If $$A =\begin{bmatrix}2 & -1 & 3\\ -5 & 3 & 1\\ -3 & 2 & 3\end{bmatrix}$$, then $$A.(Adj A)=$$

$$\left( Adj{ .A }^{ T } \right) $$

MCQ Questions for Class 12 Commerce Maths Determinants Quiz 10

Let

$A$ be a square matrix of order

$3\times 3$ then

$\left| KA \right|$ is equal to

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$K\left| A \right|$
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${K}^{2}\left| A \right|$
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${K}^{3}\left| A \right|$
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$3\left| KA \right|$

Explanation

$Using \space properties \space of \space determinant$

$|kA|=k^{n}|A|$

$where \space n \space is \space order \space determinant$

$Since \space given \space matrix \space A \space has \space order \space 3$

$hence\space\left| kA \right| ={k}^{3}\left| A \right|$

Find the equation of line joining

$(1,2)$ and

$(3,6)$ using determinants. Let

$p(x,y)$ be any point on the line joining

$(1,2)(3,6)$

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$y=x$
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$2y=x$
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$y=2x$
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$y=-2x$

Explanation

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

$y=2x$

Find the equation of the line joining

$(3,1)$ and

$(9,3)$ using determinants.

0%
$x=-3y$
0%
$y=3x$
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$y=-3x$
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$3y=x$

Explanation

Let

$p(x,y)$ be any point on the line joining

$(3,1)$ and

$(9,3)$

$\begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix}=0$

${R}_{2}\Rightarrow {R}_{2}-{R}_{1}$ ;

${R}_{3}\Rightarrow {R}_{3}-{R}_{1}$

$\begin{vmatrix} x & y & 1 \\ 3-x & 1-y & 0 \\ 9-x & 3-y & 0 \end{vmatrix}=0$

$\Rightarrow$

$1[(3-x)(3-y)-(1-y)(9-x)]=0$

$\Rightarrow$

$9-3y-3x+xy+x-9+9y-xy=0$

$\Rightarrow$

$6y-2x=0$

$\Rightarrow$

$x=3y$

Value of determinant

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

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$0$
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$1$
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$1/2$
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$-1/2$

Explanation

$\begin{vmatrix} \cos 50^\circ & \sin 10^\circ\\ \sin 50^\circ & \cos 10^\circ \end{vmatrix}$

$=\cos 50^\circ \cos 10^\circ - \sin 50^\circ \sin 10^\circ$

$= \cos (50^\circ + 10^\circ)$

$= \cos 60^\circ =\dfrac{1}{2}$
So, option (c) is correct.

Value of determinant

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

0%
$0$
0%
$1$
0%
$-1$
0%
None of these

Explanation

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

$=\cos 80^\circ \sin 10^\circ - (- \cos 10^\circ) sin 80^\circ$

$=\cos 80^\circ \sin 10^\circ + \cos 10^\circ sin 80^\circ$

$= \sin (10^\circ + 80^\circ)$

$= \sin 90^\circ = 1$

Co-factors of the first column of determinant

$\begin{vmatrix} 5 & 20 \\ 3 & -1\end{vmatrix}$

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$-1, 3$
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$-1,- 3$
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$- 1, 20$
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$- 1,- 20$

Explanation

$\begin{vmatrix} 5 & 20 \\ 3 & -1\end{vmatrix}$
Co-factor of

$a_{11}$

$F_{11} = (- 1)^2 M_{11}$

$= 1 \times (- 1) = - 1$
Co-factor of

$a_{12}$

$F_{21} = (-1)^{11}\,M_{21}$

$= (- 1) × 20 = - 20$

$a, b, c$ 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

$abx+ry+1=0$ ,

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are concurrent
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form a triangle
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are parallel
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mutually perpendicular lines

Explanation

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.

$x_{1},y_{1}$ 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

$(x_{3},y_{3})$

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are collinear
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form an equilateral triangle
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form a right angled isosceles triangle
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are concyclic

Explanation

$x^{2}+8x-20=0$

$\rightarrow (x+10)(x-2)=0$

$x=-10,x=2$

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.

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

$\displaystyle \frac{p}{p-1}+\frac{q}{q-1}+\frac{r}{r-1}=$

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$1$
0%
$-1$
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$2$
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$-2$

Explanation

$x+py+p=0$

$qx+y+q=0$

$rx+ry+1=0$

$\begin{vmatrix} 1 & p & p \\ q & 1 & q \\ r & r & 1 \end{vmatrix}$

$\left( 1-qr \right) -p\left( q-pqr \right) +p\left( qr-r \right) =0$

$1-qr-pq+pqr+pqr-pr=0$

$qr(p-1)+pr(q-1)-pq(r-1)-pqr=0$

$\Rightarrow \cfrac { p }{ p-1 } +\cfrac { q }{ q-1 } +\cfrac { r }{ r-1 } =1$

Option A

The lines

$px+qy+r=0,qx+ry+p=0 \, \,and\,\,\, rx+py+q=0$ are concurrent then

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$p+q+r=0$
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$p^{3}+q^{3}+r^{3}=3pqr$
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$p^{2}+q^{2}+r^{2}-pq-qr-rp=0$
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$p^{2}+q^{2}+r^{2}=2(pq+qr+rp)$

Explanation

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

$p(rq-p^2)-q(q^2-rp)+r(qp-r^2)=0$

$3pqr-p^3-q^3-r^3=0$

$p^3+q^3+r^3=3pqr$

$p^3+q^3+r^3-3pqr=0$

$(p+q+r)(p^2+q^2+r^2-pa-qr-pr)=0$

$\Rightarrow p+q+r=0$ or

$p^2+q^2+r^2-pq-qr-pr=0$

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

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$(4, -3)$
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$(7, -5)$
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$(10, -7)$
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$(-8, 5)$

Explanation

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

$P(7,-5)$

If the lines

$p_{1}x+q_{1}y=1,p_{2}x+q_{2}y=1$ and

$p_{3}x+q_{3}y=1$ be concurrent, then the points

$(p_{1},q_{1}),(p_{2},q_{2})$ and

$(p_{3},q_{3})$ ,

0%
are collinear
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form an equilateral triangle
0%
form a scalene triangle
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form a right angled triangle

Explanation

$p_{1}x+q_{1}y=1,\ p_{2}x+q_{2}y =1\ p_{3}x+q_{3}y=1$
Given lines are concurrent

$\Rightarrow \begin{vmatrix} p_{1} & q_{1} & 1 \\ p_{2} & q_{2}& 1 \\ p_{3} & q_{3} & 1 \end{vmatrix}=0$

$\Rightarrow p_{1}(q_{2}-q_{3})-q_{1}(p_{2}-p_{3})+(p_{2}q_{3}-p_{3}q_{2})=0$

$\Rightarrow (p_{1}q_{2}-p_{2}q_{1})+(p_{2}q_{3}-p_{3}q_{2})+(p_{3}q_{1}-p_{1}q_{3})=0$
The left hand side of the above equation is also equal to twice the area of a triangle with coordinates

$(p_1, q_1),\; (p_2, q_2),\; (p_3,q_3)$
Since it is equal to zero,

$(p_{1},q_{1}),(p_{2},q_{2}),(p_{3},q_{3})$ are collinear.

