Determinants Test

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Determinants Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

Points $$(1, 5), (2, 3)$$ and $$(-2, -11) $$ are ____

A
Non-collinear

B
Collinear

C
Vertices of equilateral triangle

D
Vertices of right angle triangle

Solution

Let $$A (1, 5), B (2, -3)$$ and $$C (2, 3)$$ be the given points.
According to distance formula, we have
$$AB=\sqrt { { (2-1) }^{ 2 }+{ (3-5) }^{ 2 } } \\ =\sqrt { 1+4 } =\sqrt { 5 } \quad \\ $$
$$BC=\sqrt { { (-2-2) }^{ 2 }+{ (-11-3) }^{ 2 } } \\ =\sqrt { 16+196 } =\sqrt { 212 } =2\sqrt { 53 }$$
AC$$=\sqrt { { (1+2) }^{ 2 }+{ (5+11) }^{ 2 } } \\ =\sqrt { 9+256 } =\sqrt { 265 } \quad \\ $$
Clearly, the given points didn't satisfy any condition of collinear points, verties of equilateral and right angle triangle.
Therefore, the given points are non-collinear points.
Question 2
1 / -0

If $$a\neq p, b\neq q, c\neq r$$ and $$\begin{vmatrix}p & b & c\\ a & q & c\\ a & b & r\end{vmatrix}=0$$, then the value of $$\dfrac{p}{p-a}+\dfrac{q}{q-b}+\dfrac{r}{r-c}$$ is

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$-2$$

Solution

$$\begin{vmatrix}p & b &c \\ a & q &c \\ a & b &r \end{vmatrix}=0$$

$$R_1\rightarrow R_1-R_3$$

$$\begin{vmatrix} p-a& 0 &c-r \\ a & q &c \\ a & b &r \end{vmatrix}=0$$

$$R_2\rightarrow R_2-R_3$$

$$\begin{vmatrix} p-a& 0 & c-r\\ 0 & q-b &c-r \\ a & b & r\end{vmatrix}$$

$$p-a(r(q-b)-b(c-r)+(c-r)(-a(q-b))$$

$$R_3\rightarrow R_1+R_3$$

$$\begin{vmatrix} p-a& 0 &c-r \\ 0 & q-b & c-r\\ p & b &c \end{vmatrix}=0$$

$$\begin{vmatrix} 1& 0 & 1\\ 0 & 1 & 1\\ \dfrac {p}{p-a} & \dfrac {b}{q-b} & \dfrac {c}{c-r}\end{vmatrix}=0$$

$$\left (\dfrac {c}{c-r}-\dfrac {b}{q-b}\right)+\left (\dfrac {-p}{p-a}\right)=0$$

$$\dfrac {p}{p-a}=\dfrac {c}{c-r}-\dfrac {b}{q-b}$$ substituting gets value $$=-2$$
Question 3
1 / -0

Find the value of $$m$$ if the points $$(5, 1), (2, 3)$$ and $$(8, 2m )$$ are collinear.

A
$${-1}$$

B
$$\dfrac{1}{2}$$

C
$$-\dfrac{1}{2}$$

D
$${2}$$

Solution

If three points are collinear, then the area of the triangle formed by taking these points as vertices should zero.
Area $$=\dfrac{1}{2}$$[$${5(3-2m)+2(2m-1)+8(1-3)}]$$......(using determinant formula of area of triangle)
$$= \dfrac{1}{2}$$[$${15-10m +4m-2+(-16)}]$$
$$= \dfrac{1}{2}$$ {$$15-18-6m$$}
$$ = \dfrac{1}{2}$$$$(-3-6m)=0$$
$$\Rightarrow (3+6m)=0$$
$$m=-\dfrac{1}{2}$$.
Question 4
1 / -0

If the points $$(0, 0), (1, 2)$$ and $$(x, y)$$ are collinear, then

A
$$x = y$$

B
$$2x = y$$

C
$$x =2 y$$

D
$$2x = -y$$

Solution

$$ {\textbf{Step - 1: Find equation}} $$

$$ {\text{Let A, B, C are the three points}} $$

$$ {\text{Since A, B, C are linear they will lie on same line}}{\text{.}} $$

$$ {\text{Given points are A}}(0,0),B(1,2)\;{\text{and C}}(x, y) $$

$$ {\text{Therefore,}} $$

$$ {\text{area }}\Delta {\text{ABC = 0}} $$

$$ \Rightarrow \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0 $$

$$ {\text{Here, }}{x_1} = 0,{x_2} = 1\;{\text{and }}{x_3} = x $$

$$ {\text{and }}{y_1} = 0,\;{y_2} = 2\;{\text{and }}{y_3} = y $$

$$ {\textbf{Step - 2: Find the relation}} $$

$$ \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0 $$

$$   \Rightarrow \dfrac{1}{2}\left[ {0\left( {2 - y} \right) + 1\left( {y - 0} \right) + x\left( {0 - 2} \right)} \right] = 0 $$

