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

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

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

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

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

and matrix \mathbf{A} doesnt have an inverse

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