MCQ Questions for Class 12 Commerce Maths Determinants Quiz 3

Adj

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

$\begin{bmatrix} 5 & a & -2\\ 1 & 1 & 0\\ -2& -2 & b \end{bmatrix}$

$\Rightarrow [a\, \, \, \, \, b]=$

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$[-4\, \, \, \, \, \, 1]$
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$[-4\, \, \, \, \, \, -1]$
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$[4\, \, \, \, \, \, 1]$
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$[4\, \, \, \, \, \, -1]$

Explanation

Co-factor of the matrix is given by,

$\begin{bmatrix} M11 & -M12 & M13 \\ -M21 & M22 & -M23 \\ M31 & -M32 & M33 \end{bmatrix}$ ,

where

$M11$

$=$

$\begin{bmatrix} 1 & -2\\ 2 & 1 \end{bmatrix} = 1+4 = 5$

$M12$

$=$

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

$M13$

$=$

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

$M21$

$=$

$\begin{bmatrix} 0 & 2 \\ 2 & 1 \end{bmatrix} = 0-4 = -4$

$M22$

$=$

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

$M23$

$=$

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

$M31$

$=$

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

$M32$

$=$

$\begin{bmatrix} 1 & -2\\ -1 &-2 \end{bmatrix} = -2+2 = 0$

$M33$

$=$

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

Substituting all the values in the Co-factor of the matrix formula we get,

$=$

$\begin{bmatrix} 5 & 1 & -2 \\ 4 & 1 & -2\\ -2 & 0 & 1\end{bmatrix}$

The adjoint of the matrix is transpose of cofactor matrix.

$\therefore$ Adjoint of the matrix is

$\begin{bmatrix} 5 & 4 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{bmatrix}$

Here

$a =4$ and

$b =1$

Hence, the answer is option C.

A determinant of second order is made with the elements

$0$ and

$1.$ The number of determinants with non-negative values is:

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3
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10
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11
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13

Explanation

There are only three determinants of second order with negative value,

$\begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix},\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix},\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}$
Number of possible determinants with elements

$0$ and

$1$ are

${ 2 }^{ 4 }=16$
therefore, number of determinants with non-negative values is

$13.$

$A=\displaystyle \int_{1}^{sin\theta}\frac{t}{1+t^{2}}dt$ and

$B=\displaystyle \int_{1}^{cosec\theta}\frac{1}{t(1+t^{2})}dt$ , then the value of determinant

$\begin{vmatrix} A & A^{2}& B\\ e^{A+B}& B^{2} &-1 \\ 1& A^{2}+B^{2} & -1 \end{vmatrix}$ is

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$\sin\theta$
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$cosec \theta$
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$0$
0%
$1$

Explanation

$A=\displaystyle \int _{ 1 }^{ sin\theta } \frac { t }{ 1+t^{ 2 } } dt$

$\displaystyle B=\int _{ 1 }^{ cosec\theta } \frac { 1 }{ t(1+t^{ 2 }) } dt$

Put

$\displaystyle z=\frac { 1 }{ t } \quad$

$\displaystyle dz=-\frac { 1 }{ { t }^{ 2 } } dt$

$\displaystyle B=\int _{ 1 }^{ \sin { \theta } } \frac { -z }{ (z^{ 2 }+1) } dz$

$\displaystyle =\int _{ 1 }^{ \sin { \theta } } \frac { -t }{ (t^{ 2 }+1) } dt$

$\Rightarrow B=-A$

Now,

$\begin{vmatrix} A & A^{ 2 } & B \\ e^{ A+B } & B^{ 2 } & -1 \\ 1 & A^{ 2 }+B^{ 2 } & -1 \end{vmatrix}$

$=\begin{vmatrix} A & A^{ 2 } & -A \\ e^{ 0 } & A^{ 2 } & -1 \\ 1 & 2A^{ 2 } & -1 \end{vmatrix}$

$=-\begin{vmatrix} A & A^{ 2 } & A \\ 1 & A^{ 2 } & 1 \\ 1 & 2A^{ 2 } & 1 \end{vmatrix}$

$=0$ (On expanding the determinant)

I. If A,B,C are angles of angle and

$\begin{vmatrix} 1 & 1 & 1\\ 1+sinA& 1+sinB&1+sinC \\ sinA+sin^{2}A & sinB+sin^{2}B & sinC+sin^{2}C \end{vmatrix}$ =0 then triangle is isosceles
II. lf

$a=1+2+4+---$ upto

$\mathrm{n}$ terms

$b=1+3+9+---$ up to

$\mathrm{n}$ terms

$c=1+5+25+----$ up to

$\mathrm{n}$ terms then

$\Delta \begin{vmatrix} a &2b &4c \\ 2& 2& 2\\ 2^{n} & 3^{n} & 5^{n} \end{vmatrix}$ =0

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I, II both are true
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only I is true
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only II is true
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neither of them are true

Explanation

$\begin{vmatrix} 1 & 1 & 1 \\ 1+sinA & 1+sinB & 1+sinC \\ sinA+sin^{ 2 }A & sinB+sin^{ 2 }B & sinC+sin^{ 2 }C \end{vmatrix}$ =0

$\Rightarrow \sin { B } \sin ^{ 2 }{ C } -\sin { C } \sin ^{ 2 }{ B+ } \sin { C } \sin ^{ 2 }{ A } -\sin { A } \sin ^{ 2 }{ C } +\sin { A } \sin ^{ 2 }{ B } -\sin { B } \sin ^{ 2 }{ A } =0$

$\Rightarrow (\sin A-\sin B)(\sin B-\sin C)(\sin C-\sin A)=0$

$\Rightarrow (\sin A-\sin B)=0$ or

$(\sin B-\sin C)=0$ or

$(\sin C-\sin A)=0$

$\Rightarrow A=B$ or

$B=C$ or

$C=A$

$a=1+2+4+---$ upto

${ n }$ terms

$\Rightarrow a={ 2 }^{ n }-1$

$b=1+3+9+---$ upto

${ n }$ terms

$\Rightarrow\displaystyle b=\dfrac{{3}^{n}-1}{2}$

$c=1+5+25+----$ upto

${ n }$ terms

$\Rightarrow\displaystyle c=b=\dfrac { { 5 }^{ n }-1 }{ 4 }$

$\Delta =\begin{vmatrix} a & 2b & 4c \\ 2 & 2 & 2 \\ 2^{ n } & 3^{ n } & 5^{ n } \end{vmatrix}$

$=\begin{vmatrix} 2^{ n }-1 & 3^{ n }-1 & 5^{ n }-1 \\ 2 & 2 & 2 \\ 2^{ n } & 3^{ n } & 5^{ n } \end{vmatrix}$

