- #1

karush

Gold Member

MHB

- 3,269

- 5

$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