Question 831410
<pre>
Don't listen to him.  You can't stop and go "PDF-ing" everytime you need 
to do a small problem.  

First get a 0 in the upper left corner by multiplying the 4th column by 3
and adding it to the first column.  
     
| 3  0  0 -1|               | 0  0  0 -1|         | 3  2 -2|    | 3  2 -2|
| 0  2 -2  1|  3C4+C1->C1   | 3  2 -2  1|  = -(-1)| 3  3 -2| =  | 3  3 -2|
|-3  3 -2  2|               | 3  3 -2  2|         | 9  1  0|    | 9  1  0|
| 6  1  0  1|               | 9  1  0  1|

When you get down to a 3×3 determinant, it's often too tedious to
always expand it by minors.  I'm sure your error was a sign or 
something.  Expand by minors until you get down to a 3×3, then
do the 3×3 this following way.  You're less likely to make a mistake.
Copy the first two columns over to the right of the determinant

| 3  2 -2| 3  2
| 3  3 -2| 3  3
| 9  1  0| 9  1

Then multiply the three diagonals that slant this way \ and add them:

(3)(3)(0) + (2)(-2)(9) + (-2)(3)(1) = 0-36-6 = -42

Then multiply the three diagonals that slant this way / and add them

(-2)(3)(9) + (3)(-2)(1) + (-2)(3)(0) = -54 - 6 - 0 = -60

Then subtract (-42)-(-60) = -42 + 60 = 18

Edwin</pre>