SOLUTION: Find the inverse of the matrix if it exists. A= [3,-2,0] [3,3,0] [0,0,4]

[3 -2  0]
[3  3  0]
[0  0  4]

Augment the matrix with the identity matrix on the right

[3 -2  0 | 1  0  0]
[3  3  0 | 0  1  0]
[0  0  4 | 0  0  1]

Get 0's everywhere except on the diagonal,
working column by column, using row operations:

[3 -2  0 | 1  0  0]
[3  3  0 | 0  1  0]
[0  0  4 | 0  0  1]

Get a 0 where the red 3 is by adding -1 times
row 1 to 1 times row 2. (It's a good idea to put
the multipliers to the left of the two rows you're 
working with so you can do the row operations in 
your head):

-1[3 -2  0 | 1  0  0]
 1[3  3  0 | 0  1  0]
  [0  0  4 | 0  0  1]

  [3 -2  0 | 1  0  0]
  [0  5  0 |-1  1  0]
  [0  0  4 | 0  0  1]

Get a 0 where the -2 is by adding 2 times
row 2 to 5 times row 1. 

 5[3 -2  0 | 1  0  0]
 2[0  5  0 |-1  1  0]
  [0  0  4 | 0  0  1]

  [15  0  0 | 3  2  0]
  [ 0  5  0 |-1  1  0]
  [ 0  0  4 | 0  0  1]

Now we have all 0's except on the diagonal,
So we now divide through every row by the
diagonal element to make them 1's, so that
we now have the identity matrix on the left:

  [15/15  0/15  0/15 | 3/15  2/15  0/15]
  [  0/5   5/5   0/5 | -1/5   1/5   0/5]
  [  0/4   0/4   4/4 |  0/4   0/4   1/4]

Reduce all the fractions that will reduce, and
we have the identity matrix on the left, and the
inverse of the original matrix is on the right:

  [    1     0     0 |  1/5  2/15     0]
  [    0     1     0 | -1/5   1/5     0]
  [    0     0     1 |    0     0   1/4]  

So the inverse is the augmented part on the right:

  [ 1/5  2/15     0]
  [-1/5   1/5     0]
  [   0     0   1/4]

Edwin