Question 274498
A neat way of doing this is as follows: write the first two columns again on the right and draw a line down the first three forward diagonals:

<tt>
3 3 4 3 3
&nbsp;\ \ \
6 1 2 6 1
&nbsp;&nbsp;&nbsp;\ \ \
3 2 2 3 2
</tt>

Multiply down each diagonal and add them up: 6 + 18 + 48 = 72

Now do the same with the last three backward diagonals:

<tt>
3 3 4 3 3
&nbsp;&nbsp;&nbsp;/ / /
6 1 2 6 1
&nbsp;/ / /
3 2 2 3 2
</tt>

You get 12 + 12 + 36 = 60

The determinant is the first (72) minus the second (60), giving 12.

For a square matrix of size n, you repeat the first (n-1) columns and use the first and last n diagonals.