Determinants Test

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Determinants Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=\begin{vmatrix} 1 & -1 & 1\\ 0 & 2 & -3\\ 2 & 1 & 0\end{vmatrix}$$ and $$B=(adj A)$$, and $$C=5A$$, then $$\displaystyle\frac{|adj B|}{|C|}$$ is?

A
$$5$$

B
$$25$$

C
$$-1$$

D
$$1$$

Solution

$$|adj(A)|=|A|^{n-1} \dots eqn (1)$$ (property of adjoint of matrix)

$$|A|=1(3)+1(6)+1(-4)=5\dots eqn(2)$$

from the eqn (1) and (2)
$$|B|= 5^{3-1}= 5^2$$
$$\dfrac{|adj(B)|}{|C|}= \dfrac{5^2}{5\times5}=1$$
Question 2
1 / -0

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

A
$$\dfrac { 1 }{ 3 }$$ or $$ 1$$

B
$$\dfrac { 1 }{ \sqrt { 2 } } $$ or $$\dfrac { \sqrt { 3 } }{ 2 } $$

C
$$0$$ or $$\dfrac { 1 }{ 2 } $$

D
None of these

Solution

Applying $${ R }_{ 2 }\rightarrow { R }_{ 2 }+4{ R }_{ 1 }$$ and $${ R }_{ 3 }\rightarrow { R }_{ 3 }+7{ R }_{ 1 }$$ we get
$$\begin{vmatrix} 3 & -2 & \sin { 3\theta } \\ 5 & 0 & \cos { 2\theta } +4\sin { 3\theta } \\ 10 & 0 & 2+7\sin { 3\theta } \end{vmatrix}=0$$
$$\Rightarrow 2\left[ 5\left( 2+7\sin { 3\theta } \right) -10\left( \cos { 2\theta } +4\sin { 3\theta } \right) \right] =0$$
$$\Rightarrow 2+7\sin { 3\theta } -2\cos { 2\theta } -8\sin { 3\theta } =0$$
$$\Rightarrow 2-2\cos { 2\theta } -\sin { 3\theta } =0$$
$$\Rightarrow \sin { \theta } \left( 4\sin ^{ 2 }{ \theta } +4\sin { \theta } -3 \right) =0$$
$$\Rightarrow \sin { \theta } =0$$ or $$\left( 2\sin { \theta } -1 \right) =0$$ or $$\left( 2\sin { \theta } +3 \right) =0$$
$$\Rightarrow \sin { \theta } =0$$ or $$\sin { \theta } =\dfrac { 1 }{ 2 } $$
Question 3
1 / -0

If the matrix $$A=\begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{bmatrix},xyz=60$$ and $$8x+4y+3z=10$$, then $$A(adj\quad A)$$ is equal to

A
$$\begin{bmatrix} 64 & 0 & 0 \\ 0 & 64 & 0 \\ 0 & 0 & 64 \end{bmatrix}$$

B
$$\begin{bmatrix} 88 & 0 & 0 \\ 0 & 88 & 0 \\ 0 & 0 & 88 \end{bmatrix}$$

C
$$\begin{bmatrix} 68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68 \end{bmatrix}$$

D
$$\begin{bmatrix} 35 & 0 & 0 \\ 0 & 35 & 0 \\ 0 & 0 & 35 \end{bmatrix}$$

Solution

$$\because A\cdot dj(A)=\left| A \right| I\quad $$
$$\therefore \left| A \right| =xyz-8x-3(z-8)+2(2-2y)$$
$$=60-20+28=68\quad $$
$$\quad \therefore A\cdot adj(A)=68\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68 \end{bmatrix}$$
Question 4
1 / -0

If $$A=\begin{bmatrix} x & x-1 \\ 2x & 1 \end{bmatrix}$$ and if $$\text{det}A=-9$$, then the values of $$x$$ are

