Can this symmetric matrix be an orthogonal matrix? So the question is to prove whether this matrix M is orthogonal?
$$
        M = \begin{bmatrix}
        a & k & k \\
        k & a & k \\
        k & k & a \\
        \end{bmatrix}
$$
My attempt to find the inverse of M :
$$det(M)=(a+2k)(a-k)(a-k)$$
The inverse of $M$ is :
$$
        \begin{matrix}
        (a+k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k))\\
        (-k)/((a-k)(a+2k)) & (a+k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k))\\
        (-k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k)) & (a+k)/((a-k)(a+2k))\\
        \end{matrix}
$$
$M$ is orthogonal if $M^T = M^{-1}$
So I end up with a system of 2 equations to solve :
$a=\dfrac{a+k}{(a-k)(a+2k)}$ and $k=\dfrac{-k}{(a-k)(a+2k)}$
and now I am stucked.
I would really appreciate feedbacks about whether my steps are all correct and if yes, what to do next ?

Since $M$ is symmetric, $M=M^T$, so to check orthogonality, we compute $MM^T=M^2$ and get
$$
\begin{bmatrix}
a^2+2k^2&2ak+k^2&2ak+k^2\\
2ak+k^2&a^2+2k^2&2ak+k^2\\
2ak+k^2&2ak+k^2&a^2+2k^2
\end{bmatrix}
$$
Therefore, you need $2ak+k^2=0$ and $a^2+2k^2=1$.  Factoring the first equation, we have $k(2a+k)=0$.  Therefore, either $k=0$ or $k=-2a$.


*

*When $k=0$, the second equation simplifies to $a^2=1$, so $a=\pm 1$ gives orthogonality.

*When $k=-2a$, the second equation simplifies to $9a^2=1$, so $a=\pm\frac{1}{3}$ with $k=\mp \frac{2}{3}$ gives orthogonality.
