Determinate of symmetric $82\times 82$ matrix. I am trying to calculate the determinant of the matrix:
$
\left(\begin{matrix} -6 & -5 & & \dots & -5\\
-5 & -6 & \dots & & -5 \\
\vdots & & \ddots & & \vdots \\
-5 & \dots & -5 & -6 & -5\\
-5 & \dots & & -5 & -6
\end{matrix}\right)\in \mathbb{R}^{82\times 82}
$
How am I supposed to do this with a matrix so large? I tried looking for a pattern of Gauss Elimination but I got stuck. For clarification, the diagonal entries are $-6$ and the rest of the entries are $-5$. 
 A: The matrix has the form $-I - 5uu^T$ where $I$ is the $82\times 82$ identify matrix and $u$ is a $82 \times 1$ matrix with all entries one. There is a matrix determinant lemma for evaluating the determinant of this sort of matrix.

For any $n \times n$ invertible matrix $A$ and $n \times 1$ column matrix $u$, $v$, one has
  $$\det(A + uv^T) = \det(A)( 1 + v^TA^{-1}u)$$

Apply this to our matrix, we find
$$\det(-I - 5uu^T) = (-1)^{82} \det(I + 5uu^T) = 1  + 5u^Tu =
1 + 5\cdot 82 = 411$$
A: First, try to substract the third row from the second row. Then the third from the second... and so on. You will see a quite nice matrix (almost triangular and lots of zeros):
$$\left(\begin{array}{cccccc}
-6&-5&-5&-5&\dots&-5 \\
0&-1&1&0&\dots&0\\
0&0&-1&1&\dots&0\\
0&0&0&-1&\dots&0\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
1&0&0&0&\dots&-1
\end{array}\right)$$
Now, you can easily remove the "-5"s. Use the second row to obtain a zero in the second element of the first row. Then the third row for the third element and so on.
I hope you can continue from here.
A: Let $M = \left(\begin{matrix} -6 & -5 & & \dots & -5\\
-5 & -6 & \dots & & -5 \\
\vdots & & \ddots & & \vdots \\
-5 & \dots & -5 & -6 & -5\\
-5 & \dots & & -5 & -6
\end{matrix}\right)$.
Then $M=\left(\begin{matrix} -5 & -5 & & \dots & -5\\
-5 & -5 & \dots & & -5 \\
\vdots & & \ddots & & \vdots \\
-5 & \dots & -5 & -5 & -5\\
-5 & \dots & & -5 & -5
\end{matrix}\right)+\left(\begin{matrix} -1 & 0 & & \dots & 0\\
0 & -1 & \dots & & 0\\
\vdots & & \ddots & & \vdots \\
0 & \dots & 0 & -1 & 0\\
0 & \dots & & 0 & -1
\end{matrix}\right)$
$=
\left(\begin{matrix} 5\\
5\\
\vdots \\
5\\
5 
\end{matrix}\right)\left(\begin{matrix} -1 & -1 & & \dots & -1\\
\end{matrix}\right)+\left(\begin{matrix} -1 & 0 & & \dots & 0\\
0 & -1 & \dots & & 0\\
\vdots & & \ddots & & \vdots \\
0 & \dots & 0 & -1 & 0\\
0 & \dots & & 0 & -1
\end{matrix}\right)$.
By the matrix determinant lemma, $\det M=\left(1+\left(\begin{matrix} -1 & -1 & & \dots & -1\\
\end{matrix}\right)\left(\begin{matrix} -1 & 0 & & \dots & 0\\
0 & -1 & \dots & & 0\\
\vdots & & \ddots & & \vdots \\
0 & \dots & 0 & -1 & 0\\
0 & \dots & & 0 & -1
\end{matrix}\right)\left(\begin{matrix} 5\\
5\\
\vdots \\
5\\
5 
\end{matrix}\right)\right)\cdot \left\vert\begin{matrix} -1 & 0 & & \dots & 0\\
0 & -1 & \dots & & 0\\
\vdots & & \ddots & & \vdots \\
0 & \dots & 0 & -1 & 0\\
0 & \dots & & 0 & -1
\end{matrix}\right\vert=(1+82\cdot5)\cdot1=411.$
A: This is not really an answer, just an advice.
Use some (free) software to check your results in cases like this.
I really like R, because it's (a) free, (b) great with vectors and matrices and (c) good for plotting.
Here's a little program to find the determinant:
N <- 82;
M <- matrix(rep(-5,N^2), nrow=N, ncol=N, byrow=TRUE);
D <- diag(rep(1,N));
det(M-D)

It gives $411$. 
Just in case, I also checked with Mathematica which is also great, but definitely not free.
