If $$A=\begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$$, then $$A.adj(a)=$$

$$\begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix}$$

$$\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$

$$\begin{bmatrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{bmatrix}$$

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

MCQ Questions for Class 12 Commerce Maths Determinants Quiz 11

$D_r = \begin{vmatrix} r & x & n(n+1)/2 \\ 2r-1 & y & n^{ 2 } \\ 3r-2 & z & n(3n-1)/2 \end{vmatrix}$ , then

$\sum^n_{r \times 1} \, D_r$ is equal to

0%
$\dfrac{1}{6} n (n + 1)(2n + 1)$
0%
$\dfrac{1}{4} n^2 (n + 1)^2$
0%
$0$
0%
None of these

$\Delta = \left| \matrix{ 1 + {a^2} + {a^4}\;\;1 + ab + {a^2}{b^2}\;\;1 + ac + {a^2}{c^2} \hfill \cr 1 + ab + {a^2}{b^2}\;\;1 + {b^2} + {b^4}\;\;1 + bc + {b^2}{c^2} \hfill \cr 1 + ac + {a^2}{c^2}\;\;1 + bc + {b^2}{c^2}\;\;1 + {c^2} + {c^4} \hfill \cr} \right|is\;equal\;to$

0%
${\left( {a + b + c} \right)^6}$
0%
${\left( {a - b} \right)^2}{\left( {b - c} \right)^2}{\left( {c - a} \right)^2}$
0%
4 (a-b)(b-c)(c-a)
0%
None of these

If in the determinant

$\Delta =\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3\end{vmatrix}$ ,

$A_1, B_1, C_1$ etc., be the co-factors of

$a_1, b_1, c_1$ etc., then which of the following relations is incorrect?

0%
$a_1A_1+b_1B_1+c_1C_1=\Delta$
0%
$a_2A_2+b_2B_2+c_2C_2=\Delta$
0%
$a_3A_3+b_3B_3+c_3C_3=\Delta$
0%
$a_1A_2+b_1B_2+c_1C_2=\Delta$

Explanation

$\textbf{Expand determinant in terms of the elements of any row.}$

$\text{Given,}$

$\Rightarrow \Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$

$\Rightarrow A_1,B_1,C_1\text{ etc. be the co-factors of }a_1,b_1,c_1\text{ etc.}$

$\text{Now,}$

$\text{Sum of the product of element of a row with their co-factor is equal to the value of the determinant.}$

$\text{So,}$

$\text{For }1st\text{ row, } a_1A_1+b_1B_1+c_1C_1 = \Delta$

$\text{Similarly, for }2nd\text{ and }3rd\text{ row, it will be}$

$\Rightarrow a_2A_2+b_2B_2+c_2C_2 = \Delta$

$\Rightarrow a_3A_3+b_3B_3+c_3C_3 = \Delta$

$\textbf{Therefore, the incorrect relation is }\boldsymbol{a_1A_2+b_1B_2+c_1C_2 = \Delta.}$

$\textbf{Hence, the correct option is (D).}$

Let

$A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \, and \, U_1, U_2, U_3$ be column matrices satisfying

$AU_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , AU_2 = \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix} , AU_3 = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$ . If U is

$3 \times 3$ matrix, whose columns are

$U_1, U_2, U_3$ . then |U| is

0%
$-11$
0%
$-3$
0%
$\dfrac{3}{2}$
0%
$2$

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

0%
$xyz\left( \dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } \right)$
0%
$xyz$
0%
$1+\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z }$
0%
$\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z }$

$\triangle_{r}=\begin{vmatrix} { 2 }^{ r-1 } & 2.{ 3 }^{ r-1 } & 4.{ 5 }^{ r-1 } \\ x & y & z \\ { 2 }^{ n-1 } & { 3 }^{ n-1 } & { 5 }^{ n-1 } \end{vmatrix}$ , then

$\sum_{r=1}^{n}(\triangle_{r})$ is equal to

0%
$xyz$
0%
$1$
0%
$-1$
0%
$0$

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

0%
$(a-1)^2$
0%
$(a-1)^3$
0%
$(a-1)^4$
0%
$2(a-1)$

Let k be a positive real number and let
A =

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

$\begin{bmatrix} 0 & 2k - 1 & \sqrt{ k} \\ 1- 2\sqrt{ k} & 0 & -2k \\ 2\sqrt{k} & -\sqrt{k} & 0 \end{bmatrix}$
If

