2x2 Inv

Inv = (1 / Det) * [d -b]
                  [-c a]

3x3 Det

Det = aei + bfg + cdh - afh - bdi - ceg

For matrix:

   [a b c]
   [d e f]
   [g h i]

3x3 Inv

   [1 2 3]
   [3 6 4] --> Det = -90 (using the previous formula)
   [9 0 1]

To find the first element (1,1) in the matrix of minors (MoM) can be determined by.

   [*    ]           [* * *] 
   [     ] where * = [* 6 4] (so using ad - bc) 6(1) - 4(0) = 6
   [     ]           [* 0 1] 

   So MoM(1,1) = 6

   [6    ] 
   [     ] 
   [     ]

   [6 *  ]           [* * *] 
   [     ] where * = [3 * 4] (so using ad - bc) 3(1) - 4(9) = -33
   [     ]           [9 * 1]

   So MoM(1,2) = -33

   [6 -33  ] 
   [       ] 
   [       ]

Using this method we can continue to solve the MoM until it is complete.

   [6  -33 -54] 
   [2  -26 -18] 
   [-10 -5   0]

Now we need to apply two transformations to the MoM. 
The first transform is a 90' counter clock-wise turn, such that:

   [a b c]         [c f i]
   [d e f] becomes [b e h]
   [g h i]         [a d g]

The second transformation is a flip over the horizontal axis, such that:

   [c f i]         [a d g]
   [b e h] becomes [b e h]
   [a d g]         [c f i]

Or in terms of our example:

   [6  -33 -54]         [6    2  -10]
   [2  -26 -18] becomes [-33 -26  -5]
   [-10 -5   0]         [-54 -18   0]

Now we take the transformed matrix and multiply it by (1/Det):

            [6    2  -10]
  (1/-90) * [-33 -26  -5]
            [-54 -18   0]

Which equals the inverse matrix of:

   [-.066' -.022' .11']
   [.366'  .288' .055']
   [.6       .2      0]

Det

Det = ad - bc

For matrix:

   [a b]
   [c d]

Orientation

For example, a 3x2 matrix would like this:

   [3 5]
   [5 1]
   [0 1]   

And a 2x3 matrix would be like this:

   [6 7 2]
   [0 1 5]