$=\begin{vmatrix} 2^{ n } & 3^{ n } & 5^{ n } \\ 2 & 2 & 2 \\ 2^{ n } & 3^{ n } & 5^{ n } \end{vmatrix}+\begin{vmatrix} -1 & -1 & -1 \\ 2 & 2 & 2 \\ 2^{ n } & 3^{ n } & 5^{ n } \end{vmatrix}$

$=0-2\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 2^{ n } & 3^{ n } & 5^{ n } \end{vmatrix}=0$

$1\mathrm{f}\mathrm{A}=\left[\begin{array}{lll} 1 & 5 & -6\\ -8 & 0 & 4\\ 3 & -7 & 2 \end{array}\right]$ then the cofactors of the elements

$3,-7,2$ are p,q,r respectively their ascending order is

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$\mathrm{p},\ \mathrm{r}, \mathrm{q}$
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$\mathrm{q},\mathrm{r},\ \mathrm{p}$
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$\mathrm{p},\mathrm{q},\mathrm{r}$
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$\mathrm{r},\mathrm{p},\ \mathrm{q}$

Explanation

$\displaystyle A= \begin{bmatrix} 1 & 5 & -6 \\ -8 & 0 & 4 \\ 3 & -7 & 2 \end{bmatrix}$

Cofactor of

$\displaystyle 3={ c }_{ 31 }=\begin{vmatrix} 5 & -6 \\ 0 & 4 \end{vmatrix}=5\times 4-0\left( -6 \right) =20=p$

Cofactor of

$\displaystyle -7={ c }_{ 32 }=-\begin{vmatrix} 1 & -6 \\ -8 & 4 \end{vmatrix}=-(1\times 4-\left( -8 \right) \left( -6 \right)) =-(4-48)=44=q$

Cofactor of

$\displaystyle 2={ c }_{ 33 }=\begin{vmatrix} 1 & 5 \\ -8 & 0 \end{vmatrix}=1\times 0-\left( -8 \right) \left( 5 \right) =40=r$

$\therefore$ In ascending order

$p,r,q$

$f(x)=$

$\begin{vmatrix} \sin\, \, x &1 &0 \\ 1& 2\sin\, \, x& 1\\ 0& 1 & 2\sin\, \, x \end{vmatrix}$ then

$\displaystyle \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}} f\left ( x \right )$ equals

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$0$
0%
$-1$
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$1$
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$\dfrac{3\pi }{2}$

Explanation

$f(x)=\begin{vmatrix} \sin\, \, x &1 &0 \\ 1& 2\sin\, \, x& 1\\ 0& 1 & 2\sin\, \, x \end{vmatrix}$
Expanding along first column we get,

$f(x)=\sin x(4\sin^2x-1)-1(2\sin x-0)=4\sin^3x-3\sin x$
Clearly

$f(-x)=f(x)\Rightarrow$ f is an odd function
Hence

$\displaystyle \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}} f\left ( x \right )=0$

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

$\begin{vmatrix} 0 & cos\alpha & cos\beta \\ cos\alpha & 0 & cos\gamma \\ cos\beta &cos\gamma & 0 \end{vmatrix}$ then

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$cos\alpha +cos\beta +cos\gamma =0$
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$cos\alpha .cos\beta .cos\gamma =0$
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$cos^{2} \, \, \alpha +cos^{2}\, \, \beta + cos^{2}\gamma =1$
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$\sum cos \: \: \alpha \, cos\, \, \beta \, \, =0$

Explanation

$\begin{vmatrix} 1 & cos\alpha & cos\beta \\ cos\alpha & 1 & cos\gamma \\ cos\beta & cos\gamma & 1 \end{vmatrix}=\begin{vmatrix} 0 & cos\alpha & cos\beta \\ cos\alpha & 0 & cos\gamma \\ cos\beta & cos\gamma & 0 \end{vmatrix}$

$\Rightarrow 1-\cos ^{ 2 }{ \gamma - } \cos { \alpha } (\cos { \alpha } -\cos { \beta } \cos { \gamma } )+\cos { \beta } (\cos { \alpha } \cos { \gamma } -\cos { \beta } )=-cos\alpha (-\cos { \beta } \cos { \gamma } )+cos\beta (cos\alpha cos\gamma )$

$\Rightarrow 1-\cos ^{ 2 }{ \gamma - } \cos ^{ 2 }{ \alpha + } \cos { \alpha \cos { \beta \cos { \gamma } } + } \cos { \alpha \cos { \beta \cos { \gamma } } - } \cos ^{ 2 }{ \beta } =\cos { \alpha \cos { \beta \cos { \gamma } } + } \cos { \alpha \cos { \beta \cos { \gamma } } }$

$\Rightarrow \cos ^{ 2 }{ \alpha + } \cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \gamma =1 }$

Match the following elements of

$\begin{vmatrix} 1 & -1 &0 \\ 0& 4 & 2\\ 3 & -4 & 6 \end{vmatrix}$ with their cofactors and choose the correct
answer
Element Cofactor
I. -1 a) -2
II. 1 b) 32
III. 3 c) 4
IV. 6 d) 6
e) -6

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$I-b,II-d,III-a,IV-c$
0%

$I-b,II-d,III-c,IV-a$
0%

$I-d,II-b,III-a,IV-c$
0%

$I-d,II-a,III-b,IV-c$

Explanation

$\displaystyle \begin{vmatrix} 1 & -1 & 0 \\ 0 & 4 & 2 \\ 3 & -4 & 6 \end{vmatrix}$

(A) Cofactor of

$\displaystyle -1={ c }_{ 12 }=-\begin{vmatrix} 0 & 2 \\ 3 & 6 \end{vmatrix}=-\left( 0\times 6-2\times 3 \right)$

$\displaystyle =-\left( 0-6 \right) =6$

(B) Cofactor of

$\displaystyle 1={ C }_{ 11 }=\begin{vmatrix} 4 & 2 \\ -4 & 6 \end{vmatrix}=\left( 4\times 6-\left( -4 \right) \times 2 \right)$

$\displaystyle =24+8=32$

$\displaystyle 3={ c }_{ 31 }=\begin{vmatrix} -1 & 0 \\ 4 & 2 \end{vmatrix}=\left( -1\times 2+4\times 0 \right) =-2+0=-2$

(D) Cofactor of

$\displaystyle 6={ c }_{ 33 }=\begin{vmatrix} 1 & -1 \\ 0 & 4 \end{vmatrix}=\left( 1\times 4-0\left( -1 \right) \right) =4-0=4.$

