How to solve an linear system with b variable?

Question

I have the following:

$$ A = \begin{bmatrix}-1 & 2 \\ 3 & -5\end{bmatrix}; b = \begin{bmatrix}0 \\ \beta \end{bmatrix} $$

and I'm thinking how to analyse the Ax = b system. As I saw I could introduce the missing variables as so:

$$ A = \begin{bmatrix} -1 & 2 & 0 \\ 3 & -5 & 0 \\ 0 & 0 & -1\end{bmatrix} x = \begin{bmatrix} \xi \\ \eta \\ \beta \end{bmatrix} b = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

Then I would be able to analyze the system starting with something like so:

$$ (i) \begin{bmatrix} -1 & 2 & 0 &| &0 \\ 3 & -5 & 0 &| &0\\ 0 & 0 & -1&| &0\end{bmatrix} $$

However it doesn't seems so good, since as I deduce I would find something like:

$$ \beta = -1 \\ \xi = 0 \\ \eta = 0 $$

How can I solve that?

Answer 1

Just solve the $2 \times 2$ system, treating $\beta$ as a constant. That will give you $x_1$ and $x_2$ in terms of $\beta$.

Answer 2

The equation $\begin{bmatrix}1 & 2 \\ 3 & -5\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}0 \\ \beta\end{bmatrix}$ is equivalent to the two equations $x+ 2y= 0$, $3x- 5y= \beta$. Subtract 3 times the first equation from the second: $-11y= \beta$ so $y= -\frac{\beta}{11}$. Then $x= -2y= -\frac{2\beta}{11}$.