Answer
Proof shown below
Work Step by Step
The $(i, j)$-entry in $(rA)B$ is $ra_{i1}b_{1j} + ... +ra_{in}b_{nj}$
The $(i, j)$-entry in $r(AB)$ is $r(a_{i1}b_{1j} + ... +a_{in}b_{nj})$
The $(i, j)$-entry in $A(rB)$ is $A(rb_{1j} + ... +rb_{nj})=(a_{i1}rb_{1j} + ... +a_{in}rb_{nj})$
Because the $(i, j)$-entries are all equal, the matrix products are also equal
