Question

If $A,B,C$ are the angles of a triangle, then the value of determinant

$\left|\begin{array}{ccc}-1+\cos B& \cos C+\cos B& \cos B\\ \cos C+\cos A& -1+\cos A& \cos A\\ -1+\cos B& -1+\cos A& -1\end{array}\right|$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$0$

Given $\left|\begin{array}{ccc}-1+\cos B& \cos C+\cos B& \cos B\\ \cos C+\cos A& -1+\cos A& \cos A\\ -1+\cos B& -1+\cos A& -1\end{array}\right|$

$=\left|\begin{array}{ccc}-1& \cos C& \cos B\\ \cos C& -1& \cos A\\ \cos B& \cos A& -1\end{array}\right|\left(\begin{array}{c}{C}_1\to C_1-C_3\\ C_2\to C_2-C_3\end{array}\right)$

$=\frac{1}{a}\left|\begin{array}{ccc}-a& \cos C& \cos B\\ a\cos C& -1& \cos A\\ a\cos B& \cos A& -1\end{array}\right|$

$=\frac{1}{a}\left|\begin{array}{ccc}0& \cos C& \cos B\\ 0& -1& \cos A\\ 0& \cos A& -1\end{array}\right|(C_1\to C_1+b{C}_2+c{C}_3)$

$=0$