Determinants Test

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Determinants Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=\begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$$, then $$\text{adj}$$ $$(3{ A }^{ 2 }+12A)$$ is equal to

A
$$\begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$$

B
$$\begin{bmatrix} 51 & 84 \\ 63 & 72 \end{bmatrix}$$

C
$$\begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$$

D
$$\begin{bmatrix} 72 & -84 \\ -63 & 51 \end{bmatrix}$$

Solution

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

$$A^2 = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} \, \begin{bmatrix} 2 & -3\\ -4 &1 \end{bmatrix}$$

$$A^2 = \begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix}$$

$$3A^2 = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix}$$

$$3A^2 + 12 A = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix} + \begin{bmatrix} 24 & -36 \\ -48 & 12 \end{bmatrix} = \begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$$

$$adj (3A^2 + 12 A) = \begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$$
Question 2
1 / -0

If $$\begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$$ is the adjoint of a $$3\times 3$$ matrix $$A$$ and $$|A|=4$$, then $$\alpha$$ is equal to:

A
$$4$$

B
$$11$$

C
$$5$$

D
$$0$$

Solution

$$\begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix} = 1 \left( 3 \times 4 - 3 \times 4 \right) - \alpha \left( 1 \times 4 - 3 \times 2 \right) + 3 \left( 1 \times 4 - 3 \times 2 \right) = 2 \alpha - 6$$
$$\begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix} = \begin{vmatrix} adj. \; A \end{vmatrix}$$
$$\Rightarrow \left( 2 \alpha - 6 \right) = \begin{vmatrix} adj. \; A \end{vmatrix}$$
Now,
$$\begin{vmatrix} adj. \; A \end{vmatrix} = {\left| A \right|}^{3 - 1} = {4}^{2} = 16$$
$$\therefore 2 \alpha - 6 = 16$$
$$\Rightarrow 2 \alpha = 16 + 6$$
$$\Rightarrow \alpha = \cfrac{22}{2} = 11$$
Question 3
1 / -0

If $$\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix}$$ = $$(A+Bx)(x-A)^2$$,
then the ordered pair $$(A , B)$$ is equal to:

A
( -4 , -5 )

B
( -4 , 3 )

C
( -4 , 5 )

D
( 4 , 5 )

Solution

$$\left|\begin{matrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{matrix}\right|=$$ $$(x-4)[(x-4)^2-4x^2]-2x[2x(x-4)-4x^2]+2x[4x^2-2x(x-4)]$$
$$=(x-4)[x^2+16-8x-4x^2]-2x[2x^2-8x-4x^2]+2x[4x^2-2x^2+8x]$$
$$=(x-4)[16-8x-3x^2]-2x[-8x-2x^2]+2x[2x^2+8x]$$
$$=(x-4)(16-8x-3x^2)+4x[2x^2+8x]$$
$$=16x-8x^2-3x^3-64+32x+12x^2+8x^3+32x^2$$
$$=5x^3+36x^2+48x-64$$

$$(A+Bx)(x-A)^2=(A+Bx)(x^2+A^2-2xA)$$
$$=Ax^2+A^3-2xA^2+Bx^3+BA^2x-2ABx^2$$
$$=Bx^3+(A-2AB)x^2+(BA^2-2x)x+A^3$$
$$B=5$$ $$A-2AB=36$$
$$A(1-2B)=36$$
$$A(1-10)=36$$
$$A=-4$$
$$\therefore (A,B)=(-4,5)$$ .
Question 4
1 / -0

If $$\left| \begin{array} { c c } { 1 } & { \sin x } & { \sin ^ { 2 } x } \\ { 1 } & { \cos x } & { \cos ^ { 2 } x } \\ { 1 } & { \tan x } & { \tan ^ { 2 } x } \end{array} \right| = 0 , x \in [ 0,2 \pi ] ,$$ then number of possible values of $$x$$ is.

A
$$4$$

B
$$5$$

C
$$6$$

D
None of these

Solution

$$\left| \overset { 1 }{ \underset { 1 }{ 1 } } \quad \overset { sinx }{ \underset { tanx }{ cosx } } \quad \overset { { \sin }^{ 2 }x }{ \underset { { \tan }^{ 2 }x }{ { \cos }^{ 2 }x } } \right| =0$$
$$\left( \underset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 } }{ { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 } } \right) $$
$$\Rightarrow \left| \overset { 0 }{ \underset { 1 }{ 0 } } \quad \quad \overset { sinx-cosx }{ \underset { tanx }{ cosx-tanx } } \quad \quad \overset { { \sin }^{ 2 }x-{ \cos }^{ 2 }x }{ \underset { { \tan }^{ 2 }x }{ { \cos }^{ 2 }x-{ \tan }^{ 2 }x } } \right| =0$$
$$\Rightarrow \underset { \left( cosx-tanx \right) }{ \left( sinx-cosx \right) } \left| \overset { 0 }{ \underset { 1 }{ 0 } } \quad \quad \overset { 1 }{ \underset { tanx }{ 1 } } \quad \quad \overset { sinx+cosx }{ \underset { { \tan }^{ 2 }x }{ cosx+tanx } } \right| =0$$
Expanding along colomn I we get
$$\Rightarrow \left( cosx-tanx \right) \left( sinx-cosx \right) \left( cosx+tanx-sinx-cosx \right) =0$$
$$\Rightarrow \left( cosx-tanx \right) \left( sinx-cosx \right) \left( tanx-sinx \right) =0$$
$$\therefore$$ $$cosx-tanx=0$$ or $$sinx=cosx$$ or $$tanx=sinx$$
or $$cosx=tanx$$ ; $$sinx=cosx$$ ; $$tanx=sinx$$
for $$x\epsilon \left[ 0,2\pi \right] $$
$$csox=tanx$$
from graph we see there are $$2$$ solutions.
for $$x\epsilon \left[ 0,2\pi \right] $$
$$sinx=cosx$$
from graph we have two solutions
for $$x\epsilon \left[ 0,2\pi \right] $$
from graph we have solutions
So a total of $$2+2+3=7$$ solutions
i.e Answer : D.
Question 5
1 / -0

