Answer
$\begin{bmatrix}
6&32\\
1&3\\
-2&-4
\end{bmatrix}$
Work Step by Step
To Find: $\begin{bmatrix}4&-2&3\\0&1&2\\-1&0&1\end{bmatrix}\begin{bmatrix}2&6\\1&-1\\0&2\end{bmatrix}$
Let the resultant matrix be matrix $M$.
$\implies M_{1,1} = (4)(2)+(-2)(1)+(3)(0) = 6$
$\implies M_{1,2} = (4)(6)+(-2)(-1)+(3)(2) = 32$
$\implies M_{2,1} = (0)(2)+(1)(1)+(2)(0) = 1$
$\implies M_{2,2} = (0)(6)+(1)(-1)+(2)(2) = 3$
$\implies M_{3,1} = (-1)(2)+(0)(1)+(1)(0) = -2$
$\implies M_{3,2} = (-1)(6)+(0)(-1)+(1)(2) = -4$
$\therefore M = \begin{bmatrix}
6&32\\
1&3\\
-2&-4
\end{bmatrix}$
