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CBSE Questions for Class 12 Commerce Maths Determinants Quiz 11 - MCQExams.com
CBSE
Class 12 Commerce Maths
Determinants
Quiz 11
If $$D_r = \begin{vmatrix} r & x & n(n+1)/2 \\ 2r-1 & y & n^{ 2 } \\ 3r-2 & z & n(3n-1)/2 \end{vmatrix} $$, then $$\sum^n_{r \times 1} \, D_r$$ is equal to
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0%
$$\dfrac{1}{6} n (n + 1)(2n + 1)$$
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$$\dfrac{1}{4} n^2 (n + 1)^2$$
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$$0$$
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None of these
$$\Delta = \left| \matrix{ 1 + {a^2} + {a^4}\;\;1 + ab + {a^2}{b^2}\;\;1 + ac + {a^2}{c^2} \hfill \cr 1 + ab + {a^2}{b^2}\;\;1 + {b^2} + {b^4}\;\;1 + bc + {b^2}{c^2} \hfill \cr 1 + ac + {a^2}{c^2}\;\;1 + bc + {b^2}{c^2}\;\;1 + {c^2} + {c^4} \hfill \cr} \right|is\;equal\;to$$
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$${\left( {a + b + c} \right)^6}$$
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$${\left( {a - b} \right)^2}{\left( {b - c} \right)^2}{\left( {c - a} \right)^2}$$
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4 (a-b)(b-c)(c-a)
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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?
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$$a_1A_1+b_1B_1+c_1C_1=\Delta$$
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$$a_2A_2+b_2B_2+c_2C_2=\Delta$$
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$$a_3A_3+b_3B_3+c_3C_3=\Delta$$
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$$a_1A_2+b_1B_2+c_1C_2=\Delta$$
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
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0%
$$-11$$
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$$-3$$
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$$\dfrac{3}{2}$$
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$$2$$
$$\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix}=$$
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$$xyz\left( \dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z } \right) $$
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$$xyz$$
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$$1+\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z }$$
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$$\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +\dfrac { 1 }{ z }$$
If $$\triangle_{r}=\begin{vmatrix} { 2 }^{ r-1 } & 2.{ 3 }^{ r-1 } & 4.{ 5 }^{ r-1 } \\ x & y & z \\ { 2 }^{ n-1 } & { 3 }^{ n-1 } & { 5 }^{ n-1 } \end{vmatrix}$$, then $$\sum_{r=1}^{n}(\triangle_{r})$$ is equal to
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$$xyz$$
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$$1$$
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$$-1$$
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$$0$$
$$\begin{vmatrix} a^2
+2a & 2 a + 1 & 1 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1
\end{vmatrix} =$$
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$$(a-1)^2$$
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$$(a-1)^3$$
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$$(a-1)^4$$
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$$2(a-1)$$
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
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0%
4
0%
6
0%
0
0%
1
If $$\theta \varepsilon R,$$ then the determinant $$\Delta =\begin{vmatrix} \sin { \theta } & \cos { \theta } & \sin { 2\theta } \\ \sin { \left( \theta +\dfrac { 2\pi }{ 3 } \right) } & \cos { \left( \theta +\dfrac { 2\pi }{ 3 } \right) } & \sin { \left( 2\theta +\dfrac { 4\pi }{ 3 } \right) } \\ \sin { \left( \theta -\dfrac { 2\pi }{ 3 } \right) } & \cos { \left( \theta -\dfrac { 2\pi }{ 3 } \right) } & \sin { \left( 2\theta -\dfrac { 4\pi }{ 3 } \right) } \end{vmatrix}=$$
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$$-\sin {\theta} - \cos {\theta}$$
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$$\sin {2\theta}$$
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$$1+\sin {2\theta} - \cos {2\theta}$$
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$$None\ of\ these$$
If $$\theta \epsilon R$$, 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}=
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$$-\sin { \theta } -\cos { \theta } $$
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$$\sin { 2\theta } $$
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$$1+\sin { 2\theta } -\cos { 2\theta } $$
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None of these
Solve $$\Delta=\begin{vmatrix} \sqrt { 13 } +\sqrt { 3 } & 2\sqrt { 5 } & \sqrt { 5 } \\ \sqrt { 15 } +\sqrt { 26 } & 5 & \sqrt { 10 } \\ 3+\sqrt { 65 } & \sqrt { 15 } & 5 \end{vmatrix}=$$
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$$15\sqrt { 2 } -25\sqrt { 3 } $$
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$$25\sqrt { 3 } -15\sqrt { 2 } $$
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$$3\sqrt { 5 } $$
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$$-15\sqrt { 2 } +7\sqrt { 3 } $$
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
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$$3$$
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$$0$$
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$$1$$
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$$2$$
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
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$$\left| {\begin{array}{*{20}{c}}
{64} & 0 & 0\\
0 & {64} & 0\\
0 & 0 & {64}
\end{array}} \right|$$
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$$\left| {\begin{array}{*{20}{c}}
{88} & 0 & 0\\
0 & {88} & 0\\
0 & 0 & {88}
