If $x,y,z$ are three real numbers and $A=\left[\begin{array}{ccc}1& \cos (x-y)& \cos (x-z)\\ \cos (y-x)& 1& \cos (y-z)\\ \cos (z-x)& \cos (z-y)& 1\end{array}\right]$
then
$A$ is

The correct option is C not invertibe matrix
$x,y,z\in R$
$A=\left[\begin{array}{ccc}1& \cos (x-y)& \cos (x-z)\\ \cos (y-x)& 1& \cos (y-z)\\ \cos (z-x)& \cos (z-y)& 1\end{array}\right]$

$\Rightarrow A=A^T\because \cos (-\theta )=\cos \theta$
$\Rightarrow A$ is symmetric

$\Rightarrow A=\left[\begin{array}{ccc}1& \cos x\cos y+\sin x\sin y& \cos x\cos z+\sin x\sin z)\\ \cos y\cos x+\sin y\sin x& 1& \cos y\cos z+\sin y\sin z\\ \cos z\cos x+\sin z\sin x& \cos z\cos y+\sin z\sin y& 1\end{array}\right]$

$\Rightarrow A=\left[\begin{array}{ccc}{\cos }^{2}x+{\sin }^{2}x& \cos x\cos y+\sin x\sin y& \cos x\cos z+\sin x\sin z)\\ \cos y\cos x+\sin y\sin x& {\cos }^{2}y+{\sin }^{2}y& \cos y\cos z+\sin y\sin z\\ \cos z\cos x+\sin z\sin x& \cos z\cos y+\sin z\sin y& {\cos }^{2}z+{\sin }^{2}z\end{array}\right]$

$\Rightarrow A=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ \cos y& \sin y& 0\\ \cos z& \sin z& 0\end{array}\right]\left[\begin{array}{ccc}\cos x& \cos y& \cos z\\ \sin x& \sin y& \sin z\\ 0& 0& 0\end{array}\right]$

$A=A_1\times A_2|A_1|=0,|A_2|=0$

$\Rightarrow A$ is singular as $|A|=|A_1|.|A_2|=0$
$\Rightarrow A$ is not invertible
$\Rightarrow A{A}^T=A.A\Rightarrow A^2\ne I$
$\Rightarrow A$ is not orthogonal