Answer
$b_{1} A_{1}+b_{2} A_{2}+b_{3} A_{3}=\left[\begin{array}{ccc}-7 & 3 & -1 \\ 6 & -1 & -4 \\ -11 & 6 & 4\end{array}\right]$
Work Step by Step
point the columns of the matrix $\mathrm{B}$ by $b_{1}, b_{2},$ and $b_{3}$
\[
A=\left[\begin{array}{ccc}
1 & 0 & -2 \\
-3 & 1 & 1 \\
2 & 0 & -1
\end{array}\right], B=\left[\begin{array}{ccc}
2 & 3 & 0 \\
1 & -1 & 1 \\
-1 & 6 & 4
\end{array}\right]
\]
and the rows of the matrix $\mathrm{A}$ by $A_{1}, A_{2},$ and $A_{3}$
\[
B A=\left[\begin{array}{lll}
b_{1} & b_{2} & b_{3}
\end{array}\right]\left[\begin{array}{l}
A_{1} \\
A_{2} \\
A_{3}
\end{array}\right]
\]
Calculate the outer products.
\[
\begin{aligned}
A_{1} &=\left[\begin{array}{c}
2 \\
1 \\
-1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & -2
\end{array}\right]=\left[\begin{array}{ccc}
2 & 0 & -4 \\
1 & 0 & -2 \\
-1 & 0 & 2
\end{array}\right] \\
A_{2} &=\left[\begin{array}{c}
3 \\
-1 \\
6
\end{array}\right]\left[\begin{array}{ccc}
-3 & 1 & 1
\end{array}\right] \\
A_{3} &=\left[\begin{array}{c}
0 \\
1
\end{array}\right]\left[\begin{array}{ccc}
-9 & 3 & 3 \\
3 & -1 & -1 \\
-18 & 6 & 6
\end{array}\right] \\
\left[\begin{array}{ccc}
0 & 0 & 0 \\
2 & 0 & -1
\end{array}\right]
\end{aligned}
\]
Add the three matrices to get the matrix BA.