${A}=\begin{bmatrix} cos\theta & sin\theta\\ -sin\theta & cos\theta \end{bmatrix}$ then

$\displaystyle \lim_{n\rightarrow\infty}\frac{1}{n}|A^{n}|=$

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$1$
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$0$
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$A$
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$\displaystyle \frac{1}{n}A$

Explanation

$A=\begin{bmatrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{bmatrix}$

${ A }^{ n }=\begin{bmatrix} \cos { n\theta } & \sin { n\theta } \\ -\sin { n\theta } & \cos { n\theta } \end{bmatrix}$

$\displaystyle \lim _{ n\rightarrow \infty }{ \cfrac { 1 }{ n } \left| { A }^{ n } \right| } =\displaystyle \lim _{ n\rightarrow \infty }{ \cfrac { 1 }{ n } \begin{bmatrix} \cos { n\theta } & \sin { n\theta } \\ -\sin { n\theta } & \cos { n\theta } \end{bmatrix} }$

$\Rightarrow \begin{bmatrix} \displaystyle \lim _{ n\rightarrow \infty }{ \cfrac { 1 }{ n } \cos { n\theta } } & \displaystyle \lim _{ n\rightarrow \infty }{ \cfrac { 1 }{ n } \sin { n\theta } } \\ -\sin { n\theta } & \cos { n\theta } \end{bmatrix}$

$=\begin{bmatrix} 0 & 0 \\ -\sin { n\theta } & \cos { n\theta } \end{bmatrix}=0$

Option B

$\mathrm{D}\mathrm{e}\mathrm{t} \left\{\begin{array}{lll} -2a & a+b & c+a\\ b+a & -2b & b+c\\ c+a & c+b & -2c \end{array}\right\}=$

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$(\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c})(\mathrm{c}+\mathrm{a})$
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$(a-b) (b-c) (c-a)$
0%
$4(\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c})(\mathrm{c}+\mathrm{a})$
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$4(a-b) (b-c) (c-a)$

Explanation

Taking

$a=b=c=1$ in the given matrix, we get

$\begin{pmatrix} -2 & 2 & 2\\ 2 & -2 & 2\\ 2 & 2 & -2 \end{pmatrix}$

$=-2(4-4)-2(-4-4)+2(4+4)$

$=-2(-8)+2(8)$

$=32=4(a+b)(b+c)(c+a)$

The sum of infinite series

$\begin{vmatrix} 1 &2 \\ 6 & 4 \end{vmatrix}+\begin{vmatrix} \frac{1}{2} &2 \\ 2& 4 \end{vmatrix}+\begin{vmatrix} \frac{1}{4} & 2\\ \frac{2}{3}& 4 \end{vmatrix}+$ ....... is

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$-10$
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$0$
0%
$10$
0%
$\infty$

Explanation

$\begin{vmatrix} 1 & 2 \\ 6 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 2 } & 2 \\ 2 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 4 } & 2 \\ \frac { 2 }{ 3 } & 4 \end{vmatrix}+.......$

$\displaystyle =4-12+2-4+1-\frac { 4 }{ 3 } +.....$

$\displaystyle=(4+2+1+...)-(12+4+\frac { 4 }{ 3 }+....)$

$\displaystyle=\frac{4}{1-\frac{1}{2}}- \frac{12}{1-\frac{1}{3}}$

$=8-18=-10$

$f(x)=\left| \begin{matrix} \sec { x } & \cos { x } \\ \cos ^{ 2 }{ x } & \cos ^{ 2 }{ x } \end{matrix} \right|$ , then

$\displaystyle \int_{0}^{\pi/2}f(x)dx=$

0%
1/2
0%
1/3
0%
0
0%
1

Explanation

$\displaystyle f(x)=\sec x \cos^2x- \cos x \times \cos^2x.$

$\displaystyle =\cos x- \cos^3x.$

$\displaystyle \int_{0}^{\pi/2}\left(cosx-cos^{3}x \right)dx=\int_{0}^{\pi/2} \cos x \sin^{2}xdx$
Substitute

$\sin x=t$

$\Rightarrow \cos x dx=dt$
The integral becomes

$\displaystyle \int_{0}^{1}t^{2}dt$

$\displaystyle =\left[ \frac{t^{3}}{3} \right]_{0}^{1}=1/3$

$a\neq b\neq c$ Then one value of

$\mathrm{x}$ which satisfies the equation

$\begin{vmatrix} 0 & x-a &x-b \\ x+a & 0 & x-c\\ x+b& x+c& 0 \end{vmatrix}$ = 0 is given by

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x = a
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x = b
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x = c
0%
x = 0

Explanation

Suppose

$x =0$

Then, the determinant becomes

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

$= 0( 0+ {c} ^2) +a(0+bc) -b(ac-0)$

$= abc -abc$

$=0$

So LHS

$=$ RHS

Hence the answer is option D

.Let

$\begin{vmatrix} x&2 & x\\ x^{2}&x & 6\\ x & x& 6 \end{vmatrix}$ =

$ax^{4}+bx^{3}+cx^{2}+dx+e$ ,then the value of 5a + 4b + 3c + 2d + e is
equal to

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

Explanation

Put x=0

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

Diff. the determinant w.r.t. x

$\begin{vmatrix}1 &0 &1 \\ x^{2}& x &6 \\ x& x &6\end{vmatrix}+\begin{vmatrix}x &2 &x \\ 2x&1 &0 \\ x& x &6 \end{vmatrix}+\begin{vmatrix}x &2 &x \\ x^{2}& x&6 \\ 1& 1 &0 \end{vmatrix}$

Put

$x=0$

$0+0+12=12=d$

Diff. again w.r.t. x

$\begin{vmatrix}0 &0 &0 \\ x^{2}& x&6 \\ x& x &6 \end{vmatrix}+\begin{vmatrix}1 &0 &1 \\ 2x& 1 &0 \\ x& x &6\end{vmatrix}+\begin{vmatrix}1 &0 &1 \\ x^{2}& x &6 \\ 1& 1 &0\end{vmatrix}+\begin{vmatrix}1 &0 &1 \\ 2x& 1 &0 \\ x& x &6\end{vmatrix}+\begin{vmatrix}x &2 &x \\ 2& 0 &0 \\ x& x &6\end{vmatrix}+\begin{vmatrix}x &2 &x \\ 2x& 1 &0 \\ 1& 1 &0\end{vmatrix}+\begin{vmatrix}1 &0 &1 \\ x^{2}& x &6 \\ 1& 1 &0\end{vmatrix}+\begin{vmatrix}x &2 &x \\ 2x& 1 &0 \\ 1& 1 &0\end{vmatrix}+\begin{vmatrix}x &2 &x \\ x^{2}& x &6 \\ 0& 0 &0\end{vmatrix}$

