## Linear Algebra and Its Applications (5th Edition)

Any such vector $\vec{v}$ in the columns of $\mathbf{A}$ will be in the set spanned by the columns of $\mathbf{B}$, regardless of one's choice of $\vec{v}$ and $\mathbf{B}$.
If there were a column of $\mathbf{A}$ not in the span of the columns of $\mathbf{B}$, then one could include it among the columns of $\mathbf{B}$ to get a larger matrix with linearly independent columns, since the order of the vectors in a set does not matter when determining linear independence. For example, given \begin{align}\mathbf{B}=\begin{bmatrix}8&-3&2\\-9&4&-7\\6&-2&4\\5&-1&10\end{bmatrix}\end{align}, we can take the third column of $\mathbf{A}$ and solve the equation $\begin{bmatrix}8&-3&2\\-9&4&-7\\6&-2&4\\5&-1&10\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}=\begin{bmatrix}0\\5\\2\\7\end{bmatrix}$ to get $\vec{x}=(3,8,0)$. This confirms that $(0,5,2,7)$ is in the span of the columns of $\mathbf{B}$. (NB: Problem 41 has only one possible answer, but in general, there may be multiple possible solutions to this type of problem. Regardless, any vectors "left over" in $\mathbf{A}$ will always lie in the span of the columns of $\mathbf{B}$.)