If the lines

${x}+{a}{y}+{a}=0,\ {b}{x}+{y}+{b}=0,\ {c}{x}+{c}{y}+1 =0 ({a}\neq{b}\neq {c}\neq1)\$ are concurrent, then the value of

$\displaystyle \frac{{a}}{{a}-1}+\frac{{b}}{{b}-1}+\frac{{c}}{{c}-1}$ , is

0%
$-1$
0%
$0$
0%
$1$
0%
$3$

Explanation

Given $x+ay+a=0$ ...(1)

$bx+y+b=0$ ...(2)

$cx+cy+1=0$ ...(3)

These lines are concurrent means they have only one intersection point

From 1 & 2,

$(ab-1)y+ab-b=0$

$y=\dfrac{b-ab}{ab-1}$ & $x=-a\left(\dfrac{b-1}{ab-1}\right)$

From 1 & 3,

$(ac-c)y+ac-1=0$

$y=\dfrac{1-ac}{ac-c}$ & $x=-a\left(\dfrac{1-c}{ac-c}\right)$

So,

$\dfrac{b-ab}{ab-1}=\dfrac{1-ac}{ac-c}$

$\Rightarrow abc-cb+abc-a^2bc=ab-a^2bc-1+ac$

$\Rightarrow 1+2abc=ab+ac+bc$ ...(4)

Now,

$\dfrac{a}{a-1}+\dfrac{b}{b-1}+\dfrac{c}{c-1}=\dfrac{ab-a+ab-b}{ab-a-b+1}+\dfrac{c}{(c-1)}$

$=\dfrac{2abc-ac-bc-2ab+a+b+abc-ac-bc+c}{abc-ac-bc+c-ab+a+b-1}$

$=\dfrac{3abc-2(ac+bc+ab)+a+b+c}{abc-(ac+bc+ab)+a+b+c-1}$

$=\dfrac{abc-(ac+bc+ab)+(a+b+c)-1}{abc-(ac+bc+ab)+(a+b+c)-1}=1$ [From (4)]

Hence, option C.

$x_1, x_2, x_3$ as well as

$y_1, y_2, y_3$ are in G.P. with same common ratio, then the points

$P(x_1, y_1), Q (x_2, y_2)$ and

$R(x_3, y_3)$

0%
lies on a straight line
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lie on an ellipse
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lie on a circle
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are vertices of a triangle

Explanation

Let

$x_1=a , x_2=ar$ and

$x_3=ar^2; y_1=b, y_2=br$ and

$y_3=br^2$
Now,

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{br-b}{ar-a}=\dfrac{b}{a}$
Similarly,

$\dfrac{y_3-y_2}{x_3-x_2}=\dfrac{br^2-br}{ar^2-ar}=\dfrac{b}{a}$
Thus, the points are collinear as they lie on same line.

The values of

$|A^{50}|$ equals

0%
0
0%
1
0%
-1
0%
25

Explanation

Given

$A= \begin{bmatrix} 1& 0 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{bmatrix}$ satisfies

$A^n = A^{n - 2} + A^2 - I$ for

$n \geq 3$

$\Rightarrow A^4=A^2+A^2-I=2A^2-I$

$\Rightarrow A^6=A^4+A^2-I=3A^2-2I$
similarly

$A^{50}=25A^2-24I$

$A^2=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$

$\therefore A^{50}=\begin{bmatrix} 25 & 0 & 0 \\ 25 & 25 & 0 \\ 25 & 0 & 25 \end{bmatrix}-\begin{bmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$

$\Rightarrow A^{50}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$

$\therefore |A^{50}|=1$
Hence, option B.

The value of

$|\cup|$ equals

0%
$0$
0%
$1$
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$2$
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$-1$

Explanation

Given

$A= \begin{bmatrix} 1& 0 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{bmatrix}$ satisfies

$A^n = A^{n - 2} + A^2 - I$ for

$n \geq 3$

$\Rightarrow A^4=A^2+A^2-I=2A^2-I$

$\Rightarrow A^6=A^4+A^2-I=3A^2-2I$
similarly

$A^{50}=25A^2-24I$

$\Rightarrow A^{50}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$
Given

$\displaystyle \bigcup_{3 \times 3}$ with its column as

$\cup_1, \cup_2, \cup_3$ such that

$A^{50} \cup_1 = \begin{bmatrix}1\\25 \\ 25\end{bmatrix}, A^{50} \cup_2 = \begin{bmatrix}0\\ 1 \\ 0\end{bmatrix}, A^{50} \cup_3 = \begin{bmatrix}0\\ 0 \\ 1\end{bmatrix}$
let

$B=A^{50}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$

$B^{-1}=\begin{bmatrix} 1 & 0 & 0 \\ -25 & 1 & 0 \\ -25 & 0 & 1 \end{bmatrix}$

$\cup_1=B^{-1}\begin{bmatrix}1\\25 \\ 25\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -25 & 1 & 0 \\ -25 & 0 & 1 \end{bmatrix}\begin{bmatrix}1\\25 \\ 25\end{bmatrix}=\begin{bmatrix}1\\0 \\ 0\end{bmatrix}$

$\cup_2=B^{-1}\begin{bmatrix}0\\1 \\ 0\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -25 & 1 & 0 \\ -25 & 0 & 1 \end{bmatrix}\begin{bmatrix}0\\1 \\ 0\end{bmatrix}=\begin{bmatrix}0\\1 \\ 0\end{bmatrix}$

$\cup_3=B^{-1}\begin{bmatrix}0\\0 \\ 1\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -25 & 1 & 0 \\ -25 & 0 & 1 \end{bmatrix}\begin{bmatrix}0\\0 \\ 1\end{bmatrix}=\begin{bmatrix}0\\0 \\ 1\end{bmatrix}$

$\therefore |\cup|=\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}=1$
Hence, option B.

