If $$A = \begin{bmatrix} a& b\\c & d\end{bmatrix}$$ then $$adj\ A = ?$$

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

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

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

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

If $$A = \begin{bmatrix}2 & 5\\ 1 & 3\end{bmatrix}$$ then $$adj\ A = ?$$

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

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

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

MCQ Questions for Class 12 Commerce Maths Determinants Quiz 9

$\left| \begin{matrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{matrix} \right|$ =

0%
$abc({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })$
0%
$abc({ a }+{ b }+{ c })$
0%
$a+b+c({ a }^{ 2 }+{ b^{ 2 } }+{ c^{ 2 } })$
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$abc\left( 1+\frac { 1 }{ a } +\dfrac { 1 }{ b } +\frac { 1 }{ c } \right)$

$A = \left[ \begin{array} { l l } { 1 } & { 2 } \\ { 2 } & { 1 } \end{array} \right]$ then

$adj (A) =?$

0%
$\left[ \begin{array} { c c } { 1 } & { - 2 } \\ { - 2 } & { 1 } \end{array} \right]$
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$\left[ \begin{array} { l l } { 2 } & { 1 } \\ { 1 } & { 1 } \end{array} \right]$
0%
$\left[ \begin{array} { c c } { 1 } & { - 2 } \\ { - 2 } & { - 1 } \end{array} \right]$
0%
$\left[ \begin{array} { c c } { - 1 } & { 2 } \\ { 2 } & { - 1 } \end{array} \right]$

Explanation

Given,

$A = \left[ \begin{array} { l l } { 1 } & { 2 } \\ { 2 } & { 1 } \end{array} \right]$

The adjoin of a square matrix of order 2 is obtained by interchanging the diagonal elements & changing signs of off diagonal elements.

adj (A) =

$A = \left[ \begin{array} { l l } { 1 } & { -2 } \\ { -2 } & { 1 } \end{array} \right]$

To solve

$x + y = 3 : 3 x - 2 y - 4 = 0$ by determinant method find

$D.$

0%
$5$
0%
$1$
0%
$-5$
0%
$-1$

Explanation

${ a }_{ 1 }x+{ b }_{ 1 }y-{ c }_{ 1 }=0\quad \Rightarrow { a }_{ 1 }x+{ b }_{ 1 }y={ c }_{ 1 }$

${ a }_{ 2 }x+{ b }_{ 2 }y-{ c }_{ 2 }=0\quad \Rightarrow { a }_{ 2 }x+{ b }_{ 2 }y={ c }_{ 2 }$

then the solution of

$x$ and

$y$ can be obtained by evaluating the following integral :

$x=\frac { \left| \underset { { c }_{ 2 } }{ { c }_{ 1 } } \quad \underset { { b }_{ 2 } }{ { b }_{ 1 } } \right| }{ \left| \underset { { a }_{ 2 } }{ { a }_{ 1 } } \quad \underset { { b }_{ 2 } }{ { b }_{ 1 } } \right| }$ and

$y=\dfrac { \left| \underset { { a }_{ 2 } }{ { a }_{ 1 } } \quad \underset { { c }_{ 2 } }{ { c }_{ 1 } } \right| }{ \left| \underset { { a }_{ 2 } }{ { a }_{ 1 } } \quad \underset { { b }_{ 2 } }{ { b }_{ 1 } } \right| }$

$\therefore$

$x+y=3$

$3x-2y=4$

can be solved using the above method

$x=\dfrac { \left| \underset { 4 }{ 3 } \quad \underset { -2 }{ 1 } \right| }{ \left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 } \right| } \quad ;\quad y=\dfrac { \left| \underset { 3 }{ 1 } \quad \underset { 4 }{ 3 } \right| }{ \left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 } \right| }$

$x=\dfrac { -6-4 }{ -2-3 } \quad ;\quad y=\dfrac { 4-9 }{ -5 }$

$x=\dfrac { 10 }{ -5 } \quad ;\quad y=\dfrac { -5 }{ -5 }$

$x=2\quad ;\quad y=1$

now the quantity

$\left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 } \right| =D$ (determinant)

$D=\left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 } \right| =-5$

So, answer is option C.

The number of distinct real roots of the equation,

$\left| \begin{matrix} cosx & sinx & sinx \\ sinx & cosx & sinx \\ sinx & sinx & cosx \end{matrix} \right| =0$ In t interval

$\left[ -\dfrac { \pi }{ 4 } \dfrac { \pi }{ 4 } \right]$ is/are:

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

If the point

$\left(\lambda+1,1\right),\left(2\lambda+1,3\right)$ and

$\left(2\lambda+2,2\lambda\right)$ are collinear then the possible value of

$\lambda$ is

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

Solve

$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix}$

0%
$1$
0%
$0$
0%
$x$
0%
$xy$

Explanation

Let

$\begin{array}{l}A= \left| { \begin{array} { *{ 20 }{ c } }1 & 1 & 1 \\ 1 & { 1+x } & 1 \\ 1 & 1 & { 1+y } \end{array} } \right| \\ { R_{ 1 } }\to { R_{ 1 } }-{ R_{ 3 } }\, \, \, \, \, \, \, \, \, \, \, { R_{ 2 } }\to { R_{ 2 } }-{ R_{ 3 } } \\ A=\left| { \begin{array} { *{ 20 }{ c } }0 & 0 & { -y } \\ 0 & x & { -y } \\ 1 & 1 & { 1+y } \end{array} } \right| \\ =-y\left( { 0-x } \right) \\ =xy \end{array}$

