Solving non square matrix equations

Let $\mathbf{a}_1,\ldots,\mathbf{a}_k$ be the columns of $\mathbf{A}$ (so $\mathbf{A}$ is $n\times k$ for some $k$). Notice that we will have $\mathbf{B}$ is $n\times m$ and $\mathbf{X}$ is $m\times k$ for some integer $m$ (for $\mathbf{BX}=\mathbf{A}$ to work out). Let $\mathbf{x}_1,\ldots,\mathbf{x}_k$ be the columns of $\mathbf{X}$.

Notice that $\mathbf{a}_i$ depends only on $\mathbf{B}$ and $\mathbf{x}_i$, since $\mathbf{Bx}_i = \mathbf{a}_i$. So to determine the $i$th column of $\mathbf{X}$, it suffices to solve the system of linear equations $$\mathbf{B}\mathbf{x}_i = \mathbf{a}_i.$$ So finding $\mathbf{X}$ is equivalent to solving $k$ systems of linear equations.

In fact, you can just do them all at the same time. Simply take a matrix that is made up of $\mathbf{B}$ followed by $\mathbf{A}$: $$\left(\mathbf{B}|\mathbf{A}\right)$$ and use Gauss-Jordan elimination on $\mathbf{B}$. The solutions you find for each column corresponding to $\mathbf{A}$ yield the columns of $\mathbf{X}$.