatrix} 9 & 4 & 8 \\ 19 & 14 & 3 \\ -4 & 1 & 2 \end{bmatrix}$
0%
$\begin{bmatrix} 9 & -19 & -4 \\ -4 & 14 & -1 \\ 8 & -3 & 2 \end{bmatrix}$
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None of these

There are 12 points in a plane of which 5 are collinear. The maximum number of distinct quadrilaterals which can be formed with vertices at these points is

0%
$2.^{ 7 }{ { { p } } }_{ 3 }$
0%
$^{ 7 }{ { { p } } }_{ 3 }$
0%
$6.^{ 7 }{ { { C } } }_{ 3 }$
0%
420

$A$ is a square matrix

$(adj \,A)' - (adj \,A')$

0%
$2A$
0%
$2 \,adj \,A$
0%
Unit matrix
0%
Null matrix

$A$ is singular matrix, then

$A.(adj\,A)$ is

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

$A$ is

$4\times 4$ matrix and if

$\left| \left| A \right| adj\left( \left| A \right| A \right) \right| ={ \left| A \right| }^{ n }$ , then

$n$ is

0%
$11$
0%
$13$
0%
$17$
0%
$19$

$\begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$ then Adj(A)=

0%
${A^T}$
0%
$3{A^T}$
0%
${A^{ - 1}}$
0%
$- {A^T}$

$A=\begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix}$ then the determinant of the matrix

$\left( {A}^{2016}-2{A}^{2015}-{A}^{2014} \right)$ is

0%
$-2016$
0%
$-25$
0%
$2016$
0%
$-175$

If adj B = A, |P| = |Q| = 1, then adj

$\left( { Q }^{ -1 }{ BP }^{ -1 } \right)$ is

0%
PQ
0%
QAP
0%
PAQ
0%
${ PA }^{ -1 }Q$

There are

$12$ points in a plane. The number of the straight lines joining any two of them when

$3$ of them are collinear is.

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$60$
0%
$62$
0%
$64$
0%
$66$

$A=\begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$ and

$A(adj\, A)=A{A}^{T}$ then

$5a+3b$ is equal to

0%
$5$
0%
$4$
0%
$11$
0%
$-1$

$\Delta_1 = \begin{vmatrix}x & \sin \theta & \cos \theta\\-\sin \theta & -x & 1\\\cos \theta & 1 &x\end{vmatrix}$ and

$\Delta_2 = \begin{vmatrix}x & \sin 2\theta & \cos 2\theta\\-\sin 2\theta & -x & 1\\ \cos 2\theta & 1 & x\end{vmatrix}, x \neq 0$ ; then for all

$\theta \in \left(0, \dfrac{\pi}{2}\right):$

0%
$\Delta_1 - \Delta_2 = x(\cos 2\theta - \cos 4\theta)$
0%
$\Delta_1 + \Delta_2 = -2x^3$
0%
$\Delta_1 - \Delta_2 = -2x^3$
0%
$\Delta_1 + \Delta_2 = -2(x^3 + x - 1)$

Two straight lines intersects at a point O. Points

$A_1, A_2,....A_n$ are taken on one line and

$B_1,B_2,....B_n$ on the other. If the point O is not to be used, the number of triangles that can be drawn using these points as vertices, is:

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$n(n-1)$
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$n(n-1)^2$
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$n^2(n-1)$
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$n^2(n-1)^2$

$f'(x)=\begin{vmatrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx+2n & mx+2n+p & mx+2n-p \end{vmatrix}$ , then

$y=f(x)$ represents

0%
a straight line parallel to x-axis
0%
a straight line parallel to y-axis
0%
parabola
0%
a straight line with negative slope

Explanation

Given

${ f }^{ ' }\left( x \right) =\left| \begin{matrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx+2n & mx+2n+p & mx+2n-p \end{matrix} \right|$

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

$\therefore { f }^{ ' }\left( x \right) =\left| \begin{matrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx+2n-2n & mx+2n+p-2\left( n+p \right) & mx+2n-p-2\left( n-p \right) \end{matrix} \right|$

$\therefore { f }^{ ' }\left( x \right) =\left| \begin{matrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx & mx+2n+p-2n-2p & mx+2n-p-2n+2p \end{matrix} \right|$

$\therefore { f }^{ ' }\left( x \right) =\left| \begin{matrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx & mx-p & mx+p \end{matrix} \right|$

Now,

${ R }_{ 1 }={ R }_{ 3 }$

$\therefore { f }^{ ' }\left( x \right) =0$

$\therefore { f }\left( x \right) =c$

$\therefore y=c$

This is the equation of straight line parallel to x axis.

