Find the inverse of each matrix A if possible.  
Check that AA^-1=I and A^-1 A=I.

[1 -1  2]
[1  2  3]
[2  1  5]

Augment the matrix with the identity matrix:

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

The idea is to get every element in the
left side 0 except the three diagonal elements,
which must not be 0.  Then we divide each row
through to make the diagonal elements on the
left side 1.

Get a 0 where the 1 is in the 2nd row 1st column by
multiplying row 1 by -1 and adding it to row 2,
restoring row 1

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

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

Get a 0 where the 2 is in the 3rd row 1st column by
multiplying row 1 by -2 and adding it to row 3,
restoring row 1

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

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

Get a 0 where the -1 is in the 1st row 2nd column by
multiplying row 2 by 1 and adding it to 3 times row 1,
restoring row 2

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

[ 3  0  7 |  2  1  0]
[ 0  3  1 | -1  1  0]
[ 0  3  1 | -2  0  1]

Get a 0 where the 3 is in the 3rd row 2nd column by
multiplying row 2 by -1 and adding it to 1 times row 3,
restoring row 2

  [ 3  0  7 |  2  1  0]
-1[ 0  3  1 | -1  1  0]
 1[ 0  3  1 | -2  0  1]

[ 3  0  7 |  2  1  0]
[ 0  3  1 | -1  1  0]
[ 0  0  0 | -2  0  1]

Oh, oh. The element in the 3rd row 2rd column is a 0
so there is no way to get 0's where the 7 and 1 are 
above it.  This matrix has no inverse.

Edwin