Answer
$(b)$
The $(i,j)$-entry of $A(B+C)$ ...
$\displaystyle \sum_{k=1}^{n}a_{ik}(b_{kj}+c_{kj})=\sum_{k=1}^{n}(a_{ik}b_{kj}+a_{ik}c_{kj})=\sum_{k=1}^{n}a_{ik}b_{kj}+\sum_{k=1}^{n}a_{ik}c_{ii}$
... equals the $(i,j)$-entry of $AB+AC$
$(c)$
The $(i,j)$-entry of $(B+C)A ...$
$\displaystyle \sum_{k=1}^{n}(b_{ik}+c_{ik})a_{kj}$=$\displaystyle \sum_{k=1}^{n}(b_{ik}a_{kj}+c_{ik}a_{kj})=\sum_{k=1}^{n}b_{ik}a_{it}+\sum_{k=1}^{n}c_{ik}a_{kj}$
... equals the $(i,j)$-entry of $BA+CA.$
Work Step by Step
proofs are given in the answer.