${ f }_{ r }\left( x \right) ,{ g }_{ r }\left( x \right) ,{ h }_{ r }\left( x \right),\ r =1,2,3$ are polynomials in

$x$ such that

${ f }_{ r }\left( a \right) = { g }_{ r }\left( a \right) = { h }_{ r }\left( a \right),\ r =1,2,3$ and

${ F }\left( x \right) = \begin{vmatrix} { f }_{ 1 }\left( x \right) & { f }_{ 2 }\left( x \right) & { f }_{ 3 }\left( x \right) \\ { g }_{ 1 }\left( x \right) & { g }_{ 2 }\left( x \right) & { g }_{ 3 }\left( x \right) \\ { h }_{ 1 }\left( x \right) & { h }_{ 2 }\left( x \right) & { h }_{ 3 }\left( x \right) \end{vmatrix}$

then

${ F }^{ ' }\left( x \right)$ at

$x = a$ is

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$1$
0%
$2$
0%
$3$
0%
None of these

If A and B are square matrices of order 3 such that

$\left | A \right |$ = -1,

$\left | B \right |$ =3, then

$\left | 3AB \right |$ equals

0%
-9
0%
-81
0%
-27
0%
81

Explanation

For any square matrix

$X$ of order

$n$ ,

$det(kX)=k^n(det X)$ where

$k$ is any constant.

Here,

$A$ and

$B$ are square matrices of order 3 such that

$|A|=-1$ and

$|B|=3$ .

Now,

$|3AB|=3^3|AB|$ (Since

$A$ and

$B$ are both of order 3 thus

$AB$ is also a matrix of order 3)

$\Rightarrow |3AB| =27|A||B|$ (Since for any square matrices

$A$ and

$B$ ,

$|AB|=|A||B|$ )

$\Rightarrow |3AB| =27(-1)(3)=-81$

Thus,

$|3AB|=-81$ .

$A = \left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{matrix} \right]$ then

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$adj(adj A ) = A$
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$| adj(adj A) | = 1$
0%
$|adj A | = 1$
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none of these

Let

$A= \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 0 & 5 \\ 0 & 2 & 1 \end{matrix} \right]$ and

$B = \left[ \begin{matrix} 0 \\ -3 \\ 1 \end{matrix} \right]$ which of the following is true ?

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$AX = B$ has a unique solution
0%
$AX = B$ has exactly three solution
0%
$AX =B$ has infinitely many solutions
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$AX = B$ is inconsistent

Let

$f\left( x \right) =\begin{vmatrix} { x }^{ 3 } & \sin { x } & \cos { x } \\ 6 & -1 & 0 \\ p & { p }^{ 2 } & { p }^{ 3 } \end{vmatrix}$ where

$p$ is a constant. Then

$\dfrac{d^2 }{d x^3}\left\{ f\left( x \right) \right\}$ at

$x=0$ is

0%
$p$
0%
$p + p^{2}$
0%
$p + p^{3}$
0%
Independent of $p$

Explanation

Given,

$f(x)=\begin{vmatrix}x^3 &\sin x &\cos x \\6 & -1 &0 \\p &p^2 &p^3\end{vmatrix}$

Since, the variable is all in one single row thus it is sufficient to just find the derivative of each element in the first row to get the derivative of the determinant. Hence we get-

$\frac{\mathrm{d} }{\mathrm{d} x}f(x)=\frac{\mathrm{d} }{\mathrm{d} x}\begin{vmatrix}x^3 &\sin x &\cos x \\6 & -1 &0 \\p &p^2 &p^3\end{vmatrix}$

$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x}\left (f(x) \right )=\begin{vmatrix}3x^2 &\cos x &-\sin x \\6 & -1 &0 \\p &p^2 &p^3\end{vmatrix}$

$\Rightarrow \frac{\mathrm{d^2} }{\mathrm{d} x^2}\left (f(x) \right )=\frac{\mathrm{d} }{\mathrm{d} x}\begin{vmatrix}3x^2 &\cos x &-\sin x \\6 & -1 &0 \\p &p^2 &p^3\end{vmatrix}$

$\Rightarrow \frac{\mathrm{d^2} }{\mathrm{d} x^2}\left (f(x) \right )=\begin{vmatrix}6x &-\sin x &-\cos x \\6 & -1 &0 \\p &p^2 &p^3\end{vmatrix}$