$$   \Rightarrow y - 2x = 0 $$

$$   \Rightarrow {\text{2x = y}} $$

$$ {\textbf{Hence, the correct answer is option B}}{\text{.}} $$
Question 5
1 / -0

$$\left| {\begin{array}{*{20}{c}}
{{{\sin }^2}x}&{{{\cos }^2}x}&1 \\
{{{\cos }^2}x}&{{{\sin }^2}x}&1 \\
{ - 10}&{12}&2
\end{array}} \right| = $$

A
$$0$$

B
$$12 cos^2x -10 sin^2x$$

C
$$12 sin^2x -10 cos^2x - 2$$

D
$$10 sin 2x$$

Solution

$$\begin{vmatrix}sin^2x & cos^2x & 1\\ cos^2 x & sin^2 x & 1\\ -10 & 12 &2 \end{vmatrix}$$
$$C_1\rightarrow C_1+C_2$$
$$=\begin{vmatrix} 1& cos^2x & 1\\ 1 & sin^2x & 1\\ 2 & 12 & 2\end{vmatrix}=0 \ (\because C_1=C_3)$$
Question 6
1 / -0

$$A=\left[\begin{matrix}1&0&0\\2&1&0\\3&2&1\end{matrix}\right], U_1, U_2$$ and $$U_3$$ are columns matrices satisfying $$AU_1=\left[\begin{matrix}1\\0\\0\end{matrix}\right], AU_2 = \left[\begin{matrix}2\\3\\0\end{matrix}\right], AU_3 = \left[\begin{matrix}2\\3\\1\end{matrix}\right]$$ and $$U$$ is $$3\times3$$ matrix whose columns are $$U_1, U_2, U_3$$ then answer the following question
The value of $$|U|$$ is

A
$$3$$

B
$$-3$$

C
$$\dfrac32$$

D
$$2$$

Solution

Given,
$$AU_1=\left[\begin{matrix}1\\0\\0\end{matrix}\right]$$ $$\Rightarrow U_{1}=A^{-1}\left[\begin{matrix}1\\0\\0\end{matrix}\right]$$ ...(1)

$$AU_2 = \left[\begin{matrix}2\\3\\0\end{matrix}\right]$$ $$\Rightarrow U_{2}=A^{-1} \left[\begin{matrix}2\\3\\0\end{matrix}\right]$$ ...(2)

$$AU_3 = \left[\begin{matrix}2\\3\\1\end{matrix}\right]$$ $$\Rightarrow U_{3}=A^{-1} \left[\begin{matrix}2\\3\\1\end{matrix}\right]$$ ...(3)

Also, given
$$A=\left[\begin{matrix}1&0&0\\2&1&0\\3&2&1\end{matrix}\right]$$
$$|A|=1$$
Now,
$$adj\ A=C^{T}=\begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}^{T}$$

$$\Rightarrow adj A=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$$

$$\Rightarrow A^{-1}= \dfrac{adj\ A}{|A|} = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$$

Put the value of $$A^{-1}$$ in (1), (2) and (3)
$$ U_{1}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\left[\begin{matrix}1\\0\\0\end{matrix}\right]$$ $$\Rightarrow U_{1}=\begin{bmatrix}1\\-2\\1\end{bmatrix}$$

$$ U_{2}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix} \left[\begin{matrix}2\\3\\0\end{matrix}\right]$$ $$\Rightarrow U_{2}=\begin{bmatrix}2\\-1\\-4\end{bmatrix}$$

$$ U_{3}=A^{-1} \left[\begin{matrix}2\\3\\1\end{matrix}\right]$$ $$\Rightarrow U_{3}=\begin{bmatrix}2\\-1\\-3\end{bmatrix}$$

$$\therefore U=\begin{bmatrix}1&2&2\\-2&-1&-1\\1&-4&-3\end{bmatrix}$$
$$\Rightarrow |U|= -1-14+18=3$$

Hence, option A.
Question 7
1 / -0

If $$(3, 2)$$, $$\left (x, \dfrac {22}{5}\right), (8, 8)$$ lie on a line, then $$x$$ is equal to

