Computing the determinant of a large matrix? How would I go about computing the determinant of large matrices, such as $6 \times 6$. I believe that I need to use multilinear maps, but I am not sure how I can go about computing the determinant in a nice and efficient way. 
Can anyone show me how I can determine the determinant of the matrix below in a simple and efficient way?
\begin{pmatrix}
0 &  0&  1&  1& 1 & 1\\ 
1 & 0 & 0 &  0&  0& 1\\ 
1 & 0 &  1& 1 & 1 &1 \\ 
0 & 1 & 1 & 1 & 0 &1 \\ 
0 & 1 &  0& 1 &  0& 0\\ 
0 &  0&  1& 0 &  0& 0
\end{pmatrix}
 A: As mentioned in the comments above, row reduction can take you a long way.  Each step in the row reduction process will change the determinant in a predefined fashion (e.g. swapping rows will multiply the determinant by negative one, adding two rows together won't change the determinant at all, etc...).
If going through the whole row reduction process is too tedious, you might try your luck at spotting whether any rows are linear combinations of others or if any columns are linear combinations of the others.  If they are, then use what you know about invertible versus noninvertible matrices.

 For your matrix, the second column plus the fifth column is equal to the fourth column.

A: You can use Laplace's expansion of the  determinant. 
More precisely, begin expanding by the last row, then some row manipulation and expanding by convenient rows/columns ends up in a $2\times2$ determinant:
\begin{align}
\begin{vmatrix}
0&0&1&1&1&1 \\ 1&0&0&0&0&1 \\ 1&0&1&1&1&1 \\
0&1&1&1&0&1 \\ 0&1&0&1&0&0 \\ 0&0&\color{red}1&0&0&0\end{vmatrix}&=
-\begin{vmatrix}
0&0&1&1&1 \\ 1&0&0&0&1 \\ 1&0&1&1&1 \\
0&1&1&0&1 \\ 0&1&1&0&0\end{vmatrix}=
-\begin{vmatrix}
0&0&1&1&1 \\ 1&0&0&0&1 \\ 1&0&1&1&1 \\
0&0&0&0&\color{red}1 \\ 0&1&1&0&0\end{vmatrix}=
+\begin{vmatrix}
0&0&1&1 \\ \color{red}1&0&0&0 \\ 1&0&1&1 \\ 0&1&1&0 \end{vmatrix}\\
&=-\begin{vmatrix}
0&1&1 \\ 0&1&1 \\ \color{red}1&1&0 \end{vmatrix}=-\begin{vmatrix}
1&1 \\ 1&1  \end{vmatrix}=0.
\end{align}
A: First, note that row 6 is mostly '$0$'s. Next note the red values in row 2. The non-zero part of the determinant has to go through that $1$, so swap rows 2 and 5,
\begin{align}
\begin{vmatrix}
0&0&1&1&1&1 \\ 
1&\color{red}0&0&\color{red}0&0&1 \\ 
1&0&1&1&1&1 \\
0&1&1&1&0&1 \\ 
0&1&0&1&0&0 \\ 
0&0&\color{red}1&0&0&0
\end{vmatrix}& = -
\begin{vmatrix}
0&0&1&1&1&1 \\ 
0&1&0&1&0&0 \\ 
1&0&1&1&1&1 \\
0&1&1&1&0&1 \\ 
1&\color{red}0&0&\color{red}0&0&1 \\ 
\color{red}0&\color{red}0&1&\color{red}0&\color{red}0&\color{red}0
\end{vmatrix}= 0
\end{align}
which shows that the determinant is zero.
The column solution is even easier:
\begin{align}
\begin{vmatrix}
0&\color{red}0&1&1&1&1 \\ 
1&\color{red}0&0&0&0&1 \\ 
1&\color{red}0&1&1&1&1 \\
\color{red}0&1&1&1&0&1 \\ 
\color{red}0&1&\color{red}0&1&0&0 \\ 
\color{red}0&0&1&0&0&0
\end{vmatrix}& = 0
\end{align}
This is a medium sized matrix at most - to find the determinant for a actual large matrix ($n>100$), look up RRQR.
A: In this particular case, the fifth column is equal to the sum of the second and fourth columns; so the determinant is zero.