\end{array}} \right|$$
0%
$$\left| {\begin{array}{*{20}{c}}
{68} & 0 & 0\\
0 & {68} & 0\\
0 & 0 & {68}
\end{array}} \right|$$
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$$\left| {\begin{array}{*{20}{c}}
{34} & 0 & 0\\
0 & {34} & 0\\
0 & 0 & {34}
\end{array}} \right|$$
The number of distinct values of a $$2 \times 2$$ determinant whose entries are from set $$\{-1, 0, 1\}$$ is
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$$4$$
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$$6$$
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$$5$$
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$$3$$
$$f\left( x \right) = \left| {\begin{array}{*{20}{c}}{x - 2}&{{{\left( {x - 1} \right)}^2}}&{{x^3}}\\{x - 1}&{{x^2}}&{{{\left( {x + 1} \right)}^3}}\\x&{{{\left( {x + 1} \right)}^2}}&{{{\left( {x + 2} \right)}^3}}\end{array}} \right|$$
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$$0$$
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$$2$$
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$$-2$$
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None of these
If $$\left( \omega \neq 1 \right)$$ 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-
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$$0$$
0%
$$1$$
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$$i$$
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$$\omega$$
If A is a square matrix of order 3, then
$$\left| Adj(Adj{ A }^{ 2 }) \right| =$$
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$${ \left| A \right| }^{ 2 }$$
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$${ \left| A \right| }^{ 4 }$$
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$${ \left| A \right| }^{ 8 }$$
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$${ \left| A \right| }^{ 16 }$$
If $$1, \omega, \omega^{2}$$ 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-
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$$0$$
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$$1$$
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$$\omega$$
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$$\omega ^{2}$$
If $$|A|$$ denotes the value of the determinant of the square matrix $$A$$ order $$3$$, then $$|-2A|=$$
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$$-8|A|$$
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$$8|A|$$
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$$-2|A|$$
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None of these
$$f(x) = \begin{vmatrix}2\cos x & 1 & 0\\ x - \dfrac {\pi}{2} & 2\cos x & 1\\ 0 & 1 & 2\cos x\end{vmatrix}\Rightarrow f'(x) =$$
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$$0$$
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$$2$$
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$$\pi/2$$
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$$\pi - 6$$
State whether the statement is true/false.
If $$\mathbf { A } ( \mathbf { x } )$$ $$= \left[ \begin{array} { c c c } { \cos x } & { - \sin x } & { 0 } \\ { \sin x } & { \cos x } & { 0 } \\ { 0 } & { 0 } & { 1 } \end{array} \right]$$, then adj $$[ \mathrm { A } ( \mathrm { x } ) ] = \mathrm { A } ( - \mathrm { x } )$$.
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True
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False
If $$a + b + c = 0$$ 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
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$$x = 1$$
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$$x = 2$$
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$$x = {a^2} + {b^2} + {c^2}$$
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$$x = 0$$
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:
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$$a+b+c=0,x+y+z=0$$
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$$a+b+c=0,x=y=z$$
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$$a=b=c,x+y+z=0$$
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None of these
If $$\left| \begin{array} { r r r } { x } & { 2 } & { x } \\ { x ^ { 2 } } & { x } & { 0 } \\ { x } & { x } & { 8 } \end{array} \right|$$ = $$A x ^ { 4 } + B x ^ { 3 } + c x ^ { 2 } + D x + E$$ , then the value of $$5 A + 4 B + 2 C + 2 D + E$$ is equal to
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$$-11$$
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$$17$$
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$$-17$$
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0
The maximum and minimum values of $$(3\times 3)$$ determinant whose elements belong to $$\left\{ 0,1,2,3 \right\} $$ is
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$$\pm 9$$
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$$\pm 15$$
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$$\pm 54$$
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$$\pm 32$$
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
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0%
$$0$$
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$$30^{x}$$
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$$30^{-x}$$
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$$None\ of\ these$$
If the points $$A(x, 2), B(-3, -4)$$ and $$C(7, -5)$$ are collinear, then the value of $$x$$ is :
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$$-63$$
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$$63$$
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$$60$$
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$$-60$$
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
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$$a,b,c$$are in A.P.
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$$a,b,c$$ are in G.P.
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$$a,b,c$$ are in H.P.
