MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Commerce Maths Determinants Quiz 12 - MCQExams.com
CBSE
Class 12 Commerce Maths
Determinants
Quiz 12
If $$A=A=\left[ \begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix} \right]$$,then $$\left| A \right| \left| AdjA \right|$$ is equal to
Report Question
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
Report Question
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%
$$\begin{bmatrix} 9 & -19 & -4 \\ -4 & 14 & -1 \\ 8 & -3 & 2 \end{bmatrix}$$
0%
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
Report Question
0%
$$2.^{ 7 }{ { { p } } }_{ 3 }$$
0%
$$^{ 7 }{ { { p } } }_{ 3 }$$
0%
$$6.^{ 7 }{ { { C } } }_{ 3 }$$
0%
420
If $$A$$ is a square matrix $$(adj \,A)' - (adj \,A')$$
Report Question
0%
$$2A$$
0%
$$2 \,adj \,A$$
0%
Unit matrix
0%
Null matrix
If $$A$$ is singular matrix, then $$A.(adj\,A)$$ is
Report Question
0%
$$singular$$
0%
$$non-singular$$
0%
$$symmetric$$
0%
$$not \,defined$$
If $$A$$ 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
Report Question
0%
$$11$$
0%
$$13$$
0%
$$17$$
0%
$$19$$
A= \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix} then Adj(A)=
Report Question
0%
$${A^T}$$
0%
$$3{A^T}$$
0%
$${A^{ - 1}}$$
0%
$$ - {A^T}$$
If $$A=\begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix}$$ then the determinant of the matrix $$\left( {A}^{2016}-2{A}^{2015}-{A}^{2014} \right) $$ is
Report Question
0%
$$-2016$$
0%
$$-25$$
0%
$$2016$$
0%
$$-175$$
If adj B = A, |P| = |Q| = 1, then adj $$\left( { Q }^{ -1 }{ BP }^{ -1 } \right) $$ is
Report Question
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.
Report Question
0%
$$60$$
0%
$$62$$
0%
$$64$$
0%
$$66$$
If $$A=\begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$$ and $$A(adj\, A)=A{A}^{T}$$ then $$5a+3b$$ is equal to
Report Question
0%
$$5$$
0%
$$4$$
0%
$$11$$
0%
$$-1$$
If $$\Delta_1 = \begin{vmatrix}x & \sin \theta & \cos \theta\\-\sin \theta & -x & 1\\\cos \theta & 1 &x\end{vmatrix}$$ 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 $$\theta \in \left(0, \dfrac{\pi}{2}\right):$$
Report Question
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:
Report Question
0%
$$n(n-1)$$
0%
$$n(n-1)^2$$
0%
$$n^2(n-1)$$
0%
$$n^2(n-1)^2$$
If $$f'(x)=\begin{vmatrix} mx & mx-p & mx+p \\ n & n+p & n-p \\ mx+2n & mx+2n+p & mx+2n-p \end{vmatrix}$$, then $$y=f(x)$$ represents
Report Question
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.
If $${ f }_{ r }\left( x \right) ,{ g }_{ r }\left( x \right) ,{ h }_{ r }\left( x \right),\ r =1,2,3$$ 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
Report Question
0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
None of these
Explanation
Hence, option (D) is the correct answer.
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
Report Question
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, $$|3AB|=-81$$.
If $$ A = \left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{matrix} \right] $$ then
Report Question
0%
$$ adj(adj A ) = A $$
0%
$$ | adj(adj A) | = 1 $$
0%
$$ |adj A | = 1 $$
0%
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 ?
Report Question
0%
$$ AX = B $$ has a unique solution
0%
$$ AX = B $$ has exactly three solution
0%
$$ AX =B $$ has infinitely many solutions
0%
$$ AX = B $$ 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
Report Question
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
Report Question
0%
$$ |A| $$
0%
$$ |A| / 2 $$
0%
$$ 2|A| $$
0%
None of these
If $$ A = \left[ \begin{matrix} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{matrix} \right] ,xyz = 80 , 3x +2y +10z = 20 $$ then $$ A adj \quad A = \left[ \begin{matrix} 81 & 0 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 81 \end{matrix} \right] $$
Report Question
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
Report Question
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 $$ \frac {1}{2} $$
Report Question
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, $${ D }_{ max }=1+sin\left( \frac { \pi }{ 4 } \right) cos\left( \frac { \pi }{ 4 } \right) $$
$$\therefore { D }_{ max }=1+\left( \frac { 1 }{ \sqrt { 2 } } \right) \left( \frac { 1 }{ \sqrt { 2 } } \right) $$
$$\therefore { D }_{ max }=1+\frac { 1 }{ 2 } $$
$$\therefore { D }_{ max }=\frac { 3 }{ 2 } $$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Commerce Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
