Determinants Test

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Determinants Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

Let $$A$$ be a square matrix of order $$3\times 3$$ then $$\left| KA \right| $$ is equal to

A
$$K\left| A \right| $$

B
$${K}^{2}\left| A \right| $$

C
$${K}^{3}\left| A \right| $$

D
$$3\left| KA \right| $$

Solution

$$Using \space properties \space of \space determinant$$
$$|kA|=k^{n}|A|$$
$$where \space n \space is \space order \space determinant$$
$$Since \space given \space matrix \space A \space has \space order \space 3$$
$$hence\space\left| kA \right| ={k}^{3}\left| A \right| $$
Question 2
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Find the equation of line joining $$(1,2)$$ and $$(3,6)$$ using determinants. Let $$p(x,y)$$ be any point on the line joining $$(1,2)(3,6)$$

A
$$y=x$$

B
$$2y=x$$

C
$$y=2x$$

D
$$y=-2x$$

Solution

Equation of line joining $$(1,2)$$ and $$(3,6)$$ can be obtained by
$$\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix}=0$$
$${R}_{2}\Rightarrow {R}_{2}-{R}_{1}$$; $${R}_{3}\Rightarrow {R}_{3}-{R}_{1}$$
$$\begin{vmatrix} x & y & 1 \\ 1-x & 2-y & 0 \\ 3-x & 6-y & 0 \end{vmatrix}$$
expanding along
$${C}_{3}=1[(1-x)(6-y)-(3-x)(2-y)]=0$$
$$=(6-y-6x+xy-6+3y+2x-xy)$$
$$2y-4x=0$$ or $$y=2x$$
Question 3
1 / -0

Value of determinant $$\begin{vmatrix} \cos 80^\circ & -\cos 10^\circ\\ \sin 80^\circ & \sin 10^\circ \end{vmatrix}$$ is:

A
$$0$$

B
$$1$$

C
$$-1$$

D
None of these

Solution

$$\begin{vmatrix} \cos 80^\circ & -\cos 10^\circ\\ \sin 80^\circ & \sin 10^\circ \end{vmatrix}$$

$$=\cos 80^\circ \sin 10^\circ - (- \cos 10^\circ) sin 80^\circ$$
$$=\cos 80^\circ \sin 10^\circ + \cos 10^\circ sin 80^\circ$$
$$= \sin (10^\circ + 80^\circ)$$
$$= \sin 90^\circ = 1$$
Question 4
1 / -0

Find the equation of the line joining $$(3,1)$$ and $$(9,3)$$ using determinants.

A
$$x=-3y$$

B
$$y=3x$$

C
$$y=-3x$$

D
$$3y=x$$

Solution

Let $$p(x,y)$$ be any point on the line joining $$(3,1)$$ and $$(9,3)$$
$$\begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix}=0$$
$${R}_{2}\Rightarrow {R}_{2}-{R}_{1}$$; $${R}_{3}\Rightarrow {R}_{3}-{R}_{1}$$
$$\begin{vmatrix} x & y & 1 \\ 3-x & 1-y & 0 \\ 9-x & 3-y & 0 \end{vmatrix}=0$$
$$\Rightarrow$$ $$1[(3-x)(3-y)-(1-y)(9-x)]=0$$
$$\Rightarrow$$ $$9-3y-3x+xy+x-9+9y-xy=0$$
$$\Rightarrow$$ $$6y-2x=0$$
$$\Rightarrow$$ $$x=3y$$
Question 5
1 / -0

Value of determinant $$\begin{vmatrix} \cos 50^\circ & \sin 10^\circ\\ \sin 50^\circ & \cos 10^\circ \end{vmatrix}$$ is:

A
$$0$$

B
$$1$$

C
$$1/2$$

D
$$-1/2$$

Solution

$$\begin{vmatrix} \cos 50^\circ & \sin 10^\circ\\ \sin 50^\circ & \cos 10^\circ \end{vmatrix}$$
$$=\cos 50^\circ \cos 10^\circ - \sin 50^\circ \sin 10^\circ$$
$$= \cos (50^\circ + 10^\circ)$$
$$= \cos 60^\circ =\dfrac{1}{2}$$
So, option (c) is correct.
Question 6
1 / -0

Co-factors of the first column of determinant
$$\begin{vmatrix} 5 & 20 \\ 3 & -1\end{vmatrix}$$