Put

$x=0$

$C = -12\ \ Similarly\ \ a=1\ \, b=-1$

$5-4-36+24=-11$

$\begin{vmatrix} 1+i &1-i &1 \\ 1-i& i&1+i \\ i & 1+i & 1-i \end{vmatrix}$ is a

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real number
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irrational number
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complex member
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Purely imaginary

Explanation

$1+i[i(1-i)-(1+i)^{2}]-(1-i)\left [ (1-i)^{2}-i(1+i) \right ]+1\left [ (1-i)(1+i)-i^{2} \right ]$

$=(1+i)\left [ 1-l^{2}-2i \right ]-(1-i)\left [ 1+l^{2}-2i-i-l^{2} \right ]+\left [ 1-l^{2}-l^{2} \right ]$

$=(1+i)[-2i^{2}-2i]-(1-i)(1-3i)+[1-2i^{2}]$

$=(1+i)[+2-2i]-(1-i) (1-3i)-1$

$2[1-l^2]-(1-i)(1-3i)-1$

$4-(1-3i-i+3i^2)-1$

$3-(1-4i-3)$

$3-(-2-4i)$

$=5+4i$

so (c) complex number

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

$Adj(A)=$

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$\mathrm{A}^{\mathrm{T}}$
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$3\mathrm{A}^{\mathrm{T}}$
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$\mathrm{A}^{-1}$
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$-\mathrm{A}^{\mathrm{T}}$

Explanation

$A_{11}=-3$

$A_{12}=-6$

$A_{13}=-6$
clearly B is the ans

$\Delta$ =

$\begin{vmatrix} cos\frac{\theta }{2} &1 &1 \\ 1&cos\frac{\theta }{2} &-cos\frac{\theta }{2} \\ -cos\frac{\theta }{2} &1 & -1 \end{vmatrix}$ the minimum of

$\Delta$ is

$m_{1}$ and maximum of

$\Delta$ is

$m_{2}$ then

$[m_{1} , m_{2}]$ is

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[ 4, 2]
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[2,4]
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[4,0]
0%
[0,2]

Explanation

$\displaystyle \triangle =\begin{vmatrix} \cos { \frac { \theta }{ 2 } } & 1 & 1 \\ 1 & \cos { \frac { \theta }{ 2 } } & -\cos { \frac { \theta }{ 2 } } \\ -\cos { \frac { \theta }{ 2 } } & 1 & -1 \end{vmatrix}$

Expanding along

${ R }_{ 1 },$ we get

$\displaystyle =\cos { \frac { \theta }{ 2 } } \left( -\cos { \frac { \theta }{ 2 } } +\cos { \frac { \theta }{ 2 } } \right) -1\left( -1-\cos ^{ 2 }{ \frac { \theta }{ 2 } } \right) +1\left( 1+\cos ^{ 2 }{ \frac { \theta }{ 2 } } \right)$

$\displaystyle =2\left( 1+\cos ^{ 2 }{ \frac { \theta }{ 2 } } \right) =2+2\cos ^{ 2 }{ \frac { \theta }{ 2 } }$

$\displaystyle -1\le \cos { x } \le 1\Rightarrow 0\le \cos ^{ 2 }{ x } \le 1$

$\therefore$ Max of

$\triangle$ i.e.,

$\displaystyle { m }_{ 2 }=4$ and min of

$\triangle$ i.e.,

$\displaystyle { m }_{ 1 }=2$

$f(x)=$

$\begin{vmatrix} \cos x &x &1 \\ 2 \sin x & x^{2} &2x \\ \tan x& x & 1 \end{vmatrix}$ then

$\displaystyle \lim _{ x\rightarrow 0 }{ f(x) } =$

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

Explanation

$f(x)=\begin{vmatrix} cos\, \, x & x & 1 \\ 2sin\, \, x & x^{ 2 } & 2x \\ tan\, \, x & x & 1 \end{vmatrix}$

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

$f(x)=\begin{vmatrix} cos\, \, x & x & 1 \\ 2sin\, \, x & x^{ 2 } & 2x \\ \tan { x } -\cos { x } & 0 & 0 \end{vmatrix}$

$=(\tan { x } -\cos { x } )x^{ 2 }$

Now,

$\lim _{ x\rightarrow 0 }{ f(x) }$

$=\lim _{ x\rightarrow 0 }{ (\tan { x } -\cos { x } )x^{ 2 } }$

$=0$

The value of

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

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$(x-6)x^{2}$
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$(x-6)x$
0%
$x^{2}(x+6)$
0%
$(x-6)$

Explanation

$\begin{vmatrix}1+x &2 &3 \\ 1& 2+x& 3\\ 1& 2 &3+x \end{vmatrix}$

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

$R_{2}\rightarrow R_{2}-R_{3}$

$\begin{vmatrix}x &-x &0 \\ 0& x& -x\\ 1& 2 &3+x \end{vmatrix}$

$=x(x^{2}+3x-(-2x))+1(x^{2})$ ....... [Expanding along

$C_{1}$ ]

$=x(x^{2}+5x)+x^{2}$

$=x^{3}+6x^{2}$

$=x^{2}(x+6)$

$[\mathrm{x}]$ stands greatest integer

$\leq \mathrm{x}$ then the value of

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

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

Explanation

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

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

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

$=\begin{vmatrix} -1 & 0 & 1 \\ 0 & 1 & -1 \\ 3 & 2 & 3 \end{vmatrix}$

$\Delta=-8$

$adjA=\left\{\begin{array}{lll} 1 & -1 & 0\\ 2 & 3 & 1\\ 2 & 1 & -1 \end{array}\right\}$ then adj 2

$\mathrm{A}=$

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$\left\{\begin{array}{lll} 2 & -2 & 0\\ 4 & 6 & 2\\ 4 & 2 & -2 \end{array}\right\}$
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$\left\{\begin{array}{ll} 4&-4 & 0\\ 8&12 & 4\\ 8&4 & -4 \end{array}\right\}$
0%
$\left\{\begin{array}{lll} 8 & -8 & 0\\ 16 & 24 & 8\\ 16 & 8 & -8 \end{array}\right\}$
0%
$\left\{\begin{array}{lll} 1 & -1 & 0\\ 2 & 3 & 1\\ 2 & 1 & -1 \end{array}\right\}$

Explanation

adj2A

$=\;2^2\times adjA$

$\therefore$ adj2A

$= 4\times \begin{Bmatrix} 1&-1 &0 \\2 &3 &1 \\ 2 & 1 &-1 \end{Bmatrix} = \begin{Bmatrix} 4& -4&0 \\8 &12 &4 \\8 &4 &-4 \end{Bmatrix}$

