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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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    @DylanMoreland So the only way will be to use the method of Gauss?2011-12-18

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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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    Looks good now!2011-12-18