Determinants Test

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Determinants Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle \begin{vmatrix} 2 & -4 \\ 9 & d-3 \end{vmatrix}=4$$ then $$d=$$

A
$$10$$

B
$$-11$$

C
$$12$$

D
$$-13$$

Solution

Given, determinant of the matrix $$ \begin{vmatrix} 2 & -4 \\ 9 & d-3 \end{vmatrix} = 4 $$
$$ => 2(d-3) - (9)(-4) = 4 $$
$$ => 2d-6+36 = 4 $$
$$ => 2d = -26 $$
$$ => d = -13 $$
Question 2
1 / -0

The value of $$\displaystyle \begin{vmatrix}a - b - c & 2a & 2a\\ 2b & b - c- a & 2b\\ 2c & 2c & c- a -b\end{vmatrix}$$ will be

A
$$(a+b+c)^2$$

B
$$(a+b+c)^3$$

C
$$(a-b-c)^2$$

D
$$(a + b -c)^2$$

Solution

$$\displaystyle \begin{vmatrix}a - b - c & 2a & 2a\\ 2b & b - c- a & 2b\\ 2c & 2c & c- a -b\end{vmatrix}$$

$$R_1 \rightarrow R_1 + (R_2 + R_3)$$

$$\begin{vmatrix}a+b+c & a+b+c & a+b+c\\2b &b - c-a & 2b\\2c & 2c & c-a-b\end{vmatrix}$$

Taking common term (a + b + c) from the first row

$$= (a + b + c) \begin{vmatrix} 1& 1 & 1\\2b & b-c-a & 2b\\ 2c & 2c & c-a-b\end{vmatrix}$$

Subtracting first column from second and third column

$$= (a + b + c) \begin{vmatrix}1 & 0 & 0\\ 2b & -b-c-a & 0\\ 2c & 0 & -c-a-b\end{vmatrix}$$

On expansion in terms of

$$R_1 = (a + b + c) \begin{vmatrix}-b-c-a & 0 \\ 0 & -c-a-b\end{vmatrix}$$
.
$$= (a + b+c)\{(-b-c-a)(-c-a -b)-0\}$$
$$= (a + b+c)^3$$
Question 3
1 / -0

If the coordinates of the vertices of a triangle are (0,0) , (0,2) and(3,1) , then area of the triangle is

A
3 sq.units

B
-3 sq. units

C
2 sq. units

D
1 sq.units

Solution

Area of triangle $$=\frac{1}{2} \begin{vmatrix} 0 &0 &1 \\ 0&2 &1 \\3 &1 &1 \end{vmatrix}= \frac{1}{2}\times|-6|=3$$
Question 4
1 / -0

Find the value of K if A(8, 1), B(K, -4), C(2, -5) are collinear :

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

Since the given points are collinear, they do not form a

triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \frac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (8,1) $$ ; $$({ x

}_{ 2 },{ y }_{ 2 }) = (k,-4) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (2,-5)$$
the area formula, we get

$$ \left| \frac { 8(-4+5) + k(-5-1) + 2(1+4) }{ 2 } \right| =

0 $$
$$ => 8 - 6k + 10 = 0 $$
$$ => 6k = 18 $$
$$ => k = 3 $$
Question 5
1 / -0

What is the value of y if $$(y, 3), (-5, 6)$$ and $$(-8, 8)$$ are collinear?

A
$$-1$$

B
$$2$$

C
$$\displaystyle \frac{1}{2}$$

D
$$\displaystyle -\frac{1}{2}$$

Solution

Let A(y,3),B(-5,6) and C(-8,8) be the given points
Condition for collinear equation-
$$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$$
where $$x_1=y,x_2=3$$
$$x_2=-5,y_2=6$$
$$x_3=-8,y_3=8$$
$$\Rightarrow y(6-8)+-5(8-3)+-8(3-6)=0$$
$$\Rightarrow 6y-8y-40+15-24+48=0$$
$$\Rightarrow -2y-1=0$$
$$\Rightarrow -2y=1$$
$$\Rightarrow y=-\frac{1}{2}$$
Question 6
1 / -0

Which of the following points are collinear?