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

$B=$

$\begin{bmatrix} 2\\ 0\\ 3\end{bmatrix}$ , then

$Adj(BA)=$

0%
$\left\{\begin{array}{lll} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right\}$
0%
$\left\{\begin{array}{lll} 8 & -4 & 10\\ 0 & 0 & 0\\ 12 & -6 & 15 \end{array}\right\}$
0%
$\left\{\begin{array}{lll} 8 & 0 & 12\\ -4 & 0 & -6\\ 10 & 0 & 5 \end{array}\right\}$
0%
None of the above

Explanation

$A=\left [ 4\ -2\ \ 5 \right ]$ ,

$B=\begin{bmatrix} 2\\ 0\\ 3 \end{bmatrix}$

$BA=\begin{bmatrix} 8 & -4 & 0\\ 0& 0 &0 \\ 12& -6 &15 \end{bmatrix}$

$M_{11}, M_{12}, M_{13}$

$M_{31},M_{32},M_{33}$ are all clearly zero
and

$M_{21},M_{22}M_{23}$ also out

So, all the co factors are also zero
thus,

$Adj\left ( BA \right )=\begin{bmatrix} 0 &0 &0 \\ 0 & 0& 0\\ 0& 0 & 0 \end{bmatrix}$ (Transpose matrix of co factors)

.Let

$\Delta$

$(x)=$

$\begin{vmatrix} x+a & x+b &x+a-c \\ x+b & x+c &x-1 \\ x+c & x+d &x-b+d \end{vmatrix}$ and

$\displaystyle \int_{0}^{2}\Delta(x) dx =-16$ , where

$\mathrm{a}$ , b,c,d are in A.

$\mathrm{P}$ . then the common difference of the A.

$\mathrm{P}$ . is

0%
$\pm 1$
0%
$\pm 2$
0%
$\pm 3$
0%
$\pm 4$

Explanation

Let

$common\ diff\ = m$

$R_1\rightarrow R_1-R_2$

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

$R_3\rightarrow R_3-R_2$

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

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

Applying

$C_2\rightarrow C_2-C_1$

$\begin{vmatrix} m & 0 & -2m+1\\ x+b & m &x-1 \\ m & 0 & 2m+1 \end{vmatrix}$

$=m(2m^2+m+2m^2-m)$

$=-4m^3$

$\int_{0}^{2}\Delta=-m^3(2)=-16$

$m=\pm2$

$\begin{bmatrix} b^{2}c^{2}& bc & b+c\\ c^{2}a^{2}& ca &c+a \\ a^{2}b^{2}& ab & a+b \end{bmatrix}$ then

$\left | A \right |$ =?

0%
$abc$
0%
$abc-1$
0%
$abc+1$
0%
$0$

Explanation

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

$=b^2c^2[a^2c+abc-abc-a^2b]-bc[a^3c^2+a^2bc^2-a^2b^2c-a^3b^2]+(b+c)[a^3bc^2-a^3b^2c]$

$=a^2b^2c^3 - a^2b^3c^2 -a^3bc^3-a^2b^2c^3+a^2b^3c^2+a^3b^3c+a^3b^2c^2 + a^3bc^3-a^3b^3c-a^3b^3c-a^3b^2c^2$

$=0$

The straight lines

$\mathrm{x}+2\mathrm{y}-9=0,3\mathrm{x}+5\mathrm{y}-5=0$ and

$\mathrm{a}\mathrm{x}+\mathrm{b}\mathrm{y}-1=0$ are concurrent if the straight line

$22\mathrm{x}-35\mathrm{y}-1=0$ passes through the point

0%
(a, b)
0%
(b,a)
0%
(-a,b)
0%
(-a, -b)

Explanation

$x+2y-9=0$ ---(1)

$3x+5y-5=0$ ---(2)

$ax+by-1=0$ ---(3)

Solving (1) and (2) simultaneously we get

$y=22, x=-35$

Now, equation (1), (2) and (3) will be concurrent, that is they will pass through one point if $y=22, x=-35$ satisfy the third equation $ax+by-1=0$ .

Substituting the values of 'x' and 'y' in this equation we get $-35a+22b-1=0$ ---(4)

And another equation given is $22x-35y-1=0$ ---(5)

Equation (5) will be of the form of equation (4), if we substitute

$x=b$ & $y=a$

That is $22x-35y-1=0$ passes through $(b,a)$

If maximum and minimum values of the determinant

$\begin{vmatrix} 1+\sin ^{ 2 }{ x } & \cos ^{ 2 }{ x } & \sin { 2x } \\ \sin ^{ 2 }{ x } & 1+\cos ^{ 2 }{ x } & \sin { 2x } \\ \sin ^{ 2 }{ x } & \cos ^{ 2 }{ x } & 1+\sin { 2x } \end{vmatrix}$ are

$\alpha$ and

$\beta$ , then

0%
$\alpha +{ \beta }^{ 99 }=4$
0%
${ \alpha }^{ 3 }-{ \beta }^{ 17 }=26$
0%
$\left( { \alpha }^{ 2n }-{ \beta }^{ 2n } \right)$ is always an even integer for $n\in N$
0%
a triangle can be constructed having its sides as $\alpha -\beta ,\alpha +\beta$ and $\alpha+3\beta$

Explanation

Given

$A=\begin{vmatrix} 1+\sin ^{ 2 }{ x } & \cos ^{ 2 }{ x } & \sin { 2x } \\ \sin ^{ 2 }{ x } & 1+\cos ^{ 2 }{ x } & \sin { 2x } \\ \sin ^{ 2 }{ x } & \cos ^{ 2 }{ x } & 1+\sin { 2x } \end{vmatrix}$

Apply

${ c }_{ 1 }\rightarrow { c }_{ 1 }+{ c }_{ 2 }$ to get

$A=\begin{vmatrix} 2 & \cos ^{ 2 }{ x } & \sin { 2x } \\ 2 & 1+\cos ^{ 2 }{ x } & \sin { 2x } \\ 1 & \cos ^{ 2 }{ x } & 1+\sin { 2x } \end{vmatrix}$

Now apply

${ r }_{ 2 }\rightarrow { r }_{ 2 }-{ r }_{ 1 }$ and

${ r }_{ 3 }\rightarrow { r }_{ 3 }-{ r }_{ 1 }$

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

Since, the

$\sin { 2x }$ has maximum value of

$1$ and minimum value of

$-1$

$\alpha =3,\beta =1$

$\alpha -\beta =2,\alpha +\beta =4,\alpha =3\ , \beta =6$

Thus

$\left( \alpha -\beta \right) +\left( \alpha +\beta \right) =\alpha +3\beta =6$

$\left( \alpha -\beta \right) ,\left( \alpha +\beta \right) ,\left( \alpha +3\beta \right)$ cannot form a triangle

All other options are correct.