$(k,2-2k),(-k+1,2k),(-4-k,6-2k)$ are collinear, then

$k=$

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

Explanation

$\begin{array}{l} where\, \, more\, than\, \, three\, \, po{ { int } }s\, are\, collinear\, slope\, of\, the\, line\, are\, same \\ So, \\ slope:\, \, \left[ { \left( { k,2-2k } \right) ,\left( { -k+1,2k } \right) } \right] =\left[ { \left( { -k+1,\, 2k } \right) ,\left( { -4-k,6-2k } \right) } \right] \\ \Rightarrow \dfrac { { 2-2k-2k } }{ { k+k-1 } } =\dfrac { { 2k-6+2k } }{ { -k+1+4+k } } \\ \Rightarrow \dfrac { { 2-4k } }{ { 2k-1 } } =\dfrac { { 4x-6 } }{ 5 } \\ \Rightarrow \left( { 1-2k } \right) 5=\left( { 2k-3 } \right) \left( { 2k-1 } \right) \end{array}$

$5-10k=4k^2-2k-6k+3$

$4k^2+2k-2=0$

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

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

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

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

$k=-1\, or\, \dfrac{1}{2}$

Hence,option

$B$ is the correct answer.

${ D }_{ P }=\left| \begin{matrix} P & 15 & 8 \\ { P }^{ 2 } & 35 & 9 \\ { P }^{ 3 } & 25 & 10 \end{matrix} \right| ,$ then

${ D }_{ 1 }+{ D }_{ 2 }+{ D }_{ 3 }+{ D }_{ 4 }+{ D }_{ 5 }$ is equal to -

0%
$-29000$
0%
$-25000$
0%
$25000$
0%
none of these

Explanation

Given

$D_p=\begin{vmatrix} p & 15 & 8\\ p^2 & 35 & 9\\ p^3 & 25 & 10\end{vmatrix}$

$=p(350-225)-15(10p^2-9p^3)+8(25p^2-35p^3)$

$D_1=1(350-225)-15(10(1)^2-9(1)^3)+8(25(1)^2-35(1)^3)$

$=125-15.(1)+8(-10)$

$=125-80-15$

$=125-95$

$=30$

$D_2=2(350-225)-15(10(2)^2-9(2)^3)+8(25(2)^2-35(2)^3)$

$=2.(125)-15(40-72)+8(100-280)$

$=250+15(32)-1440$

$=730-1440$

$=-710$

$D_3=3(350-225)-15(10(3)^2-9(3)^3)+8(25(3)^2-35(3)^3)$

$=3.(125)-15(90-243)+8(225-945)$

$=375+15(153)-8(720)$

$=375+2,295-5760$

$=-3090$ .

$D_4=4(350-225)-15(10(4)^2-9(4)^3)+8(25(4)^2-35(4)^3)$

$=4.(125)-15(160-576)+8(400-2240)$

$=500+6240-8.(1840)$

$=5740-14720$

$=-8980$

$D_5=5(350-225)-15(10(5)^2-9(5)^3)+8(25(5)^2-35(5)^3)$

$=5(125)-15(250-1125)+8(625-4375)$

$=625+13125+(-30,000)$

$=-16250$ .

$\therefore D_1+D_2+D_3+D_4+D_5=30-710-3090-8980-16250$

$=-29000$ .

1191686_1333483_ans_d5df9a6197ad497ea5c09d6f6dd37567.jpg

The value of determinant

$\left| \begin{matrix} { bc-a }^{ 2 } & { ac-b }^{ 2 } & ab-c^{ 2 } \\ { ac-b }^{ 2 } & { ab-c }^{ 2 } & { bc-a }^{ 2 } \\ { ab-c }^{ 2 } & { bc-a }^{ 2 } & ac-b^{ 2 } \end{matrix} \right|$ is

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always non-negative
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always non-positive
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always zero
0%
can't say anything

$A=\begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$ , then

$\text{adj}$

$(3{ A }^{ 2 }+12A)$ is equal to

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$\begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$
0%
$\begin{bmatrix} 51 & 84 \\ 63 & 72 \end{bmatrix}$
0%
$\begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$
0%
$\begin{bmatrix} 72 & -84 \\ -63 & 51 \end{bmatrix}$

Explanation

$A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$ then

$A^2 = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} \, \begin{bmatrix} 2 & -3\\ -4 &1 \end{bmatrix}$

$A^2 = \begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix}$

$3A^2 = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix}$

$3A^2 + 12 A = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix} + \begin{bmatrix} 24 & -36 \\ -48 & 12 \end{bmatrix} = \begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$

$adj (3A^2 + 12 A) = \begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$

If a,b,c are distinct and

$\left| \begin{matrix} a & { a }^{ 2 } & { a }^{ 3 }-1 \\ b & { b }^{ 2 } & { b }^{ 3 }-1 \\ c & { c }^{ 2 } & { c }^{ 3 }-1 \end{matrix} \right| =0$ then

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$a+b+c=1$
0%
$ab+bc+ca=0$
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$a+b+c=0$
0%
$abc=1$