$det(Adj(A)) + det(Adj(B))$ = 2 then [k] is equal to

0%
4
0%
6
0%
0
0%
1

$\theta \varepsilon R,$ then the determinant

$\Delta =\begin{vmatrix} \sin { \theta } & \cos { \theta } & \sin { 2\theta } \\ \sin { \left( \theta +\dfrac { 2\pi }{ 3 } \right) } & \cos { \left( \theta +\dfrac { 2\pi }{ 3 } \right) } & \sin { \left( 2\theta +\dfrac { 4\pi }{ 3 } \right) } \\ \sin { \left( \theta -\dfrac { 2\pi }{ 3 } \right) } & \cos { \left( \theta -\dfrac { 2\pi }{ 3 } \right) } & \sin { \left( 2\theta -\dfrac { 4\pi }{ 3 } \right) } \end{vmatrix}=$

0%
$-\sin {\theta} - \cos {\theta}$
0%
$\sin {2\theta}$
0%
$1+\sin {2\theta} - \cos {2\theta}$
0%
$None\ of\ these$

$\theta \epsilon R$ , then the

$determinant$

$\Delta$

$\begin{vmatrix} \sin { \theta } & \cos { \theta } & \sin { 2\theta } \\ \sin { \left( \theta +\cfrac { 2\pi }{ 3 } \right) } & \cos { \left( \theta +\cfrac { 2\pi }{ 3 } \right) } & \sin { \left( 2\theta +\cfrac { 2\pi }{ 3 } \right) } \\ \sin { \left( \theta -\cfrac { 2\pi }{ 3 } \right) } & \cos { \left( \theta -\cfrac { 2\pi }{ 3 } \right) } & \sin { \left( 2\theta -\cfrac { 2\pi }{ 3 } \right) } \end{vmatrix}$ =

0%
$-\sin { \theta } -\cos { \theta }$
0%
$\sin { 2\theta }$
0%
$1+\sin { 2\theta } -\cos { 2\theta }$
0%
None of these

Solve

$\Delta=\begin{vmatrix} \sqrt { 13 } +\sqrt { 3 } & 2\sqrt { 5 } & \sqrt { 5 } \\ \sqrt { 15 } +\sqrt { 26 } & 5 & \sqrt { 10 } \\ 3+\sqrt { 65 } & \sqrt { 15 } & 5 \end{vmatrix}=$

0%
$15\sqrt { 2 } -25\sqrt { 3 }$
0%
$25\sqrt { 3 } -15\sqrt { 2 }$
0%
$3\sqrt { 5 }$
0%
$-15\sqrt { 2 } +7\sqrt { 3 }$

Let

$\left[ \begin{matrix} \cos ^{ -1 }{ x } & \cos ^{ -1 }{ y } & \cos ^{ -1 }{ z } \\ \cos ^{ -1 }{ y } & \cos ^{ -1 }{ z } & \cos ^{ -1 }{ x } \\ \cos ^{ -1 }{ z } & \cos ^{ -1 }{ x } & \cos ^{ -1 }{ y } \end{matrix} \right]$ such that

$|A|=0$ , then maximum value of

$x+y+z$ is

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

Matrix

$A = \left| {\begin{array}{*{20}{c}}x & 3 & 2\\1 & y & 4\\2 & 2 & z\end{array}} \right|$ , if

$xyz=60$ and

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

$a(adjA)$ is equal to

0%
$\left| {\begin{array}{*{20}{c}} {64} & 0 & 0\\ 0 & {64} & 0\\ 0 & 0 & {64} \end{array}} \right|$
0%
$\left| {\begin{array}{*{20}{c}} {88} & 0 & 0\\ 0 & {88} & 0\\ 0 & 0 & {88} \end{array}} \right|$
0%
$\left| {\begin{array}{*{20}{c}} {68} & 0 & 0\\ 0 & {68} & 0\\ 0 & 0 & {68} \end{array}} \right|$
0%
$\left| {\begin{array}{*{20}{c}} {34} & 0 & 0\\ 0 & {34} & 0\\ 0 & 0 & {34} \end{array}} \right|$