$\mathrm{y}=\sin \mathrm{x},\ y_{n}=\displaystyle \frac{d^{n}(\sin x)}{dx^{n}}$
then

$\begin{vmatrix} y& y_{1} & y_{2}\\ y_{3}& y_{4} &y_{5} \\ y_{6}& y_{7} & y_{8} \end{vmatrix}$ =?

0%
-sin x
0%
0
0%
sin x
0%
cos x

Explanation

Given

$y=sin x$ and

$Y_n=\dfrac{\delta ^n(sin x)}{dx^{n}}$

Let

$A=\begin{vmatrix}y& y_{1} & y_{2}\\ y_{3}& y_{4} &y_{5} \\ y_{6}& y_{7} & y_{8}\end{vmatrix}$

$\begin{vmatrix} sin x & cos x & -sin x\\ - cos x & sin x & cos x\\ -sin x & -cos x & sin x \end{vmatrix}$

$= sin x(sin^2 x + cos ^2 x)-cos x[0]-sin x[cos^2x + sin ^2 x]$

$=0$

If f (x) = tan x and A, B, C are the angles of

$\Delta ABC$ , then

$\begin{vmatrix} f(A) & f(\pi /4) & f(\pi /4)\\ f(\pi /4) &f(B) & f(\pi /4)\\ f(\pi /4) &f(\pi /4) & f(C) \end{vmatrix}$

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

Explanation

Given

$f(x)= \tan x,$ So,
Let,

$A=\begin{vmatrix} f(A) & f(\pi/4) & f(\pi/4)\\ f(\pi/4) & f(B) & f(\pi/4)\\ f(\pi/4) & f(\pi/4) & f(c) \end{vmatrix}$

$A=\begin{vmatrix} \tan A & 1 & 1\\ 1 & \tan B & 1\\ 1 & 1 & \tan c \end{vmatrix}$
So,

$A=\tan A(\tan B \tan c-1)-1(\tan c-1)+1(1-\tan B)$

$=\tan A \tan B \tan C +2 -\tan A-\tan C -\tan B$
By property of triangle

$\Delta ABC$

$\angle (A+B+C) =180 ^{\circ}$
So,

$\tan (A+B+C)=0$
So,

$\tan A \tan B \tan C=\tan A + \tan C +\tan B$

$\therefore A=2+0=2$

If the lines

$2\mathrm{x}-\mathrm{a}\mathrm{y}+1 =0$ ,

$\ 3\mathrm{x}-\mathrm{b}\mathrm{y}+1 =0$ ,

$\ 4\mathrm{x}-\mathrm{c}\mathrm{y}+1 =0$ are concurrent then

$a,b,c$ are in ?

.

0%
G.P.
0%
A.P.
0%
H.P.
0%
A.G.P.

Explanation

$2x-ay+1 =0,\ 3x-by+1 =0,\ 4x-cy+1 =0$

Given lines are concurrent

$\Rightarrow \begin{vmatrix} 2 & -a & 1 \\ 3 & -b & 1 \\ 4 & -c & 1 \end{vmatrix}=0$

$\Rightarrow 2b-a-c=0$

$\Rightarrow 2b=a+c$

Hence, a,b,c are in A.P.

$\mathrm{a}\neq b\neq \mathrm{c}$ and if

$ax+by+\mathrm{c}=0\ bx+cy+\mathrm{a}=0$ and

$cx+ay+b=0$ are concurrent,

then find the value of

$2^{\mathrm{a}^{2}b^{-1}\mathrm{c}^{-1}}2^{b^{2}\mathrm{c}^{-1}\mathrm{a}^{-1}}2^{\mathrm{c}^{2}\mathrm{a}^{-1}b^{-1}}$

0%
1
0%
4
0%
8
0%
16

Explanation

Give,

$ax + by + c = 0$

$bx + cy + a = 0$

$cx + ay + b = 0$ are concurrent.

i.e., their determinant is equal to zero.

$\Rightarrow \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}=0$

$\Rightarrow a(bc-a^2)-b(b^2-ac) + c(ab-c^2)=0$

$\Rightarrow abc - a^3 - b^3 + abc + abc - c^3 = 0$

$\Rightarrow 3abc - (a^3+b^3+c^3)=0$

$\Rightarrow a^3+b^3+c^3 = 3abc$ ....(1)

We need to find

$2^{a^2b^{-1}c^{-1}} \cdot 2^{b^2c^{-1}a^{-1}} \cdot 2^{c^2a^{-1}b^{-1}}$

$\Rightarrow 2^{(a^2b^{-1}c^{-1}+b^2c^{-1}a^{-1} + c^2a^{-1}b^{-1})}$

$= \Rightarrow 2^{\left( \dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab}\right)}$

$= 2^{\left(\dfrac{a^3 + b^3+c^3}{abc}\right)}$

$= 2 ^{\left(\dfrac{3abc}{abc}\right)}$ [From (1)]

$=2^3 = 8$

lf the lines

$3\mathrm{x}+2\mathrm{y}-5=0,\ 2\mathrm{x}-5\mathrm{y}+3=0,\ 5\mathrm{x}+\mathrm{b}\mathrm{y}+\mathrm{c}=0$ are concurrent then

$\mathrm{b}+\mathrm{c}=$

0%
7
0%
-5
0%
6
0%
9

Explanation

We know that concurrent lines have same intersection points.Then

$\therefore 3x+2y-5=0$ --(1)

$2x-5y+3=0$ ---(2)

$5x+by+c=0$ ---(3)

