How do I solve this matrices question

Given that

A =
[3 0]
[0 4]

B =
[1 -2]
[-1 1]

C =
[2 -2]
[3 1]

Hence find the Matrix X which satisfies AXB = C

AXB = C
A'AXB = A'C . . . I'm using A' to denote the multiplicative inverse of A.
XB = A'C
XBB' = A'CB'
X = A'CB'
X =
[1/3 0][2 - 2][-1 -2]
[0 1/4 ][3 1][-1 -1]
=
[ 2/3 -2/3[-1 -2]
[3/4 1/4 ][-1 -1]
=
[0 -2/3]
[-1 -7/4]