# Solving a system of linear equations that depends on parameters $\alpha, \beta \in \Bbb R$

Solve the system depending on $\alpha, \beta \in \Bbb R$: $$x_1+2x_2-2x_3=0$$ $$3x_1-2x_2+2x_3=4$$ $$x_1+\alpha x_2+2x_3=2\beta$$

My attempt: $$\left(\begin{array}{ccc|c} 1 &2 &-2 &0\\ 3 &-2 &2 &4\\ 1 &\alpha &2 &2\beta \end{array}\right)\sim\left(\begin{array}{ccc|c} 1 &2 &-2 &0\\ 0 &-8 &8 &4\\ 0 &\alpha-2 &4 &2\beta \end{array}\right)\sim\left(\begin{array}{ccc|c} 1 &0 &0 &1\\ 0 &1 &-1 &-\frac{1}{2}\\ 0 &0 &\alpha+2 &\frac{\alpha+4\beta -2}{2} \end{array}\right)$$

$$\Rightarrow x_1=1 \\ x_2-x_3=-\frac{1}{2} \\ x_3(\alpha+2)=\frac{\alpha+4\beta -2}{2}$$

If $\alpha=-2\Rightarrow \beta=1\Rightarrow X=\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}=\begin{pmatrix} 1\\ -\frac{1}{2}+x_3\\ x_3 \end{pmatrix}$

If $\alpha\neq -2 \Rightarrow X=\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}=\begin{pmatrix} 1\\ 2\frac{\beta-1}{\alpha+2}\\ \frac{\alpha+4\beta-2}{2\alpha+4} \end{pmatrix}, \beta\in \Bbb R$

Is this correct?

• There is at least one case missing, if $\alpha=-2$ and $\beta \ne 1$ then the system has no solutions. – dxiv Feb 1 '17 at 7:19
• @dxiv Oh, you're right, but what do you mean at least? – lmc Feb 1 '17 at 7:36
• That meant I'll leave it to you to check if there are any other special cases left (I don't think there are). – dxiv Feb 1 '17 at 7:40
• Am I the only one who gets the feeling that I have seen this before? – polfosol Feb 1 '17 at 8:30