Determinants Test

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Determinants Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of $$k$$ for which the points $$ (2, 3), (3, k)$$ and $$(3, 7)$$ are collinear.

A
5

B
6

C
8

D
7

Solution

If three points are collinear, then the area of the triangle is zero.
Area of the triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is:
$$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$$
Here the points are $$(2,3)$$, $$(3,k)$$ and $$(3,7)$$ and it is given that these points are collinear therefore the area is zero that is:
$$0=\dfrac { 1 }{ 2 } \left| 2(k-7)+3(7-3)+3(3-k) \right| \\ \Rightarrow 0=\left| 2k-14+3(4)+9-3k \right| \\ \Rightarrow 0=\left| -k-5+12 \right| \\ \Rightarrow 0=\left| -k+7 \right| \\ \Rightarrow k-7=0\\ \Rightarrow k=7$$
Hence, $$k=7$$
Question 2
1 / -0

If the points $$(a,\,b),\,(3,\,-5)$$ and $$(-5,\,-2)$$ are collinear. Then find the value of $$3a+8b$$

A
$$20$$

B
$$-31$$

C
$$10$$

D
None

Solution

If three points are collinear then the area of the triangle is zero.
Area of the triangle passing through the points $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is given by:
$$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$$
Here the points are $$(a,b)$$, $$(3,-5)$$ and $$(-5,-2)$$ and it is given that these points are collinear therefore the area is zero that is:
$$A=\dfrac { 1 }{ 2 } \left| a(-5-(-2))+3(-2-b)+(-5)(b-(-5)) \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| a(-5+2)+3(-2-b)-5(b+5) \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| -3a-6-3b-5b-25 \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| -3a-8b-31 \right| \\ \Rightarrow 0=-3a-8b-31\\ \Rightarrow 3a+8b=-31$$
Hence, $$3a+8b=-31$$.
Question 3
1 / -0

Find the correct option regarding the given points $$(2, -1), (0, 2)$$ and $$(3, 2)$$.

A
Collinear

B
Non-collinear

C
Exists on parallel lines

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:
$$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$$
Here the points are $$(2,-1)$$, $$(0,2)$$ and $$(3,2)$$, therefore, the area of the triangle is:
$$A=\dfrac { 1 }{ 2 } \left| 2(2-2)+0(2+1)+3(-1-2) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| 2(0)+0(3)+3(-3) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| -9 \right| \\ \Rightarrow A=\dfrac { 9 }{ 2 } \neq 0$$
Since the area of the triangle is not zero,
Hence, the points are non-collinear.
Question 4
1 / -0

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

A
Collinear

B
Non-collinear

C
Exists on parallel lines

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:
$$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$$
Here the points are $$(2,2)$$, $$(1,2)$$ and $$(3,1)$$, therefore the area of the triangle is:
$$A=\dfrac { 1 }{ 2 } \left| 2(2-1)+1(1-2)+3(2-2) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| 2(1)+1(-1)+3(0) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| 2-1 \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \neq 0$$
Since the area of the triangle is not zero,
Hence, the points are non-collinear.
Question 5
1 / -0

Consider the three collinear points $$(3, p), (4, 4)$$ and $$(5, 6)$$. Find the value of $$p$$.

A
1

B
2

C
3

D
4

Solution

If three points are collinear then the area of the triangle is zero.
Area of the triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is:
$$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$$
Here the points are $$(3,p)$$, $$(4,4)$$ and $$(5,6)$$ and it is given that these points are collinear, therefore the area is zero that is:
$$0=\dfrac { 1 }{ 2 } \left| 3(4-6)+4(6-p)+5(p-4) \right| \\ \Rightarrow 0=\left| 3(-2)+24-4p+5p-20 \right| \\ \Rightarrow 0=\left| -6+p+4 \right| \\ \Rightarrow 0=\left| p-2 \right| \\ \Rightarrow p-2=0\\ \Rightarrow p=2$$
Hence, $$p=2$$
Question 6
1 / -0

The graph of $$f(x)$$ is shown above in the $$xy$$-plane. The points $$(0,3), (5b, b)$$ and $$(10b, -b)$$ are on the line described by $$f(x)$$. If $$b$$ is a positive constant, find the coordinates of point $$C$$.

A
$$(5, 1)$$

B
$$(10, -1)$$

C
$$(15, -0.5)$$

D
$$(20, -2)$$

Solution

In given figure the line goes to point $$A(0,3) ,B(5b,b)$$ and $$C(10b,-b)$$
Then slope of line goes at point $$A(0,3)$$ and $$B(5b,b)=$$ $$\dfrac{b-3}{5b-0}=\dfrac{b-3}{5b}$$
And slope of line goes at point $$B(5b,b)$$ and $$C(10b,-b)=$$ $$\dfrac{-b-b}{10b-5b}=\dfrac{-2b}{5b}$$
Then given line is a st reline then both slopes are equal
$$\therefore \dfrac{b-3}{5b}=\frac{-2b}{5b }$$
$$\Rightarrow -2b=b-3$$
$$\Rightarrow -3b=-3$$
$$\Rightarrow b=1$$
Put the value of $$b=1$$ in point C
We get $$C=(10,-1)$$
Question 7
1 / -0

If $$A = \begin{bmatrix}3 &5 \\ 2 & 0\end{bmatrix}$$ and $$B = \begin{bmatrix}1 &17 \\ 0 & -10\end{bmatrix}$$, then $$|AB|$$ is equal to

