Solution:
Solution
We have,
$$\quad A' = \begin{bmatrix}1 & -4 \\ -2 & 2 \\ 3 & 5\end{bmatrix}$$
and $$\quad B' = \begin{bmatrix}1& -1 & 2 \\ 3 & 0 & 4 \end{bmatrix}$$
$$\therefore\quad
AB = \begin{bmatrix}1 & -2 & 3 \\ -4 & 2 &
5\end{bmatrix}\times\begin{bmatrix}1 & 3 \\ -1 & 0 \\ 2 &
4\end{bmatrix}$$
$$\quad =
\begin{bmatrix}1+2+6 & 3+0+12 \\ -4-2+10 & -12+0+20\end{bmatrix}
= \begin{bmatrix}9& 15 \\ 4 & 8 \end{bmatrix}$$
$$\therefore\quad (AB)' = \begin{bmatrix}9& 4 \\ 15 & 8 \end{bmatrix}\quad ...(i)$$
and
$$B'A' = \begin{bmatrix}1& -1 & 2 \\ 3 & 0 &
4 \end{bmatrix}\times\begin{bmatrix}1 & -4 \\ -2 & 2 \\ 3 &
5\end{bmatrix}$$
$$\quad = \begin{bmatrix}1+2+6 & -4-2+10 \\ 3+0+12 & -12+0+20\end{bmatrix}$$
$$\quad = \begin{bmatrix}9 & 4 \\ 15 & 8\end{bmatrix} \quad ...(ii)$$
From $$(i)$$ and $$(ii)$$, we get
$$\quad (AB)' = B'A'$$