Answer
The $(i,j)$-entry of $(AB)^{T}$ is the $(j, i)$-entry of $AB$,
which is the inner product of (row j in A) and (column i in B)
$[(AB)^{T}]_{ij}=a_{j1}b_{1i}+\cdots+a_{jn}b_{ni}$
On the other hand, the $(i,j)$-entry of $B^{T}A^{T}$
equals the inner product of (row i in $B^{T}$) and (column j in $A^{T}$),
equals the inner product of (column i in $B$) and (row j in $A$),
equals the inner product of (row j in $A$) and (column i in $B$)
or, $a_{j1}b_{1i}+\cdots+a_{jn}b_{ni}$
This proves that $(AB)^{T}=B^{T}A^{T}$
Work Step by Step
The answer contains the proof of: $(AB)^{T}=B^{T}A^{T}$
