Question

If $a^2+b^2+c^2=1$, then
$\begin{vmatrix}a^{2}+\left(b^{2}+c^{2}\right)\cos \theta&ab\left(1-\cos\theta\right)&ac\left(1-\cos\theta\right)\\ ba\left(1-\cos \theta\right)&b^{2}+\left(c^{2}+a^{2}\right)\cos\theta&bc\left(1-\cos \theta\right)\\ ca\left(1-\cos\theta\right)&cb\left(1-\cos \theta\right)&c^{2}+\left(a^{2}+b^{2}\right)\cos \theta\end{vmatrix}$equals

• A. $\cos ^2 \theta$
• B. 0
• C. 1
• D. $\sin ^2 \theta$

Answer 1

Answer: (A)

First multiply $C_1$ by $a, C_2$ by $b, C_3$ by $c$, followed by
multiplying $R_1$ by $1 / a, R_2$ by $1 / b$ and $R_3$ by $1 / c$, we get $\begin{vmatrix}a^{2}+\left(b^{2}+c^{2}\right)\cos \theta&b^{2}\left(1-\cos \theta\right)&c^{2}\left(1-\cos\theta\right)\\ a^{2}\left(1-\cos\theta\right)&b^{2}+\left(c^{2}+a^{2}\right)\cos\theta&c^{2}\left(1-\cos\theta\right)\\ a^{2}\left(1-\cos \theta\right)&b^{2}\left(1-\cos \theta\right)&c^{2}+\left(a^{2}+b^{2}\right)\cos \theta\end{vmatrix}$
Using $C_1 \rightarrow C_1+C_2+C_3$ and $a^2+b^2+c^2=1$, we get $\Delta=\begin{vmatrix} 1 & b^2(1-\cos \theta) & c^2(1-\cos \theta) \\ 1 & b^2+\left(1-b^2\right) \cos \theta & c^2(1-\cos \theta) \\ 1 & b^2(1-\cos \theta) & c^2+\left(1-c^2\right) \cos \theta \end{vmatrix}$
Using $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$, we get
$\Delta=\begin{vmatrix} 1 & b^2(1-\cos \theta) & c^2(1-\cos \theta) \\ 0 & \cos \theta & 0 \\ 0 & 0 & \cos \theta \end{vmatrix}=\cos ^2 \theta$