A
$$-5$$

B
$$2$$

C
$$4$$

D
$$5$$

Solution

The area of the triangle made by the points $$\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 3,2 \right) , \left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( x,\dfrac { 22 }{ 5 } \right) , \left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 8,8 \right)$$ will be $$ =0$$, if they lie on the same straight line.
Therefore, the area of the $$\Delta =\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] $$
In this case, the area of the triangle
$$ = \dfrac { 1 }{ 2 } \left[ 3\left( \dfrac { 22 }{ 5 } -8 \right) +x\left( 8-2 \right) +8\left( 2-\dfrac { 22 }{ 5 } \right) \right] =0$$
$$\Rightarrow -\dfrac { 54 }{ 5 } +6x-\dfrac { 96 }{ 5 } =0$$
$$ \Rightarrow 30x=150$$
$$ \Rightarrow x=5$$
Question 8
1 / -0

If the points $$(0, 4), (4, 0)$$ and $$(5, p)$$ are collinear, then value of $$p$$ is

A
$$- 1$$

B
$$7$$

C
$$6$$

D
$$4$$

Solution

The given points are $$ A=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 0,4 \right) , B=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 4,0 \right) , C=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 5,p \right) $$.
They are collinear.
$$\therefore ar\Delta ABC=0$$
$$ \Rightarrow \dfrac { 1 }{ 2 } \left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { 1 } \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix} \right| =0 $$
$$\Rightarrow \left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { 1 } \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix} \right| =0$$
$$ \Rightarrow \left| \begin{matrix} 0 & 4 & 1 \\ 4 & 0 & 1 \\ 5 & p & 1 \end{matrix} \right| =0$$
$$\Rightarrow 0\left( 0-p \right) +4\left( p-4 \right) +5\left( 4-0 \right) =0$$
$$ \Rightarrow p=-1$$
Question 9
1 / -0

P, Q, R are three collinear points. The coordinates of P and R are (3, 4) and (11, 10) respectively and PQ is equal to 2.5 units. Coordinates of Q are-

A
(5, 11/2)

B
(11, 5/2)

C
(5, -11/2)

D
(-5, 11/2)

Solution

Given that PQR are three collinear points and

PQ=2.5. Let us now find the distance of PR
Recall Distance Formula is =$$ \sqrt { { \left( { x}_{ 2 }{ { x }_{ 1 } } \right) }^{ 2}+{ \left( { y }_{ 2 }{ { y }_{ 1 } } \right) }^{ 2 } } \\$$
Distance PR=$$\sqrt { (113)^{ 2 })+( 104) ^{ 2 } }\\=\sqrt{100}=10$$
Let us now consider S is the midpoint of PR,
then we have the co-ordinates of S as $$\dfrac{11+3}{2},\dfrac{10+4}{2}=(7, 7)$$.
$$ \therefore$$ it is clear that Q is the
midpoint of PS as we have know PQ=2.5 units.
Thus, to get the co-ordinates of Q take
midpoint of PS which is $$\dfrac{3+7}{2},\dfrac{4+7}{2}=(5,11/2)$$.
Question 10
1 / -0

If the points $$(a, 0), (0, b)$$ and $$(1, 1)$$ are collinear, then $$\displaystyle \frac{1}{a} + \frac{1}{b}$$ equal to -

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Given points are $$(a,0),(0,b)$$ and $$(1,1)$$
$$x_1 = a, y_1 = 0, x_2 =0, y_2 = b$$ and $$x-3=1, y_3=1$$
Condition for collinearity
$$x_1y_2+x_2y_3+x_3y_1=x_2y_1+x_3y_2+x_1y_3$$ gives
$$ab+0+0=0 +1.b+a.1 \Rightarrow ab = a+b$$
$$\Rightarrow \displaystyle 1 = \frac{1}{b}+ \frac{1}{a}$$
$$ \Rightarrow \dfrac{1}{a} + \dfrac{1}{b}=1$$

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Determinants Test - 31

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Determinants Test - 31

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SHARING IS CARING

Question 1

Non-collinear

Collinear

Vertices of equilateral triangle

Vertices of right angle triangle

Solution

Question 2

$$0$$

$$1$$

$$-1$$

$$-2$$

Solution

Question 3

$${-1}$$

$$\dfrac{1}{2}$$

$$-\dfrac{1}{2}$$

$${2}$$

Solution

Question 4

$$x = y$$

$$2x = y$$

$$x =2 y$$

$$2x = -y$$

Solution

Question 5

$$0$$

$$12 cos^2x -10 sin^2x$$

$$12 sin^2x -10 cos^2x - 2$$

$$10 sin 2x$$

Solution

Question 6

$$3$$

$$-3$$

$$\dfrac32$$

$$2$$

Solution

Question 7

$$-5$$

$$2$$

$$4$$

$$5$$

Solution

Question 8

$$- 1$$

$$7$$

$$6$$

$$4$$

Solution

Question 9

(5, 11/2)

(11, 5/2)

(5, -11/2)

(-5, 11/2)

Solution

Question 10

$$1$$

$$2$$

$$3$$

$$4$$

Solution

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