Explanation

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

$\Rightarrow a(b^2c^3-b^2-c^2b^3+c^2)-a^2(bc^3-b-cb^3+c)+(a^3-1)(bc^2-cb^2)=0$

$\Rightarrow a (b^2c^2(c-b)+c^2-b^2)-a^2(bc(c^2-b^2)+c-b)+(a^3-1)bc(c-b)=0$

$\Rightarrow a(b^2c^2(c-b)+(c-b)(c+b))-a^2(bc(c-b)(c+b)+(c-b))+bc(a^3-1)(c-b)=0$

$\Rightarrow a(c-b)(b^2c^2+c+b)-a^2(c-b)(bc(c+b)+1)+bc(a^3-1)(c-b)=0$

$\Rightarrow (c-b)\left \{ a(b^2c^2+c+b)-a^2(bc(c+b)+1)+bc(a^3-1) \right \}=0$

$\Rightarrow(c-b)\left \{ ab^2c^2+ac+ab-a^2bc^2-a^2b^2c-a^2+a^3bc-bc \right \}=0$

$\Rightarrow (c-b)\left \{ abc (bc-ac-ab+b^2)-(bc-ac-ab+a^2) \right \}=0$

$\Rightarrow(c-b)(bc-ac-ab+a^2)(abc-1)=0$

$\Rightarrow (c-b)(b(c-a)-a(c-a))(abc-1)=0$

$\Rightarrow (c-b)(c-a)(b-a)(abc-1)=0$

$\Rightarrow c-b=0$ or

$c-a =0$ or

$b-a =0$

$\Rightarrow c=b$

$c=a$

$b=a$

which is not possible since

$a, b, c$ are distinct.

$\therefore abc -1=0$

$\Rightarrow abc=1$

$\begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$ is the adjoint of a

$3\times 3$ matrix

$A$ and

$|A|=4$ , then

$\alpha$ is equal to:

0%
$4$
0%
$11$
0%
$5$
0%
$0$

Explanation

$\begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix} = 1 \left( 3 \times 4 - 3 \times 4 \right) - \alpha \left( 1 \times 4 - 3 \times 2 \right) + 3 \left( 1 \times 4 - 3 \times 2 \right) = 2 \alpha - 6$

$\begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix} = \begin{vmatrix} adj. \; A \end{vmatrix}$

$\Rightarrow \left( 2 \alpha - 6 \right) = \begin{vmatrix} adj. \; A \end{vmatrix}$

Now,

$\begin{vmatrix} adj. \; A \end{vmatrix} = {\left| A \right|}^{3 - 1} = {4}^{2} = 16$

$\therefore 2 \alpha - 6 = 16$

$\Rightarrow 2 \alpha = 16 + 6$

$\Rightarrow \alpha = \cfrac{22}{2} = 11$

If f(x), g(x), h(x) are polynomials in x of degree 2 and F(x)=

$\left| \begin{matrix} f & g & h \\ { f' } & g' & h' \\ f" & g" & h" \end{matrix} \right| ,$ , then F(x) is equal to

0%
1
0%
0
0%
-1
0%
$f(x).g(x).h(x)$

Explanation

$\begin{array}{l} Let, \\ f\left( x \right) =a{ x^{ 2 } }+bx+c \\ f'\left( x \right) =2ax+b \\ f''\left( x \right) =2a \\ g\left( x \right) ={ a_{ 1 } }{ x^{ 2 } }+{ b_{ 1 } }x+{ c_{ 1 } } \\ g'\left( x \right) =2{ a_{ 1 } }x+{ b_{ 1 } } \\ g''\left( x \right) =2{ a_{ 1 } } \\ and, \\ h\left( x \right) ={ a_{ 2 } }{ x^{ 2 } }+{ b_{ 2 } }x+{ c_{ 2 } } \\ h'\left( x \right) =2{ a_{ 2 } }x+{ b_{ 2 } } \\ f\left( x \right) =\left| { \begin{array} { *{ 20 }{ c } }{ a{ x^{ 2 } }+bx+c } & { { a_{ 1 } }{ x^{ 2 } }+{ b_{ 1 } }x+{ c_{ 1 } } } & { { a_{ 2 } }{ x^{ 2 } }+{ b_{ 2 } }x+{ c_{ 2 } } } \\ { 2ax+b } & { 2{ a_{ 1 } }x+{ b_{ 1 } } } & { 2{ a_{ 2 } }x+{ b_{ 2 } } } \\ { 2a } & { 2{ a_{ 1 } } } & { 2{ a_{ 2 } } } \end{array} } \right| \\ f'\left( x \right) =\left| { \begin{array} { *{ 20 }{ c } }{ 2ax+b } & { 2{ a_{ 1 } }x+{ b_{ 1 } } } & { 2{ a_{ 2 } }x+{ b_{ 2 } } } \\ { 2ax+b } & { 2{ a_{ 1 } }x+{ b_{ 1 } } } & { 2{ a_{ 2 } }x+{ b_{ 2 } } } \\ { 2a } & { 2{ a_{ 1 } } } & { 2{ a_{ 2 } } } \end{array} } \right| +0+0 \\ =0 \end{array}$

Hence, the option

$B$ is the correct answer.

