Question
If $A=\left[\begin{array}{ccc}2 & 1 & -3 \\ 0 & 2 & 6\end{array}\right], B=\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 4\end{array}\right]$, find $A B^{\top}$ and $A^{\top} B$.

Answer

$\begin{aligned} & A=\left[\begin{array}{ccc}2 & 1 & -3 \\ 0 & 2 & 6\end{array}\right] \text { and } B=\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 4\end{array}\right] \\ & A^T=\left[\begin{array}{cc}2 & 0 \\ 1 & 2 \\ -3 & 6\end{array}\right] \text { and } B^{\mathrm{T}}=\left[\begin{array}{cc}1 & 3 \\ 0 & -1 \\ -2 & 4\end{array}\right]\end{aligned}$

$\begin{aligned} A B^T & =\left[\begin{array}{ccc}2 & 1 & -3 \\ 0 & 2 & 6\end{array}\right]\left[\begin{array}{cc}1 & 3 \\ 0 & -1 \\ -2 & 4\end{array}\right] \\ & =\left[\begin{array}{cc}2+0+6 & 6-1-12 \\ 0+0-12 & 0-2+24\end{array}\right] \\ & =\left[\begin{array}{cc}8 & -7 \\ -12 & 22\end{array}\right]\end{aligned}$

and $A^{\mathrm{T}} B=\left[\begin{array}{cc}2 & 0 \\ 1 & 2 \\ -3 & 6\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 4\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{ccc}2+0 & 0+0 & -4+0 \\ 1+6 & 0-2 & -2+8 \\ -3+18 & 0-6 & 6+24\end{array}\right] \\ & =\left[\begin{array}{ccc}2 & 0 & -4 \\ 7 & -2 & 6 \\ 15 & -6 & 30\end{array}\right]\end{aligned}$

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