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

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

Answer 1

Answer: (A), (C), (D)

The correct options are
A $A$ is symmetric
C $A$ is singular
D $A$ is not invertible.

As $\cos (-x)=\cos x$

we can write $\cos (\alpha -\beta )=\cos (\beta -\alpha )$

So clearly $A=A^T$

$A$ is symmetric

We can write $A$ as

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

so $|A|=0$

means $A$ is singular and non-invertible.