For which value(s) of parameter m is there a solution for this system

Imagine a system with one parameter $m$:

\begin{cases} mx + y = m\\ mx + 2y = 1\\ 2x + my = m + 1 \end{cases}

Now the question is: when does this system of equations have a solution?

I know how to do it with the Gaussian method, but how can I do this without the Gaussian method, let's say with Cramer's rule?

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Careful: Cramer's rule doesn't tell you much when the system has more than one solution. – Dylan Moreland Dec 18 '11 at 20:34
@DylanMoreland So the only way will be to use the method of Gauss? – user21385 Dec 18 '11 at 20:38

Compute the values of $x$ and $y$ dependent on $m$ for the following system, then solve $2x + my = m + 1$ (the last equation) to find the values of parameter $m$ for $x$ and $y$:

\begin{cases} mx + y = m\\ mx + 2y = 1\\ \end{cases}

So,

\begin{cases} 2mx + 2y =2 m\\ mx + 2y = 1\\ \end{cases}

Subtracting two equations, will have:

$$mx=2m-1$$

• If $m \neq 0$, we may divide by $m$ and get $x = (2m-1)/m$ and $y = 1-m$.

• If $m = 0$, the system has no solution.

Putting $x$ and $y$ in the last equation ($m\neq 0$), we'll have:

$$m^3-3m+2=0$$

$$(m^3-1)-3m+3=0$$ $$(m-1)(m^2+m+1)-3(m-1)=0$$ $$(m-1)(m^2+m-2)=0$$

Thus the values of parameter $m$ are $m=1$ or $m=-2$.

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Be careful: If $m = 0$ then the equation $mx = 2m - 1$ gives you $0 = -1$, so you have no solution. Ruling out this case you have $x = \frac{2m - 1}{m}$ or $(2m-1)/m$ which is not th esame as $2m - 1/m$. – t.b. Dec 18 '11 at 22:03
@t.b.: Edited, thank you. – Gigili Dec 18 '11 at 22:09
Looks good now! – Dylan Moreland Dec 18 '11 at 22:44