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 $\mathbf{x_2}$?