From 1 & 2,

$6x+4y-10=0$

$6x-15y+9=0$

$\Rightarrow 19y-19=0$

$y=1$ & $x=1$

So, $substituting\ \ x=1$ & $y=1$ , in equation (3),

$5x+by+c=0$

$5+b+c=0$

$\therefore b+c=-5$

$\alpha,\ \beta$ are the roots of

$\begin{vmatrix} x & 1 & 2\\ 0 & 1& 1\\ 1 & x & 2 \end{vmatrix}$ = 0 then

$\alpha^{n}+\beta^{n}=?$

0%
0
0%
1
0%
2
0%
2n

Explanation

$\begin{vmatrix} x & 1 &2 \\ 0 & 1 & 1\\ 1 & x & 2 \end{vmatrix}=0$
So,

$x(2-x)-1(-1)+2(-1)=0$

$2x-x^2-1=0$

$x^2-2x+1=0$

$x=1$
So, Given that

$\alpha , \beta$ are roots of

$x^2-2x+1=0$
So,

$\alpha=\beta=1$
So,

$\alpha^{n} + \beta^{n}=2$

If the points

$A (1, 2), O (0, 0)$ and

$C (a, b)$ are collinear, then

0%
$a=b$
0%
$a=2b$
0%
$2a=b$
0%
$a=-b$

Explanation

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

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

${\text{Given points are A}}(1,2),{\text{O}}(0,0)\;{\text{and C}}(a,b)$

${\text{Therefore,}}$

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

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

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

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

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

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

$\Rightarrow \dfrac{1}{2}\left[ {1\left( {0 - b} \right) + 0\left( {b - 2} \right) + {\text{a}}\left( {2 - 0} \right)} \right] = 0$

$\Rightarrow - {\text{b + 2a}} = 0$

$\Rightarrow {\text{2a = b}}$

${\textbf{Hence, the correct answer is option C}}{\text{.}}$

Relation between

$x$ and

$y$ , if the points

$(x, y), (1, 2)$ and

$(7, 0)$ are collinear is _____

0%
$x+3y=7$
0%
$x+3y=14$
0%
$-x+3y=7$
0%
$-x+3y=14$

Explanation

For the points $AB =(x, y), BC =(1,2)$ and $CD=(7,0)$ to be collinear they must have area $=0$ .

Thus, area of a triangle ABC = $\frac { 1 }{ 2 } \left( x_{ 1 }\left( y_{ 2 }-y_{ 3 } \right) +x_{ 2 }\left( y_{ 3 }-y_{ 1 } \right) +x_{ 3 }\left( y_{ 1 }-y_{ 2 } \right) \right)= 0$

$=\dfrac{1}{2}(2x\times 2+6y-14)=0$

$=(2x+6y-14)=0$

Thus, we have the relation $x+3y=7$ for the points to be collinear.

Points

$(1, 5), (2, 3)$ and

$(-2, -11)$ are ____

0%
Non-collinear
0%
Collinear
0%
Vertices of equilateral triangle
0%
Vertices of right angle triangle

Explanation

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.

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-

0%
(5, 11/2)
0%
(11, 5/2)
0%
(5, -11/2)
0%
(-5, 11/2)

Explanation

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 $\dfrac{3+7}{2},\dfrac{4+7}{2}=(5,11/2)$ .

Find the value of

$m$ if the points

$(5, 1), (2, 3)$ and

$(8, 2m )$ are collinear.

0%
${-1}$
0%
$\dfrac{1}{2}$
0%
$-\dfrac{1}{2}$
0%
${2}$

Explanation

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)

$= \dfrac{1}{2}$ [

${15-10m +4m-2+(-16)}]$

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

$15-18-6m$ }

$= \dfrac{1}{2}$

$(-3-6m)=0$

$\Rightarrow (3+6m)=0$

$m=-\dfrac{1}{2}$ .

$a\neq p, b\neq q, c\neq r$ 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

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

Explanation

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

$R_1\rightarrow R_1-R_3$

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

$R_2\rightarrow R_2-R_3$

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

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

$R_3\rightarrow R_1+R_3$

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

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

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

$\dfrac {p}{p-a}=\dfrac {c}{c-r}-\dfrac {b}{q-b}$ substituting gets value

$=-2$

If the points

$(0, 0), (1, 2)$ and

$(x, y)$ are collinear, then

0%
$x = y$
0%
$2x = y$
0%
$x =2 y$
0%
$2x = -y$

Explanation

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

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

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

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

${\text{Therefore,}}$

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

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

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

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

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

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

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

$\Rightarrow y - 2x = 0$

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

${\textbf{Hence, the correct answer is option B}}{\text{.}}$

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

0%
$0$
0%
$12 cos^2x -10 sin^2x$
0%
$12 sin^2x -10 cos^2x - 2$
0%
$10 sin 2x$

Explanation

$\begin{vmatrix}sin^2x & cos^2x & 1\\ cos^2 x & sin^2 x & 1\\ -10 & 12 &2 \end{vmatrix}$

$C_1\rightarrow C_1+C_2$

$=\begin{vmatrix} 1& cos^2x & 1\\ 1 & sin^2x & 1\\ 2 & 12 & 2\end{vmatrix}=0 \ (\because C_1=C_3)$

$\mathrm{If}\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=-2$ and

$\mathrm{f}(\mathrm{x})=$

$\begin{vmatrix}1+a^{2}x &(1+b^{2})x &(1+c^{2})x \\(1+a^{2})x&1+b^{2}x &(1+c^{2})x\\(1+a^{2})x&(1+b^{2})x & 1+c^2x\end{vmatrix}$ then

$f(x)$ is a polynomial of degree

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

Explanation

Applying

$C_{1}\rightarrow C_{1}+ C_{2}+C_{3}$

$\begin{vmatrix} 1+(a^{2}+b^{2}+c^{2}+2)x&(1+b^{2})x &(1+c^{2})x \\ 1+(a^{2}+b^{2}+c^{2}+2)x& 1+b^{2}x & (1+c^{2})x\\ 1+(a^{2}+b^{2}+c^{2}+2)x& (1+b^{2})x & 1+c^{2}x \end{vmatrix}$ ,

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

$(\because a^{2}+b^{2}+c^{2}+2=0)$

Applying

$R_{1}\rightarrow R_{1}-R_{2}$ ;

$R_{2}\rightarrow R_{2}-R_{3}$

$F(x) = \begin{vmatrix} 0 & x-1 &0 \\ 0& 1-x & x-1\\ 1& (1+B^{2})x & 1+C^{2}x \end{vmatrix}$ ;

$\Rightarrow F(x) = (x-1)^{2}$
Hence degree

$= 2.$

Find the values of

$k,$ if the points

$A (k + 1, 2k), B (3k, 2k + 3)$ and

$C (5k +1, 5k)$ are collinear.

0%
$1$
0%
$\dfrac{1}{2}$
0%
$2$
0%
$2.5$

Explanation

Let us assume that the points

$A(x_{1}, y_{1}) =(k+1, 2k), B(x_{2}, y_{2})=(3k, 2k+3) \ \& \ C(x_{3}, y_{3}) =(5k+1, 5k)$ form a triangle.

