If $C = 2\cos {\rm{\theta }}$, then the value of the determinant $\Delta = \left| {\begin{array}{*{20}{c}} C&1&0\\ 1&C&1\\ 6&1&C \end{array}} \right|$ is

$\frac{{\sin 4{\rm{\theta }}}}{{\sin {\rm{\theta }}}}$

$\frac{{2{{\sin }^2}2{\rm{\theta }}}}{{{\rm{sin\theta }}}}$

$4{\cos ^2}{\rm{\theta }}(2\cos {\rm{\theta }} - 1)$

JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Application +4 | -1

C = 2\cos \mathrm{\theta }

, then the value of the determinant

\Delta = \left| {\begin{array}{cccccccccccccccccccc} C&1&0\\ 1&C&1\\ 6&1&C \end{array}} \right|

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the determinant simplifies to

8{\cos ^3}{\theta } - 4\cos \\,\mathrm{\theta } + 6

, which doesn't match any of the other options.

Practice All Chapter MCQs →

\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

a, b, c

a x+b y+c=0

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

\omega

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

A

2 \times 2

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

|A|