For what value of x, will the points (-1,x),(-3,2) and (-4,4) lie on a line?

A
-3

B
3

C
-2

D
2

Solution

$$A(-1, x), B(-3, 2), C(-4, 4)$$ lie on a line
$$\therefore \dfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|=0$$
$$\dfrac{1}{2}\left|-1(2-4)+-3(4-x)+-4(x-2)\right|=0$$
$$\dfrac{1}{2}\left|-1(-2)-3(4-x)-4(x-2)\right|=0$$
$$\dfrac{1}{2}\left|2-3\times 4+3x-4x+4\times 2\right|=0$$
$$\dfrac{1}{2}\left|2-12+3x-4x+8\right|=0$$
$$\dfrac{1}{2}\left|-2-x\right|=0$$
$$-2-x=0$$
$$x=-2$$.
Question 6
1 / -0

If $$\left|( {adj\,A}) \right| = 81,$$ for $$3 \times 3$$ matrix, then det $$A$$ is equal to

A
$$1$$

B
$$2$$

C
$$4$$

D
$$9$$

Solution

Given for a square matrix $$A$$ of order $$3$$.

$$|adjA|=|A|^{n-1}$$, where $$n$$ is the order of matrix.

We have, $$|adj(A)|=|A|^{(3-1)}$$

or, $$|A|^2=81$$

or, $$|A|=9$$. [ Taking the positive value]
Question 7
1 / -0

The point $$(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{1}, y_{2})$$ & $$(x_{2}, y_{1})$$ are always

A
Collinear

B
Concyclic

C
Vertices of a square

D
Vertices of a rhombus.

Solution

Let the coordinates be denoted as $$A(x_1,y_2)$$, $$B(x_2,y_2)$$, $$C(x_2,y_1)$$ and $$D(x_1,y_1)$$
Plot the given points on a graph as above,
It is not necessary that
$$|x_2-x_1|=|y_2-y_1|$$
With $$(x_2,y_1)$$ and $$(x_1,y_2)$$ as ends of diameter $$\angle ABC=90^{\circ}$$ and $$\angle ADC=90^{\circ}$$
$$\therefore ABCD$$ are concyclic.
So, $$\text{B}$$ is the correct option.
Question 8
1 / -0

$$|A|=6$$ and $$A$$ is $$3\times 3$$ matrix then $$det(2\ adj(2(A^{-1})^{T}))=$$

A
$$162$$

B
$$36$$

C
$$216$$

D
$$512$$
Question 9
1 / -0

Find the value of $$\begin{vmatrix} 5 & 3 \\ -7 & -4 \end{vmatrix}$$

A
-1

B
-41

C
41

D
1

Solution

$$\begin{vmatrix} 5 & 3 \\ -7 & -4 \end{vmatrix}$$ $$= 5 \times (-4) - 3 \times (-7) = -20 + 21 = 1$$
Hence, the correct answer is 1.
Question 10
1 / -0

If a determinant of order $$3\times 3$$ is formed by using the numbers $$1$$ or $$-1$$, then the minimum value of the determinant is?

A
$$-2$$

B
$$-4$$

C
$$0$$

D
$$-8$$

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Re-Attempt Weekly Quiz Competition

Determinants Test - 57

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Determinants Test - 57

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SHARING IS CARING

Question 1

$$\begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$$

$$\begin{bmatrix} 51 & 84 \\ 63 & 72 \end{bmatrix}$$

$$\begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$$

$$\begin{bmatrix} 72 & -84 \\ -63 & 51 \end{bmatrix}$$

Solution

Question 2

$$4$$

$$11$$

$$5$$

$$0$$

Solution

Question 3

( -4 , -5 )

( -4 , 3 )

( -4 , 5 )

( 4 , 5 )

Solution

Question 4

$$4$$

$$5$$

$$6$$

None of these

Solution

Question 5

-3

3

-2

2

Solution

Question 6

$$1$$

$$2$$

$$4$$

$$9$$

Solution

Question 7

Collinear

Concyclic

Vertices of a square

Vertices of a rhombus.

Solution

Question 8

$$162$$

$$36$$

$$216$$

$$512$$

Question 9

-1

-41

41

1

Solution

Question 10

$$-2$$

$$-4$$

$$0$$

$$-8$$

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