The number of distinct values of a

$2 \times 2$ determinant whose entries are from set

$\{-1, 0, 1\}$ is

0%
$4$
0%
$6$
0%
$5$
0%
$3$

$f\left( x \right) = \left| {\begin{array}{*{20}{c}}{x - 2}&{{{\left( {x - 1} \right)}^2}}&{{x^3}}\\{x - 1}&{{x^2}}&{{{\left( {x + 1} \right)}^3}}\\x&{{{\left( {x + 1} \right)}^2}}&{{{\left( {x + 2} \right)}^3}}\end{array}} \right|$

0%
$0$
0%
$2$
0%
$-2$
0%
None of these

$\left( \omega \neq 1 \right)$ is a cubic root of unity then

$\left| \begin{matrix} 1 & 1+i+{ \omega }^{ 2 } & { { \omega } }^{ 2 } \\ 1-i & -1 & { \omega }^{ 2 }-1 \\ -i & -1+\omega -i & -1 \end{matrix} \right|$ equals-

0%
$0$
0%
$1$
0%
$i$
0%
$\omega$

If A is a square matrix of order 3, then

$\left| Adj(Adj{ A }^{ 2 }) \right| =$

0%
${ \left| A \right| }^{ 2 }$
0%
${ \left| A \right| }^{ 4 }$
0%
${ \left| A \right| }^{ 8 }$
0%
${ \left| A \right| }^{ 16 }$

$1, \omega, \omega^{2}$ are the roots of unity then

$\triangle =\left| \begin{matrix} 1 & { \omega }^{ n } & { \omega }^{ 2n } \\ { \omega }^{ n } & { \omega }^{ 2n } & 1 \\ { \omega }^{ 2n } & 1 & { \omega }^{ n } \end{matrix} \right|$ is equal to-

0%
$0$
0%
$1$
0%
$\omega$
0%
$\omega ^{2}$

$|A|$ denotes the value of the determinant of the square matrix

$A$ order

$3$ , then

$|-2A|=$

0%
$-8|A|$
0%
$8|A|$
0%
$-2|A|$
0%
None of these

$f(x) = \begin{vmatrix}2\cos x & 1 & 0\\ x - \dfrac {\pi}{2} & 2\cos x & 1\\ 0 & 1 & 2\cos x\end{vmatrix}\Rightarrow f'(x) =$

0%
$0$
0%
$2$
0%
$\pi/2$
0%
$\pi - 6$

State whether the statement is true/false.

$\mathbf { A } ( \mathbf { x } )$

$= \left[ \begin{array} { c c c } { \cos x } & { - \sin x } & { 0 } \\ { \sin x } & { \cos x } & { 0 } \\ { 0 } & { 0 } & { 1 } \end{array} \right]$ , then adj

$[ \mathrm { A } ( \mathrm { x } ) ] = \mathrm { A } ( - \mathrm { x } )$ .

0%
True
0%
False

$a + b + c = 0$ one root of

$\left| {\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}} \right|$ =0 is

0%
$x = 1$
0%
$x = 2$
0%
$x = {a^2} + {b^2} + {c^2}$
0%
$x = 0$

If a matrix

$\begin{bmatrix} { \left( x-a \right) }^{ 2 } & { \left( x-b \right) }^{ 2 } & { \left( x-c \right) }^{ 2 } \\ { \left( y-a \right) }^{ 2 } & { \left( y-b \right) }^{ 2 } & { \left( y-c \right) }^{ 2 } \\ { \left( z-a \right) }^{ 2 } & { \left( z-b \right) }^{ 2 } & { \left( z-c \right) }^{ 2 } \end{bmatrix}$ is a zero matrix, then

$a,b,c,x,y,z$ are connected by:

0%
$a+b+c=0,x+y+z=0$
0%
$a+b+c=0,x=y=z$
0%
$a=b=c,x+y+z=0$
0%
None of these

$\left| \begin{array} { r r r } { x } & { 2 } & { x } \\ { x ^ { 2 } } & { x } & { 0 } \\ { x } & { x } & { 8 } \end{array} \right|$ =

