<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] The LDU-decomposition gives
$$ L=\small \begin{bmatrix} 
 1 & . & . & . \\ 
 1/a & 1 & . & . \\ 
 1/a & 1/(a+1) & 1 & . \\ 
 1/a & 1/(a+1) & 1/(a+2) & 1 \\ 
 \end{bmatrix} \\ D= \small \begin{bmatrix}
 a & . & . & . \\ 
 . & (a^2-1)/a & . & . \\ 
 . & . & (a^2+a-2)/(a+1) & . \\ 
 . & . & . & (a^2+2a-3)/(a+2) \\ 
 \end{bmatrix} \\ U=\small \begin{bmatrix}
 1 & 1/a & 1/a & 1/a \\ 
 . & 1 & 1/(a+1) & 1/(a+1) \\ 
 . & . & 1 & 1/(a+2) \\ 
 . & . & . & 1
 \end{bmatrix} $$
Here L and U are simply transposed (which reflects the symmetry of the original matrix), and all of the determinant-compuation happens in the diagonal of D.     
The determinant is $$ a {a^2-1\over a} {a^2+a-2\over a+1}{a^2+2a-3\over a+2} $$
Expanding gives
$$ a\cdot  {(a-1)(a+1)\over a} \cdot {(a-1)(a+2)\over a+ 1}\cdot {(a-1)(a+3)\over a+2}$$ and cancelling over the diagonal gives $$
 1{a-1\over 1}\cdot  {a-1\over 1}\cdot {(a-1)(a+3)\over 1}$$
which is the same as that in the answer of triplesec above    
This representation has also the advantage that it is easily extensible to larger matrices 
</s>