Easy way to calculate the determinant of a big matrix? Given this matrix:
\begin{matrix}
2 & 3 & 0 & 9 & 0 & 1 & 0 & 1 & 1 & 2 & 1 \\
1 & 1 & 0 & 3 & 0 & 0 & 0 & 9 & 2 & 3 & 1 \\
1 & 4 & 0 & 2 & 8 & 5 & 0 & 3 & 6 & 1 & 9 \\
0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0 \\
2 & 2 & 4 & 1 & 1 & 2 & 1 & 6 & 9 & 0 & 7 \\
0 & 0 & 0 & 6 & 0 & 7 & 0 & 1 & 0 & 0 & 0 \\
2 & 5 & 0 & 7 & 0 & 4 & 6 & 8 & 5 & 1 & 3 \\
0 & 0 & 0 & 1 & 0 & 4 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 8 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 2 & 1 & 1 \\
2 & 6 & 0 & 1 & 0 & 30 & 0 & 2 & 3 & 2 & 1 \\
        \end{matrix}
Is there an efficient way of calculating its determinant using the matrix minors? it just seems that there's lots of zeroes inside so it has to ease up the operation and I'm missing something.
Thanks.
 A: It is possible to notice that the third column has only one non-zero value, $4$. Hence we cancel out the third column and the fifth row. Now the (original) fourth row has only one non-zero value, $5$, hence we cancel out the fourth row and the sixth column and continue this way. After six steps, i.e. after using the elements in positions $(5,3),(4,6),(3,5),(7,7),(9,4),(8,8)$ as "pivots", we can notice that no non-zero element on the sixth row remains, so the determinant is zero.
A: There are three ways to attack this.  One is just the slow standard algorithm.  This is a bad idea.
The next is to look for rows or columns with just a few nonzero entries.  As you do this, you may notice that rows 6, 8, and 9 each have only 2 nonzero entries, both of which are in the same 2 columns.  It is impossible for them to be linearly independent.  So the determinant is 0.
If you aren't lucky enough to notice this, then you'll notice that either row 4 or column 3 have only one nonzero entry.  You can do your expansion based on either of these.  If you start with row 4, then you'll maybe notice that those rows I mentioned above now only have 1 nonzero entry, so you're done.  But let's say you start with column 3.  You'll again end up with other rows/columns with just 1 nonzero entry.  Just keep doing this, and eventually you'll find a row/column that has all 0s.
More generally, usually these sorts of matrices don't have determinant 0, but by expanding on rows and columns with a minimal number of nonzero entries, you can keep your work very minimal.
