MHB Find $(AB)^T$: Calculate Matrix Product & Transpose

[karush]

Let
$A=\left[\begin{array}{c}1 & 2 & -3 \\ 2 & 0 & -1 \end{array}\right] \textit { and }
B=\left[\begin{array}{c}3&2 \\ 1 & -1 \\ 0 & 2 \end{array}\right]$
Find $(AB)^T$$AB=\left[ \begin{array}{cc}(1\cdot 3)+(2\cdot1)+(-3\cdot0) & (1\cdot2)+(2\cdot-1)+(-3\cdot2) \\
(2\cdot3)+(0\cdot1)+ (-1\cdot0) & (2\cdot2)+(0\cdot-1)+(-1\cdot2) \end{array} \right]=\left[\begin{array}{c}5 & -6 \\ 6 & 2 \end{array}\right]$
then transpose I think this is just a diagonal reflection?
$\left[\begin{array}{c}5 & -6 \\ 6 & 2 \end{array}\right]^T=\left[\begin{array}{c}5 & 6 \\ -6 & 2 \end{array}\right]$
ok think this is correct but would like comments .. if any

 

[joypav]

Looks good to me!
However, I wouldn't refer to it as the "diagonal reflection", but that is only my opinion. I only say this because a matrix need not be a square matrix to take the transpose.

I think of it as, the rows of the original matrix become the columns of the transposed matrix.

Just as an example...

$M =\left[\begin{array}{c}2 & 3 & -1 \\ 3 & 1 & 0 \end{array}\right] $
$M^T =\left[\begin{array}{c}2 & 3 \\ 3 & 1 \\ -1 & 0 \end{array}\right] $