$\left| \begin{array} { c c } { 1 } & { \sin x } & { \sin ^ { 2 } x } \\ { 1 } & { \cos x } & { \cos ^ { 2 } x } \\ { 1 } & { \tan x } & { \tan ^ { 2 } x } \end{array} \right| = 0 , x \in [ 0,2 \pi ] ,$ then number of possible values of

$x$ is.

0%
$4$
0%
$5$
0%
$6$
0%
None of these

Explanation

$\left| \overset { 1 }{ \underset { 1 }{ 1 } } \quad \overset { sinx }{ \underset { tanx }{ cosx } } \quad \overset { { \sin }^{ 2 }x }{ \underset { { \tan }^{ 2 }x }{ { \cos }^{ 2 }x } } \right| =0$

$\left( \underset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 } }{ { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 } } \right)$

$\Rightarrow \left| \overset { 0 }{ \underset { 1 }{ 0 } } \quad \quad \overset { sinx-cosx }{ \underset { tanx }{ cosx-tanx } } \quad \quad \overset { { \sin }^{ 2 }x-{ \cos }^{ 2 }x }{ \underset { { \tan }^{ 2 }x }{ { \cos }^{ 2 }x-{ \tan }^{ 2 }x } } \right| =0$

$\Rightarrow \underset { \left( cosx-tanx \right) }{ \left( sinx-cosx \right) } \left| \overset { 0 }{ \underset { 1 }{ 0 } } \quad \quad \overset { 1 }{ \underset { tanx }{ 1 } } \quad \quad \overset { sinx+cosx }{ \underset { { \tan }^{ 2 }x }{ cosx+tanx } } \right| =0$

Expanding along colomn I we get

$\Rightarrow \left( cosx-tanx \right) \left( sinx-cosx \right) \left( cosx+tanx-sinx-cosx \right) =0$

$\Rightarrow \left( cosx-tanx \right) \left( sinx-cosx \right) \left( tanx-sinx \right) =0$

$\therefore$

$cosx-tanx=0$ or

$sinx=cosx$ or

$tanx=sinx$

$cosx=tanx$ ;

$sinx=cosx$ ;

$tanx=sinx$

for

$x\epsilon \left[ 0,2\pi \right]$

$csox=tanx$

from graph we see there are

$2$ solutions.

for

$x\epsilon \left[ 0,2\pi \right]$

$sinx=cosx$

from graph we have two solutions

for

$x\epsilon \left[ 0,2\pi \right]$

from graph we have solutions

So a total of

$2+2+3=7$ solutions

i.e Answer : D.

1218131_1309929_ans_07d0bed09b444c1d9e5614daf4d836cf.jpg

$\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix}$ =

$(A+Bx)(x-A)^2$ ,
then the ordered pair

$(A , B)$ is equal to:

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( -4 , -5 )
0%
( -4 , 3 )
0%
( -4 , 5 )
0%
( 4 , 5 )

Explanation

$\left|\begin{matrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{matrix}\right|=$

$(x-4)[(x-4)^2-4x^2]-2x[2x(x-4)-4x^2]+2x[4x^2-2x(x-4)]$

$=(x-4)[x^2+16-8x-4x^2]-2x[2x^2-8x-4x^2]+2x[4x^2-2x^2+8x]$

$=(x-4)[16-8x-3x^2]-2x[-8x-2x^2]+2x[2x^2+8x]$

$=(x-4)(16-8x-3x^2)+4x[2x^2+8x]$

$=16x-8x^2-3x^3-64+32x+12x^2+8x^3+32x^2$

$=5x^3+36x^2+48x-64$

$(A+Bx)(x-A)^2=(A+Bx)(x^2+A^2-2xA)$

$=Ax^2+A^3-2xA^2+Bx^3+BA^2x-2ABx^2$

$=Bx^3+(A-2AB)x^2+(BA^2-2x)x+A^3$

$B=5$

$A-2AB=36$

$A(1-2B)=36$

$A(1-10)=36$

$A=-4$

$\therefore (A,B)=(-4,5)$ .

$A=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}\\$ then

$\left|A\right|=$

0%
0
0%
10
0%
12
0%
60

Explanation

$A=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}\\$

$=10\times 6-30\times 2$

$=60-60=0$

$\left|( {adj\,A}) \right| = 81,$ for

$3 \times 3$ matrix, then det

$A$ is equal to

0%
$1$
0%
$2$
0%
$4$
0%
$9$

Explanation

Given for a square matrix

$A$ of order

$3$ .

$|adjA|=|A|^{n-1}$ , where

$n$ is the order of matrix.

We have,

$|adj(A)|=|A|^{(3-1)}$

or,

$|A|^2=81$

or,

$|A|=9$ . [ Taking the positive value]

For what value of x, will the points (-1,x),(-3,2) and (-4,4) lie on a line?

0%
-3
0%
3
0%
-2
0%
2

Explanation

$A(-1, x), B(-3, 2), C(-4, 4)$ lie on a line

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

$\dfrac{1}{2}\left|-1(2-4)+-3(4-x)+-4(x-2)\right|=0$

$\dfrac{1}{2}\left|-1(-2)-3(4-x)-4(x-2)\right|=0$

$\dfrac{1}{2}\left|2-3\times 4+3x-4x+4\times 2\right|=0$

$\dfrac{1}{2}\left|2-12+3x-4x+8\right|=0$

$\dfrac{1}{2}\left|-2-x\right|=0$

$-2-x=0$

$x=-2$ .

