If $C= 2 cos θ,$ then the value of the determinant
$Δ=\begin{vmatrix}C & 1 & 0\\1 & C & 1\\6 & 1 & C\end{vmatrix}$, is
Answer & explanation
Correct answer: option 4
The correct answer is option (4) : none of these
We have,
$Δ=\begin{vmatrix}C & 1 & 0\\1 & C & 1\\6 & 1 & C\end{vmatrix}$
$⇒Δ=\begin{vmatrix}0 & 1 & 0\\1-C^2 & C & 1\\6-C & 1 & C\end{vmatrix}$ [Applying $C_1→C_1-CC_2$]
$⇒Δ=\begin{vmatrix}1-C^2 & 1\\6-C& C\end{vmatrix} $ [Expanding along $R_1$]
$⇒Δ = -(C-C^3-6+C)$
$⇒Δ=C^3-2C+6$
$⇒Δ=8cos^3\theta - 4 cos \theta + 6 $
Hence, option (4) is correct.