Question

If $A$, $B$ and $C$ are angles of a triangle, then the determinant
$\left|\begin{matrix}-1&cos\,C&cos\,B\\ cos\,C&-1&cos\,A\\ cos\,B&cos\,A&-1\end{matrix}\right|=$

• A. $0$
• B. $-1$
• C. $1$
• D. None of these

Answer 1

Answer: (A)

Let $\Delta=\left|\begin{matrix}-1&cos\,C&cos\,B\\ cos\,C&-1&cos\,A\\ cos\,B&cos\,A&-1\end{matrix}\right|$

Applying $C_{2}\rightarrow C_{2}+\left(cos\,C\right)C_{1}$ and $C_{3} \rightarrow C_{3}+\left(cos\,B\right)C_{1}$, we get
$\Delta=\left|\begin{matrix}-1&0&0\\ cos\,C&-1+cos^{2}C&cos\,A+cos\,C\,cos\,B\\ cos\,B&cos\,A+cos\,B\,cos\,C&-1+cos^{2}\,B\end{matrix}\right|$

$=\left|\begin{matrix}-1&0&0\\ cos\,C&-sin^{2}\,C&cos\,A+cos\,C+cos\,B\\ cos\,B&cos\,A+cos\,B\,cos\,C&-sin^{2}\,B\end{matrix}\right|$

Expanding along $R_{1}$ we get
$\Delta=-1\left[sin^{2}\,C \,sin^{2}\,B-\left(cos\,A+cos\,C\,cos\,B\right)\left(cos\,A+cos\,B\,cos\,C\right)\right]$

$=[-sin^{2}\,C sin^{2}\,B-cos^{2}\,A-cos\,A cos\,B cos\,C-cos\,A-cos\,B cos\,C-cos^{2}\,B cos^{2}\,C]$

$=-\left[(1-cos^{2}\,C\right)\left(1-cos^{2}\,B\right)-cos^{2}A-2cos \,A cos\,B cos\,C-cos^{2}B cos^{2}C]$

$=-[1-cos^{2}\,C-cos^{2}\,B+cos^{2}\,B cos^{2}\,C-cos^{2}\,B cos^{2}C]$

$=-[1-cos^{2}C-cos^{2}B-cos^{2}A-2cosA cosB cosC]$

$=-[1+cos^{2}A+cos^{2}B+cos^{2}C+2cosA cosB cosC]$

Now, in a triangle
$cos^{2}A+cos^{2}B+cos^{2}C+2cosA cosB cosC=1$
$\Rightarrow \quad\Delta=-1+1=0$