1225369_1501943_ans_bfa8b3076f1b41179c89f03bf4102bda.jpeg

$\left| \begin{matrix} \frac { 1 }{ a } & { a }^{ 2 } & bc \\ \frac { 1 }{ b } & { b }^{ 2 } & ca \\ \frac { 1 }{ c } & { c }^{ 2 } & ab \end{matrix} \right| =$

0%
abc
0%
a+b+c
0%
0
0%
4abc

Explanation

Given,

$\begin{vmatrix}\frac{1}{a}&a^2&bc\\ \frac{1}{b}&b^2&ca\\ \frac{1}{c}&c^2&ab\end{vmatrix}$

$=\dfrac{1}{a}\det \begin{pmatrix}b^2&ca\\ c^2&ab\end{pmatrix}-a^2\det \begin{pmatrix}\frac{1}{b}&ca\\ \frac{1}{c}&ab\end{pmatrix}+bc\det \begin{pmatrix}\frac{1}{b}&b^2\\ \frac{1}{c}&c^2\end{pmatrix}$

$=\dfrac{1}{a}\left(ab^3-ac^3\right)-a^2\cdot \:0+bc\left(\dfrac{c^2}{b}-\dfrac{b^2}{c}\right)$

$=b^3-c^3-0+bc\left(\dfrac{c^2}{b}-\dfrac{b^2}{c}\right)$

$=b^3+c^3-b^3-c^3$

$=0$

The point

$(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{1}, y_{2})$ &

$(x_{2}, y_{1})$ are always

0%
Collinear
0%
Concyclic
0%
Vertices of a square
0%
Vertices of a rhombus.

Explanation

Let the coordinates be denoted as

$A(x_1,y_2)$ ,

$B(x_2,y_2)$ ,

$C(x_2,y_1)$ and

$D(x_1,y_1)$

Plot the given points on a graph as above,

It is not necessary that

$|x_2-x_1|=|y_2-y_1|$

With

$(x_2,y_1)$ and

$(x_1,y_2)$ as ends of diameter

$\angle ABC=90^{\circ}$ and

$\angle ADC=90^{\circ}$

$\therefore ABCD$ are concyclic.

So,

$\text{B}$ is the correct option.

1940934_1415967_ans_cfacbeab028b474f8369d893f1f639f2.png

Find the value of

$\begin{vmatrix} 5 & 3 \\ -7 & -4 \end{vmatrix}$

0%
-1
0%
-41
0%
41
0%
1

Explanation

$\begin{vmatrix} 5 & 3 \\ -7 & -4 \end{vmatrix}$

$= 5 \times (-4) - 3 \times (-7) = -20 + 21 = 1$
Hence, the correct answer is 1.

$a,b,c$ are

$pth$ ,

$qth$ and

$rth$ terms of a GP, then

$\begin{vmatrix} \log { a } & p & 1 \\ \log { b } & q & 1 \\ \log { c } & r & 1 \end{vmatrix}$ is equal to

0%
$0$
0%
$1$
0%
$\log { abc }$
0%
none of these

Explanation

It is given that the given terms are in G.P

$\therefore {p}^{th}$ term

$=a,{q}^{th}$ term

$=b$ and

${r}^{th}$ term

$=c$

$\Rightarrow a=A{R}^{p-1}\Rightarrow \log{a}=\log{A{\left(R\right)}^{p-1}}=\log{A}+\left(p-1\right)\log{R}$

$\Rightarrow b=A{R}^{q-1}\Rightarrow \log{b}=\log{A{\left(R\right)}^{q-1}}=\log{A}+\left(q-1\right)\log{R}$

$\Rightarrow c=A{R}^{q-1}\Rightarrow \log{c}=\log{A{\left(R\right)}^{r-1}}=\log{A}+\left(r-1\right)\log{R}$

Now,

$\Delta=\begin{vmatrix} \log{a} & p & 1 \\ \log{b} & q & 1 \\\log{c} & r & 1 \end{vmatrix}$

$=\begin{vmatrix} \log{A}+\left(p-1\right)\log{R} & p & 1 \\ \log{A}+\left(q-1\right)\log{R} & q & 1\\\log{A}+\left(r-1\right)\log{R} & r & 1 \end{vmatrix}$

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

$=\begin{vmatrix} \log{A}+\left(p-1\right)\log{R} & p-1 & 1 \\ \log{A}+\left(q-1\right)\log{R} & q-1 & 1\\\log{A}+\left(r-1\right)\log{R} & r-1 & 1 \end{vmatrix}$

${C}_{1}\rightarrow {C}_{1}-\left(\log{A}\right){C}_{3}-\left(\log{R}\right){C}_{2}$

$=\begin{vmatrix} 0 & p-1 & 1 \\0 & q-1 & 1 \\0 & r-1 & 1\end{vmatrix}$

$=0$

$|A|=6$ and

$A$ is

$3\times 3$ matrix then

$det(2\ adj(2(A^{-1})^{T}))=$

0%
$162$
0%
$36$
0%
$216$
0%
$512$

$\Delta{(x)} = \begin{vmatrix} e^x & \sin 2x & \tan x^2\\ ln(1 + x) & \cos x & \sin x\\ \cos x^2 & e^x - 1 & \sin x^2 \end{vmatrix} = A + Bx + Cx^2 + .....$ then B is equal to

0%
0
0%
1
0%
2
0%
4

$A=\begin{bmatrix} 5 & 5x & x\\ 0 & x & 5x\\ 0 & 0 & 5\end{bmatrix}$ and

$|A^2|=25$ , then

$|x|$ is equal to?