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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
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0%
$$1+x$$
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$$(1+x)^{2}$$
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$$x^{2}$$
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$$none\ of\ these$$
If $$A=\begin{bmatrix} 2 & 1 & -1 \\ 0 & 1 & 4 \\ 0 & 0 & 3 \end{bmatrix}$$, then $$tr(adj(adj\ A))$$ is equal to
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0%
$$18$$
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$$24$$
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$$36$$
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$$48$$
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
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$$(i)$$
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$$(ii)$$
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$$(iii) and (iv)$$
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$$None$$ $$of$$ $$these$$
If $$A=\begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix}$$, then the value of $$|A||adj A|$$ is
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$$a^{9}$$
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$$a^{5}$$
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$$a^{6}$$
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$$a^{27}$$
If $$A=\left[ \begin{matrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{matrix} \right]$$, then $$det(adj(adj A))$$
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$$(14)^{4}$$
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$$(14)^{3}$$
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$$(14)^{2}$$
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$$(14)^{1}$$
$$A=\begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix}$$ and A (adj A)=KI, then the value of 'K' is ...
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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
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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
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$$f(x)= \frac {\pi}{2}$$ wherever defined
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domain of $$f(x)$$ is $$x= n \pi \pm \frac {\pi}{4}, n \in 1$$
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period $$f(x)$$ is $$\frac {\pi}{2}$$
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$$f(x)$$ in many one function
If $$A = \begin{bmatrix} 4 & 2 \\ 3 & 4 \end{bmatrix}$$ then |adj A| is equal to
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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
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$$\dfrac{A}{K}$$
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$$K^{n-1}\left(adj A\right)$$
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$$K^{n-2}A$$
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$$KA$$
Which of the following values of $$\alpha $$ satisfy the equation
$$\left| \begin{array}{ll} { (1+\alpha )^{ { 2 } } } & { (1+2\alpha )^{ { 2 } } } & { (1+3\alpha )^{ { 2 } } } \\ { (2+\alpha )^{ { 2 } } } & { (2+2\alpha )^{ { 2 } } } & { (2+3\alpha )^{ { 2 } } } \\ { (3+\alpha )^{ { 2 } } } & { (3+2\alpha )^{ { 2 } } } & { (3+3\alpha )^{ { 2 } } } \end{array} \right| =-648\alpha $$ ?
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$$-4$$
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
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0%
$$5$$
0%
$$10$$
0%
$$20$$
0%
$$40$$
If $$A = \begin{bmatrix}1 & -1 & 2\\ 3 & 0 & -2\\ 1 & 0 & 3\end{bmatrix}$$, value of $$|A(adj \,A)|$$:
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0%
$$11$$
0%
$$11^2$$
0%
$$11^3$$
0%
$$-11$$
If $$\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{matrix} \right]$$ then $$|adj\ (adj\ A)|$$ is equal to
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0%
$$18^{3}$$
0%
$$18^{2}$$
0%
$$18^{4}$$
0%
$$18^{6}$$
If $$A=\begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$$, then $$A.adj(a)=$$
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$$\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$
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%
$$\begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix}$$
If $$A$$ is a square matrix of order $$n$$ and $$|A|=D$$ and $$|adj A|=D^{\prime}$$, then
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$$DD^{\prime}=D^{2}$$
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$$DD^{-1}=D^{-1}$$
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$$DD^{\prime}=D^{n-1}$$
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$$DD^{\prime}=D^{n}$$
If the points (k, 2 - 2k) (1 - k, 2k) and (-k -4, 6 -2x) be collinear the possible values of k are
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-$$\dfrac{1}{2}$$
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
if
$$\Rightarrow \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] =0\quad \longrightarrow \left( 1 \right) $$
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]
If $$A=\begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$$, then $$adj(3A^{2}+12A)$$ is equal to:
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$$\begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$$
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$$\begin{bmatrix} 72 & -84 \\ -63 & 51 \end{bmatrix}$$
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$$\begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$$
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$$\begin{bmatrix} 51 & 84 \\ 63 & 72 \end{bmatrix}$$
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
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$${\Delta _0}^{2n}\;$$
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$${\Delta _0}^{{2^n}}\;$$
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$${\Delta _0}^{{n^2}}\;$$
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$${\Delta ^2}_0\;$$
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
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$$4$$
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If $$A = \left( \begin{array} { l l } { 1 } & { 2 } \\ { 3 } & { 5 } \end{array} \right), $$ then the value of the determinant $$\left| A ^ { 2009 } - 5 A ^ { 2008 } \right|$$ is
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If $$A=\begin{bmatrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{bmatrix}$$ and $$A_i,B_i,C_i$$ are cofactors of $$a_i,b_i,c_i$$ then $$a_1B_1+a_2B_2+a_3B_3=$$
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$$|A|^2$$
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2|A|
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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