$A x ^ { 4 } + B x ^ { 3 } + c x ^ { 2 } + D x + E$ , then the value of

$5 A + 4 B + 2 C + 2 D + E$ is equal to

0%
$-11$
0%
$17$
0%
$-17$
0%
0

The maximum and minimum values of

$(3\times 3)$ determinant whose elements belong to

$\left\{ 0,1,2,3 \right\}$ is

0%
$\pm 9$
0%
$\pm 15$
0%
$\pm 54$
0%
$\pm 32$

The value of

$\begin{vmatrix} 1 & 1 & 1 \\ { \left( { 2 }^{ x }+{ 2 }^{ -x } \right) }^{ 2 } & { \left( { 3 }^{ x }+{ 3 }^{ -x } \right) }^{ 2 } & { \left( { 5 }^{ x }+{ 5 }^{ -x } \right) }^{ 2 } \\ { \left( { 2 }^{ x }-{ 2 }^{ -x } \right) }^{ 2 } & { \left( { 3 }^{ x }-{ 3 }^{ -x } \right) }^{ 2 } & { \left( { 5 }^{ x }-{ 5 }^{ -x } \right) }^{ 2 } \end{vmatrix}$ is equal to

0%
$0$
0%
$30^{x}$
0%
$30^{-x}$
0%
$None\ of\ these$

If the points

$A(x, 2), B(-3, -4)$ and

$C(7, -5)$ are collinear, then the value of

$x$ is :

0%
$-63$
0%
$63$
0%
$60$
0%
$-60$

Explanation

Given points are

$A(x,2)\: B(-3,-4) \: C(7, -5)$

$(x_{1},y_{1})=(x,2),(x_{2},y_{2})= (-3,-4)(x_{3},y_{3})=(7,-5)$

$x_{1}[y_{2}-y_{3}]+x_{2}[y_{3}-y_{1}]+x_{3}[y_{1}-y_{2}]=0$

$x[(-4)-(-5)]+(-3)[(-5)-2]+7[2-(-4)]=0$

$x(-4+5)-3(-5-2)+7(2+4)=0$

$x(1)-3(-7)+7(6)=0$

$x+21+42=0$

$x+63=0$

$x=-63$

$\therefore$ The value of x is

$-63$ .

The determinant

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

0%
$a,b,c$ are in A.P.
0%
$a,b,c$ are in G.P.
0%
$a,b,c$ are in H.P.
0%
None of these

The determinant

$\Delta=\left| \begin{matrix} { a }^{ 2 }\left( 1+x \right) & ab & ac \\ ab & { b }^{ 2 }\left( 1+x \right) & bc \\ ac & bc & { c }^{ 2 }\left( 1+x \right) \end{matrix} \right|$ is divisible by

0%
$1+x$
0%
$(1+x)^{2}$
0%
$x^{2}$
0%
$none\ of\ these$

$A=\begin{bmatrix} 2 & 1 & -1 \\ 0 & 1 & 4 \\ 0 & 0 & 3 \end{bmatrix}$ , then

$tr(adj(adj\ A))$ is equal to

0%
$18$
0%
$24$
0%
$36$
0%
$48$

Which of the following is/are true ?
(i) Adjoint of a symmetric matrix is symmetric
(ii) Adjoint of a unit matrix is a unit matrix
(iii) A(adj A)=(adj A) A= [A]f and
(iv) Adjoint of a diagonal matrix is a diagonal matrix

0%
$(i)$
0%
$(ii)$
0%
$(iii) and (iv)$
0%
$None$ $of$ $these$

$A=\begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix}$ , then the value of

$|A||adj A|$ is

0%
$a^{9}$
0%
$a^{5}$
0%
$a^{6}$
0%
$a^{27}$

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

$det(adj(adj A))$

0%
$(14)^{4}$
0%
$(14)^{3}$
0%
$(14)^{2}$
0%
$(14)^{1}$

$A=\begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix}$ and A (adj A)=KI, then the value of 'K' is ...