A
(2a,0), (3a,0), (a,2a)

B
(3a,0), (0,3b), (a,2b)

C
(3a,b), (a,2b), (-a,b)

D
(a,-6), (-a,3b), (-2a,-2b)

Solution

Ans option B
Let A(3a,0),B(0,3b) and C(a,2b) be the given points
Condition for collinear equation-
$$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$$
where $$x_1=3a,y_1=0$$
$$x_2=0,y_2=3b$$
$$x_3=a,y_3=2b$$
$$\Rightarrow 3a(3b-2b)+0(2b-0)+a(0-3b)=0$$
$$\Rightarrow 3ab+0-3ab=0$$
Question 7
1 / -0

Find the value of K if (2, 3), (4, K) and (6, -3) are collinear :

A
0

B
1

C
2

D
3

Solution

Since the given points are collinear, they do not form a

triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \dfrac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$
Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (2,3) $$ ; $$({ x

}_{ 2 },{ y }_{ 2 }) = (4.k) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (6,-3)$$

in the area formula, we get

$$ \left| \dfrac { 2 (k+3) + 4(-3-3) + 6(3-k) }{ 2 } \right| =

0 $$
$$ => 2k + 6 -24 + 18 - 6k = 0 $$
$$ => 4k = 0 $$
$$ => k = 0 $$
Question 8
1 / -0

If $$\displaystyle \left[ \begin{matrix} \cos { \theta } \\ -\sin { \theta } \\ 0 \end{matrix}\begin{matrix} \sin { \theta } \\ \cos { \theta } \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] $$ then the value of $$\displaystyle { a }_{ 11 }{ A }_{ 11 }+{ a }_{ 12 }{ A }_{ 12 }+{ a }_{ 13 }{ A }_{ 13 }=$$ where $$\displaystyle { A }_{ 11 },{ A }_{ 12 }{ A }_{ 13 }$$ are cofactors of $$\displaystyle { a }_{ 11 },{ a }_{ 12 },{ a }_{ 13 }$$ respectively

A
$$-1$$

B
$$1$$

C
$$0$$

D
$$\displaystyle \frac { 1 }{ 2 } $$

Solution

$$A_{11} = \begin{vmatrix} \cos{\theta} & 0 \\ 0 & 1 \end{vmatrix} = \cos{\theta}$$

$$A_{12} = - \begin{vmatrix} -\sin{\theta} & 0 \\ 0 & 1 \end{vmatrix} = \sin{\theta}$$

$$A_{13} = \begin{vmatrix} -\sin{\theta} & \cos{\theta} \\ 0 & 0 \end{vmatrix} = 0$$

Now,
$$a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = \cos^2{\theta} + \sin^2{\theta} = 1$$
Question 9
1 / -0

If $$z=\begin{vmatrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{vmatrix} $$, then $$i=\left( \sqrt { -1 } \right) $$

A
$$z$$ is purely real

B
$$z$$ is purely imaginary

C
$$z+\overline { z } =0$$

D
$$\left( z-\overline { z } \right) i$$ is purely imaginery

Solution

Given $$z=\begin{vmatrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{vmatrix}$$
$$=1(-21-(25+9))-(1+2i)(7-14i-(25i-15))-5i[(-1-13i)+15i]$$
$$-55-(1+2i)(22-39i)-5i(-1+2i)$$
$$=-55-100-5i+5i+10$$
$$=-145$$
Hence, $$z$$ is purely real.
Question 10
1 / -0

Find the correct option regarding given points $$(1, 2), (2, 4)$$ and $$(3, 6)$$

A
Non-colllinear

B
Exists on parallel lines

C
Collinear

D
Exists on perpendicular lines

Solution

Area of the triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is:
$$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 2 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$$
Here the points are $$(1,2)$$, $$(2,4)$$ and $$(3,6)$$, therefore, the area of the triangle is:
$$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right| \\ A=\dfrac { 1 }{ 2 } \left| 1(4-6)+2(6-2)+3(2-4) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| 1(-2)+2(4)+3(-2) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| -2+8-6 \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| -8+8 \right| \\ \Rightarrow A=0$$
Since the area of the triangle is zero,
Hence, the points are collinear.

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Determinants Test - 42

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Determinants Test - 42

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Question 1

$$10$$

$$-11$$

$$12$$

$$-13$$

Solution

Question 2

$$(a+b+c)^2$$

$$(a+b+c)^3$$

$$(a-b-c)^2$$

$$(a + b -c)^2$$

Solution

Question 3

3 sq.units

-3 sq. units

2 sq. units

1 sq.units

Solution

Question 4

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 5

$$-1$$

$$2$$

$$\displaystyle \frac{1}{2}$$

$$\displaystyle -\frac{1}{2}$$

Solution

Question 6

(2a,0), (3a,0), (a,2a)

(3a,0), (0,3b), (a,2b)

(3a,b), (a,2b), (-a,b)

(a,-6), (-a,3b), (-2a,-2b)

Solution

Question 7

0

1

2

3

Solution

Question 8

$$-1$$

$$1$$

$$0$$

$$\displaystyle \frac { 1 }{ 2 } $$

Solution

Question 9

$$z$$ is purely real

$$z$$ is purely imaginary

$$z+\overline { z } =0$$

$$\left( z-\overline { z } \right) i$$ is purely imaginery

Solution

Question 10

Non-colllinear

Exists on parallel lines

Collinear

Exists on perpendicular lines

Solution

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