Matrix Multiplication In general, it does not always hold that if
$$AB=AC \implies B=C.\tag{1}$$
Is there any specific case where $(1)$ holds?
 A: Well, let $A$ not be a zero divisor. This happens when $A$ is invertible, since units are not zero divisors.
Then $AB=AC$ implies $A(B-C)=0$. By hypothesis, $B-C = 0$ and so $B=C$.
A: Here is a necessary and sufficient condition.

Given an $m \times n$ matrix $A$, the following are equivalent:
  
  
*
  
*$AB = AC \implies B = C$ for all $n \times k$ matrices $B,C$,
  
*$A$ has a trivial nullspace,
  
*The columns of $A$ are linearly independent,
  
*The matrix $A$ has rank $n$ (i.e. "full column-rank"),
  
*There exists a matrix $M$ such that $MA = I$.
  

Proof: 
$1 \implies 2$: Suppose that $A$ satisfies condition 1.  Note that $A 0 = 0$, where $0$ here denotes the $n \times 1$ zero-vector.  For any $n \times 1$ column vector $x$, we have
$$
Ax = A 0 \implies x = 0
$$
So, $A$ has a trivial nullspace.
That $2 \iff 3 \iff 4$ is something covered in most introductory linear algebra texts, so I won't dwell on it here.
$4 \implies 5$: This is quick using pseudoinverses or the fact that $A^TA$ and $A$ have the same rank, but I'll avoid using that here.
Suppose that $A$ has full column-rank; it necessarily holds that $m \geq n$.  Because row-rank and column-rank are equal, $A$ has $n$ linearly independent rows.  
Let $J$ denote the $n \times m$ matrix that we get by taking the $m \times m$ identity matrix and removing all rows except for the ones corresponding to the rows that form the linearly independent rows of $A$. Then $(JA)$ is a square, invertible matrix (in particular, $JA$ is the matrix attained by removing all the rows of $A$ except for those in our chosen linearly independent set).  
So, we have $(JA)^{-1}JA = I$.  Thus, $M = (JA)^{-1}J$ satisfies $MA = I$.
$5 \implies 1$:  Suppose that $AB = AC$.  Multiplying both sides by $M$, we have 
$$
(MA)B = (MA)C \implies B = C.
$$
So, condition 1 holds. 
