Chapter 2 - Matrices - 2.3 The Inverse of a Matrix - 2.3 Exercises - Page 73: 68

See the proof below.

Let $C$ be invertible, such that $CA=CB$. Then $C$ has an inverse matrix. Thus we can multiply the last relation by $C^{-1}$ to get $C^{-1} (CA)=C^{-1} (CB)$ This implies that $IA=IB$ and then $A=B$.