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?
 A: 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 $(1+5n(-\frac{1}{6}))(-1)^{\frac{n(n-1)}{2}}(-6)^{n}$.
A: Add all the rows to the last row and in the last row all entries will be $99 \cdot 5 - 1 = 494$. Now you can take $494$ outside of the determinant and now start subtracting $-5$ times the last row from each other row. Then the entries in the second diagonal will be $-6$, the last row will be $1$'s , while all other entries will be $0$. Then you can start changing the first row with the last one, the second with the second to last and so on. As there will be $50$ changes (even number), the value of the determinant will not change. Eventually you will get a triangular number and hence the determinant is:
$$494 \cdot (-6)^{99}$$
A: If we reverse the order of the rows, we end up with a matrix of the form $M = A-6I$, where $A$ has $5$ for every entry.  This requires $50$ transpositions, so the determinant of this new matrix is the same.  To find this determinant, we can multiply the eigenvalues of the matrix $A - 6I$, which can be found to be 
$$
\{\overbrace{-6,-6,\dots,-6}^{99 \text{ times}},494\}
$$
in the method outlined here.
A: Starting from
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
0 &0 &0 &\ldots &0 &-6 &6\\
0 &0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &6\\\
0 &-6 &0 &\ldots &0 &0 &6\\
-6 &0 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
you can add each column to the last column to obtain
$$ D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1+99\times5\\
0 &0 &0 &\ldots &0 &-6 &0\\
0 &0 &0 &\ldots &-6 &0 &0\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &0\\\
0 &-6 &0 &\ldots &0 &0 &0\\
-6 &0 &0 &\ldots &0 &0 &0 \end{vmatrix}.
$$
Now the matrix is upper-tridiagonal (along the not so conventional diagonal I have to admit). By Leibniz formula (or Laplace along the last column), it can be seen that the determinant is the product of the diagonal elements. This yields
$$D=  (-1+99\times5)(-6)^{99} =-494\times 6^{99}.$$
A: 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 $494.(6)^{50}.(-6)^{49}$