$A =\begin{bmatrix}2 & -1 & 3\\ -5 & 3 & 1\\ -3 & 2 & 3\end{bmatrix}$ , then

$A.(Adj A)=$

0%
$\left( Adj{ .A }^{ T } \right)$
0%
$(Adj. A) . A$
0%
$|A| . A$ .
0%
None of these

Explanation

${ A }^{ -1 }=\dfrac { adj\quad A }{ |A| }$

${ AA }^{ -1 }=\dfrac { A\left( adj\quad A \right) }{ |A| }$

$\therefore \quad A\left( Adj\quad A \right) =|A|I$

Also

$\left( Adj\ A \right)A=A\left( Adj\quad A \right) =|A|I$

Option B satisfies

If A is a square matrix of order 3, then

$|(A - A^T)^{105}|$ is equal to

0%
$105|A|$
0%
$105|A|^2$
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$105$
0%
none of these

Explanation

$\left |\left (A-A^{T} \right )^{105} \right |=\left |-\left (A^{T}-A \right ) \right |^{105}$

$=-\left |A^{T}-A \right |^{105}$

$=-\left |\left (A^{T}-A \right )^{T} \right |^{105}$ (

$\because$ determinant of matrix and its transpose is equal)

$=-\left |A-A^{T} \right |^{105}$

$\Rightarrow \left |A-A^{T} \right |^{105}=-\left |A-A^{T} \right |^{105}=0 (\because$ determinant of skew-symmetric matrix of odd order is 0.)

$\left |\left (A-A^{T} \right )^{105} \right |=0$
Hence, option 'D' is correct.

Let

$\begin{vmatrix} 1+x & x & x^{2}\\ x & 1+x & x^{2} \\ x^{2} & x & 1+x \end{vmatrix} = ax^{5} + bx^{4} + cx^{3} + dx^{2} + \lambda x + \mu$ be an identity in x, where a,b,c,d,

$\lambda, \mu$ are independent of x. Then the value of

$\lambda$ is

0%
$3$
0%
$2$
0%
$4$
0%
$1$

Explanation

Given,

$\begin{vmatrix} 1+x & x & x^{2}\\ x & 1+x & x^{2} \\ x^{2} & x & 1+x \end{vmatrix} = ax^{5} + bx^{4} + cx^{3} + dx^{2} + \lambda x + \mu$
Put

$x=0$

$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}=\mu$

$\Rightarrow \mu=1$
Differentiating both sides,

$\begin{vmatrix} 1 & 1 & 2x \\ x & 1+x & x^{ 2 } \\ x^{ 2 } & x & 1+x \end{vmatrix}+\begin{vmatrix} 1+x & x & x^{ 2 } \\ 1 & 1 & 2x \\ x^{ 2 } & x & 1+x \end{vmatrix}+\begin{vmatrix} 1+x & x & x^{ 2 } \\ x & 1+x & x^{ 2 } \\ 2x & 1 & 1 \end{vmatrix}=5ax^{ 4 }+4bx^{ 3 }+3cx^{ 2 }+2dx+\lambda$
Put

$x=0$

$\lambda =\begin{vmatrix} 1&1&0 \\ 0&1&0 \\ 0&0&1 \end{vmatrix}+\begin{vmatrix} 1&0&0 \\ 1&1&0 \\ 0&0&1 \end{vmatrix}+\begin{vmatrix} 1&0&0 \\ 0&1&0 \\ 0&1&1 \end{vmatrix}$

$=1+1+1=3$

The value of |B| is equal to

0%
|A|
0%
|A|/2
0%
2|A|
0%
none of these

Let f (n)=

$\displaystyle \left | \begin{matrix}n &n+1 &n+2 \\^{n}P_{n} &^{n+1}P_{n+1} &^{n+2}P_{n+2} \\^{n}C_{n} &^{n+1}C_{n+1} &^{n+2}C_{n+2} \end{matrix} \right |,$ where the symbols have their usual meanings. The

$f(n)$ is divisible by

0%
$n^{2}+n+1$
0%
$(n+1)!$
0%
$n!$
0%
none of these

Explanation

$\displaystyle \left | \begin{matrix}n &n+1 &n+2 \\^{n}P_{n} &^{n+1}P_{n+1} &^{n+2}P_{n+2} \\^{n}C_{n} &^{n+1}C_{n+1} &^{n+2}C_{n+2} \end{matrix} \right |,$

$=\begin{vmatrix}n & n+1 & n+2\\ n! & (n+1)! & (n+2)!\\ 1 & 1 & 1\end{vmatrix}$

Doing,

$C_{1}=C_{1}-C_{2}$ and

$C_{2}=C_{2}-C_{3}$ ,

$=n!\begin{vmatrix}-1 & -1 & n+2\\ -n & -n^{2}-2n-1 & (n+2)(n+1)\\ 0 & 0 & 1\end{vmatrix}$

$=-1(-n^2-2n-n-0)+(-n-0)+(0)$

$=n!(n^{2}+2n+1-n)$

$=n!(n^{2}+n+1)$

So, it is divisible by

$n!$ and

$(n^{2}+n+1)$

Consider the points

$P=(-\sin (\beta -\alpha ), -\cos \beta )$ ,

$Q=(\cos (\beta -\alpha ), \sin \beta )$ and

$R=(\cos (\beta -\alpha +\theta ), \sin (\beta -\theta ))$ , where

$0< \alpha , \beta < \dfrac{\pi }{4}$ then

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$P$ lies on the line segment $RQ$
0%
$Q$ lies on the line segment $PR$
0%
$R$ lies on the line segment $QP$
0%
$P, Q, R$ are non-collinear.

Explanation

$\Delta=\displaystyle \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$

Put

$\beta -\alpha =\phi$ and consider the determinant

$\Delta =\displaystyle \begin{vmatrix} -\sin \phi & -\cos \beta & 1 \\ \cos \phi & \sin \beta & 1 \\ \cos (\phi +\theta ) & \sin (\beta -\theta ) & 1 \end{vmatrix}$
Using

$R_{3}\rightarrow R_{3}-\cos \theta R_{2}-\sin \theta R_{1}$

$\Delta =\displaystyle \begin{vmatrix} -\sin \phi & -\cos \beta & 1 \\ \cos \phi & \sin \beta & 1 \\ 0 & 0 & 1-\cos \theta -\sin \theta \end{vmatrix}$

$=(1-\cos \theta -\sin \theta )\cos (\phi +\beta )$

$=(1-\cos \theta -\sin \theta )\cos (2\beta -\alpha )$

$=\left [ 1-\sqrt{2}\:\sin (\theta +\dfrac{\pi}4) \right ]\cos (2\beta -\alpha )$
As

$0< \theta < \pi /4\Rightarrow \dfrac{\pi}4< \theta +\dfrac{\pi}4< \dfrac{\pi}2$

$\Rightarrow \displaystyle \frac{1}{\sqrt{2}}< \sin (\theta +\dfrac{\pi}4)< 1$

$\Rightarrow \displaystyle \cos (2\beta -\alpha )\neq 0$
Thus

$\Delta \neq 0$ and the points

$P, Q, R$ are non-collinear.