$\Rightarrow \frac{\mathrm{d^3} }{\mathrm{d} x^3}\left (f(x) \right )=\frac{\mathrm{d} }{\mathrm{d} x}\begin{vmatrix}6x &-\sin x &-\cos x \\6 & -1 &0 \\p &p^2 &p^3\end{vmatrix}$

$\Rightarrow \frac{\mathrm{d^3} }{\mathrm{d} x^3}\left (f(x) \right )=\begin{vmatrix}6 &-\cos x &\sin x \\6 & -1 &0 \\p &p^2 &p^3\end{vmatrix}$

Thus, at

$x=0$ we have-

$\left [ \frac{\mathrm{d^3} }{\mathrm{d} x^3}\left (f(x) \right ) \right ]_{x=0}=\begin{vmatrix}6 &-\cos 0 &\sin 0 \\6 & -1 &0 \\p &p^2 &p^3\end{vmatrix}=\begin{vmatrix}6 &-1 & 0 \\6 & -1 &0 \\p &p^2 &p^3\end{vmatrix}$

Since two rows are identical the value of the determinant is 0.

Thus,

$\left [ \frac{\mathrm{d^3} }{\mathrm{d} x^3}\left (f(x) \right ) \right ]_{x=0}=0$ , which is independent of p.

Hence, the correct answer is option D.

Consider an arbitary

$3 \times 3$ matrix

$A = [a_{ij} ]$ , a matrix

$B = [ b_{ij} ]$ is formed such that

$b_{ij}$ i sthe sum of all the elements expect

$a_{ij}$ in the

$i^{th}$ row of A . answer the following questions.

The value of

$to |B|$ is equal

0%
$|A|$
0%
$|A| / 2$
0%
$2|A|$
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None of these

$A = \left[ \begin{matrix} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{matrix} \right] ,xyz = 80 , 3x +2y +10z = 20$ then

$A adj \quad A = \left[ \begin{matrix} 81 & 0 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 81 \end{matrix} \right]$

0%
True
0%
False

Explanation

Given

$A=\left[ \begin{matrix} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{matrix} \right]$

1) Minors Matrix of A-

${ M }_{ A }=\left[ \begin{matrix} yz-3 & 2z-3 & 2-y \\ 5z-2 & xz-2 & x-5 \\ 15-2y & 3x-4 & xy-10 \end{matrix} \right]$

2) Cofactors of the matrix A-

${ C }_{ A }=\left[ \begin{matrix} yz-3 & -\left( 2z-3 \right) & 2-y \\ -\left( 5z-2 \right) & xz-2 & -\left( x-5 \right) \\ 15-2y & -\left( 3x-4 \right) & xy-10 \end{matrix} \right]$

3) Adjoint of matrix A is,

$Adj.\quad A=\left[ \begin{matrix} yz-3 & -\left( 5z-2 \right) & 15-2y \\ -\left( 2z-3 \right) & xz-2 & -\left( 3x-4 \right) \\ 2-y & -\left( x-5 \right) & xy-10 \end{matrix} \right]$

$\therefore Adj.\quad A=\left[ \begin{matrix} yz-3 & -5z+2 & 15-2y \\ -2z+3 & xz-2 & -3x+4 \\ 2-y & -x+5 & xy-10 \end{matrix} \right]$

${ R }_{ 1 }\times x$ ,

${ R }_{ 2 }\times 5$ and

${ R }_{ 3 }\times 2$

$\therefore Adj.\quad A=\left[ \begin{matrix} xyz-3x & -5xz+2x & 15x-2xy \\ -10z+15 & 5xz-10 & -15x+20 \\ 4-2y & -2x+10 & 2xy-20 \end{matrix} \right]$

${ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }\rightarrow { R }_{ 1 }$

$\therefore Adj.\quad A=\left[ \begin{matrix} xyz-3x-10z+15+4-2y & -5xz+2x+5xz-10-2x+10 & 15x-2xy-15x+20+2xy-20 \\ -10z+15 & 5xz-10 & -15x+20 \\ 4-2y & -2x+10 & 2xy-20 \end{matrix} \right]$

$\therefore Adj.\quad A=\left[ \begin{matrix} xyz-\left( 3x+2y+10z \right) +19 & 0 & 0 \\ -10z+15 & 5xz-10 & -15x+20 \\ 4-2y & -2x+10 & 2xy-20 \end{matrix} \right]$

