Chapter 2.1, Problem 4P

Explanation of Solution

Proving ${(AB)}^T=B^T{A}^T$:

Consider a matrix A of order $m\times n$ and another matrix B of order $n\times k$, then the product AB is defined because the columns in A are equal to the rows in B.

Thus, the matrix AB will be of the order $m\times k$.

The matrix A has elements $a_{li}$ and the matrix B has elements $b_{ij}$.

Then, the element of D=AB will be of the form given below:

$d_{lj}={\displaystyle \sum _{i=1}^n{a}_{li}{b}_{ij}}\mathrm{......}(1)$

Transpose of this matrix is ${(AB)}^T$. The matrix is of the order $k\times m$.

And the element of the matrix $D^T={(AB)}^T$ is of the form $d_{jl}$.

Now the transpose of the matrix B is $B^T$ and it will be of the order $k\times n$.

The element in $B^T$ is of the form $b_{ji}$