0%
$\dfrac{1}{5}$
0%
$5$
0%
$5^2$
0%
$1$

Explanation

Given

$A=\begin{bmatrix} 5 & 5x & x\\ 0 & x & 5x\\ 0 & 0 & 5\end{bmatrix}$ and

$|A^2|=25$

$|A^2|=|A|^2=25$

$|A|=5[5x-0]+0=25x$

$\Rightarrow (25x)^2=25$

$\Rightarrow x^2 =\dfrac {1}{25}$

$\Rightarrow |x|=\dfrac{1}{5}$ .

If a determinant of order

$3\times 3$ is formed by using the numbers

$1$ or

$-1$ , then the minimum value of the determinant is?

0%
$-2$
0%
$-4$
0%
$0$
0%
$-8$

$\alpha, \beta$ are the roots of

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

$\begin{vmatrix} y+1 & \beta & \alpha\\ \beta & y+\alpha & 1\\ \alpha & 1 & y+\beta\end{vmatrix}=?$

0%
$y^2-1$
0%
$y(y^2-1)$
0%
$y^2-y$
0%
$y^3$

$\begin{vmatrix} \sin ^{ 2 }{ \theta } & \cos ^{ 2 }{ \theta } \\ -\cos ^{ 2 }{ \theta } & \sin ^{ 2 }{ \theta } \end{vmatrix}=$

0%
$\cos { 2\theta }$
0%
$\cfrac { 1 }{ 2 } \left( 1+\cos ^{ 2 }{ 2\theta } \right)$
0%
$\cfrac { 1 }{ 2 } \left( 1-\sin ^{ 2 }{ 2\theta } \right)$
0%
$\cfrac { 1 }{ 2 } \sin ^{ 2 }{ 2\theta }$

Explanation

we know that

$(sin^2x+cos^2x)^2=sin^4x+cos^4x+2sin^2xcos^2x$

$\begin{vmatrix} \sin ^{ 2 }{ \theta } & \cos ^{ 2 }{ \theta } \\ -\cos ^{ 2 }{ \theta } & \sin ^{ 2 }{ \theta } \end{vmatrix}=$

$\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } =1-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta }$

$=1-\cfrac { 1 }{ 2 } \sin ^{ 2 }{ 2\theta }$

$=1-\cfrac { 1 }{ 2 } \left( 1-\cos ^{ 2 }{ 2\theta } \right)$

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

The sum of the real roots of the equation

$\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}=0$ is equal to

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

Explanation

Given

$\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}=0$

By expansion, we get

$x(-3x^2+6x)-(-6)(2x+4-3x+3)+(-1)(4x+9x)$

$\Rightarrow -5{ x }^{ 3 }+30x-30+5x=0\Rightarrow -5{ x }^{ 3 }+35x-30=0\Rightarrow { x }^{ 3 }-7x+6=0$

$\Rightarrow ({x-1})({x-2})({x-3})$

$\Rightarrow {x}=1,2,-3$

Therefore, All roots are real

So, sum of roots

$=0$

$\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix}=$ ?

0%
$(a^2+b^2-c^2-d^2)$
0%
$(a^2-b^2+c^2-d^2)$
0%
$(a^2+b^2+c^2+d^2)$
0%
$none\ of\ these$

For square matrices

$A$ and

$B$ of the same order, we have

$adj (AB) = ?$

0%
$(adj\ A)(adj\ B)$
0%
$(adj\ B)(adj\ A)$
0%
$|AB|$
0%
None of these

$\begin{vmatrix} \cos { { 70 }^{ o } } & \sin { { 20 }^{ o } } \\ \sin { { 70 }^{ o } } & \cos { { 20 }^{ o } } \end{vmatrix}=$ ?

0%
$1$
0%
$0$
0%
$\cos 50^o$
0%
$\sin 50^o$

Explanation

Let us consider

$\Delta=\begin{vmatrix} \cos { { 70 }^{ o } } & \sin { { 20 }^{ o } } \\ \sin { { 70 }^{ o } } & \cos { { 20 }^{ o } } \end{vmatrix}$

$\Delta =\cos 70^0\cos 20^0-\sin 70^0\sin 20^0$

$=\cos(70^0+20^0)$

$=\cos 90^0$

$\Delta =0$

Option

$B$ .

Evaluate :

$\begin{vmatrix} \sin { { 23 }^{ o } } & -\sin { { 7 }^{ o } } \\ \cos { { 23 }^{ o } } & \cos { { 7 }^{ o } } \end{vmatrix}$

0%
$\dfrac {\sqrt 3}{2}$
0%
$\dfrac {1}{2}$
0%
$\sin 16^o$
0%
$\cos 16^o$

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

$adj\ A = ?$

0%
$\begin{bmatrix} d& -c\\ -b & a\end{bmatrix}$
0%
$\begin{bmatrix} -d& b\\ c & -a\end{bmatrix}$
0%
$\begin{bmatrix} d& -b\\ -c & a\end{bmatrix}$
0%
$\begin{bmatrix} -d& -b\\ c & a\end{bmatrix}$

$A$ is a

$3-rowed$ square matrix and

$|A| = 5$ then

$|adj\ A| = ?$

0%
$5$
0%
$25$
0%
$125$
0%
None of these

Explanation

$|adj\ A| = |A|^{(n - 1)} = |A|^{2} = 5^{2} = 25$ .