A
$$-1, 3$$

B
$$-1,- 3$$

C
$$- 1, 20$$

D
$$- 1,- 20$$

Solution

$$\begin{vmatrix} 5 & 20 \\ 3 & -1\end{vmatrix}$$
Co-factor of $$a_{11}$$
$$F_{11} = (- 1)^2 M_{11}$$
$$= 1 \times (- 1) = - 1$$
Co-factor of $$a_{12}$$
$$F_{21} = (-1)^{11}\,M_{21}$$
$$= (- 1) × 20 = - 20$$
Question 7
1 / -0

If $$a, b, c$$ are the $$p^{th}, q^{th}$$ and $$r^{th}$$ terms of an $${H}.{P}$$, then the lines $$bcx+py+1=0,\ cax+ qy+1=0$$ and $$abx+ry+1=0$$,

A
are concurrent

B
form a triangle

C
are parallel

D
mutually perpendicular lines

Solution

Given $$a,b,c$$ are $$p^{th} , q^{th} , r^{th}$$ terms of H.P. respectively.
$$\Rightarrow \displaystyle \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$$ are $$p^{th} , q^{th} , r^{th}$$ terms of A.P. respectively.

Let $$A$$ be the first term and d be the common difference
$$T_p=A+(p-1)d$$
$$\displaystyle \frac{1}{a}=A+(p-1)d$$ .....(1)

$$T_q=A+(q-1)d$$
$$\displaystyle \frac{1}{b}=A+(q-1)d$$ .....(2)

$$T_r=A+(r-1)d$$
$$\displaystyle \frac{1}{c}=A+(r-1)d$$ .....(3)

Subtracting (2) from (1), we get
$$\displaystyle \frac{1}{a}-\frac{1}{b}=(p-q)d$$
$$\Rightarrow \displaystyle ab(p-q)=\frac{b-a}{d}$$ .....(4)

Subtracting (3) from (2), we get
$$\displaystyle \frac{1}{b}-\frac{1}{c}=(q-r)d$$
$$\Rightarrow \displaystyle bc(q-r)=\frac{c-b}{d}$$ .....(5)

Subtracting (3) from (1), we get
$$\displaystyle \frac{1}{a}-\frac{1}{c}=(p-r)d$$
$$\Rightarrow \displaystyle ac(p-r)=\frac{c-a}{d}$$ .....(6)

Now, consider $$\begin{vmatrix} bc & p & 1 \\ ca & q & 1 \\ ab & r & 1 \end{vmatrix}$$
(The above matrix is the matrix representation of the given equations)

Expanding along first column, we get
$$=bc(q-r)-ca(p-r)+ab(p-q)$$
$$=\displaystyle \frac{c-b}{d}-\frac{c-a}{d}+\frac{b-a}{d}$$
$$=0$$

Hence, the given lines are concurrent.

Hence, option A.
Question 8
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If $$x_{1},y_{1}$$ are the roots of $$x^{2}+8x-20=0$$ and $$x_{2},y_{2}$$ are the roots of $$4x^{2}+32x-57=0$$ and $$x_{3},y_{3}$$ are the roots of $$9x^{2}+72x-112=0$$ such that $$y_{i}<0,$$ then the points $$(x_{1},y_{1}),(x_{2},y_{2})$$ and $$(x_{3},y_{3})$$

A
are collinear

B
form an equilateral triangle

C
form a right angled isosceles triangle

D
are concyclic

Solution

$$x^{2}+8x-20=0$$
$$\rightarrow (x+10)(x-2)=0$$
$$x=-10,x=2$$
Therefore
$$(x_{1},y_{1})=(2,-10)$$...(i)

Similarly
$$4x^{2}+32x-57=0$$
$$(x-\dfrac{3}{2})(x+\dfrac{19}{2})=0$$

$$x=\dfrac{3}{2},\dfrac{-19}{2}$$
Hence
$$(x_{2},y_{2})=(\dfrac{3}{2},\dfrac{-19}{2})$$...(ii)

$$9x^{2}+72x-112=0$$

$$(x-\dfrac{4}{3})(x+\dfrac{28}{3})=0$$

$$x=\dfrac{4}{3},\dfrac{-28}{3}$$
Hence
$$(x_{3},y_{3})=(\dfrac{4}{3},\dfrac{-28}{3})$$...(iii)
Therefore the three points are
$$A=(x_{1},y_{1})=(2-10)$$

$$B=(x_{2},y_{2})=(\dfrac{3}{2},\dfrac{-19}{2})$$

$$C=(x_{3},y_{3})=(\dfrac{4}{3},\dfrac{-28}{3})$$

Thus, there are two possibilities
I) The points are co-linear.
II) The points are non-co-linear (forming a triangle).
Hence, first we check for co-linearity.
If the points are co-linear, then the must lie on a single straight line.