If the points

$(a, 1), (1, b)$ and

$(a -1, b -1)$ are collinear,

$\alpha ,\beta$ are respectively the arithmetic and geometric means of

$a$ and

$b$ , then

$4\alpha -\beta^{2}$ is equal to

0%
$-1$
0%
$0$
0%
$3$
0%
$2$

Explanation

Since, The given points are collinear

$\displaystyle \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}=0$

so,

$\displaystyle \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\a-1& b-1 & 1 \end{vmatrix}=0$
On expanding,we get

$a(b-b+1)-1(1-a+1)+1(b-1-ab+b)=0$

$2(a+b)-ab=3$ -----------1)
Now

$\alpha=\dfrac{a+b}{2} and\ \beta=\sqrt{ab}$ ---------2)

From 1) and 2)
Now

$4\alpha-{\beta}^{2}=3$

$2(a+b)-ab{=}3$

If the points

$\displaystyle(-2,0),(-1,\dfrac{1}{\sqrt{3}})$ and

$\displaystyle(\cos\theta,\sin \theta)$ are collinear, then the number of values of

$\displaystyle \theta \in [0,2\pi]$ :

0%
$0$
0%
$1$
0%
$2$
0%
infinite

Explanation

Given points

$\displaystyle(-2,0),\left(-1,\dfrac1{\sqrt{3}}\right)$ and

$\displaystyle(\cos\theta,\sin \theta)$
For collinearity,

$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}=0$

$\begin{vmatrix} -2 & 0 & 1 \\ -1 & \dfrac { 1 }{ \sqrt { 3 } } & 1 \\ \cos { \theta } & \sin { \theta } & 1 \end{vmatrix}=0$

$\displaystyle \sin { \theta } -\frac { 1 }{ \sqrt { 3 } } \cos { \theta } =\frac { 2 }{ \sqrt { 3 } }$

$\Rightarrow \displaystyle \frac { \sqrt { 3 } }{ 2 } \sin { \theta } -\frac { 1 }{ 2 } \cos { \theta } =1$

$\Rightarrow \displaystyle \sin { (\theta -\frac { \pi }{ 6 } ) } =\sin { \frac { \pi }{ 2 } }$

$\Rightarrow \displaystyle \theta -\frac { \pi }{ 6 } =n \pi + (-1)^n \frac { \pi }{ 2 }$ (

$n \in Z$ )

$\Rightarrow \displaystyle \theta=\frac{2\pi}{3}$ (when

$\theta \in [0, 2 \pi ], n=1$ )

$A=\begin{bmatrix}a&b &c \\x &y &z \\p &q &r \end{bmatrix}$ ,

$B=\begin{bmatrix}q&-b &y \\-p &a &-x \\r &-c &z \end{bmatrix}$ then

0%
$|A| = |B|$
0%
$|A| = -|B|$
0%
$|A| =2|B|$
0%
$A$ is invertible if and only if $B$ is invertible.

Explanation

$|B|=\begin{vmatrix}q&-b &y \\-p &a &-x \\r &-c &z \end{vmatrix}=(-1)\begin{vmatrix}q&-b &y \\p &-a &x \\r &-c &z \end{vmatrix}$

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

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

$=-|A|$ [

$\because |A^{T}|=|A|$ ]

$\therefore$ From here, we get

$|A|\neq 0$ if and only if

$|B|\neq 0$ .
i.e

$A$ is invertible if and only if

$B$ is invertible.

Hence, options B and D.

Let

$\begin{vmatrix} x& 2 & x\\ x^2 & x & 6\\ x & x & 6\end{vmatrix} = \alpha x^4 + \beta x^3 + \gamma x^2 + \delta x + \lambda$ then the value of

$5 \alpha + 4 \beta + 3\gamma + 2 \delta + \lambda =$

0%
$-11$
0%
$0$
0%
$-16$
0%
$16$

Explanation

Given,

$\begin{vmatrix} x& 2 & x\\ x^2 & x & 6\\ x & x & 6\end{vmatrix} = \alpha x^4 + \beta x^3 + \gamma x^2 + \delta x + \lambda$

Put

$x=0$

$\begin{vmatrix} 0 & 2 & 0 \\ 0 & 0 & 6 \\ 0 & 0 & 6 \end{vmatrix}=\lambda$

$\Rightarrow \lambda=0$

So,

$\begin{vmatrix} x& 2 & x\\ x^2 & x & 6\\ x & x & 6\end{vmatrix} = \alpha x^4 + \beta x^3 + \gamma x^2 + \delta x$
Differentiating w.r.t x,

$\begin{vmatrix} 1 & 0 & 1 \\ x^{ 2 } & x & 6 \\ x & x & 6 \end{vmatrix}+\begin{vmatrix} x & 2 & x \\ 2x & 1 & 0 \\ x & x & 6 \end{vmatrix}+\begin{vmatrix} x & 2 & x \\ x^{ 2 } & x & 6 \\ 1 & 1 & 0 \end{vmatrix}=4\alpha x^{ 3 }+3\beta x^{ 2 }+2\gamma x+\delta$
Put

$x=0$

$\Rightarrow \begin{vmatrix} 1 & 0 & 1 \\ 0 & 0 & 6 \\ 0 & 0 & 6 \end{vmatrix}+\begin{vmatrix} 0 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 6 \end{vmatrix}+\begin{vmatrix} 0 & 2 & 0 \\ 0 & 0 & 6 \\ 1 & 1 & 0 \end{vmatrix}=\delta$

$\Rightarrow \delta=-12$

If [x] stands greatest integer

$\leq x$ then the value of

$\begin{vmatrix} \left [ e \right ] & \left [ \pi \right ] & \left [ \pi ^{2}-6 \right ]\\ \left [ \pi \right ] & \left [ \pi ^{2}-6 \right ] & \left [ e \right ]\\ \left [ \pi ^{2}-6 \right ] & \left [ e \right ] & \left [ \pi \right ] \end{vmatrix}$ equals to=?

