Answer
$\text{col}_{1}(B)=\begin{bmatrix}7\\4\end{bmatrix}$
$\text{col}_{2}(B)=\begin{bmatrix}-8\\-5\end{bmatrix}$
Work Step by Step
The definition of matrix multipliucation yields the following:
$AB=\begin{bmatrix}A\vec{b}_{1}&A\vec{b}_{2}&A\vec{b}_{3}\end{bmatrix}=\begin{bmatrix}-1&2&-1\\6&-9&3\end{bmatrix}\implies
\begin{bmatrix}1&-2\\-2&5\end{bmatrix}\begin{bmatrix}b_{11}\\b_{21}\end{bmatrix}=\begin{bmatrix}-1\\6\end{bmatrix},
\begin{bmatrix}1&-2\\-2&5\end{bmatrix}\begin{bmatrix}b_{21}\\b_{22}\end{bmatrix}=\begin{bmatrix}2\\-9\end{bmatrix}
$
Note that we are not asked to find $\vec{b}_{3}$, the third column of $B$.
We solve by Gaussian elimination, converting the augmented matrices to reduced row-echelon form:
$\begin{bmatrix}1&-2&-1\\-2&5&6\end{bmatrix}\sim \begin{bmatrix}1&0&7\\0&1&4\end{bmatrix}$
$\begin{bmatrix}1&-2&2\\-2&5&-9\end{bmatrix}\sim \begin{bmatrix}1&0&-8\\0&1&-5\end{bmatrix}$
Hence, the solutions are given by the final columns of the rref matrices.
