Trying to prove this but I'm stuck. I'm assuming A $\neq$ B.
$$ (AB)_{i,j}=\sum^{n}_{k=1}a_{i,k}b_{k,j} $$ $$ (BA)_{i,j}=\sum^{n}_{k=1}a_{k,j}b_{i,k} $$ So we need $$ (AB)_{i,j} = (BA)_{i,j} $$ $$ \sum^{n}_{k=1}a_{i,k}b_{k,j} = \sum^{n}_{k=1}a_{k,j}b_{i,k} $$
If you write out the first few terms $$ a_{i,1}b_{1,j}+a_{i,2}b_{2,j}+a_{i,3}b_{3,j}+\cdots+a_{i,n}b_{n,j} = a_{1,j}b_{i,1}+a_{2,j}b_{i,2}+a_{3,j}b_{i,2}+\cdots+a_{n,j}b_{i,n} $$
and I don't know quite how to move forward. You can notice that if you choose some position in $AB$( or $BA$) say $(AB)_{2,3}$ then in the last equation the two $b_{2,3}$ terms are paired with $a_{2,2}$ and $a_{3,3}$ (which we are trying to show must be 1 as we want to show A is the identity matrix) but I don't know how to get there.
Thanks for any help or hints!
