Question

If $\alpha ,\beta ,\gamma$ are three real numbers and $A=\left[\begin{array}{ccc}1& cos(\alpha -\beta )& cos(\alpha -\gamma )\\ cos(\beta -\alpha )& 1& cos(\beta -\gamma )\\ cos(\gamma -\alpha )& cos(\gamma -\beta )& 1\end{array}\right]$, then which of the following is/are true?

• A. A is singular
• B. A is symmetric
• C. A is orthogonal
• D. A is not invertible

Answer 1

Answer: (A), (B), (D)

The correct options are
A A is singular
B A is symmetric
D A is not invertible

$A=\left[\begin{array}{ccc}1& \cos (\alpha -\beta )& \cos (\alpha -\gamma )\\ \cos (\beta -\alpha )& 1& \cos (\beta -\gamma )\\ \cos (\gamma -\alpha )& \cos (\gamma -\beta )& 1\end{array}\right]\therefore A^T=\left[\begin{array}{ccc}1& \cos (\beta -\alpha )& \cos (\gamma -\alpha )\\ \cos (\alpha -\beta )& 1& \cos (\gamma -\beta ))\\ \cos (\alpha -\gamma )& \cos (\beta -\gamma )& 1\end{array}\right]$

$\therefore A=A^T$ So $A$ is symmetric (Option $B$)

Now by using formula we have, $\cos (A+B)=\cos A\cos B+\sin A\sin B$

$\therefore \left|A\right|$ can be represented as:

$\left|A\right|=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ \cos \beta & \sin \beta & 0\\ \cos \gamma & \sin \gamma & 0\end{array}\right]\cdot \left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ \cos \beta & \sin \beta & 0\\ \cos \gamma & \sin \gamma & 0\end{array}\right]=0$

$\therefore A$ is singular (option $A$)

$A^{-1}=\frac{adjA}{\left|A\right|}$ but $\left|A\right|=0$

$\therefore A$ is non-invertible (option $D$)