Find the determinant of order $100$ Find the determinant of order $100$: 
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}$$
I think I should be using recurrence relations here but I'm not entirely sure how that method works. I tried this:
Multiplying the first row by $(-1)$ and adding it to all rows: 
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
0 &0 &0 &\ldots &0 &-6 &6\\
0 &0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &6\\\
0 &-6 &0 &\ldots &0 &0 &6\\
-6 &0 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
Applying Laplace's method to the first column
$$D=5\begin{vmatrix}
0 &0  &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}+6\begin{vmatrix}
5 &5 &\ldots &5 &5 &-1\\
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
I can see that this one is $D$ but of order $99$...Is this leading anywhere? How would you solve this?

Using this matrix determinant lemma I am getting $(1+5\cdot100\cdot(-\frac{1}{6}))(-6)^{100}$.
Note about general size, one needs to think about the determinant of anti-diagonal matrix to compute it. My guess is that it equals $(1+5n(-\frac{1}{6}))(-1)^{\frac{n(n-1)}{2}}(-6)^{n}$.
