Associative law of scalar and matrix multiplication 
Prove that $$c(AB)= (cA)B= A(cB)$$ where $A,B$ are matrices with $AB$ is defined  and $c$ is a scalar.

I don't know where to start, so I guess we start there. This is a theorem that is used in my textbook, and my professor stated that we should practice proving some of these for the test. I can see that this is true but I can't put it into a proof. Obviously it's true just thinking about it, but how do you even start to prove this. 
 A: In order to do this properly, you're going to need proper definitions for scalar and matrix multiplication.
Start with scalar multiplication first, as it's easier. If $A$ is an $n \times m$ matrix, and $c$ is a scalar, then
$$(cA)_{ij} := c(A)_{ij}.$$
This is to say, the new matrix which we are calling $cA$, has an entry in its $i$th row and $j$th column equal to the entry of $A$ in the same position, just multiplied by $c$. More briefly, we form $cA$ by multiplying each entry by $c$.
A little more complicated is the definition of matrix multiplication. If $A$ is $n \times m$ and $B$ is $m \times k$, then we define
$$(AB)_{ij} := \sum_{t = 1}^m (A)_{it} (B)_{tj}.$$
You should convince yourself that this is indeed the definition of matrix multiplication that you're familiar with: each entry is the dot product between the appropriate row of $A$ and the appropriate column of $B$.
Now we have what we need to actually do a proof. We have,
\begin{align*}
(c(AB))_{ij} &= c(AB)_{ij} & \ldots \text{ by definition of scalar multiplication} \\
&= c \sum_{t = 1}^m (A)_{it} (B)_{tj} & \ldots \text{ by definition of matrix multiplication} \\
&= \sum_{t = 1}^m c(A)_{it} (B)_{tj} & \ldots \text{ by the distributive property of real numbers} \\
&= \sum_{t = 1}^m (cA)_{it} (B)_{tj} & \ldots \text{ by definition of scalar multiplication} \\
&= ((cA)B)_{ij} & \ldots \text{ by definition of matrix multiplication.}
\end{align*}
Since they are equal in each entry, we therefore have
$$c(AB) = (cA)B$$
as required. Try showing they are equal to $A(cB)$ yourself!
