How to compute determinant of $n$ dimensional matrix? I have this example:
$$\left|\begin{matrix}
-1 &amp; 2 &amp; 2 &amp; \cdots &amp; 2\\
2 &amp; -1 &amp; 2 &amp; \cdots &amp; 2\\
\vdots &amp; \vdots &amp; \ddots &amp; \ddots &amp; \vdots\\
2 &amp; 2 &amp; 2 &amp; \cdots &amp; -1\end{matrix}\right|$$

When first row is multiplied by $2$ and added to second, to $nth$ row, determinant is:
$$\left|\begin{matrix}
-1 &amp; 2 &amp; 2 &amp; \cdots &amp; 2\\
0 &amp; 3 &amp; 6 &amp; \cdots &amp; 6\\
\vdots &amp; \vdots &amp; \ddots &amp; \ddots &amp; \vdots\\
0 &amp; 6 &amp; 6 &amp; \cdots &amp; 3\end{matrix}\right|$$

Now using laplace expansion on first column:
$$-\left|\begin{matrix}
3 &amp; 6 &amp;  \cdots &amp; 6\\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
6 &amp; 6 &amp;  \cdots &amp; 3\end{matrix}\right|$$

Is it possible to get recursive relation?
What to do next?

Thanks for replies.
Hint: the matrix $M = e e^T$ (where $e$ is a column vector consisting of $n$  $1$'s) satisfies $M^2 = n M$, so its eigenvalues are ...
You can row reduce the matrix to an upper triangular matrix then take the product of all the elements of the diagonal.