MCQ Questions for Class 12 Commerce Maths Determinants Quiz 2

If two rows of a determinant are identical, then what is the value of the determinant ?

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0
0%
1
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-1
0%
Can be any real value.

Explanation

Let determinant of this matrix is

$x$ , if we interchange the two identical rows of the matrix then by property the determinant of the new matrix is

$-x$ , but overall the matrix will be same as we have interchanged only the two identical rows.

So,

$x=-x$ , we have

$x=0$ .

Hence, the determinant is zero.

The value of k for which

$kx+3y-k+3=0$ and

$12x+ky=k$ , have infinite solutions, is?

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

Explanation

For infinite many solution

$\displaystyle\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

$\Rightarrow \displaystyle\dfrac{k}{12}=\dfrac{3}{k}=\dfrac{-(k-3)}{-k}, \dfrac{3}{k}=\dfrac{(k-3)}{k}$

$k^2=36$

$k-6=0$

$k=\pm 6$

$k=6$ .

The number of line segments possible with three collinear points is ________.

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

Explanation

Let three collinear points be

$A , B , C$

They can represent three line segments namely,

$AB, BC, AC$ .

Thus namely

$3$ line segments are possible with three collinear points.

For positive numbers

$x, y$ and

$z$ the numerical value of the determinant

$\begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix}$ is

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$0$
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$1$
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$\log_e xyz$
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$-\log_{e} xyz$

Explanation

Let

$D=\begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix}$

$=1(1\times 1-\log_yz\times \log_zy)-\log_xy(\log_yx\times 1-\log_zx\times \log_yz)+\log_xz(\log_yx\times \log_zy-1\times \log_zx)$

Since,

$\log_ab=\dfrac{\log a}{\log b}$

So simplifying above further we get

$\log_yz\times \log_zy=\dfrac{\log z}{\log y}\times \dfrac{\log y}{\log z}=1$

$\Rightarrow D=1(1-1)-\log_xy(\log_yx-\log_yx)+\log_xz(\log_zx-\log_zx)$

$\Rightarrow D=0$

If any two adjacent rows or columns of a determinant are interchanged in position, the value of the determinant :

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Becomes zero
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Remains the same
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Changes its sign
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Is doubled

Explanation

If two rows or columns are interchanged

then the value of determinant changes its sign

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

Interchanging column

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

Similarly, interchanging rows will give

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

Hence, C is correct.

$a, b, c$ are non-zero and different from

$1$ , then the value of

$\begin{vmatrix}\log_a 1 & \log_a b & \log_ac\\ \log_a \left( \dfrac{1}{b} \right ) & \log_b 1 &\log_a \left( \dfrac{1}{c} \right ) \\ \log_a \left( \dfrac{1}{c}\right ) & \log_a c & \log_c 1\end{vmatrix}$ is

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$0$
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$1 + \log_a (a +b + c)$
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$\log_a (ab + bc + ca)$
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$1$
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$\log_a(a + b + c)$

Explanation

$\Delta =\left| \begin{matrix} \log _{ a }{ 1 } & \log _{ a }{ b } & \log _{ a }{ c } \\ \log _{ a }{ \dfrac { 1 }{ b } } & \log _{ b }{ 1 } & \log _{ a }{ \dfrac { 1 }{ c } } \\ \log _{ a }{ \dfrac { 1 }{ c } } & \log _{ a }{ c } & \log _{ c }{ 1 } \end{matrix} \right| \\ \Rightarrow \Delta =\left| \begin{matrix} 0 & \log _{ a }{ b } & \log _{ a }{ c } \\ -\log _{ a }{ b } & 0 & -\log _{ a }{ c } \\ -\log _{ a }{ c } & \log _{ a }{ c } & 0 \end{matrix} \right| \\ \Rightarrow \Delta =0-\log _{ a }{ b } \{ 0-{ (\log _{ a }{ c) } }^{ 2 }\} +\log _{ a }{ c } \{ -\log _{ a }{ b } \log _{ a }{ c } -0\} \\ \Rightarrow \Delta =\log _{ a }{ b } { (\log _{ a }{ c) } }^{ 2 }-\log _{ a }{ b } { (\log _{ a }{ c) } }^{ 2 }=0$

So, option A is correct.

The points (2, -3), (4,3) and (5, k/2) are on the same straight line. The value(s) of k is (are):

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$12$
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$-12$
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$\pm 12$
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$12$ or $6$

Explanation

Given (2, -3), (4, 3) and (5,

$\cfrac{k}{2}$ ) are on same straight line.

$\therefore \cfrac { 3-(-3) }{ 4-2 } =\cfrac { \cfrac { k }{ 2 } -3 }{ 5-4 } \Longrightarrow \cfrac { 6 }{ 2 } =\cfrac { \cfrac { k }{ 2 } -3 }{ 1 } \Longrightarrow 3=\cfrac { k }{ 2 } -3\\ \therefore k=12$

Let a be the square matrix of order 2 such that

$A^2 - 4A + 4I =0$ where I is an identify matrix of order .If

$B = A ^5 - 4A^4 + 6 A^3 + 4A^2 + A$ then Det (B) is equal to

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162
0%
$(162)^2$
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256
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$(256)^2$

Explanation

$A^2 - 4A + 4I = O \Rightarrow ( A- 2 I)^2 = O \Rightarrow A = 2 I$

$B = A (A^4 \, + \, 4A^3\, + \, 6A^2 \, + \, 4A +I) = A(A \, + \, I)^4 = 2I (3I)^4=162 I$
det

$(B) = (162)^2$

$A$ is a skew symmetric matrix, then

$\left| A \right|$ is

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$1$
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$-1$
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$0$
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none