A
$$\dfrac { 3 }{ 2 } ,-3$$

B
$$\dfrac { -2 }{ 3 } ,3$$

C
$$\dfrac { 2 }{ 3 } ,3$$

D
$$\dfrac { -3 }{ 2 } ,3$$

E
$$\dfrac { -3 }{ 2 } ,-3$$

Solution

Given, $$A=\begin{bmatrix} x & x-1 \\ 2x & 1 \end{bmatrix}$$, $$\text{det} A=-9$$
$$\Rightarrow \left| A \right| =\begin{vmatrix} x & x-1 \\ 2x & 1 \end{vmatrix}$$
$$\Rightarrow -9=x-2x\left( x-1 \right) $$
$$\Rightarrow -9=x-2{ x }^{ 2 }+2x$$
$$\Rightarrow 2{ x }^{ 2 }-3x-9=0$$
$$\Rightarrow 2{ x }^{ 2 }-6x+3x-9=0$$
$$\Rightarrow 2x\left( x-3 \right) +3\left( x-3 \right) =0$$
$$\Rightarrow \left( x-3 \right) \left( 2x+3 \right) =0$$
$$\Rightarrow x=3,-\dfrac { 3 }{ 2 } $$
Question 5
1 / -0

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

A
$$-3$$

B
$$3$$

C
$$2$$

D
$$-8$$

E
$$-2$$

Solution

Given, $$A=\begin{bmatrix} x & 1 & -x \\ 0 & 1 & -1 \\ x & 0 & 7 \end{bmatrix}$$. Then
det $$(A) = x7-1(0+x)-x(0-x)$$
$$=7x-x+x^2=x^2+6x$$ $$...(i)$$
Also $$det(A)=\begin{vmatrix} 3 & 0 & 1 \\ 2 & -1 & 0 \\ 0 & 6 & 7 \end{vmatrix}$$
$$=3(-7-0)-0+1(12)=-9$$ $$...(ii)$$
From Eqs(i) and (ii), we get
$$\quad { x }^{ 2 }+6x=-9$$
$$\Rightarrow { x }^{ 2 }+6x+9=0\Rightarrow { (x+3) }^{ 2 }=0$$
$$x+3=0$$
$$x=-3$$
Question 6
1 / -0

If $$A=\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix}$$, then the value of $$\left| A \right| \left| adj\left( A \right) \right| $$ is

A
$${ a }^{ 3 }$$

B
$${ a }^{ 6 }$$

C
$${ a }^{ 9 }$$

D
$${ a }^{ 27 }$$

Solution

$$A=\left| \begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix} \right| =\\ \Rightarrow |A|={ a }^{ 3 }\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right| ={ a }^{ 3 }\\ \left| adj(A) \right| ={ \left| A \right| }^{ n-1 }$$
Here $$n=3$$
$$\Rightarrow \left| adj(A) \right| ={ \left| A \right| }^{ 3-1 }={ \left| A \right| }^{ 2 }={ a }^{ 6 }\\ \Rightarrow |A|.\left| adj(A) \right| ={ a }^{ 3 }.{ a }^{ 6 }={ a }^{ 9 }$$
So option $$C$$ is correct
Question 7
1 / -0

If $$A=\begin{bmatrix} 1 & 2 & -1 \\ -1& 1 & 2 \\ 2 & -1 & 1\end{bmatrix}$$, then $$\text{det} (\text{adj}(\text{adj} A))$$ is equal to.

A
$$14^4$$

B
$$14^3$$

C
$$14^2$$

D
$$14$$

Solution

We have, $$A=\begin{bmatrix} 1 & 2 & -1\\ -1 & 1 & 2 \\ 2 & -1 & 1\end{bmatrix}$$
$$\Rightarrow |A|=1(1+2)-2(-1-4)-1(1-2)$$
$$=3+10+1=14$$
We know that, for a square matrix of order $$n$$,
$$\text{adj}(\text{adj} A)=|A|^{n-2}A,$$ if $$|A|\neq 0$$
$$\Rightarrow \text{det}(\text{adj} (\text{adj} A))=||A|^{n-2}A|$$
$$\Rightarrow \text{det}(\text{adj} (\text{adj} A))=(|A|^{n-2})^n|A|$$
$$\Rightarrow \text{det}(\text{adj} (\text{adj} A))=|A|^{n^2-2n+1}$$
Here, $$n=3$$ and $$|A|=14$$
Therefore, $$ \text{det}(\text{adj}(\text{adj} A))=(14)^{3^2-2\times 3+1}=14^4$$.
Question 8
1 / -0

If $$3x+2y=I$$ and $$2x-y=O$$, where $$I$$ and $$O$$ are unit and null matrices of order $$3$$ respectively, then

