<s>[INST] How to find the determinant of this matrix I have the following matrix:
$
\begin{bmatrix}
 a & 1 & 1 & 1 \\
 1 & a & 1 & 1 \\
 1 & 1 & a & 1 \\
 1 & 1 & 1 & a \\
\end{bmatrix}
$
My approach is to rref the matrix so that i can find the determinant by multiplying along the diagonals.
I attempted to do an rref and ended up with 
$
\begin{bmatrix}
 1 & a & 1 & 1 \\
 1-a & a-1 & 0 & 0 \\
 0 & 1-a & a-1 & 1 \\
 0 & 0 & 1-a & a-1 \\
\end{bmatrix}
$
I then factored out (1-a) to get this
$
\begin{bmatrix}
 1 & a & 1 & 1 \\
 1 & -1 & 0 & 0 \\
 0 & 1 & -1 & 1 \\
 0 & 0 & 1 & -1 \\
\end{bmatrix}
$
which will then make my determinant a multipile of $(1-a)^3$: 
$\det(A) = (1-a)^3x$
But now here I don't know what to do. I have a feeling that my approach is wrong. Any help or guide please?
 [/INST] How about the LU decomposition (modulo a rearrangements of rows)?
\begin{align*}
&\left[\begin{array}{cccc}0&1&0&0\\0&0&0&1\\0&0&1&0\\1&0&0&0\end{array}\right]\times\left[\begin{array}{cccc}
a&1&1&1\\
1&a&1&1\\
1&1&a&1\\
1&1&1&a\end{array}\right]\\=&\left[\begin{array}{cccc}1&0&0&0\\1&1&0&0\\1&1&1&0\\a&1+a&-1&1\end{array}\right]\times\left[\begin{array}{cccc}1&a&1&1\\0&1-a&0&a-1\\0&0&a-1&1-a\\0&0&0&(3+a)(1-a)\end{array}\right].
\end{align*}
It follows that the determinant is $(a-1)^3(3+a)$.
</s>