Find the determinant of the following $5\times 5$ real matrix: 
Let $A\in\mathbb{R^{5\times5}}$ be the matrix: $\left(\begin{array}{l}a&a&a&a&b\\a&a&a&b&a\\a&a&b&a&a\\a&b&a&a&a\\b&a&a&a&a\end{array}\right)$
Find the determinant of $A$.

Hey everyone. What I've done so far: $det\left(\begin{array}{l}a&a&a&a&b\\a&a&a&b&a\\a&a&b&a&a\\a&b&a&a&a\\b&a&a&a&a\end{array}\right)=det\left(\begin{array}{l}b&a&a&a&a\\a&b&a&a&a\\a&a&b&a&a\\a&a&a&b&a\\a&a&a&a&b\end{array}\right)$ (since switching two pairs of rows does not change the determinant)
$= det\left(\begin{array}{l}b-a&0&0&0&a-b\\0&b-a&0&0&a-b\\0&0&b-a&0&a-b\\0&0&0&b-a&a-b\\a&a&a&a&b\end{array}\right)$ (since adding a multiple of one row to another does not change the determinant) for all $1\le i\le 4 \rightarrow R_i-R_5$
Now I am quite stuck. I wanted to obtain a triangular matrix so I can calculate its determinant by the diagonal entries, but I don't know what to do with the last row. I've tried some column operations as well, but have had no success.
Would be happy to get your help, thank you :)
 A: Note that the first $4$ vectors
\begin{eqnarray*}
\begin{bmatrix}
1  \\0 \\0 \\0 \\-1 \\
\end{bmatrix} ,
\begin{bmatrix}
0  \\1 \\0 \\0 \\-1 \\
\end{bmatrix} ,
\begin{bmatrix}
0  \\0 \\1 \\0 \\-1 \\
\end{bmatrix} ,
\begin{bmatrix}
0  \\0 \\0 \\1 \\-1 \\
\end{bmatrix} ,
\begin{bmatrix}
1  \\1 \\1 \\1 \\1 \\
\end{bmatrix} 
\end{eqnarray*}
are linearly independent & have eigenvalues $a-b$ and the final vector has eigen value $4a+b$. So ...
A: ...now take out common factor from rows $1$ to $4$ :
$$(b-a)^4\begin{vmatrix}1&0&0&0&-1\\0&1&0&0&-1\\0&0&1&0&-1\\0&0&0&1&-1\\a&a&a&a&b\end{vmatrix}\;\;(*)$$
and now take $\;aR_1\;$ from row $\;R_5\;$ , then $\;aR_2\;$ for $\;R_5\;$, etc.
$$(*)=(b-a)^4\begin{vmatrix}1&0&0&0&-1\\
0&1&0&0&-1\\
0&0&1&0&-1\\
0&0&0&1&-1\\
0&0&0&0&b+4a\end{vmatrix}=(b-a)^4(b+4a)$$
A: The eigenvalues of
$$ \left(\begin{array}{l}a&a&a&a&a\\a&a&a&a&a\\a&a&a&a&a\\a&a&a&a&a\\a&a&a&a&a\end{array}\right) $$
are $$ 5a,0,0,0,0 $$
with eigenvectors as the columns (pairwise perpendicular) of 
$$    
 \left(  \begin{array}{rrrrr}
  1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  1  &  -1  &  -1  &  -1     \\
  1  &  0  &  2  &  -1  &  -1     \\
  1  &  0  &  0  &  3  &  -1     \\
  1  &  0  &  0  &  0  &  4    \\  
\end{array}
  \right).
  $$
After adding $(b-a)I,$ the eigenvalues of 
$$ \left(\begin{array}{l}b&a&a&a&a\\a&b&a&a&a\\a&a&b&a&a\\a&a&a&b&a\\a&a&a&a&b\end{array}\right) $$
are $$ b+ 4a, \; b-a, \; b-a, \; b-a, \; b-a $$