The point

$(-a,-b),(0,0),(a,b)$ and

$(a^{2},ab)$ are-

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collinear
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concyclic
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vertices of a rectangle
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vertices of a parallelogram

Explanation

The equation of line passing through points

$\left( {0,0} \right)$ and

$\left( {a,b} \right)$ is given by

$\begin{array}{l} L:\left( { y-0 } \right) =\frac { { b-0 } }{ { a-0 } } \left( { x-0 } \right) \\ \Rightarrow y=\frac { b }{ a } x \\ \Rightarrow ay=bx \\ \Rightarrow L:bx-ay=0 \end{array}$

Putting

$\left( { - a, - b} \right)$ in

$R.H.S$

we get

$L: - ab - a\left( { - b} \right)$ Satisfied

$L$

$\left( { - a, - b} \right)$ lies on

$L$

Now Putting

$\left( { { a^{ 2 } },ab } \right)$ in

$R.H.S$

we get

$L:b\left( { { a^{ 2 } } } \right) -a\left( { ab } \right) -{ a^{ 2 } }b=0$

Therefore,

$\left( { { a^{ 2 } },ab } \right)$ Satisfies

$L$

$\left( { { a^{ 2 } },ab } \right)$ lies on

$L$

Hence The points

$\left( { 0,0 } \right) ,\left( { a,b } \right) ,\left( { -a,-b } \right) ,\left( { { a^{ 2 } },ab } \right)$ are Collinear. So the option

$(A)$ is the correct answer.

Let

$\omega\neq{1}$ be a cube root of unity and

$S$ be the set of all non-singular matrices of the form

$\begin{bmatrix} 1 & a & b \\ \omega & 1 & c \\ { \omega }^{ 2 } & \omega & 1 \end{bmatrix}$ Where each of

$a,\ b$ and

$c$ is either

$\omega$ or

${\omega}^{2}$ . Then the number of distinct matrices in the set

$S$ is

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$2$
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$6$
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$4$
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$8$

Explanation

For the given matrix to be non-singular

$1-(a+c)w+acw^2 \ne 0$

$(1-aw)(1-cw) \ne 0$

$a \ne w^2$ and

$c \ne w^2$

where

$w$ is complex cube root of unity

And

$a,b,c$ are complex cube root of unity

therefore,

$a$ and

$c$ can take two values

$w$ and

$w^2$

Hence, total numbers of distinct matrices

$=1 \times 1\times 2=2$

$A=\begin{bmatrix} 5 & 5a & a \\ 0 & a & 5a \\ 0 & 0 & 5 \end{bmatrix}$ If

$\left| A^{ 2 } \right| =25$ then

$|a|=$

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

Explanation

$A=\left[ { \begin{array} { *{ 20 }{ c } }5 & { 5a } & a \\ 0 & a & { 5a } \\ 0 & 0 & 5 \end{array} } \right]$

$\left| A \right| =\left[ { \begin{array} { *{ 20 }{ c } }5 & { 5a } & a \\ 0 & a & { 5a } \\ 0 & 0 & 5 \end{array} } \right] =25a$

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

$(25a)^2=25$

$a=\pm \frac{1}{5}$

$|a|=\frac{1}{5}$

$a\neq6,b,c$ satisfy

$\begin{vmatrix} a&2b&2c \\ 3&b&c \\ 4&a&b \end{vmatrix}=0,$ then

$abc =$

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$a+b+c$
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$0$
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$b^{3}$
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$ab +b -c$

Explanation

Expanding the Determinant we get,

$a(b^2-ac)-2b(3b-4c)+2c(3a-4b)=0\Rightarrow ab^2-a^2c-6b^2+8bc+6ca-8bc=0\Rightarrow ab^2-6b^2=a^2c-6ca\\\Rightarrow (a-6)b^2=(a-6)ca\Rightarrow b^2=ca$

Multilpying

$b$ both sides we get

$b^3=abc$

The value of (adj

$A$ ) is equal to

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$2A$
0%
$4A$
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$8A$
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$16A$

Explanation

The value of (adj A) is equal to

$2A$ .

Option

$A$ is correct answer.

Two points

$(a, 0)$ and

$(0, b)$ are joined by a straight line. Another point on this line is

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$(3a, -2b)$
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$({a}^{2}, ab)$
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$(-3a, 2b)$
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$(a, b)$

Explanation

Given that

$(a,0)$ and

$(0,b)$ lie on a straight line.

We know that a straight line is represented by

$y=mx+c$

Substituting co-ordinates in equation, we get

$(1) 0 = am + c$

$(2) b = a(0) + c = c$

$\Rightarrow m = \dfrac{-b}{a} , c = b$

$\therefore$ Equation of line is

$ay = ab - bx$

Substituting options we see that

$(3a,-2b)$ lies on this line

$\begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} =xyz(x-y)(y-z)(z-x)$

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True
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False

Explanation

Let

$A=\begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix}$

Taking

$x,y,z$ common from

$C_{1},C_{2},C_{3}$ respectively, we get,

$A=xyz\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix}$

Applying

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

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

$A=xyz\begin{vmatrix} 1 & 0 & 0 \\ x & y-x & z-x \\ x^2 & y^2-x^2 & z^2-x^2 \end{vmatrix}$

Taking

$(y-x)$ and

$(z-x)$ common from

$C_{2}$ and from

$C_{3}$

$A=xyz(y-x)(z-x)\begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ x^2 & y+x & z+x \end{vmatrix}$

Expanding along

$R_{1}$ ,

$A=xyz(y-x)(z-x)(z+x-y-x)$

$A=xyz(x-y)(y-z)(z-x)$

$\begin{vmatrix} 2^3 & 3^3 & 3.2^2+3.2+1\\ 3^3 & 4^3 & 3.3^2+3.3+1\\ 4^3 & 5^3 & 3.4^2+3.4+1\end{vmatrix}$ is equal to?