The equation of the line passing through $$A=(2,-10)$$ and $$B=(\dfrac{3}{2},\dfrac{-19}{2})$$ is

$$\dfrac{y+10}{x-2}=\dfrac{y+\dfrac{19}{2}}{x-\dfrac{3}{2}}$$

$$\rightarrow x+y=-8$$.
Now if C is collinear with A and B, then it must satisfy the above equation of the straight line.
Substituting $$C=(x_{3},y_{3})$$ in the given line gives us

$$\dfrac{4}{3}-\dfrac{28}{3}$$

$$=\dfrac{-24}{3}$$

$$=-8$$

$$=RHS$$

Hence the points are collinear.
Question 9
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If the lines $$\mathrm{x}+\mathrm{p}\mathrm{y}+\mathrm{p}=0,\ \mathrm{q}\mathrm{x}+\mathrm{y}+\mathrm{q}=0$$ and $$\mathrm{r}\mathrm{x}+\mathrm{r}\mathrm{y}+1 =0 (\mathrm{p},\mathrm{q}, \mathrm{r}$$ being distinct and $$ \neq$$ 1) are concurrent, then the value of
$$\displaystyle \frac{p}{p-1}+\frac{q}{q-1}+\frac{r}{r-1}=$$

A
$$1$$

B
$$-1$$

C
$$2$$

D
$$-2$$

Solution

$$x+py+p=0$$
$$qx+y+q=0$$
$$rx+ry+1=0$$
$$\begin{vmatrix} 1 & p & p \\ q & 1 & q \\ r & r & 1 \end{vmatrix}$$
$$\left( 1-qr \right) -p\left( q-pqr \right) +p\left( qr-r \right) =0$$
$$1-qr-pq+pqr+pqr-pr=0$$
$$qr(p-1)+pr(q-1)-pq(r-1)-pqr=0$$
$$\Rightarrow \cfrac { p }{ p-1 } +\cfrac { q }{ q-1 } +\cfrac { r }{ r-1 } =1$$
Option A
Question 10
1 / -0

The coordinates of the point $$P$$ on the line $$2x+3y+1=0$$ such that $$|PA-PB|$$ is maximum, where $$A(2, 0)$$ and $$B(0, 2)$$ is

A
$$(4, -3)$$

B
$$(7, -5)$$

C
$$(10, -7)$$

D
$$(-8, 5)$$

Solution

Given Point $$P(x_{1},y_{1})$$ lies on
$$2x+3y+1=0$$-----(1)
Given points $$A(2,0),B(0,2)$$
maximum value of $$\left | PA-PB \right | $$ is equal to the $$AB$$
consider PAB is triangle then
from Triangular inequalities
$$PB+AB>PA$$
$$\left | PA-PB \right |<PA$$-------(2)
Now $$PA=\sqrt{(0-2)^2+(2-0)^2}$$
$$PA=\sqrt{8}$$
$$PA=2\sqrt{2}$$
from eq (2)
$$2\sqrt{2}>\left | PA-PB \right |$$ is similar to
$$2\sqrt{2}>$$eq of line $$AB$$
on solving above eq we get
$$y_{1}=-1(x_{1}-2)$$
$$y_{1}=-x_{1}+2$$
$$x_{1}+y_{1}=2$$----(3)
Point P lies in line (1)
$$2x_{1}+3y_{1}=-1$$----(4)
from eq (3) and (4)
$$x_{1}=7,y_{1}=-5$$
Point $$P(7,-5)$$

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Determinants Test - 60

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Determinants Test - 60

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SHARING IS CARING

Question 1

$$K\left| A \right| $$

$${K}^{2}\left| A \right| $$

$${K}^{3}\left| A \right| $$

$$3\left| KA \right| $$

Solution

Question 2

$$y=x$$

$$2y=x$$

$$y=2x$$

$$y=-2x$$

Solution

Question 3

$$0$$

$$1$$

$$-1$$

None of these

Solution

Question 4

$$x=-3y$$

$$y=3x$$

$$y=-3x$$

$$3y=x$$

Solution

Question 5

$$0$$

$$1$$

$$1/2$$

$$-1/2$$

Solution

Question 6

$$-1, 3$$

$$-1,- 3$$

$$- 1, 20$$

$$- 1,- 20$$

Solution

Question 7

are concurrent

form a triangle

are parallel

mutually perpendicular lines

Solution

Question 8

are collinear

form an equilateral triangle

form a right angled isosceles triangle

are concyclic

Solution

Question 9

$$1$$

$$-1$$

$$2$$

$$-2$$

Solution

Question 10

$$(4, -3)$$

$$(7, -5)$$

$$(10, -7)$$

$$(-8, 5)$$

Solution

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