Determinants Test

Result

Determinants Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

If $$\begin{vmatrix} a + b& b + c & c +a \\ c + a & a + b & b + c\\ b + c & c + a & a + b\end{vmatrix} = t\times det$$ of circulant matrix whose elements of first column are $$a, b, c$$ then $$'t'$$ equals.

A
$$5$$

B
$$6$$

C
$$-2$$

D
$$2$$

Solution
Question 2
1 / -0

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}=

A
$$-\sin { \theta } -\cos { \theta } $$

B
$$\sin { 2\theta } $$

C
$$1+\sin { 2\theta } -\cos { 2\theta } $$

D
None of these
Question 3
1 / -0

If $$|A|$$ denotes the value of the determinant of the square matrix $$A$$ order $$3$$, then $$|-2A|=$$

A
$$-8|A|$$

B
$$8|A|$$

C
$$-2|A|$$

D
None of these

Solution
Question 4
1 / -0

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

A
$$0$$

B
$$2$$

C
$$\pi/2$$

D
$$\pi - 6$$

Solution
Question 5
1 / -0

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

A
$$3$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution
Question 6
1 / -0

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

A
$$x = 1$$

B
$$x = 2$$

C
$$x = {a^2} + {b^2} + {c^2}$$

D
$$x = 0$$

Solution
Question 7
1 / -0

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-

A
$$0$$

B
$$1$$

C
$$i$$

D
$$\omega$$

Solution
Question 8
1 / -0

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

A
$$-11$$

B
$$17$$

C
$$-17$$

D
0

Solution
Question 9
1 / -0

If A is a square matrix of order 3, then $$\left| Adj(Adj{ A }^{ 2 }) \right| =$$

A
$${ \left| A \right| }^{ 2 }$$

B
$${ \left| A \right| }^{ 4 }$$

C
$${ \left| A \right| }^{ 8 }$$

D
$${ \left| A \right| }^{ 16 }$$

Solution
Question 10
1 / -0

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

A
$$0$$

B
$$30^{x}$$

C
$$30^{-x}$$

D
$$None\ of\ these$$

Solution