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$0$
0%
$1$
0%
$92$
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None of these

Explanation

$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }{ { 2^{ 3 } } } & { { 3^{ 3 } } } & { 3\cdot { 2^{ 2 } }+3\cdot 2+1 } \\ { { 3^{ 3 } } } & { { 4^{ 3 } } } & { 3\cdot { 3^{ 2 } }+3\cdot 3+1 } \\ { { 4^{ 3 } } } & { { 5^{ 3 } } } & { 3\cdot { 4^{ 2 } }+3\cdot 4+1 } \end{array} } \right| \\ \left| { \begin{array} { *{ 20 }{ c } }{ { 2^{ 3 } } } & { { 3^{ 3 } } } & { { 3^{ 3 } }-{ 2^{ 3 } } } \\ { { 3^{ 3 } } } & { { 4^{ 3 } } } & { { 4^{ 3 } }-{ 3^{ 3 } } } \\ { { 4^{ 3 } } } & { { 5^{ 3 } } } & { { 5^{ 3 } }-{ 4^{ 3 } } } \end{array} } \right| \\ { C_{ 3 } }\to { C_{ 3 } }+{ C_{ 1 } } \\ \left| { \begin{array} { *{ 20 }{ c } }{ { 2^{ 3 } } } & { { 3^{ 3 } } } & { { 3^{ 3 } } } \\ { { 3^{ 3 } } } & { { 4^{ 3 } } } & { { 4^{ 3 } } } \\ { { 4^{ 3 } } } & { { 5^{ 3 } } } & { { 5^{ 3 } } } \end{array} } \right| =0 \end{array}$

The determinant

$\left| {\begin{array}{*{20}{c}}a&b&{a\alpha + b}\\b&c&{b\alpha + c}\\{a\alpha + b}&{b\alpha + c}&0\end{array}} \right|$ is equal to zero, if-

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$a, b, c$ are in AP
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$a, b, c$ are in GP
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$\alpha$ is a root of the equation $a{x^2} + bx + c = 0$
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$\left( {x - \alpha } \right)$ is a factor of $a{x^2} + 2bx + c$

Explanation

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

By expanding matrix,

$a(c(0)-(b\alpha+c)(b\alpha+c))-b(b(0)-(a\alpha+b)(b\alpha+c))+(a\alpha+b)(b(b\alpha+c)-c(a\alpha+b))=0$

$(a\alpha+b)(b^2\alpha+bc-ac\alpha-bc)=a(b\alpha+c)^2-b(a\alpha+b)(b\alpha+c)$

$(a\alpha+b)(b^2\alpha-ac\alpha)=(a(b\alpha+c)-b(a\alpha+b))(b\alpha+c)$

$(a\alpha+b)((b^2-ac)\alpha)=(ab\alpha+ac-ba\alpha+b^2))(b\alpha+c)$

$(a\alpha+b)(b^2-ac)\alpha=(ac-b^2)(b\alpha+c)$

$(a\alpha+b)(-1)\alpha=(b\alpha+c)$

$-a\alpha^2-b\alpha=b\alpha+c$

$a\alpha^2+2b\alpha+c=0$

Hence ,

$(x-\alpha)$ is a factor of

$ax^2+2bx+c$

$A = \left[ \begin{array}{l}1\,\,\,\,\,\,\,\,1\\3\,\,\,\,\,\,\,4\end{array} \right]$ and

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

$'K'$

0%
$1$
0%
$-2$
0%
$10$
0%
$-10$

$D=\begin{vmatrix} 18 & 40 & 89 \\ 40 & 89 & 198 \\ 89 & 198 & 440 \end{vmatrix}=$

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

Explanation

We have,

$D=\left| \begin{matrix} 18 & 40 & 89 \\ 40 & 89 & 198 \\ 89 & 198 & 440 \end{matrix} \right|$

$D = 18\times (89\times 440 - 198^{2})- 40\times (40\times 440-198\times 89) + 89 \times(40\times 198 - 89^{2})$

$D = 18\times (39160 - 39204)- 40\times (17600-17622) + 89 \times(7920 - 7921)$

$D = 18\times (-44)- 40\times (-22) + 89 \times(-1)$

$D = -792+880 -89$

$D =-1$

Therefore, the value of the determinant is

$-1$ .

