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\begin{align*} x - \alpha y &= 1\\ \alpha x - y &= 1 \end{align*}

For which values of alpha does the system have an infinite number of solutions, no solutions and one solution.

Find the solution when it is unique.

My attempt:

$-\alpha \cdot \mathrm{eqn}_1 + \mathrm{eqn}_2$ resulting in $(\alpha^2 - 1)y = 1-\alpha$.

then we get $y = (1-\alpha)/(\alpha^2-1)$

so, $y = -1/(1+\alpha)$, but I am trying to proceed

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Is there a second half to your question? It looks like it got cut off mid sentence –  Willie Wong May 3 '12 at 9:30
    
thanks for noticing & editing –  mary May 3 '12 at 9:39
    
@mary: Note that you can't divide by $\alpha^2 - 1$ if this term is 0. –  martini May 3 '12 at 9:39
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2 Answers

up vote 1 down vote accepted

\begin{align*} x - \alpha y &= 1\\ \alpha x - y &= 1 \end{align*}

By simplifying, we have: $$\frac{x-1}{\alpha}=\alpha x-1$$

Which is:

$$(\alpha^2-1)x-(\alpha+1)=0$$

$$x=\frac{\alpha+1}{\alpha^2-1}$$

Assuming $\alpha \neq \pm 1$, $x=\frac{1}{\alpha-1}$ is your unique solution. For $\alpha=1$ and $\alpha=-1$ you have many and no solution, respectively.

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What you did looks good so far. If every single step you made is legitimate you have found your unique solution: $y=-1/(1+\alpha)$ and $x=1-\alpha/(1+\alpha)$. What do I mean by legitimate? When you are dealing with this kind of exercise what usually "goes wrong" is that you want to divide by $0$ somewhere. You divided by $\alpha^2-1$ at one point, this doesn't work when $\alpha=\pm1$. What happens in these two cases?

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