Question

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

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

Answer 1

Answer: (A)

Let $\Delta =\left|\begin{array}{ccc}-1& cosC& cosB\\ cosC& -1& cosA\\ cosB& cosA& -1\end{array}\right|$

Applying $C_2\to C_2+\left(cosC\right)C_1$ and $C_3\to C_3+\left(cosB\right)C_1$, we get
$\Delta =\left|\begin{array}{ccc}-1& 0& 0\\ cosC& -1+co{s}^{2}C& cosA+cosCcosB\\ cosB& cosA+cosBcosC& -1+co{s}^{2}B\end{array}\right|$

$=\left|\begin{array}{ccc}-1& 0& 0\\ cosC& -si{n}^{2}C& cosA+cosC+cosB\\ cosB& cosA+cosBcosC& -si{n}^{2}B\end{array}\right|$

Expanding along $R_1$ we get
$\Delta =-1\left[si{n}^{2}Csi{n}^{2}B-\left(cosA+cosCcosB\right)\left(cosA+cosBcosC\right)\right]$

$=[-si{n}^{2}Csi{n}^{2}B-co{s}^{2}A-cosAcosBcosC-cosA-cosBcosC-co{s}^{2}Bco{s}^{2}C]$

$=-\left[(1-co{s}^{2}C\right)\left(1-co{s}^{2}B\right)-co{s}^{2}A-2cosAcosBcosC-co{s}^{2}Bco{s}^{2}C]$

$=-[1-co{s}^{2}C-co{s}^{2}B+co{s}^{2}Bco{s}^{2}C-co{s}^{2}Bco{s}^{2}C]$

$=-[1-co{s}^{2}C-co{s}^{2}B-co{s}^{2}A-2cosAcosBcosC]$

$=-[1+co{s}^{2}A+co{s}^{2}B+co{s}^{2}C+2cosAcosBcosC]$

Now, in a triangle
$co{s}^{2}A+co{s}^{2}B+co{s}^{2}C+2cosAcosBcosC=1$
$\Rightarrow \Delta =-1+1=0$