A
$$x=\dfrac { 1 }{ 7 } , y=\dfrac { 2 }{ 7 } $$

B
$$x=\dfrac { 2 }{ 7 } , y=\dfrac { 1 }{ 7 } $$

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

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

Solution

We have,
$$3x+2y=I$$ ...(i)
$$2x-y=O$$ ....(ii)
On multiplying equation (ii) by $$2$$, we get
$$4x-2y=2\cdot O=O$$ ...(iii)
On adding equations (i) and (iii), we get
$$3x+2y=I$$
$$4x-2y=O$$
____________
$$7x=I$$
$$\Rightarrow x=\dfrac { I }{ 7 } $$ or $$\dfrac { 1 }{ 7 } I$$
From equation (i), we get
$$2y=I-\dfrac { 3 }{ 7 } I=\dfrac { 4 }{ 7 } I$$
$$\Rightarrow y=\dfrac { 2 }{ 7 } I$$
Question 9
1 / -0

If $$A = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix}$$ and $$A\ adj\ A = k\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$$ then the value of $$k$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

We know that,
$$A(adj\ A) = |A|I$$
Given, $$A = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix}$$
$$|A| = \cos^{2}x + \sin^{2}x = 1$$
$$|A| = \cos^{2} x + \sin^{2} x = 1$$
$$\therefore A(adj A) = 1\cdot I=kI$$
$$k = 1$$.
Question 10
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The value of $$a$$, for which the points $$(9, 5), (1, 2)$$ and $$(a, 8)$$ are collinear, is

A
$$17$$

B
$$8$$

C
$$7$$

D
$$71$$

E
$$1$$

Solution

Given, points $$(9, 5), (1, 2)$$ and $$(a, 8)$$ are collinear.
$$\begin{vmatrix}9 & 5 & 1\\ 1 & 2 & 1\\ a & 8 & 1\end{vmatrix} = 0$$
Applying operation $$R_{2}\rightarrow R_{2} - R_{1}, R_{3}\rightarrow R_{3} - R_{1}$$ we get
$$\begin{vmatrix}9& 5 & 1\\ -8 & -3 & 0\\ a - 9 & 3 & 0\end{vmatrix} = 0$$
Expanding along $$C_{3}$$, we get
$$-24 + 3(a - 9) = 0$$
$$\Rightarrow 3a - 51 = 0\Rightarrow a = 17$$.

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Re-Attempt Weekly Quiz Competition

Determinants Test - 47

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Determinants Test - 47

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SHARING IS CARING

Question 1

$$5$$

$$25$$

$$-1$$

$$1$$

Solution

Question 2

$$\dfrac { 1 }{ 3 }$$ or $$ 1$$

$$\dfrac { 1 }{ \sqrt { 2 } } $$ or $$\dfrac { \sqrt { 3 } }{ 2 } $$

$$0$$ or $$\dfrac { 1 }{ 2 } $$

None of these

Solution

Question 3

$$\begin{bmatrix} 64 & 0 & 0 \\ 0 & 64 & 0 \\ 0 & 0 & 64 \end{bmatrix}$$

$$\begin{bmatrix} 88 & 0 & 0 \\ 0 & 88 & 0 \\ 0 & 0 & 88 \end{bmatrix}$$

$$\begin{bmatrix} 68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68 \end{bmatrix}$$

$$\begin{bmatrix} 35 & 0 & 0 \\ 0 & 35 & 0 \\ 0 & 0 & 35 \end{bmatrix}$$

Solution

Question 4

$$\dfrac { 3 }{ 2 } ,-3$$

$$\dfrac { -2 }{ 3 } ,3$$

$$\dfrac { 2 }{ 3 } ,3$$

$$\dfrac { -3 }{ 2 } ,3$$

$$\dfrac { -3 }{ 2 } ,-3$$

Solution

Question 5

$$-3$$

$$3$$

$$2$$

$$-8$$

$$-2$$

Solution

Question 6

$${ a }^{ 3 }$$

$${ a }^{ 6 }$$

$${ a }^{ 9 }$$

$${ a }^{ 27 }$$

Solution

Question 7

$$14^4$$

$$14^3$$

$$14^2$$

$$14$$

Solution

Question 8

$$x=\dfrac { 1 }{ 7 } , y=\dfrac { 2 }{ 7 } $$

$$x=\dfrac { 2 }{ 7 } , y=\dfrac { 1 }{ 7 } $$

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

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

Solution

Question 9

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 10

$$17$$

$$8$$

$$7$$

$$71$$

$$1$$

Solution

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