$\therefore Adj.\quad A=\left[ \begin{matrix} 80-20+19 & 0 & 0 \\ -10z+15 & 5xz-10 & -15x+20 \\ 4-2y & -2x+10 & 2xy-20 \end{matrix} \right]$

$\therefore Adj.\quad A=\left[ \begin{matrix} 79 & 0 & 0 \\ -10z+15 & 5xz-10 & -15x+20 \\ 4-2y & -2x+10 & 2xy-20 \end{matrix} \right]$

${ R }_{ 2 }\times y$ and

${ R }_{ 3 }\times z$

$\therefore Adj.\quad A=\left[ \begin{matrix} 79 & 0 & 0 \\ -10yz+15y & 5xyz-10y & -15xy+20y \\ 4z-2yz & -2xz+10z & 2xyz-20z \end{matrix} \right]$

The value of

$| \cup |$ equals

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

Explanation

$A^n - A^{n-2} =A^2 -I \Rightarrow A^{50} = A^{48} +A^2 - I$

Further,

$A^{48} = A^{46} +A^2 -I$

$A^{46} = A^{44} + A^2 - I$

$A^4 = A^2 + A^2 - I$

____________________

$A^{50} = 25A^2 - 24 I$

Here,

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

$\Rightarrow A^{50} = \left[ \begin{matrix} 25 & 0 & 0 \\ 25 & 25 & 0 \\ 25 & 0 & 25 \end{matrix} \right] -24\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{matrix} \right]$

$\therefore |A^{50}| = 1$

Also

$tr(A^{50}) = 1 +1 + 1 =3$ further,

$\left[ \begin{matrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 25 \\ 25 \end{matrix} \right] \Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \cup _{ 2 }=\left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right]$ similarly

$\cup _{ 2 }=\left[ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right] \quad and\quad \cup _{ 3 }=\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] \Rightarrow \cup \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] ,i.e.|\cup |=1$

The maximum value of

$\left| \begin{matrix}1 &1 &1 \\ 1 & (1+sin \theta) &1 \\ 1 & 1 & 1 +cos \theta \end{matrix} \right|$ is

$\frac {1}{2}$

0%
True
0%
False

Explanation

Let

$D=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1+sin\theta & 1 \\ 1 & 1 & 1+cos\theta \end{matrix} \right|$

$\therefore D=1\left[ \left( 1+sin\theta \right) \left( 1+cos\theta \right) \right] -1\left[ \left( 1+cos\theta \right) -1 \right] +1\left[ 1-\left( 1+sin\theta \right) \right]$

$\therefore D=1+cos\theta +sin\theta +sin\theta cos\theta -1\left[ cos\theta \right] +1\left[ 1-1-sin\theta \right]$

$\therefore D=1+cos\theta +sin\theta +sin\theta cos\theta -cos\theta -sin\theta$

$\therefore D=1+sin\theta cos\theta$

For maximum value of

$D$ , differentiate w.r.t.

$\theta$ and equate to zero.

$\therefore \frac { dD }{ d\theta } =sin\theta \frac { d }{ d\theta } \left( cos\theta \right) +cos\theta \frac { d }{ d\theta } \left( sin\theta \right)$

$\therefore \frac { dD }{ d\theta } =sin\theta \left( -sin\theta \right) +cos\theta \left( cos\theta \right)$

$\therefore \frac { dD }{ d\theta } =-{ sin }^{ 2 }\theta +{ cos }^{ 2 }\theta$

For maximum value of

$\theta$ ,

$\frac { dD }{ d\theta } =0$

$\therefore { cos }^{ 2 }\theta -{ sin }^{ 2 }\theta =0$

$\therefore cos\left( 2\theta \right) =0$

$\therefore 2\theta ={ cos }^{ -1 }\left( 0 \right)$

$\therefore 2\theta =\frac { \pi }{ 2 }$

$\therefore \theta =\frac { \pi }{ 4 }$

Thus,

${ D }_{ max }=1+sin\left( \frac { \pi }{ 4 } \right) cos\left( \frac { \pi }{ 4 } \right)$

$\therefore { D }_{ max }=1+\left( \frac { 1 }{ \sqrt { 2 } } \right) \left( \frac { 1 }{ \sqrt { 2 } } \right)$

$\therefore { D }_{ max }=1+\frac { 1 }{ 2 }$

$\therefore { D }_{ max }=\frac { 3 }{ 2 }$

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

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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