0%
-8
0%
8
0%
-1
0%
1

Explanation

$\left[ x \right]$ Greatest integer

$=\begin{vmatrix} \left[ e \right] & \left[ \pi \right] & \left[ { \pi }^{ 2 }-6 \right] \\ \left[ \pi \right] & \left[ { \pi }^{ 2 }-6 \right] & \left[ e \right] \\ \left[ { \pi }^{ 2 }-6 \right] & \left[ e \right] & \left[ \pi \right] \end{vmatrix}$

$=\begin{vmatrix} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 3 & 2 & 3 \end{vmatrix}$

$=2(5)-3(3)+3(-3)$

$=-8$

$\Delta =\begin{vmatrix} x+1 & x+2 & x+a\\ x+2 & x+3 & x+b\\ x+3 & x+4 & x+c \end{vmatrix}=0$ , then
the family of lines

$ax+by+c=0$ passes through

0%
$(1, -1)$
0%
$(1, -2)$
0%
$(2, -3)$
0%
$(0, 0)$

Explanation

$\Delta =\begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix}=0$

Applying

${ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 },{ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }$

$\Delta =\begin{vmatrix} x+1 & 1 & a-1 \\ x+2 & 1 & b-2 \\ x+3 & 1 & c-3 \end{vmatrix}=0$

Applying

${ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }$

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

Expanding along

${ C }_{ 2 }$

$1\left( c-a-2 \right) -2\left( b-a-1 \right) =0\\ \Rightarrow a-2b+c=0$
From options for point

$(1,-2)$ lies on the line

$ax+by+c=0$

Two

$n \times n$ square matrices

$A$ and

$B$ are said to be similar if there exists a non-singular matrix

$P$ such that

$P^{-1}A\: P=B$

$A$ and

$B$ are two similar matrices, then

0%
$det \: (A) = det \: (B)$
0%
$det \: (A) + det \: (B)=0$
0%
$det (AB)\neq 0$
0%
none of these

Explanation

$A$ and

$B$ are similar matrices there exists a non-singular matrix

$P$ such that

$A=P^{-1}\:BP$

$\Rightarrow det \: (A) = det\: (P^{-1}\:BP)$

$=det\: (P^{-1})\: det \: (B) \: det \: (P)$

$\displaystyle =\frac{1}{det\: (P)}det \: (B) \: (det P)$

$= det \: B$

Hence, option A.

Let

$0< \theta < \pi /2$ and

$\Delta \left ( x, \theta \right )=\begin{vmatrix} x & \tan \theta & \cot \theta \\ -\tan \theta & -x & 1\\ \cot \theta & 1 & x \end{vmatrix}$
then

0%
$\Delta \left ( 0, \theta \right )=0$
0%
$\Delta \left ( x, \dfrac {\pi }{4} \right )=x-x^3$
0%
$Min_{0< \theta < \pi /2}\Delta \left ( 1, \theta \right )=0$
0%
$\Delta \left ( x, \theta \right )$ is independent of x

Explanation

$\Delta \left( 0,\theta \right) =\begin{vmatrix} 0 & \tan \theta & \cot \theta \\ -\tan \theta & 0 & 1 \\ \cot \theta & 1 & 0 \end{vmatrix}=\tan \theta \cot \theta \quad -\cot \theta \tan \theta =0$

$\Delta \left( x,{ \pi }/{ 4 } \right) =\begin{vmatrix} x & 1 & 1 \\ -1 & -x & 1 \\ 1 & 1 & x \end{vmatrix}=x\left( -{ x }^{ 2 }-1 \right) +1\left( 1+x \right) +1\left( x-1 \right) \\$

$=-{ x }^{ 3 }-x+1+x+x-1=x-{ x }^{ 3 }$

$\Delta \left( 1,\theta \right) =\begin{vmatrix} 1 & \tan \theta & \cot \theta \\ -\tan \theta & -1 & 1 \\ \cot \theta & 1 & 1 \end{vmatrix}=-2\quad +\tan \theta \left( \cot \theta +\tan \theta \right) +\cot \theta \left( \cot \theta -\tan \theta \right) \\ =-2+\tan ^{ 2 }{ \theta } +\cot ^{ 2 }{ \theta }$

$\tan ^{ 2 }{ \theta } +\cot ^{ 2 }{ \theta } \ge 2$

Hence,

$\Delta \left( 1,\theta \right) _{min.}=0$

$\Delta =\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}$ and

$c_{ij}=\left ( -1 \right )^{i+j}$ (determinant obtained by deleting ith row and jth column), then

$\begin{vmatrix} c_{11} & c_{12} & c_{13}\\ c_{21} & c_{22} & c_{23}\\ c_{31} & c_{32} & c_{33} \end{vmatrix}=\Delta ^{2}$

$\begin{vmatrix} 1 & x & x^{ 2 } \\ x & x^{ 2 } & 1 \\ x^{ 2 } & 1 & x \end{vmatrix}=7$ and

$\Delta =\begin{vmatrix} x^{3}-1 & 0 & x-x^{4}\\ 0 & x-x^{4} & x^{3}-1\\ x-x^{4} & x^{3}-1 & 0 \end{vmatrix}$ , then

0%
$\Delta =7$
0%
$\Delta =343$
0%
$\Delta =-49$
0%
$\Delta =49$

Explanation

For

$\begin{vmatrix} 1 & x & x^{ 2 } \\ x & x^{ 2 } & 1 \\ x^{ 2 } & 1 & x \end{vmatrix}=7$

$\begin{vmatrix} c_{ 11 } & c_{ 12 } & c_{ 13 } \\ c_{ 21 } & c_{ 22 } & c_{ 23 } \\ c_{ 31 } & c_{ 32 } & c_{ 33 } \end{vmatrix}=\begin{vmatrix} x^{ 3 }-1 & 0 & x-x^{ 4 } \\ 0 & x-x^{ 4 } & x^{ 3 }-1 \\ x-x^{ 4 } & x^{ 3 }-1 & 0 \end{vmatrix}$

$\Delta ={ 7 }^{ 2 }=49$

The determinant

$\begin{vmatrix} \sin \alpha & \cos \alpha & 1\\ \sin \beta & \cos \beta & 1\\ \sin \gamma & \cos \gamma & 1 \end{vmatrix}$ is equal to

0%
$\displaystyle -4\sin \frac{\alpha -\beta }{2}\sin \frac{\alpha -\gamma }{2}\sin \frac{\gamma -\alpha }{2}$
0%
$\sin \alpha +\sin \beta +\sin \gamma$
0%
$\sin \left ( \alpha -\beta \right )+\sin \left ( \beta -\gamma \right )+\sin \left ( \gamma -\alpha \right )$
0%
none of these

