Matrix Right-Inverse: Find x_1,x_2 for Ax=b

[TsAmE]

Homework Statement

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}.

Homework Equations

None.

The Attempt at a Solution

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 don't understand

 

[HallsofIvy]

If your textbook talks about "right inverses" then it must have a definition of "right inverse"! I suggest you look up your text's definition specifically but it probably is referring to "multiplication from the right". That is, A is a "right inverse" of B if and only if BA= I where I is the identity matrix.

So look at AR_1 and AR_2. What are they?

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

Then your understanding is wrong. An equation Ax= b has a unique solution if and only if A has an inverse and in that case x= A^{-1}b (though that is not the only way to solve the equation). If A is not invertible, then Ax= b may still have solutions, though they would not be unique.

 

[TsAmE]

Oh I see.

Would this be right?

\mathbf{Ax_1} = \mathbf{b}

\mathbf{AR_1x_1} = \mathbf{bR_1}

\mathbf{I_2x_1} = \mathbf{bR_1}

\mathbf{x_1} = \mathbf{bR_1}

and the same calculations for \mathbf{x_2}?