Question

If a matrix $A=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]$ and matrix $B=\left[\begin{array}{cc}4& 6\\ 5& 9\end{array}\right]$, then the transpose of product of these two matrices I.e., $(AB{)}^T$ is equal to

• A. $\left[\begin{array}{cc}28& 19\\ 34& 47\end{array}\right]$
• B. $\left[\begin{array}{cc}19& 34\\ 47& 28\end{array}\right]$
• C. $\left[\begin{array}{cc}48& 33\\ 28& 19\end{array}\right]$
• D. $\left[\begin{array}{cc}28& 19\\ 48& 33\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{cc}28& 19\\ 48& 33\end{array}\right]$

We will use Reversal Law

i.e. $(AB{)}^T$ = $B^T{A}^T$
= $\left[\begin{array}{cc}4& 5\\ 6& 9\end{array}\right]$ $\left[\begin{array}{cc}2& 1\\ 4& 3\end{array}\right]$ = $\left[\begin{array}{cc}28& 19\\ 48& 33\end{array}\right]$