First row of the matrix

$A$ is

$\begin{bmatrix}1& 3 & 2\end{bmatrix}$ . If

$adj (A)$ =

$\begin{bmatrix}-2 & 4 & a\\ -1& 2 & 1\\ 3a& -5 &-2 \end{bmatrix}$ then a possible value of

$det(A)$ is

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

Explanation

$|A|=a_{11}A_{11}+a_{12}A_{21}+a_{13}A_{31}$

$|A|=1(-2)+3(-1)+2(3\alpha)$

$|A|=-5+6\alpha$

$adj \ A =\begin{bmatrix} -2 & 4 & \alpha \\ -1 & 2 & 2 \\ 3\alpha & -5 & -2 \end{bmatrix}$

$|adj \ A|=-2(-4+5)-4(2-3\alpha)+\alpha(5-6\alpha)$

$=-2-8+12\alpha +5\alpha - 6\alpha^2$

$|A|^2=-6\alpha^2 +17\alpha-10$

$(6\alpha-5)^2=-6\alpha^2 +17\alpha-10$

$36\alpha^2 +25-60\alpha = -6\alpha^3 +17\alpha -10$

$42 \alpha^2 -77\alpha +35=0$

$6\alpha^2-11\alpha+5=0$

$(6\alpha-5)(\alpha-1)=0$

$\alpha=\dfrac{5}{6}$ or

$\alpha=1$

$|A|=-5+6\alpha$

When,

$\alpha=\dfrac{5}{6}, |A|=0$

When,

$\alpha=1, |A|=1$

$P=\begin{bmatrix} 1 & \alpha & 3\\ 1 & 3 & 3\\ 2 & 4 & 4\end{bmatrix}$ is a

$3\times 3$ matrix A and

$|A|=4$ , then

$\alpha$ is equal to?

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

The value of determinant

$\begin{vmatrix} 19 & 6 & 7 \\ 21 & 3 & 15 \\ 28 & 11 & 6 \end{vmatrix}$ is :

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$150$
0%
$-110$
0%
$0$
0%
None of these

Explanation

$\begin{vmatrix} 19 & 6 & 7 \\ 21 & 3 & 15 \\ 28 & 11 & 6 \end{vmatrix}$

$= 19(3\times 6-11\times 15 )$

$-6(21\times 6-15\times 28)$

$+7(21\times 11-3\times 28)$

$= 19(-147)-6(-294)$

$+7(147)$

$= 0$

1083699_1134625_ans_d45ce02a062e4a99a9948c1d95605b19.jpeg

$\left| \begin{matrix} 1+i & 1-i & i \\ 1+i & i & 1+i \\ i & 1+i & 1-i \end{matrix} \right|$ (where

$i=\sqrt {-1}$ ) equals.

0%
$5i-2$
0%
$7 4i$
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$4 7i$
0%
$48i$

Explanation

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

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

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

$(1+i)(1-i) - (1+i)(1-3i) + 3i$

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

$1+1-1+3i-i+3i^{2}+3i$

$1-3+6i$

$5i-2$

$A=\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right]$ then determinant of

$[A]$ is

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

Explanation

Determinants of a

$2\times 2$ matrix is given by:

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

$Det(A)=ad-bc$

Hence using this above property:

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

$Det(A)=0\times0-1\times (-1)=0+1=1$

`If

$(8,1),(k,-4),(2,-5)$ are collinaer, then

$k=$

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

Explanation

Let the given point be

$A(8,1),B(K, - 4),C(2, - 5)$

if the above point are collinear,

they will lie on same line

i.e. they will not form triangle,

Area of

$\Delta$ ABC

$=0$

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

Here,

$\begin{array}{l} { x_{ 1 } }=8,{ y_{ 1 } }=1 \\ { x_{ 2 } }=k,{ y_{ 2 } }=-4 \\ { x_{ 3 } }=2,{ y_{ 3 } }=-5 \end{array}$

putting value,

$\begin{array}{l} \Rightarrow \dfrac { 1 }{ 2 } \left[ { 8\left( { -4+5 } \right) +k\left( { -5-1 } \right) +2\left( { 1-4 } \right) } \right] =0 \\ \Rightarrow 8\left( { -4+5 } \right) +k\left( { -5-1 } \right) +2\left( { 1+4 } \right) =0\times 2 \\ \Rightarrow -32+40-5k-k+2+8=0 \\ \Rightarrow 8+10-6k=0 \\ \Rightarrow -6k=-18 \\ \Rightarrow 6k=18 \\ \Rightarrow k=\dfrac { { 18 } }{ 6 } \\ \Rightarrow k=3 \end{array}$

hence,we get

$k=3$

so, option

$C$ is correct.

$A=\begin{bmatrix} 5 & 1 \\ 2 & 3 \end{bmatrix}$ , the determinant of matrix

$A$ is

0%
$13$
0%
$12$
0%
$17$
0%
$-13$

Explanation

Determinants of a

$2\times 2$ matrix is given by:

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

$Det(A)=ad-bc$

Hence using this above property:

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

$Det(A)=5\times3-1\times 2=15-2=13$

Find the determinant:

$\begin{vmatrix} 1 & 2 & 1 \\ 2 & 2 & 2 \\ 3 & 1 & 4 \end{vmatrix}$

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

Explanation

$\begin{array}{l} Let\, \, the\, \, er\min ant\, \, be\, \Delta \\ Now, \\ \Delta =\left| { \begin{array} { *{ 20 }{ c } }1 & 2 & 1 \\ 2 & 2 & 2 \\ 3 & 1 & 4 \end{array} } \right| \\ \Delta =1\left| { \begin{array} { *{ 20 }{ c } }2 & 2 \\ 1 & 4 \end{array} } \right| -2\left| { \begin{array} { *{ 20 }{ c } }2 & 2 \\ 3 & 4 \end{array} } \right| +1\left| { \begin{array} { *{ 20 }{ c } }2 & 2 \\ 3 & 1 \end{array} } \right| \\ \Delta =1\left( { 8-2 } \right) -2\left( { 8-6 } \right) +1\left( { 2-6 } \right) \\ \Delta =6-4-4 \\ \Delta =-2 \\ Hence, \\ option\, \, A\, \, is\, \, correct\, \, answer. \end{array}$

