Question

If A, B and C are angles of a triangle, then the determinant
$\left|\begin{array}{ccc}-1& cosC& cosB\\ cosC& -1& cosA\\ cosB& cosA& -1\end{array}\right|$ is equal to

(a) 0
(b) -1
(c) 1
(d) None of these

Answer 1

(a) We have, $\left|\begin{array}{ccc}-1& cosC& cosB\\ cosC& -1& cosA\\ cosB& cosA& -1\end{array}\right|$
Applying $C_1\to a{C}_1+b{C}_2+c{C}_3$
$\left|\begin{array}{ccc}-a+bcosC+ccosB& cosC& cosB\\ acosC-b+ccosA& -1& cosA\\ acosB+bcosA-c& cosA& -1\end{array}\right|$
Also, by projection rule in a triangle, we know that
$a=bcosC+ccosB,b=ccosA+acosC$ and $c=acosB+bcosA$
Using above equation in column first, we get
$\left|\begin{array}{ccc}-a+a& cosC& cosB\\ b-b& -1& cosA\\ c-c& cosA& -1\end{array}\right|=\left|\begin{array}{ccc}0& cosC& cosB\\ 0& -1& cosA\\ 0& cosA& -1\end{array}\right|=0$
[since, determinant having all elements of any column or row gives value of determinant as zero]