TITLE: What values of $a$ make this matrix not invertible?
QUESTION [2 upvotes]: So I'm given this matrix:
$$\left(\begin{array}{c} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a\end{array}\right)$$
and am told to find the values of a which make it not invertible.  
I know that $a = 0$ means our matrix is invertible (since the column vectors span $\mathbb{R}^3$) but I'm not sure how to go about finding all values of $a$ which make the matrix not invertible.
My thought was to row reduce it and find values of $a$ for which rref isn't the identity.  The row reduction with $a$ instead of numbers is tripping me up.
Any help?
Thanks,
Mariogs

REPLY [0 votes]: In this answer it is shown that
$$
\det(\lambda I_n-AB)=\lambda^{n-m}\det(\lambda I_m-BA)\tag{1}
$$
We can write your matrix as
$$
\begin{pmatrix}a&1&1\\1&a&1\\1&1&a\end{pmatrix}
=(a-1)I_3+\begin{pmatrix}1\\1\\1\end{pmatrix}\begin{pmatrix}1&1&1\end{pmatrix}\tag{2}
$$
Applying $(1)$ to $(2)$, with $\lambda=a-1$, yields
$$
\begin{align}
\det\begin{pmatrix}a&1&1\\1&a&1\\1&1&a\end{pmatrix}
&=(a-1)^{3-1}\det\left((a-1)I_1+\begin{pmatrix}1&1&1\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix}\right)\\
&=(a-1)^2\det((a+2)I_1)\\[9pt]
&=(a-1)^2(a+2)\tag{3}
\end{align}
$$
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