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

0%
$(a-b)(b-c)(c-a)$
0%
$(a+b)(c-a)$
0%
$(a+b+c)^2$
0%
$2(a+b+c)^2$

Explanation

$\begin{array}{l} Let\, \, the\, \, given\, \quad determinant\, be\, \Delta .Then \\ \Delta =\left| { \begin{array} { *{ 20 }{ c } }1 & a & { { a^{ 2 } } } \\ 1 & b & { { b^{ 2 } } } \\ 1 & c & { { c^{ 2 } } } \end{array} } \right| \\ =\left| { \begin{array} { *{ 20 }{ c } }1 & a & { { a^{ 2 } } } \\ 0 & { b-a } & { { b^{ 2 } }-{ a^{ 2 } } } \\ 0 & { c-a } & { { c^{ 2 } }-{ a^{ 2 } } } \end{array} } \right| \, \, \left[ { applying\, \, { R_{ 2 } }\to \left( { { R_{ 2 } }-{ R_{ 1 } } } \right) and\, { R_{ 3 } }\to \left( { { R_{ 3 } }-{ R_{ 1 } } } \right) } \right] \\ =\left( { b-a } \right) \left( { c-a } \right) .\left| { \begin{array} { *{ 20 }{ c } }1 & a & { { a^{ 2 } } } \\ 0 & 1 & { b-a } \\ 0 & 1 & { c+a } \end{array} } \right| \\ \left[ { taking\, \, \left( { b-a } \right) common\, \, from\, \, { R_{ 2 } }\, \, and\, \, \left( { c-a } \right) common\, \, from\, \, { R_{ 3 } } } \right] \\ =\left( { b-a } \right) \left( { c-a } \right) \times 1.\left| { \begin{array} { *{ 20 }{ c } }1 & { b+a } \\ 1 & { c+a } \end{array} } \right| \, \, \left[ { anded\, \, by\, \, { C_{ 1 } } } \right] \\ =\left( { b-a } \right) \left( { c-a } \right) \left\{ { \left( { c+a } \right) -\left( { b+a } \right) } \right\} \\ =\left( { b-a } \right) \left( { c-a } \right) \left( { c-b } \right) =\left( { a-b } \right) \left( { b-c } \right) \left( { c-a } \right) . \\ Hence, \\ \Delta =\left( { a-b } \right) \left( { b-c } \right) \left( { c-a } \right) . \end{array}$

So, option

$A$ is correct answer.

If the points

$(a, 1), (2, -1)$ and

$\left(\dfrac{1}{2}, 2\right)$ are collinear, then

$a$ is equal to:

0%
$1$
0%
$0$
0%
$2$
0%
$\dfrac{1}{4}$

Explanation

Slope of

$AB=$ slope of

$BC$

$\Rightarrow \dfrac{{ - 1 - 1}}{{2 - a}} = \dfrac{{2 - \left( { - 1} \right)}}{{\dfrac{1}{2} - 2}}$

$\Rightarrow \dfrac{{ - 2}}{{2 - a}} = \dfrac{{3 \times 2}}{{ - 3}}$

$\Rightarrow 6 = 12 - 6a$

$\Rightarrow 6a = 6$

$a=1$

Hence, option

$A$ is correct answer.

$\displaystyle A = \begin {vmatrix} \dfrac{1}{2}\left ( e^{\alpha} + e^{\alpha} \right ) & \dfrac{1}{2}\left ( e^{\alpha} e^{\alpha} \right ) \\ \dfrac{1}{2}\left ( e^{\alpha} e^{\alpha} \right ) & \dfrac{1}{2}\left ( e^{\alpha} + e^{\alpha} \right ) \end {vmatrix}$ then

$A^{1}$ exists

0%
For all real $\alpha$
0%
For positive real $\alpha$ only
0%
For negative real $\alpha$ only
0%
None of these

Explanation

$\Rightarrow \displaystyle A =\frac{1}{2} \begin {vmatrix} \left ( e^{\alpha} + e^{\alpha} \right ) & \left ( e^{\alpha} e^{\alpha} \right ) \\ \left ( e^{\alpha} e^{\alpha} \right ) & \left ( e^{\alpha} + e^{\alpha} \right ) \end {vmatrix}$

$\Rightarrow A=(e^{\alpha}+e^{\alpha})^2-(e^{\alpha}e^{\alpha})^2$

$\Rightarrow A=4e^{2\alpha}-e^{4\alpha}$

So,

$A$ is correct option.

What is the value of the determinant

$\begin{vmatrix} 1!& 2! & 3!\\ 2! & 3! & 4! \\ 3!& 4!& 5!\end{vmatrix}$

$?$

0%
$0$
0%
$12$
0%
$24$
0%
$36$

Explanation

Given,

$\begin{pmatrix}1!&2!&3!\\ 2!&3!&4!\\ 3!&4!&5!\end{pmatrix}$

$=1!\cdot \det \begin{pmatrix}3!&4!\\ 4!&5!\end{pmatrix}-2!\cdot \det \begin{pmatrix}2!&4!\\ 3!&5!\end{pmatrix}+3!\cdot \det \begin{pmatrix}2!&3!\\ 3!&4!\end{pmatrix}$

$=1!\cdot \:144-2!\cdot \:96+3!\cdot \:12$

$=144-192+72$

$=24$

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

0%
44
0%
43
0%
40
0%
39

Explanation

By property of cofactor of matrix,

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

$-7=\begin{vmatrix} 1 & -6\\ -8 & 4 \end{vmatrix}=4-48=-44$
So, cofactor of (-7)=minor of

