Question

If $\alpha, \beta, \gamma$ are three real numbers such that $\alpha+\beta+\gamma=0$, then
$\Delta=\begin{vmatrix} 1 & \cos \gamma & \cos \beta \\ \cos \gamma & 1 & \cos \alpha \\ \cos \beta & \cos \alpha & 1 \end{vmatrix} \text { equals }$

• A. -1
• B. 0
• C. 1
• D. $\cos \alpha \cos \beta \cos \gamma$

Answer 1

Answer: (B)

Let $A, B$ and $C$ be three real numbers such that $\alpha=B-C, \beta=C-A$ and $\gamma=A-B$, clearly $\alpha+\beta+\gamma$ $=0$. We have
$\Delta=\begin{vmatrix}1 & \cos (A-B) & \cos (C-A) \\\cos (A-B) & 1 & \cos (B-C) \\\cos (C-A) & \cos (B-C) & 1\end{vmatrix}$
$\begin{vmatrix}\cos^{2}A +\sin ^{2 }A&\cos A \cos B +\sin A \sin B&\cos A \cos A +\sin A \sin C\\ \cos A \cos B +\sin A \sin B&\cos^{2}B +\sin^{2 }B&\cos B \cos C +\sin B \sin C\\ \cos C \cos A +\sin C \sin A&\cos B \cos C +\sin B \sin C&\cos^{2 }C +\sin ^{2}C\end{vmatrix}$
$=\begin{vmatrix}\cos A & \sin A & 0 \\ \cos B & \sin B & 0 \\ \cos C & \sin C & 0\end{vmatrix}\begin{vmatrix}\cos A & \sin A & 0 \\ \cos B & \sin B & 0 \\ \cos C & \sin C & 0\end{vmatrix}=(0)(0)=0$.