Show that $\displaystyle \mathbf{R_1} = \begin{pmatrix}-1 & 1 \\ 6& -4 \\ 4& -3\end{pmatrix}$ and $\displaystyle \mathbf{R_2} = \begin{pmatrix}1 & -1 \\ -4& 6 \\ -4& 5\end{pmatrix}$ are both right-inverses of the matrix $\displaystyle \mathbf{A} = \begin{pmatrix}1 &1 &-1 \\ 4&0 &1 \end{pmatrix}$.

Use the right-inverses $\displaystyle \mathbf{R_1}$ and $\displaystyle \mathbf{R_2}$ to find two solutions $\displaystyle \mathbf{x_1}$ and $\displaystyle \mathbf{x_2}$ of the equation $\displaystyle \mathbf{Ax = b}$, where $\displaystyle \mathbf{b} =\begin{pmatrix}0\\ 8\end{pmatrix}$.

By what I understand, the only way to solve Ax = b is with an inverse:

$\displaystyle \mathbf{A^{-1}Ax = A^{-1}b}$

$\displaystyle \mathbf{x = A^{-1}b}$

and matrix $\displaystyle \mathbf{A}$ doesnt have an inverse

but the question asks to use the right-inverse and this is what I dont understand