A
$$80$$

B
$$100$$

C
$$-110$$

D
$$92$$

Solution

Given, $$A = \begin{bmatrix}3 &5 \\ 2 & 0\end{bmatrix}$$ and $$B = \begin{bmatrix}1 &17 \\ 0 & -10\end{bmatrix}$$
Therefore, $$ AB = \begin{bmatrix}3 &5 \\ 2 & 0\end{bmatrix}\begin{bmatrix}1 &17 \\ 0 & -10\end{bmatrix}$$
$$= \begin{bmatrix}3 + 0 &51 - 50 \\ 2 + 0 & 34 - 0\end{bmatrix} = \begin{bmatrix}3 &1 \\ 2 & 34\end{bmatrix}$$
$$\Rightarrow |AB| = \begin{vmatrix}1 & 1\\ 2 & 34\end{vmatrix}$$
$$= 102 - 2$$
$$= 100$$
Question 8
1 / -0

If $$\Delta = \begin{vmatrix} -a & 2b & 0 \\ 0 & -a & 2 b \\ 2b & 0 &-a \end{vmatrix}=0$$, then

A
$$\dfrac{1}{b}$$ is a cube root of unity

B
$$a$$ is one of the cube roots of unity

C
$$b$$ is one of the cube roots of $$8$$

D
$$\dfrac{a}{b}$$ is a cube root of $$8$$

Solution

$$\begin{vmatrix} -a & 2b & 0 \\ 0 & -a & 2 b \\ 2b & 0 &-a \end{vmatrix}=0$$

$$\Rightarrow -a(a^2-0)-2b(0-4b^2)=0$$, Expand along 1st row

$$\Rightarrow -a^2+8b^3=0\Rightarrow \left(\dfrac{a}{b}\right)^3=8$$

$$\Rightarrow \dfrac{a}{b}$$ is a cube root of $$8$$
Question 9
1 / -0

If C = $$2 \cos \theta$$, then the value of the determinant to $$\Delta = \left |\begin{matrix} c & 1 & 0 \\ 1 & c & 1 \\ 6 & 1 & c \end{matrix}\right |$$ is

A
$$\dfrac{2\,\sin^22\theta}{\sin\,\theta}$$

B
$$8\, \cos^3\theta-4\,cos\theta+6$$

C
$$\dfrac{2\,\sin\,2\theta}{\sin\, \theta}$$

D
$$8\,\cos^3\theta+4\,\cos\theta+6$$

Solution

$$C=2\cos { \theta } $$
$$\triangle =\begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix}$$
$$\triangle =C\left( { C }^{ 2 }-1 \right) -1\left( C-6 \right) $$
$$\triangle ={ C }^{ 3 }-2C+6$$
$$C=2\cos { \theta } $$
$$\Rightarrow \triangle ={ \left( 2\cos { \theta } \right) }^{ 3 }-\left( 2\cos { \theta } \right) 2+6$$
$$\Rightarrow \triangle =8\cos ^{ 2 }{ \theta } -4\cos { \theta } +6$$
Question 10
1 / -0

If A = $$\begin{bmatrix}-8 & 5\\ 2 & 4\end{bmatrix}$$ satisfies the equation $$x^2\, +\, 4x\, -\, p\, =\, 0$$, then p =

A
64

B
42

C
36

D
24

Solution

$$A= \begin{bmatrix}-8&5\\2&4\end{bmatrix}$$

$$A^2=\begin{bmatrix}-8&5\\2&4\end{bmatrix}\begin{bmatrix}-8&5\\2&4\end{bmatrix}=\begin{bmatrix}64+10&-40+20\\-16+8&10+16\end{bmatrix}=\begin{bmatrix}74&-20\\-8&26\end{bmatrix}$$

$$4A=\begin{bmatrix}-32&20\\8&16\end{bmatrix}$$

$$P=x^2+4x$$ and $$A$$ satisfies this equation

$$So, P=\begin{bmatrix}74&-20\\-8&26\end{bmatrix}+\begin{bmatrix}-32&20\\8&16\end{bmatrix}$$

$$=\begin{bmatrix}42&0\\0&42\end{bmatrix}$$

$$=42\begin{bmatrix}1&0\\0&1\end{bmatrix}$$

$$=42I$$

$$=42$$

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Re-Attempt Weekly Quiz Competition

Determinants Test - 43

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Determinants Test - 43

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SHARING IS CARING

Question 1

5

6

8

7

Solution

Question 2

$$20$$

$$-31$$

$$10$$

None

Solution

Question 3

Collinear

Non-collinear

Exists on parallel lines

Exists on perpendicular lines

Solution

Question 4

Collinear

Non-collinear

Exists on parallel lines

Exists on perpendicular lines

Solution

Question 5

1

2

3

4

Solution

Question 6

$$(5, 1)$$

$$(10, -1)$$

$$(15, -0.5)$$

$$(20, -2)$$

Solution

Question 7

$$80$$

$$100$$

$$-110$$

$$92$$

Solution

Question 8

$$\dfrac{1}{b}$$ is a cube root of unity

$$a$$ is one of the cube roots of unity

$$b$$ is one of the cube roots of $$8$$

$$\dfrac{a}{b}$$ is a cube root of $$8$$

Solution

Question 9

$$\dfrac{2\,\sin^22\theta}{\sin\,\theta}$$

$$8\, \cos^3\theta-4\,cos\theta+6$$

$$\dfrac{2\,\sin\,2\theta}{\sin\, \theta}$$

$$8\,\cos^3\theta+4\,\cos\theta+6$$

Solution

Question 10

64

42

36

24

Solution

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