Explanation

Expanding along with

$C_{ 3 }$

$\begin{vmatrix} \sin \alpha & \cos \alpha & 1 \\ \sin \beta & \cos \beta & 1 \\ \sin \gamma & \cos \gamma & 1 \end{vmatrix}\\$

$=1\left( \sin { \beta } \cos { \gamma } -\sin { \gamma } \cos { \beta } \right) -1\left( \sin { \alpha } \cos { \gamma } -\sin { \gamma } \cos { \alpha } \right) +1\left( \sin { \alpha } \cos { \beta } -\sin { \beta } \cos { \alpha } \right) \\$

$=\sin { \left( \beta -\gamma \right) } +\sin { \left( \gamma -\alpha \right) } +\sin { \left( \alpha -\beta \right) }$

Now let

$A=\left( \beta -\gamma \right) ,B=\left( \gamma -\alpha \right) ,C=\left( \alpha -\beta \right)$

we have

$A+B+C=0$

Now

$\sin { \left( \beta -\gamma \right) } +\sin { \left( \gamma -\alpha \right) } +\sin { \left( \alpha -\beta \right) } =\sin { A } +\sin { B } +\sin { C } \\$

$=2\sin { \cfrac { A+B }{ 2 } } \cos { \cfrac { A-B }{ 2 } } +2\sin { \cfrac { C }{ 2 } } \cos { \cfrac { C }{ 2 } } \\$

$=2\sin { \cfrac { -C }{ 2 } } \cos { \cfrac { A-B }{ 2 } } +2\sin { \cfrac { C }{ 2 } } \cos { \cfrac { -\left( A+B \right) }{ 2 } } \\$

$=-2\sin { \cfrac { C }{ 2 } } \left( \cos { \cfrac { A-B }{ 2 } } -\cos { \cfrac { A+B }{ 2 } } \right)$

$=-2\sin { \cfrac { C }{ 2 } } \left( 2\sin { \cfrac { A }{ 2 } } \sin { \cfrac { B }{ 2 } } \right) \\$

$=-4\sin { \cfrac { A }{ 2 } } \sin { \cfrac { B }{ 2 } } \sin { \cfrac { C }{ 2 } }$

$-4\sin { \cfrac { \left( \beta -\gamma \right) }{ 2 } } \sin { \cfrac { \left( \gamma -\alpha \right) }{ 2 } } \sin { \cfrac { \left( \alpha -\beta \right) }{ 2 } }$

The number of distinct real roots of

$\begin{vmatrix} \sin\, x&\cos\, x&\cos \,x \\ \cos\, x &\sin\, x&\cos\, x \\ \cos\, x&\cos\, x&\sin \, x\end{vmatrix}=0$ in the interval

$-\dfrac{\pi}{4} < x \le \dfrac{\pi}{4}$ is

0%
$0$
0%
$2$
0%
$1$
0%
$> 2$

Explanation

Given,

$\begin{vmatrix} \sin\, x&\cos\, x&\cos \,x \\\cos\, x &\sin\, x&\cos\, x \\\cos\, x&\cos\, x&\sin \, x\end{vmatrix}=0$

$\Rightarrow \sin \,x(\sin^2x-\cos^2x)-\cos\,x(\sin\,x\,\cos\,x-\cos^2x)+\cos\,x(\cos^2x-\sin\,x\,\cos\,x)=0$

$\Rightarrow \sin \,x(\sin\,x-\cos\,x)(\sin\,x+\cos\,x)-\cos^2\,x(\sin\,x-\cos\,x)-\cos^2\,x(\sin\,x-\cos\,x)=0$

$\Rightarrow (\sin\,x-\cos\,x)(\sin^2x+\sin\,x\,\cos\,x-2\cos^2\,x)=0$

$\Rightarrow (\sin\,x-\cos\,x)(\sin^2\,x+2\,\sin\,x\,\cos\,x-\sin\,x\,\cos\,x-2\,\cos^2\,x)=0$

$\Rightarrow (\sin\,x-\cos\,x)^2(\sin\,x+2\,\cos\,x)=0$

$\Rightarrow \sin\,x-\cos\,x=0$ or

$\sin\,x+2\,\cos\,x=0$

$\Rightarrow \tan\, x=1$ or

$\tan\, x=-2$

But

$x \in\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]$

$\Rightarrow$ Only one solution i.e

$x=\dfrac{\pi}{4}$

Say true or false:

Points

$P(10,\,-6),\,Q(6,\,-4)$ and

$C(-8,\,3)$ are collinear.

0%
True
0%
False

Explanation

Since the slope of the line with points

$(x_1,y_1)$ and

$(x_2,y_2)$ is:

$m=\cfrac { y_{ 2 }-y_{ 1 } }{ x_{ 2 }-x_{ 1 } }$

We find the slope of the line

$PQ$ with points

$P(10,-6)$ and

$Q(6,-4)$ is:

$m=\cfrac { -4-(-6) }{ 6-10 } =\cfrac { -4+6 }{ -4 } =-\cfrac { 1 }{ 2 }$

Now the slope of the line

$QC$ with points

$Q(6,-4)$ and

$C(-8,3)$ is:

$m=\cfrac { 3-(-4) }{ -8-6 } =\cfrac { 7 }{ -14 } =-\cfrac { 1 }{ 2 }$

Finally the slope of the line

$PC$ with points

$P(10,-6)$ and

$C(-8,3)$ is:

$m=\cfrac { 3-(-6) }{ -8-10 } =\cfrac { 9 }{ -18 } =-\cfrac { 1 }{ 2 }$

Since the slope of

$PQ$ ,

$QC$ and

$PC$ are equal, therefore, the points are collinear.

If the points

$(-1,3), (2,p)$ and

$(5,-1)$ are collinear, the value of

$p$ is

0%
$1$
0%
$-1$
0%
$0$
0%
$\displaystyle \sqrt{2}$

Explanation

Let

$A(-1,3)$ ,

$B(2,p)$ and

$C(5,-1)$ be the given points.

Three points are collinear if

${ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 })=0$

Therefore,

${ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 })=0\\ \Rightarrow -1(p-(-1))+2(-1-3)+5(3-p)=0\\ \Rightarrow -1(p+1)+2(-4)+5(3-p)=0\\ \Rightarrow -p-1-8+15-5p=0\\ \Rightarrow -6p+6=0\\ \Rightarrow -6p=-6\\ \Rightarrow p=1$

Hence

$p=1$

$(1,6), (3.-2)$ and

$(-2,K)$ are collinear points. What is

$K$ ?

