Prove the Theorem 3(d) i.e., \({\left( {AB} \right)^T} = {B^T}{A^T}\).

Let A and B be the matrices of sizes \(m \times n\) and \(n \times k\), respectively.

Note that the ij^th entry of \({\left( {AB} \right)^T}\) is the ji^th entry of \(AB\). That is,

\({a_{j1}}{b_{1i}} + {a_{j2}}{b_{2i}} + ... + {a_{jn}}{b_{ni}}\).

The i^th row of \({B^T}\) is \(\left( {\begin{aligned}{*{20}{c}}{{b_{1i}}}&{{b_{2i}}}&{...}&{{b_{ni}}}\end{aligned}} \right)\), and the j^th column of \({A^T}\) is \(\left( {\begin{aligned}{*{20}{c}}{{a_{j1}}}\\{{a_{j2}}}\\ \vdots \\{{a_{jn}}}\end{aligned}} \right)\). Then ij^th entry of \({B^T}{A^T}\) is \({b_{1i}}{a_{j1}} + {b_{2i}}{a_{j2}} + ... + {b_{ni}}{a_{jn}} = {a_{j1}}{b_{1i}} + {a_{j2}}{b_{2i}} + ... + {a_{jn}}{b_{ni}}\).

Hence, the ij^th entry of \({\left( {AB} \right)^T}\) and \({B^T}{A^T}\) is equal. This implies, \({\left( {AB} \right)^T} = {B^T}{A^T}\).