0%
2
0%
-2
0%
10
0%
-10

Let

$P\left( x \right) =\begin{vmatrix} x & -3+4i & 3-4i \\ x & -7i & 5+6i \\ x & 7-2i & -7-2i \end{vmatrix}$
The number of values of x for which

$P\left( x \right) =0$ is

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

Let

$f\left( x \right) = {\sin ^{ - 1}}\left( {\tan x} \right) + {\cos ^{ - 1}}\left( {\cot x} \right)$ then

0%
$f(x)= \frac {\pi}{2}$ wherever defined
0%
domain of $f(x)$ is $x= n \pi \pm \frac {\pi}{4}, n \in 1$
0%
period $f(x)$ is $\frac {\pi}{2}$
0%
$f(x)$ in many one function

$A = \begin{bmatrix} 4 & 2 \\ 3 & 4 \end{bmatrix}$ then |adj A| is equal to

0%
16
0%
10
0%
6
0%
none of these

Let

$A$ be a non-singular matrix of order

$n$ nad

$\left|A\right|=K$ , then

$\left(adj A\right)^{-1}$ is

0%
$\dfrac{A}{K}$
0%
$K^{n-1}\left(adj A\right)$
0%
$K^{n-2}A$
0%
$KA$

Which of the following values of

$\alpha$ satisfy the equation

$\left| \begin{array}{ll} { (1+\alpha )^{ { 2 } } } & { (1+2\alpha )^{ { 2 } } } & { (1+3\alpha )^{ { 2 } } } \\ { (2+\alpha )^{ { 2 } } } & { (2+2\alpha )^{ { 2 } } } & { (2+3\alpha )^{ { 2 } } } \\ { (3+\alpha )^{ { 2 } } } & { (3+2\alpha )^{ { 2 } } } & { (3+3\alpha )^{ { 2 } } } \end{array} \right| =-648\alpha$ ?

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

Let

$A =[a_{ij}]$ be a

$3 \times 3$ matrix whose determinant is

$5$ . Then the determinant of the matrix

$B = [ 2^{i-j} a_{ij} ]$ is

0%
$5$
0%
$10$
0%
$20$
0%
$40$

$A = \begin{bmatrix}1 & -1 & 2\\ 3 & 0 & -2\\ 1 & 0 & 3\end{bmatrix}$ , value of

$|A(adj \,A)|$ :

0%
$11$
0%
$11^2$
0%
$11^3$
0%
$-11$

$\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{matrix} \right]$ then

$|adj\ (adj\ A)|$ is equal to

0%
$18^{3}$
0%
$18^{2}$
0%
$18^{4}$
0%
$18^{6}$

$A=\begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$ , then

$A.adj(a)=$

0%
$\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
0%
$\begin{bmatrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{bmatrix}$
0%
$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
0%
$\begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix}$

$A$ is a square matrix of order

$n$ and

$|A|=D$ and

$|adj A|=D^{\prime}$ , then

0%
$DD^{\prime}=D^{2}$
0%
$DD^{-1}=D^{-1}$
0%
$DD^{\prime}=D^{n-1}$
0%
$DD^{\prime}=D^{n}$

If the points (k, 2 - 2k) (1 - k, 2k) and (-k -4, 6 -2x) be collinear the possible values of k are

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

Explanation

Given three points

$(K,2-2K)$

$(1-K,2K)$

$(-K-4,6-2K)$

Noe three points

$\left( { x }_{ 1 },{ y }_{ 1 } \right) \left( { x }_{ 2 },{ y }_{ 2 } \right) \left( { x }_{ 3 },{ y }_{ 3 } \right)$ will be collinear

$\Rightarrow \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] =0\quad \longrightarrow \left( 1 \right)$

Putting

${ x }_{ 1 }=K;{ y }_{ 1 }=2-2K$

${ x }_{ 2 }=1-K;{ y }_{ 2 }=2K$

${ x }_{ 3 }=-K-4;{ y }_{ 3 }=6-2K$

in the equation

$(1)$ we get

$\Rightarrow K\left( 2K-6+2K \right) +\left( 1-K \right) \left( 6-2K-2+2K \right) +\left( K-4 \right) \left( 2-2K-2K \right) =0$

$\Rightarrow K\left( 4K-6 \right) +\left( 1-K \right) \left( 4 \right) +\left( -K-4 \right) \left( 2-4K \right) =0$