$(-7) \times (-1)^{3+2}=-[minor \;of \; (-7)]$

$=-(-44)$

$=44$

The vectorial angle of a point

$P$ on the line joining the points

$(r_{1}, \theta _{1})$ and

$(r_{2},\theta _{2})$ is

$\dfrac{\theta_{1} +\theta _{2}}{2}$ then the length of radius vector of

$P$ is

0%
$\displaystyle \frac{r_{1} - r_{2}}{r_{1} + r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$
0%
$\displaystyle \frac{r_{1} + r_{2}}{r_{1} r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$
0%
$\displaystyle \frac{r_{1} r_{2}}{r_{1} + r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$
0%
None of these

Explanation

Given points are

$A(r_{1},\theta_{1}),B(r_{2},\theta_{2})$

Eq of line passing through two given points is given by:

$y-y_{1}=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$

$\therefore y-\theta_{1}=\cfrac{\theta_{2}-\theta_{1}}{r_{2}-r_{1}}(x-r_{1})$

Now, angle between point P and line is

$\cfrac{\theta_{1}+\theta_{2}}{2}$

Let point P meet at point Q on the line.

The radius will be

$R=PQ\cos\theta$

$\therefore R=PQ\cos(\cfrac{\theta_{1}+\theta_{2}}{2})$

If the entries in a

$3\times 3$ determinant are either

$0$ or 1, then the greatest value of their determinats is:

0%
1
0%
2
0%
3
0%
9

Explanation

Case I: If all the entries in one row are 0 , then the determinant is 0.
Case II: If all the entries in one row are 1 , then the maximum value of determinant is 1.

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

Case II:If exactly one 1 is in one row, then the maximum value is 1.

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

Case III:If exactly two 1 is in one row, then the maximum value is 2.

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

Hence, the largest value of the determinant with entries either 0 or 1 is 2.

$A+B+C= \pi$ , then

$\displaystyle \left| \begin{matrix} \tan { \left( A+B+C \right) } & \tan { B } & \tan { C } \\ \tan { (A+C) } & 0 & \tan { A } \\ \tan { (A+B) } & -\tan { A } & 0 \end{matrix} \right|$ is equal to

0%
$0$
0%
$1$
0%
$\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{A}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{B}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{C}$
0%
$-2$

Explanation

Let

$P=\begin{vmatrix} tan(A+B+C) & tan B & tan C\\ tan (A+C) & 0 & tan A\\ tan(A+B) & -tan A & 0 \end{vmatrix}$

$=\tan(A+B+C) \tan^2A-\tan B[-\tan A \tan (A+B)]+\tan C[-\tan A \tan(A+C)]$

$=\tan (A+B+C) \tan^2 A+\tan A \tan B \tan(A+B)-\tan A \tan C \tan (A+C)$

Given

$A+B+C=\pi \implies A+B=\pi-C$ or

$A+C=\pi-B$

$\therefore P=\tan (\pi) \tan^2 A + \tan A \tan B(-\tan C) -\tan A \tan C(-\tan B)$

$=0\cdot \tan^2 A - \tan A \tan B \tan C + \tan A \tan B \tan C$

$=0$

The value of

$\left|\begin{array}{ll} 2+i & 2-i\\ 1+i & 1-i \end{array}\right|$ is:

0%
$\mathrm{A}$ complex quantity
0%
real quantity
0%
$0$
0%
cannot be determined

Explanation

Let

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

$det A=(2+i)(1-i)-(2-i)(1+i)$

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

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

$\Rightarrow det A= -2i\Rightarrow A$ complex quantity.

$I\mathrm{f}\mathrm{A}=\left[\begin{array}{ll} 1 & 3\\ 2 & 1 \end{array}\right]$ , then the determinant

$\mathrm{A}^{2}-2\mathrm{A}$ :

0%
$5$
0%
$25$
0%
$-5$
0%
$-25$

Explanation

Given

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

$A^2=\begin{bmatrix} 1 & 3\\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 3\\ 2 & 1 \end{bmatrix}$
By operation of matrix (4),

$A^2=\begin{bmatrix} 7 & 6\\ 4 & 7 \end{bmatrix}$
So,

$A^2-2A=\begin{bmatrix} 7 & 6\\ 4 & 7 \end{bmatrix}-2\begin{bmatrix} 1 & 3\\ 2 & 1 \end{bmatrix}=\begin{bmatrix} 7 & 6\\ 4 & 7 \end{bmatrix}-\begin{bmatrix} 2 & 6\\ 4 & 2 \end{bmatrix}$

$=\begin{bmatrix} 5 & 0\\ 0 & 5 \end{bmatrix}$
So,

$det A^2-2A=|A^2-2A|=25$

$\left|\begin{array}{llll} 1 & & \mathrm{l}\mathrm{o}\mathrm{g}_{b}a\\ \mathrm{l}\mathrm{o}\mathrm{g}_{a}b & 1 & \end{array}\right|$ =....

0%
$ab$
0%
$\mathrm{b}\mathrm{a}$
0%
$\mathrm{a}\mathrm{b}$
0%
$0$

Explanation

Let

$A=\begin{vmatrix} 1 & \log_ba\\ \log_ab & 1 \end{vmatrix}$

Expanding the determinant,

$A=1-\log_ab \times \log_ba=1-1=0$
(because

$\log_ab=\frac{1} {\log_ba}$ )

$\left|\begin{array}{lll} a+x & a-x & a-x\\ a-x & a+x & a-x\\ a-x & a-x & a+x \end{array}\right|=0$ then the non-zero value of x=............

