How to compute an equation if a matrix is singular?

Question

as homework, I was supposed to find $X$ in this equation: $AX=B$, where $$ A = \left( \begin{matrix} 1 & 2 & 5 \\ 2 & 4 & 7 \\ 4 & 8 & 9 \end{matrix} \right) $$ and $$ B = \left( \begin{matrix} 31 & 53 & 22 \\ 50 & 85 & 35 \\ 80 & 135 & 55 \end{matrix} \right) $$ Knowing that $A$ is a singular matrix that is therefore not invertible, I cannot do the ordinary equation $X=A^{-1}B$. This equation would not have a solution if $B$ had a determinant that was different from 0. But $B$ is also a singular matrix, so there is a possibility of a solution. I know there is a solution, which is $$ X = \left( \begin{matrix} 1 & 2 & 3 \\ 5 & 8 & 2 \\ 4 & 7 & 3 \end{matrix} \right) $$ but I don't know, how to arrive to it.

Answer 1

This matrix equation is actually three systems of equations in disguise. Let's denote the columns of $B$ as $\vec{b}_1, \vec{b}_2, \vec{b}_3$, from left to right. To find such an $X$, we need to find vectors $\vec{x}_1, \vec{x}_2, \vec{x}_3$ where $$A \vec{x}_1 = \vec{b}_1, ~ A \vec{x}_2 = \vec{b}_2, ~ A \vec{x}_3 = \vec{b}_3$$ because indeed if this is the case, then $$A \begin{bmatrix} \vec{x}_1 & \vec{x}_2 & \vec{x}_3 \end{bmatrix} = \begin{bmatrix} A\vec{x}_1 & A\vec{x}_2 & A\vec{x}_3 \end{bmatrix} = \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \vec{b}_3 \end{bmatrix} = B$$ Can you proceed from here?