$\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { { 15 }^{ o } } & \cos { { 15 }^{ o } } \end{vmatrix}=$ ?

0%
$1$
0%
$\dfrac {1}{2}$
0%
$\dfrac {\sqrt 3}{2}$
0%
$none\ of\ these$

Explanation

Let us consider

$\Delta=\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { { 15 }^{ o } } & \cos { { 15 }^{ o } } \end{vmatrix}=$

$\Delta=\cos^215^0-\sin^2 15^0$

$=\cos 30^0$ .........

$(\because \cos 2\theta=\cos^2 \theta-\sin^2 \theta)$

$=\dfrac{\sqrt{3}}{2}$

Hence

$\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { { 15 }^{ o } } & \cos { { 15 }^{ o } } \end{vmatrix}=\dfrac{\sqrt{3}}{2}$

Option

$C$ .

The roots of the equation

$\begin{vmatrix} 3{ x }^{ 2 } & { x }^{ 2 }+x\cos { \theta } +\cos ^{ 2 }{ \theta } & { x }^{ 2 }+x\sin { \theta } +\sin ^{ 2 }{ \theta } \\ { x }^{ 2 }+x\cos { \theta } +\cos ^{ 2 }{ \theta } & 3\cos ^{ 2 }{ \theta } & 1+\dfrac { \sin { 2\theta } }{ 2 } \\ { x }^{ 2 }+x\sin { \theta } +\sin ^{ 2 }{ \theta } & 1+\dfrac { \cos { 2\theta } }{ 2 } & 3\sin ^{ 2 }{ \theta } \end{vmatrix}=0$ are

0%
$\sin {\theta},\ \cos {\theta}$
0%
$\sin^2 {\theta},\ \cos^2 {\theta}$
0%
$\sin {\theta},\ \cos^2 {\theta}$
0%
$\sin^2 {\theta},\ \cos {\theta}$

When the determinant

$\begin{vmatrix} \cos { 2x } & \sin ^{ 2 }{ x } & \cos { 4x } \\ \sin ^{ 2 }{ x } & \cos { 2x } & \cos ^{ 2 }{ x } \\ \cos { 4x } & \cos ^{ 2 }{ x } & \cos { 2x } \end{vmatrix}$ is expanded in powers of

$\sin x$ , then the constant term in that expression is

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

Explanation

$f(x)=\begin{vmatrix} \cos { 2x } & \sin ^{ 2 }{ x } & \cos { 4x } \\ \sin ^{ 2 }{ x } & \cos { 2x } & \cos ^{ 2 }{ x } \\ \cos { 4x } & \cos ^{ 2 }{ x } & \cos { 2x } \end{vmatrix}$

the requried constant term is

$\quad f(0)=\begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}=\quad \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix}=1(0-1)=-1$

If the determinant

$\begin{vmatrix} \cos 2x & \sin ^{ 2 } x & \cos 4x \\ \sin ^{ 2 } x & \cos 2x & \cos ^{ 2 } x \\ \cos 4x & \cos ^{ 2 } x & \cos 2x \end{vmatrix}$ is expanded in powers of

$\sin x$ then the constant term in the expansion is

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

Explanation

Let

$\begin{vmatrix}\cos 2x & \sin^2 x &\cos 4x \\ \sin^2 x &\cos 2x &\cos^2x \\\cos 4x &\cos^2x &\cos 2x\end{vmatrix}= a_{1} +a_{2}\sin ^2 x +a_{3}\sin ^3 x + ....$ . . . . . . . . . . .

$(i)$

The constant term in the expansion is clearly the term

$a_{1}$ .

Putting

$x=0$ on both sides of

$(i)$ we get-

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

$\Rightarrow a_{1}=\begin{vmatrix}1 & 0 & 1\\ 0 & 1 &1 \\ 0 & 1 &0\end{vmatrix}$ (

$R_{3} \rightarrow R_{3} - R_{1}$ )

$\Rightarrow a_{1} = 1(0 - 1) = -1$ (Expanding along

$C_{1}$ )

Thus, the constant term is

$-1$ .

The correct answer is option C.

$|A| = 3$ and

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

$adj\ A = ?$

0%
$\begin{bmatrix} 9 & 3\\ -5 & -2 \end{bmatrix}$
0%
$\begin{bmatrix} 9 & -3\\ -5 & 2 \end{bmatrix}$
0%
$\begin{bmatrix} -9 & 3\\ 5 & -2 \end{bmatrix}$
0%
$\begin{bmatrix} 9 & -3\\ 5 & -2 \end{bmatrix}$

Explanation

$A^{-1} = \dfrac {1}{|A|}\cdot adj\ A \rightarrow adj\ A = |A| \cdot A^{-1} = 3A^{-1} = \begin{bmatrix} 9& -3\\ -5 & 2\end{bmatrix}$ .