0%
$a$
0%
$3a$
0%
$2a$
0%
$4a$

Explanation

By operation of matrix (5),

$A=\begin{vmatrix} a+x & a-x & a-x\\ a-x & a+x & a-x\\ a-x & a-x & a+x \end{vmatrix}$

$=(a+x)[(a+x)^2-(a-x)^2]-(a-x)[a^2-x^2-(a-x)^2]+(a-x)[(a-x)^2-(a^2-x^2)]$

$=(a+x)(4ax)-(a-x)(a^2-x^2)+2(a-x)^3-(a-x)(a^2-x^2)$

$=(a+x)(4ax)+2(a-x)[a^2+x^2-2ax-a^2+x^2]$

$=(a+x)4ax+2(a-x)(2x^2-2ax)$

$A=4a^2x+4ax^2+4ax^2-4a^2x-ax^3+4ax^2$

$A=8ax^2+4ax^2-4x^3$

$A=12ax^2-4x^3$
Given,

$A= 12 ax^2-4x^3=0$

$x^2(12 a-x)=0$

$x=3a$

$\left|\begin{array}{lll} 1 & 2 & x\\ 4 & -1 & 7\\ 2 & 4 & -6 \end{array}\right|$ is a singular matrix, then

$x$ is equal to

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

Explanation

Given

$A=\begin{vmatrix} 1 & 2 & x\\ 4 & -1 & 7\\ 2 & 4 & -6 \end{vmatrix}$ is a singular matrix

So,

$detA=0$

$\Rightarrow 1(6-28)-2(-24-14)+x(16+2)=0$

$\Rightarrow -22+76+18x=0$

$\Rightarrow 18x=-54$

$\therefore x=-3$

$A=\left[\begin{array}{lll} 1^{2} & 2^{2} & 3^{2}\\ 2^{2} & 3^{2} & 4^{2}\\ 3^{2} & 4^{2} & 5^{2} \end{array}\right]$ , then the minor of

$\mathrm{a}_{22}$ is

0%
$-56$
0%
$51$
0%
$-43$
0%
$41$

Explanation

Given,

$A=\begin{bmatrix} 1^2 & 2^2 & 3^2\\ 2^2 & 3^2 & 4^2\\ 3^2 & 4^2 & 5^2 \end{bmatrix}$
So, By property of minor (1),
minor of

$a_{22}=M_{22}(-1)^{2+2}=M_{22}$

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

$=25 - 9 \times 9$

$=-56$

$\begin{vmatrix} 1 &4 &20 \\ 1 & -2& 5\\ 1 &2x & 5x^{2} \end{vmatrix}=0$ find x

0%
-1, 2
0%
0,1
0%
1, 3
0%
2, 0

Explanation

$\begin{vmatrix} 1 & 4 & 20\\ 1 & -2 & 5\\ 1 & 2x & 5x^2 \end{vmatrix}=0$

$r_1 \rightarrow r_1-r_2$ and

$r_3\rightarrow r_3-r_2$

$\begin{vmatrix} 0 & 6 & 15\\ 1 & -2 & 5\\ 0 & 2x+2 & 5x^2-5 \end{vmatrix}$

$=-(6(5x^2-5)-15(2x+2))$

$=-(30x^2-30-30x-30)$

$=30(-x^2+x+2)$
So,

$x^2-x-2=0$

$x=-1,2$

If a, b, c are all positive and not all equal then the value of the determinant

$\begin{bmatrix} a & b & c\\ b & c &a \\ c & a & b \end{bmatrix}$ is

0%
0
0%
< 0
0%
> 0
0%
cannot be determined

Explanation

Let

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

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

$|A|=\begin{vmatrix} a+b+c &b &c \\ a+b+c & c& a\\ a+b+c & a & b\end{vmatrix}$

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

$R_{1}\rightarrow R_{1}-R_{2} , R_{2}\rightarrow R_{2}-R_{3}$

$=(a+b+c)\begin{vmatrix}0 &b-a &c-a \\ 0& c-a& a-b\\ 1&a & b\end{vmatrix}$

$=(a+b+c)[(b-a)(a-b)-(c-a)^{2}]$

$\therefore |A|=-(a+b+c)[(a-b)^{2}+(c-a)^{2}]$

Now,

$(a-b)^{2}, (c-a)^{2}>0$ and

$a+b+c>0$ as

$a,b,c>0$

$\therefore |A|<0$

Hence,

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

$\left|\begin{array}{lll} \mathrm{a}+\mathrm{b} & \mathrm{a} & \mathrm{b}\\ \mathrm{a} & \mathrm{a}+\mathrm{c} & \mathrm{c}\\ \mathrm{b} & \mathrm{c} & \mathrm{b}+\mathrm{c} \end{array}\right|=$

0%
4 abc
0%
abc
0%
$2\mathrm{a}^{2}\mathrm{b}^{2}\mathrm{c}^{2}$
0%
$4\mathrm{a}^{2}\mathrm{b}^{2}\mathrm{c}^{2}$

