Answer
$$AB
=\left[\begin{array}{ccc}{b_1-2b_3} \\ {-3b_1+b_2+b_3} \\ {2b_1-b_3} \end{array}\right] .$$
Work Step by Step
Let $$
A=\left[\begin{array}{ccc}{1} & {0} & {-2} \\ {-3} & {1} & {1} \\ {2} & {0} & {-1}\end{array}\right], B=\left[\begin{array}{ccc}{2} & {3} & {0} \\ {1} & {-1} & {1} \\ {-1} & {6} & {4}\end{array}\right].
$$
Suppose that $B=\left[\begin{array}{ccc}{b_1} \\ {b_2} \\ {b_3} \end{array}\right]$, where $b_1,b_2,b_3$ are the rows of $B$.
Now, we have
$$AB=\left[\begin{array}{ccc}{1} & {0} & {-2} \\ {-3} & {1} & {1} \\ {2} & {0} & {-1}\end{array}\right] \left[\begin{array}{ccc}{b_1} \\ {b_2} \\ {b_3} \end{array}\right]
=\left[\begin{array}{ccc}{b_1-2b_3} \\ {-3b_1+b_2+b_3} \\ {2b_1-b_3} \end{array}\right] .$$