$\begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & a+b & c \end{vmatrix}=0$ , then the line

$ax+by+c=0$ passes through the fixed point whcih is

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

Explanation

applying

${C}_{1}\rightarrow a{C}_{!}$ and then

${C}_{1}\rightarrow {C}_{1}+b{C}_{2}+c{C}_{3}$ and taking

$\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)$ common from

${C}_{1}$ we get

$\Delta =\cfrac { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }{ a } \begin{vmatrix} 1 & b-c & c+b \\ 1 & b & c-a \\ 1 & b+a & c \end{vmatrix}=\cfrac { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }{ a } \begin{vmatrix} 1 & b-c & c+b \\ 0 & c & -a-b \\ 0 & a+c & -b \end{vmatrix}$

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

$=\cfrac { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }{ a } (-bc+{ a }^{ 2 }+ab+ac+bc)=\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) (a+b+c)$

hence

$\Delta =0\Rightarrow a+b+c=0$

therefore line

$ax+by+c=0$ passes through the fixed point

$(1,1)$

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

$adj\ A = ?$

0%
$\begin{bmatrix}3 & -5\\ -1 & 2\end{bmatrix}$
0%
$\begin{bmatrix}3 & -1\\ -5 & 2\end{bmatrix}$
0%
$\begin{bmatrix}-1 & 2\\ 3 & -5\end{bmatrix}$
0%
None of these

If A is singular matrix , then adj A is

0%
singular
0%
non-singular
0%
symmetric
0%
not defined

Explanation

$A adj A = |A| I$

$\Rightarrow |A adj A | = |A|^n$ [ if A is of order

$n \times n ]$

$\Rightarrow |A| |adj A| = |A|^n$

$\Rightarrow |adj A| = |A|^{n-1}$

Now, A is singular .

$\therefore |A| = 0$

$\Rightarrow |adj A | = 0$

Hence adj A is singular.

$A = \left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -1/3 \end{matrix} \right]$ , then

0%
$| A | = -1$
0%
$adj A = \left[ \begin{matrix} -1 & 1 & -2 \\ 0 & -3 & -1 \\ 0 & 0 & -1/3 \end{matrix} \right]$
0%
$A = \left[ \begin{matrix} 1 & 1/3 & 7 \\ 0 & 1/3 & 1 \\ 0 & 0 & -3 \end{matrix} \right]$
0%
$A= \left[ \begin{matrix} 1 & -1/3 & -7 \\ 0 & -3 & 0 \\ 0 & 0 & 1 \end{matrix} \right]$

Explanation

$|A^{-1}| = -1 \Rightarrow |A|= -1$
Now, use

$adj A=|A|A^{-1} and A=(A^{-1})^{-1}$

There are two values of a which makes determinant

$\Delta =\left| \begin{matrix} 1\quad & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{matrix} \right| =86$ then sum of these number is

0%
$4$
0%
$5$
0%
$- 4$
0%
$9$

Explanation

we have

$\Delta =\left| \begin{matrix} 1\quad & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{matrix} \right| =86$

$\Rightarrow 1(2a^2 +4) -2(-4a -20) + 0 = 86$ [expanding along first column ]

$\Rightarrow 2a^2 +4 +8a +40 = 86$

$\Rightarrow 2a^2 + 8a +44 -86 = 0$

$\Rightarrow a^2 +4a -21 = 0$

$\Rightarrow a^2 +7a -3a-21 =0$

$\Rightarrow (a +7)(a-3) = 0$

$a = -7$ and

$3$

$\therefore$ Required sum =

$-7 +3 = -4$

$\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix}$ then value of

$x$ is

0%
$3$
0%
$\pm 3$
0%
$\pm 6$
0%
$6$

Explanation

$\because \begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix}$

$\Rightarrow 2x^2 -40 = 18 +14$

$\Rightarrow 2x^2 = 32 +40$

$\Rightarrow x^2 = \frac {72}{2} = 36$

$\therefore x = \pm 6$

Choose the correct answer from the given alternatives in the following question:
If

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} , {\text{adj}} A = \begin{bmatrix} 4 & a \\ -3 & b \end{bmatrix}$ , then the values of

$a$ and

$b$ are

0%
$a = -2 , b = 1$
0%
$a = 2 , b = 4$
0%
$a = 2 , b = -1$
0%
$a = 1 , b = -2$

Choose the correct answer from the given alternatives in the following question:
If

$A = \begin{bmatrix} 2 & -4 \\ 3 & 1 \end{bmatrix}$ , then the adjoint of matrix

$A$ is

0%
$\begin{bmatrix} -1 & 3 \\ -4 & 1 \end{bmatrix}$
0%
$\begin{bmatrix} 1 & 4 \\ -3 & 2 \end{bmatrix}$
0%
$\begin{bmatrix} 1 & 3 \\ 4 & -2 \end{bmatrix}$
0%
$\begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix}$

State whether true or false:
If the value of a third order determinant is

$12$ , then the value of determinant formed by replacing each element by its co-factor will be

$144$

0%
True
0%
False

Explanation

Let A is determinant

$\therefore |A| = 12$

Also we know that if

$A$ is a square matrix of order

$n$ then

$| adj A| = |A|^{n-1}$

For

$n = 3 , |Adj A| = |A|^{3-1} =|A|^2$

$=(12)^2 = 144$

Choose the correct answer from the given alternatives in the following question:
If

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

$A ({\text{adj}} A) = KI$ , then the value of

$k$ is

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

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 9 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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