Explanation

$\begin{vmatrix} a+b & a & b\\ a & a+c & c\\ b & c & b+c \end{vmatrix}$

$=(a+b)[(a+c)(b+c)-c^2]-a[ab+ac-bc]+b[ac-ab-bc]$

$=(a+b)[ab+ac+cb]-a^2b-a^2c+abc+abc-ab^2-b^2c$

$=a^2b+a^2c+abc+ab^2+abc+cb^2+abc+cb^2-a^2b-a^2c+abc+abc-ab^2-b^2c$

$=4abc$

Adj

$\left ( Adj\begin{bmatrix} 2 &-3 \\ 4& 6 \end{bmatrix} \right )=$

0%
$\begin{bmatrix} 2 & -3\\ 4& 6 \end{bmatrix}$
0%
$\begin{bmatrix} 6& 3\\ -4& 2 \end{bmatrix}$
0%
$\begin{bmatrix} -6& 3\\ -4& -2 \end{bmatrix}$
0%
$\begin{bmatrix} -6& -3\\ 4& -2 \end{bmatrix}$

Explanation

Adjoint of matrix is given by

$\begin{bmatrix} M11 & -M12 \\ -M21 & M22 \end{bmatrix}$

Here

$M11 = 6$

$M12 = 4$

$M21 = -3$

$M22 = 2$

So adjoint of the matrix is

$\begin{bmatrix} 6& -4 \\ 3 & 2 \end{bmatrix}$

Again adjoint of this matrix is

$\begin{bmatrix} 2&-3\\ 4&6\end{bmatrix}$

Hence the answer is option A.

$\begin{bmatrix} 3 & 0 & 0\\ 0& 3 & 0\\ 0& 0 & 3 \end{bmatrix}$ ,then Adj ( A)

0%
3A
0%
6A
0%
$9A^{T}$
0%
$2A^{T}$

Explanation

We have,

$A=\begin{bmatrix} 3 & 0 & 0\\ 0& 3 & 0\\ 0& 0 & 3 \end{bmatrix}$
co-factor matrix of

$A$ is,

$B = \begin{bmatrix} 9 & 0 & 0\\ 0& 9 & 0\\ 0& 0 & 9 \end{bmatrix}$
Thus

$\therefore$ Adj(A)

$=B^T=\begin{bmatrix} 9 & 0 & 0\\ 0& 9 & 0\\ 0& 0 & 9 \end{bmatrix}=3\begin{bmatrix} 3 & 0 & 0\\ 0& 3 & 0\\ 0& 0 & 3 \end{bmatrix}=3A$

$\left[\begin{array}{llll} \mathrm{c}\mathrm{o}\mathrm{s}\alpha+\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}\alpha & \mathrm{c}\mathrm{o}\mathrm{s}\beta+\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}\beta\\ \mathrm{s}\mathrm{i}\mathrm{n}\beta+\mathrm{i}\mathrm{c}\mathrm{o}\mathrm{s}\beta\ & \mathrm{s}\mathrm{i}\mathrm{n}\alpha+\mathrm{i}\mathrm{c}\mathrm{o}\mathrm{s}\alpha & \end{array}\right]$ is

0%
2 $\cos\alpha$
0%
2 $\sin\beta$
0%
$0$
0%
$1$

Explanation

Let

$A=\begin{bmatrix} \cos \alpha+i \sin \alpha & \cos \beta + i sin \beta\\ \sin \beta + i \cos \beta & \sin \alpha + i \cos \alpha \end{bmatrix}$

$|A|=(\cos \alpha + i \sin \alpha)(\sin \alpha+ i \cos \alpha)-(\cos \beta + i \sin \beta)(\sin \beta + i \cos \beta)$

$= \cos \alpha \sin \alpha + i - \cos \alpha \sin \alpha - [\cos \beta \sin \beta + i \sin \beta \cos \beta]$

$=0$

$\mathrm{A}$ is an unitary matrix then

$|A|$ is equal to:

0%
$1$
0%
$-1$
0%
$\pm 1$
0%
$2$

Explanation

Since, A is a unitary matrix

$AA^{*}=I$

$det A \times det A^{*}=det I$

$\Rightarrow (det A)^{2}=1$ (

$\because |A|=|\overline{A^{T}}|$ )

$\Rightarrow det A=\pm 1$

If A =

$\begin{bmatrix} 0 &1 & 2\\ 1& 2 & 3\\ 3 & 1 & 1 \end{bmatrix}$ then Adj (A) =

0%
$\begin{bmatrix}-1 &+8 & -5\\ 1& -6 & 3\\ -1 & 2 & -1\end{bmatrix}$
0%
$\begin{bmatrix}-1 &+1 & -1\\ 8& -6 & 2\\ -5 & 3 & -1\end{bmatrix}$
0%
$\begin{bmatrix}1 &-1 & 1\\ 8& -6 & 2\\ -5 & 3 & -1\end{bmatrix}$
0%
$\begin{bmatrix}-1 &-8 & 5\\ -1& 6 & -3\\ 1 & -2 & 1\end{bmatrix}$

Explanation

Cofactor of the matrix is given by

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

where

$M11$ =

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

$M12$ =

$\begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix} = 1-9 = -8$

$M13$ =

$\begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} = 1-6 = -5$

$M21$ =

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

$M22$ =

$\begin{bmatrix} 0 & 2 \\ 3 & 1 \end{bmatrix} = 0-6 = -6$

$M23$ =

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

$M31$ =

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

$M32$ =

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

$M33$ =

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

substituting all the values in the Cofactor of the matrix formula we get,

$\begin{bmatrix} -1 & 8 & -5 \\ 1& -6 & 3\\ -1 & 2 & -1\end{bmatrix}$

Hence the adjoint of the matrix is transpose of the cofactor of the given matrix

$\therefore$ adjoint of the given matrix is

$\begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}$

Hence the answer is option B.

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 2 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 2 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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