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?

you start with a $2×2$ matrix and see the eigenvalues they are $4,6$.
Then see for $3×3$ matrix the eigenvalues are $9,6,-6$
for $4×4$ they are $14,6,-6,6.$.
Hence whenever the order is even the eigenvalues 6 exceeds the eigenvalue $-6$ by 1 in multiplicity.
and hence for even $n$ the snswer is
$det= (5(n-1)-1).6^{n/2}.(-6)^{\frac{n}{2}-1}$
In your case the answer is $494.(6)^{50}.(-6)^{49}$