0%
$-6$
0%
$2$
0%
$8$
0%
$10$
0%
$18$

Explanation

If three points are collinear then the area of triangle is zero.

Therefore, if the points

$(1,6)$ ,

$(3,-2)$ and

$(-2,K)$ are collinear, then their area must be zero that is:

$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| 1(-2-K)+3(K-6)-2(6+2) \right| \\ \Rightarrow 0=\left| -2-K+3K-18-16 \right| \\ \Rightarrow 0=\left| 2K-36 \right| \\ \Rightarrow 2K-36=0\\ \Rightarrow 2K=36\\ \Rightarrow K=18$

Hence,

$K=18$

$a_{1}, a_{2}, ...., a_{n}$ , ..... are in G.P. then

$\begin{vmatrix}\log a_{n} & \log a_{n + 1} & \log a_{n + 2}\\ \log a_{n + 3} & \log a_{n + 4} & \log a_{n + 5}\\ \log a_{n + 6} & \log a_{n + 7} & \log a_{n + 8}\end{vmatrix}$ is

0%
$0$
0%
$1$
0%
$-1$
0%
None of these

Explanation

We know that

$\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=.......\dfrac{a_{n}}{a_{n-1}}=r$

which means that

$a_n,a_{n+1}$ and

$a_{n+2}$ are in G.P

So,

$a^{2}_{n+1}=a_n \times a_{n+2}$

Taking log on both the sides, we get

$2 log\ {a_{n+1}}=log\ {a_{n}}+log{a_{n+2}}$

$2 log\ {a_{n+1}}-log\ {a_{n}}-log\ {a_{n+2}}=0$ --------(1)

Similarly,

$2 log\ {a_{n+4}}-log\ {a_{n+3}}-log\ {a_{n-5}}=0$ -------(2)

$2 log\ {a_{n+7}}-log\ {a_{n+6}}-log\ {a_{n+8}}=0$ ---------(3)

Apply the operation on column, we get

$C_{1}\rightarrow 2C_2-C_1-C_3$

$det(A)=0$

If the determinant

$\begin{vmatrix} a+p & 1+x & u+f \\ b+q & m+y & v+g \\ c+r & n+z & w+h \end{vmatrix}$ splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is

0%
9
0%
8
0%
24
0%
12

The coefficient of

${x}^{2}$ in the expansion of the determinant

$\begin{vmatrix} { x }^{ 2 } & { x }^{ 3 }+1 & { x }^{ 5 }+2 \\ { x }^{ 3 }+3 & { x }^{ 2 }+x & { x }^{ 3 }+{ x }^{ 4 } \\ x+4 & { x }^{ 3 }+{ x }^{ 4 } & { 2 }^{ 3 } \end{vmatrix}$ is

0%
$-10$
0%
$-8$
0%
$-2$
0%
$-6$
0%
$8$

Explanation

For the coefficient of

${x}^{2}$ on expanding along

$R$ we get

$\Delta ={ x }^{ 2 }\left[ 8{ x }^{ 2 }+8x-{ x }^{ 6 }-2{ x }^{ 7 }-{ x }^{ 8 } \right]-\left( { x }^{ 2 }+1 \right)$

$\left[ 8{ x }^{ 3 }+24-{ x }^{ 4 }-{ x }^{ 6 }-4{ x }^{ 3 }-4{ x }^{ 4 } \right] +\left( { x }^{ 5 }+2 \right)$

$\left[ { x }^{ 6 }+{ x }^{ 7 }+3{ { x }^{ 3 }-3{ x }^{ 4 }-{ x }^{ 3 }-x }^{ 2 }-4{ x }^{ 2 }-4x \right]$

$= 8x^4+8x^3-x^5-2x^9-x^{10}-8x^6-24x^3+x^7+x^6+4x^6+4x^7-8x^3-24$

$-x^4+x^5+4x^3+4x^4+x^{11}+x^{12}+3x^8+3x^9-x^8-x^7-4x^7-4x^7-4x^6+2x^6$

$+2x^7+6x^3+6x^4-2x^3-2x^2-8x^2-8x$

Coefficient of

${x}^{2}=-2-8=-10$

The Value of the determinant

$\begin{vmatrix} { b }^{ 2 }-ab & \quad b-c & \quad bc-ac \\ ab-{ a }^{ 2 } & \quad a-b & { \quad b }^{ 2 }-ab \\ bc-ac & \quad c-a & \quad ab-{ a }^{ 2 } \end{vmatrix}$ =

0%
abc
0%
a + b + c
0%
0
0%
ab + bc + ca

Points (a, 0), (0, b) and (1, 1)are collinear, if:

0%
$\displaystyle \frac{1}{a} + \frac{1}{b} = 1$
0%
$\displaystyle \frac{1}{a} - \frac{1}{b} = 1$
0%
$\displaystyle a+b = 1$
0%
$\displaystyle a-b = ab$

Explanation

$\textbf{Step 1: Calculating}$

$\text{The given points are }(a,0),(0,b),(1,1)$

$\implies \text{Slope of line connecting }(a,0)\text{ and }(0,b)=\text{Slope of line connecting }(0,b) \text{ and }(1,1)$

$\textbf{Step 2: Calculating}$

$\text{Slope }=\dfrac{y_2-y_1}{x_2-x_1}$

$\text{Consider the points }(a,0)\text{ and }(0,b)$

$\implies \text{Slope}=\dfrac{0-b}{a-0}=\dfrac{-b}{a}$

$\text{Consider the points }(0,b)\text{ and }(1,1)$

$\implies \text{Slope}=\dfrac{b-1}{0-1}=-(b-1)$

$\implies \dfrac{-b}{a}=-(b-1)$

$\implies ab-a=b$

$\implies a+b=ab$

$\implies \dfrac1a+\dfrac1b=1$

$\textbf{Hence, The expression }\mathbf{\dfrac1a+\dfrac1b=1 \text{ is true for the given conditions}}$

The value of

$x$ satisfying the equation

$\begin{vmatrix}\cos 2x & \sin 2x & \sin 2x\\ \sin 2x & \cos 2x & \sin 2x\\ \sin 2x & \sin 2x & \cos 2x\end{vmatrix} = 0$ and

$x\epsilon \left [0, \dfrac {\pi}{4}\right ]$ is

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$\dfrac {\pi}{4}$
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$\dfrac {\pi}{2}$
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$\dfrac {\pi}{16}$
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$\dfrac {\pi}{3}$
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$\dfrac {\pi}{8}$