$\Rightarrow { 4K }^{ 2 }-6K+4-4K-\left( 2K-{ 4K }^{ 2 }+8-16K \right) =0$

$\Rightarrow { 4K }^{ 2 }-6K+4-4K-2K+{ 4K }^{ 2 }-8+16K=0$

$\Rightarrow { 8K }^{ 2 }+4K-4=0$

$\Rightarrow { 2K }^{ 2 }+K-1=0$

$\Rightarrow { 2K }^{ 2 }+2K-K-1=0$

$\Rightarrow 2K\left( K+1 \right) -1\left( K+1 \right) =0$

$\Rightarrow \left( 2K-1 \right) \left( K+1 \right) =0$

$\therefore$

$K=1/2,-1$

[option B, option D]

1179949_1305526_ans_4b9eca5386bf4a55ac099834c44b6a5c.jpg

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

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

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

Let

${\Delta _{\text{o}}} =$

$\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right]$ and let

${\Delta _1}$ denote the determinant formed by the cofactors of elements of

${\Delta _0}$ and

${\Delta _2}$ denote the determinant formed by the cofactor of

${\Delta _1},$ similarly

${\Delta _n}$ denotes the determinant formed by the cofactors of

${\Delta _{n - 1}}$ then the determinant value of

${\Delta _n}$ is

0%
${\Delta _0}^{2n}\;$
0%
${\Delta _0}^{{2^n}}\;$
0%
${\Delta _0}^{{n^2}}\;$
0%
${\Delta ^2}_0\;$

Explanation

Let

$A$ be a matrix of order

$m\times m$ and

$C$ be its co-factor matrix

We know that

$A.adj{\left(A\right)}=\left|A\right|.I$

$\Rightarrow\,\left|A\right|\left|adj{\left(A\right)}\right|={\left|A\right|}^{m}.\left|I\right|$

$\Rightarrow\,\left|A\right|\left|adj{\left(A\right)}\right|={\left|A\right|}^{m}$ ......

$(1)$ since

$\left|I\right|=1$

But

$adj{\left(A\right)}={C}^{T}$

$\therefore\,\left|adj{\left(A\right)}\right|=\left|{C}^{T}\right|=\left|C\right|$ .......

$(2)$ since

$\left|C\right|=\left|{C}^{T}\right|$

From

$(1)$ and

$(2)$ we have

$\left|A\right|\left|C\right|={\left|A\right|}^{m}$

$\Rightarrow\,\left|C\right|={\left|A\right|}^{m-1}$

Let

${\Delta}_{i}$ be the cofactor matrix then

${\Delta}_{i}={\left({\Delta}_{i-1}\right)}^{m-1}$ where

$m=3$

$\Rightarrow\,{\Delta}_{i}={\left({\Delta}_{i-1}\right)}^{3-1}$

$\Rightarrow\,{\Delta}_{i}={\left({\Delta}_{i-1}\right)}^{2}$

$\Rightarrow\,{\Delta}_{n}={\left({\Delta}_{n-1}\right)}^{2}={\left({\left({\Delta}_{n-2}\right)}^{2}\right)}^{2}={\left({\left({\left({\Delta}_{n-3}\right)}^{2}\right)}^{2}\right)}^{2}$ and so on.

$\Rightarrow\,{\Delta}_{n-1}={\left({\Delta}_{n-2}\right)}^{2}$

$\Rightarrow\,{\Delta}_{n-2}={\left({\Delta}_{n-3}\right)}^{2}$

and so on.

Hence

${\Delta}_{0}={\left({\Delta}_{0}\right)}^{{2}^{n}}$

$P = \left[ {\begin{array}{*{20}{c}}1&\alpha &3\\1&3&3\\2&4&4\end{array}} \right]$ is the adjoint of a

$3 \times 3$ matrix A and

$\left| A \right| = 4,$ then

$\alpha$ is equal to

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

$A = \left( \begin{array} { l l } { 1 } & { 2 } \\ { 3 } & { 5 } \end{array} \right),$ then the value of the determinant

$\left| A ^ { 2009 } - 5 A ^ { 2008 } \right|$ is

0%
$- 6$
0%
$- 5$
0%
$- 4$
0%
$4$
0%
$6$

$A=\begin{bmatrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{bmatrix}$ and

$A_i,B_i,C_i$ are cofactors of

$a_i,b_i,c_i$ then

$a_1B_1+a_2B_2+a_3B_3=$

0%
0
0%
|A|
0%
$|A|^2$
0%
2|A|

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 11 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Determinants Quiz 11 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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