Proof

Suppose that A is an \(n\times m\) matrix and B is a \(m\times \ell\) matrix as follows

$$A=\begin{bmatrix}a_{11} &a_{12}&\cdots&a_{1m}\\a_{21} &a_{22}&\cdots&a_{2m}\\\vdots &\vdots & \ddots & \vdots\\a_{n1} &a_{n2}&\cdots&a_{nm}\\\end{bmatrix},\ \ \ B=\begin{bmatrix}b_{11} &b_{12}&\cdots&b_{1\ell}\\b_{21} &b_{22}&\cdots&b_{2\ell}\\\vdots &\vdots & \ddots & \vdots\\b_{m1} &b_{m2}&\cdots&b_{m\ell}\\\end{bmatrix},$$

Let us denote the \((i,j)\)-entry in A and B by \([A]_{ij}\) and \([B]_{ij}\).

Then we have

\begin{eqnarray*}&&[A]_{ij}=a_{ij}\ \ \text{and }\ \ [A^{t}]_{ij}=a_{ji}=:A’_{ij}\\&&[B]_{ij}=b_{ij}\ \ \text{and }\ \ [B^{t}]_{ij}=b_{ji}=:B’_{ij}\\&&[AB]_{ij}=\displaystyle\sum_{k=1}^{n}a_{ik}b_{kj}\ \ \text{and }\ \ \ [(AB)^{t}]_{ij}=[AB]_{ji}=\displaystyle\sum_{k=1}^{n}a_{jk}b_{ki}\end{eqnarray*}

Therefore, we obtain

$$[AB^{t}]_{ij}=[AB]_{ji}=\displaystyle\sum_{k=1}^{n}a_{jk}b_{ki}=\displaystyle\sum_{k=1}^{n}b_{ki}a_{jk}=\displaystyle\sum_{k=1}^{n}B’_{ik}A’_{kj}=[B^{t}A^{t}]_{ij}$$