There is another simpler (simple as in easier to remember, it is quite slow for big matrices and if you aren't very fast performing linear operations) to invert matrices. If you are familiar with solving linear equations using matricial notation, maybe you'll find this method easy.

**The method:**

1. Put an NxN identity matrix next to your NxN matrix, so you get an Nx2N matrix.

2. Use elemental row transforms to turn the original part of the matrix into an identity matrix.

3. The part of the matrix that started as an identity matrix is now the inverse of your matrix.

Say we've got a square matrix like this one:

+ 3 2 1 +
| 2 1 0 |
+ 0 2 2 +

Now we'll put next to it an identity matrix:

+ 3 2 1 | 1 0 0 +
| 2 1 0 | 0 1 0 |
+ 0 2 2 | 0 0 1 +

Our task now is to turn the left part of this matrix (that is, the matrix we are trying to invert) into an identity matrix, using linear row operations.

As a first step, we will divide the first row by three so we have a one in the left hand corner:

+ 1 2/3 1/3 | 1/3 0 0 +
| 2 1 0 | 0 1 0 |
+ 0 2 2 | 0 0 1 +

Now we substract the first row multiplied by two to the second row:

+ 1 2/3 1/3 | 1/3 0 0 +
| 0 -1/3 -2/3 | -2/3 1 0 |
+ 0 2 2 | 0 0 1 +

We multiply the second row by -3 to get a one

+ 1 2/3 1/3 | 1/3 0 0 +
| 0 1 2 | 2 -3 0 |
+ 0 2 2 | 0 0 1 +

Now we substract the second row doubled to the third

+ 1 2/3 1/3 | 1/3 0 0 +
| 0 1 2 | 2 -3 0 |
+ 0 0 -2 | -4 6 1 +

We divide the last row by -2 to obtain our traditional one

+ 1 2/3 1/3 | 1/3 0 0 +
| 0 1 2 | 2 -3 0 |
+ 0 0 1 | 2 -3 -1/2 +

Now we substract from the first row one third of the last and from the second one, double the last.

+ 1 2/3 0 | -1/3 1 1/6 +
| 0 1 0 | -2 3 1 |
+ 0 0 1 | 2 -3 -1/2 +

And we only have to substract 2/3 of the second to the first to end:

+ 1 0 0 | 1 -1 -1/2 +
| 0 1 0 | -2 3 1 |
+ 0 0 1 | 2 -3 -1/2 +

Let's check the result:

+ 3 2 1 + + 1 -1 -1/2 + + 1 0 0 +
| 2 1 0 | x | -2 3 1 | = | 0 1 0 |
+ 0 2 2 + + 2 -3 -1/2 + + 0 0 1 +

Thank god! I have made no mistakes!!

As an exercise to the reader, you can apply the method to general matrices like these:

+ a b c +
+ a b + | d e f |
+ c d + , + g h i + , ...

Doing this will (hopefully) give you the formulas shown on Wntrmute's writeup (I have never had patience to do a 3x3 general matrix, let alone anything bigger, but the 2x2 case is not very complicated).