Explanation

Given,

$\begin{vmatrix}\cos 2x & \sin 2x & \sin 2x\\ \sin 2x & \cos 2x & \sin 2x\\ \sin 2x & \sin 2x & \cos 2x\end{vmatrix} = 0; x\epsilon \left [0, \dfrac {\pi}{4}\right ]$
Applying operation:

$R_{1}\rightarrow R_{1} + R_{2} + R_{3}$ , we get

$\begin{vmatrix}\cos 2x + 2\sin 2x & 2 \sin 2x + \cos 2x & 2 \sin 2x + \cos 2x \\ \sin 2x & \cos 2x & \sin 2x\\ \sin 2x & \sin 2x & \cos 2x\end{vmatrix} = 0$

$\Rightarrow (2\sin 2x + \cos 2x) \begin{vmatrix}1 & 1 & 1\\ \sin 2x & \cos 2x & \sin 2x\\ \sin 2x & \sin 2x & \cos 2x\end{vmatrix} = 0$
Now, apply operation

$C_{2}\rightarrow C_{2} - C_{1}$ and

$C_{3} \rightarrow C_{3} - C_{1}$ , we get

$(2\sin 2x + \cos 2x)$

$\begin{vmatrix}1 & 0 & 0\\ \sin 2x & \cos 2x - \sin 2x & 0\\ \sin 2x & 0 & \cos 2x - \sin 2x\end{vmatrix} = 0$
Expanding along

$R_{1}$ , we get

$(2\sin 2x + \cos 2x) (\cos 2x - \sin 2x)^{2} = 0$

$\Rightarrow 2\sin 2x + \cos 2x = 0$
or

$\cos 2x - \sin 2x = 0$

$\Rightarrow \tan 2x = -\dfrac {1}{2}$
or

$\tan 2x = 1 = \tan \dfrac {\pi}{4}$

$\because x\neq \dfrac {1}{2} \tan^{-1}\left (\dfrac {-1}{2}\right )$

$\therefore x = \dfrac {\pi}{8}\epsilon \left [0, \dfrac {\pi}{4}\right ]$ .

Consider the following statements:

$1$ . Determinant is a square matrix.

$2$ . Determinant is a number associated with a square matrix.
Which of the above statements is/are correct?

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$1$ only
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$2$ only
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Both $1$ and $2$
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Neither $1$ nor $2$

Explanation

Example :

A = $\begin{bmatrix} a&b \\ c&d \end{bmatrix}$

$det$ A = $\begin{vmatrix} a& b\\ c& d \end{vmatrix}$ =

$ad - bc.$

From the above one it is verified that Determinant is a square matrix but It is not associated with absolute value at all except that they both use vertical lines.

Hence the statements

$1$ only correct.

$A$ is

$3*3$ show symmetric matrix then

$|A| = ?: -$

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$0$
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$1$
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$-1$
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N.O.T.

The value of the determinant

$\begin{vmatrix} b^2-ab & b-c & bc-ac \\ ab -a^2 & a-b & b^2-ab \\ bc-ac & c-a & ab -a^2 \end{vmatrix}$ =

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abc
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a+b+c
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0
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ab+bc+ca

$B$ is a square matrix of order 4 such that

$|B|= 24$ ,then the value of

$|adj B|$ is equal to

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$24$
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${24}^{2}$
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${24}^{3}$
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${24}^{4}$

$a^{2},b^{2}+c^{2}+ab+bc+ca \le 0 \forall a,b,c \epsilon R$ , then value of the determinant

$\left| \begin{matrix} \left( a^{ 2 }+b^{ 2 }+c^{ 2 } \right) ^{ 2 } & a^{ 2 }+b^{ 2 } & 1 \\ 1 & (b+c+2) & b^{ 2 }+c^{ 2 } \\ c^{ 2 }+a^{ 2 } & 1 & (c+a+2)^{ 2 } \end{matrix} \right|$ equals

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$65$
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$a^{2}+b^{2}+c^{2}+31$
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$4(a^{2}+b^{2}+c^{2})$
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$0$

$A$ is a square matrix of order

$3$ such that

$A(adjA) =$

$\left [ \begin {array}{111} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end {array} \right ]$ then

$|adj A|=$

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$2$
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$4$
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$8$
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$16$

Adj

$\begin{bmatrix} 1 & 0 & 2 \\ -1 & 5 & -2 \\ 0 & 2 & 1 \end{bmatrix}=\begin{bmatrix} 9 & a & -2 \\ -1 & 1 & 0 \\ -2 & 2 & b \end{bmatrix}\Rightarrow \left[ \begin{matrix} a & b \end{matrix} \right] =$

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$\left[ \begin{matrix} -4 & 5 \end{matrix} \right]$
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$\left[ \begin{matrix} -4 & -1 \end{matrix} \right]$
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$\left[ \begin{matrix} 4 & 1 \end{matrix} \right]$
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$\left[ \begin{matrix} 4 & -1 \end{matrix} \right]$

Explanation

$\begin{array}{l} A=\left( { \begin{array} { *{ 20 }{ c } }1 & 0 & 2 \\ { -1 } & 5 & { -2 } \\ 0 & 2 & 1 \end{array} } \right) \\ \Rightarrow cofactor\, \, of\, \, A \\ \Rightarrow { C_{ 11 } }=9,\, \, { C_{ 12 } }=-1 ,\, \, { C_{ 13 } }=-2 \\\Rightarrow{C_{21}}=-4,\ {C_{22}}=1,\ {C_{23}}=2\\ \Rightarrow { C_{ 31 } }=-10,\, \, { c_{ 32 } }=0.\, \, { c_{ 33 } }=5 \\ \Rightarrow adJ\left( { \begin{array} { *{ 20 }{ c } }1 & 0 & 2 \\ { -1 } & 5 & { -2 } \\ 0 & 2 & 1 \end{array} } \right) =\left( { \begin{array} { *{ 20 }{ c } }9 & -4 & { -10 } \\ -1 & 1 & 0 \\ { -2 } & { 2 } & 5 \end{array} } \right) =\left( { \begin{array} { *{ 20 }{ c } }9 & a & { -10 } \\ -1 & 1 & 0 \\ { -2 } & { 2 } & b \end{array} } \right) \end{array}$

hence by compearing

$a = -4,\,\,b = 5$

x - 4 is factor of

$\begin{vmatrix}x - 2& 2x -3& 3x -4\\ x -4 &2x-9& 3x -16\\ x-8&2x-27&3x-64\end{vmatrix}$

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True
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False

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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