Now,

$area(\Delta ABC)$ with vertices

$A(x_{1}, y_{1}), B(x_{2}, y_{2} )$ and

$C(x_{3}, y_{3})$ is given by

$Area(\Delta ABC)=\dfrac{1}{2}\left[x_{1}\left(y_{2}-y_{3} \right) +x_{2}\left(y_{3}-y_{1} \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right]$

$\therefore Area(\Delta ABC)=\dfrac { 1 }{ 2 } \left[ \left( k+1 \right) \left( 2k+3-5k \right) +3k\left( 5k-2k \right) +\left( 5k+1 \right) \left( 2k-2k-3 \right) \right]$

$\therefore Area(\Delta ABC)=2{ k }^{ 2 }-5k+2$

Since,

$A,B,C$ are collinear

$\implies Area(\Delta ABC)=0$

$\implies 2k^2-5k+2=0$

$\implies (k-2)(2k-1)=0$

$\implies k=2, \dfrac{1}{2}$

If A is a square matrix, then

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

0%
2 |A|
0%
2 |A| I
0%
null matrix
0%
unit matrix

Explanation

$adj A^{T} - (adj A)^T$

$=|A^{T}|(A^{T})^{-1} -(|A|{A}^{-1})^{T}$

$=|A|^{ T }(A^{ -1 })^{ T }-(A^{ -1 })^{ T }|A|^{ T }$

$=|A|^{T}((A^{-1})^{T}-(A^{ -1 })^{ T })$

$=O$

$B=A$ ,

$|P|=|Q|=1$ , then

$adj({ Q }^{ -1 }{ BP }^{ -1) }$ is

0%
$P.Q$
0%
$Q.adj(A).P$
0%
$P.adj(A).Q$
0%
$P.{ A }^{ -1 }.Q$

Explanation

Given,

$B=A, |P|=|Q|=1$

Consider,

$adj({ Q }^{ -1 }{ BP }^{ -1) }$

$=(adj {P^{-1}}) (adj B) (adj Q^{-1}) (\because adj (ABC)=(adjC)(adj B)(adj A)$

$\displaystyle =\frac{P}{|P|} (adj B) \frac{Q}{|Q|} (\because adj P^{-1}=\frac{P}{|P|})$

$\displaystyle =P (adj A)Q$ (

$\because A=B$ )

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

0%
$\begin{bmatrix}-3 & -2\\ -1 & -5\end{bmatrix}$
0%
$\begin{bmatrix}3 & -2\\ -1 & 5\end{bmatrix}$
0%
$\begin{bmatrix}5 & 1\\ 2 & 3\end{bmatrix}$
0%
$\begin{bmatrix}3 & 2\\ 1 & 5\end{bmatrix}$

Explanation

Given

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

$C=\begin{bmatrix} -3 & -1 \\ -2 & -5 \end{bmatrix}$

$adj A= C^{T}=\begin{bmatrix} -3 & -2 \\ -1 & -5 \end{bmatrix}$

$P$ is a non singular matrix, then value of

$adj({ P }^{ -1 })$ in terms of

$P$ is

0%
${ P }/{ |P| }$
0%
${ P }{ |P| }$
0%
$P$
0%
none of these

Explanation

Given P is non-singular

$|P|\ne 0$

Now,

$adj({ P }^{ -1 })=(adj P)^{-1}$

$=(|P|P^{-1})^{-1}$

$=(P^{-1})^{-1}|P|^{-1}$

$=P/|P|$

The adjoint of

$\begin{bmatrix}1 & 1 & 1\\ 1 & 2 & -3\\ 2 & -1 & 3\end{bmatrix}$ is

0%
$\begin{bmatrix}3 & -9 & -5\\ -4 & 1 & 3\\ -5 & 4 & 1\end{bmatrix}$
0%
$\begin{bmatrix}3 & -4 & -5\\ -9 & 1 & 4\\ -5 & 3 & 1\end{bmatrix}$
0%
$\begin{bmatrix}-3 & 4 & 5\\ 9 & -1 & -4\\ 5 & -3 & -1\end{bmatrix}$
0%
None of these

Explanation

Given,

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

$C=\begin{bmatrix} 3 & -9 & -5 \\ -4 & 1 & 3 \\ -5 & 4 & 1 \end{bmatrix}$

$adj A=C^{T}=\begin{bmatrix}3 & -4 & -5\\ -9 & 1 & 4\\ -5 & 3 & 1\end{bmatrix}$

$(3, 2)$ ,

$\left (x, \dfrac {22}{5}\right), (8, 8)$ lie on a line, then

$x$ is equal to

0%
$-5$
0%
$2$
0%
$4$
0%
$5$

Explanation

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

$= \dfrac { 1 }{ 2 } \left[ 3\left( \dfrac { 22 }{ 5 } -8 \right) +x\left( 8-2 \right) +8\left( 2-\dfrac { 22 }{ 5 } \right) \right] =0$

$\Rightarrow -\dfrac { 54 }{ 5 } +6x-\dfrac { 96 }{ 5 } =0$

$\Rightarrow 30x=150$

$\Rightarrow x=5$

Given the matrix

$A=\begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{bmatrix}$ . If

$xyz=60$ and

$8x+4y+3z=20$ , then

$A(adj\ A)$ is equal to

0%
$A=\begin{bmatrix} 64 & 0 & 0 \\ 0 & 64 & 0 \\ 0 & 0 & 64 \end{bmatrix}$
0%
$A=\begin{bmatrix} 88 & 0 & 0 \\ 0 & 88 & 0 \\ 0 & 0 & 88 \end{bmatrix}$
0%
$A=\begin{bmatrix} 68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68 \end{bmatrix}$
0%
$A=\begin{bmatrix} 34 & 0 & 0 \\ 0 & 34 & 0 \\ 0 & 0 & 34 \end{bmatrix}$

Explanation

Given,

$A=\begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{bmatrix}$

$|A|=xyz-(8x+4y+3z)+28$

$\Rightarrow |A|=68$

Now,

$A(adj A)=|A|I$

$=68I$

$=\begin{bmatrix} 68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68 \end{bmatrix}$

If the points

$(0, 4), (4, 0)$ and

$(5, p)$ are collinear, then value of

$p$ is

0%
$- 1$
0%
$7$
0%
$6$
0%
$4$

Explanation

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.

$\therefore ar\Delta ABC=0$

$\Rightarrow \dfrac { 1 }{ 2 } \left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { 1 } \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix} \right| =0$

$\Rightarrow \left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { 1 } \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix} \right| =0$

$\Rightarrow \left| \begin{matrix} 0 & 4 & 1 \\ 4 & 0 & 1 \\ 5 & p & 1 \end{matrix} \right| =0$

$\Rightarrow 0\left( 0-p \right) +4\left( p-4 \right) +5\left( 4-0 \right) =0$

$\Rightarrow p=-1$

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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