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Lets say I have these three equations and let m, a real parameter :
$$x - y + z = 1$$ $$3x + my + 2z = 3$$ $$mx - 3y + z = 2$$

While I do know how to write those equations in a matrix form and use gauss jordan elimination when there is no m parameter present, I don't know how I should do when working with m, from what I remember you need to treat specific cases but I don't remember at all how to do that.

Thank you.

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    @Moo I can't find them unfortunately... – 2017-01-05
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    is there a typo in your first equation? – 2017-01-05
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    @Dr.SonnhardGraubner Yup there was, I corrected it thank you ! – 2017-01-05
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    @Moo Thank you, I don't understand why this gauss jordan elimination doesn't end up with an identity matrix as it usually does... – 2017-01-05

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The matrix to do elimination on is $$ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 3 & m & 2 & 3 \\ m & -3 & 1 & 2 \end{array}\right] $$ Let's do it, being careful when we need to normalize a pivot: \begin{align} \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 3 & m & 2 & 3 \\ m & -3 & 1 & 2 \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 0 & m+3 & -1 & 0 \\ 0 & -3+m & 1-m & 2-m \end{array}\right] &&\begin{aligned}R_2&\gets R_2-3R_1\\R_3&\gets R_3-mR_1\end{aligned} \\ \color{red}{(m+3\ne0)}\quad&\to \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 0 & 1 & -1/(m+3) & 0 \\ 0 & -3+m & 1-m & 2-m \end{array}\right] &&R_2\gets\tfrac{1}{m+3}R_1 \\ &\to \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 0 & 1 & -1/(m+3) & 0 \\ 0 & 0 & -(m^2+m)/(m+3) & 2-m \end{array}\right] &&R_3\gets R_3+(3-m)R_2 \end{align} Thus we see that, for $m=0$ or $m=-1$, the system has no solution.

For $m\notin\{-3,-1,0\}$, the system has a single solution.

It remains to see what happens for $m=-3$. We restart from the second step: $$ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & -6 & 4 & 5 \end{array}\right] $$ where a simple row switch shows the system has a single solution.

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    Thank you for your answer ! I don't understand how you concluded that for m = 0 and m = -1 there is no solution ? – 2017-01-05
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    @TrevörAnneDenise For $m=0$ or $m=-1$, the last column has a pivot, so no solution. If you prefer, the last equation becomes $0=1$. Just substitute. – 2017-01-05
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from the first equation we get $$x=1+y-z$$ plugging this in the second and third equation we obtain: $$y(m+3)-z=0$$ (I) $$y(m-3)+z(1-m)=2-m$$ (II) with $$z=y(m+3)$$ and (II) we have $$y\left(m-3+(m+3)(1-m)\right)=2-m$$ can you proceed? the last equation can written as $$ym(-1-m)=2-m$$ and now you must do case work! if $m=0$ then we get $$0=2$$ contradiction if $m=-1$ then we get $$0=3$$ contradiction! thus we get $$y=\frac{m-2}{m(m+1)}$$

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The coefficient matrix $$ \pmatrix{1 & -1 & 1\cr 3 & m & 2\cr m & -3 & 1\cr}$$ has determinant $-m^2-m = -m(m+1)$, so it is singular iff $m = 0$ or $-1$. It turns out that for $m=0$ or $-1$, the vector $[1,3,2]^T$ on the right side is not in the column space, so the system has no solutions in those cases.

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    The advantage of Gauss-Jordan elimination is that it works for systems of *any* size. It might be overkill in this case, but the idea is to teach it for use with larger systems that cannot be treated with Cramer's rule in a decent amount of time. – 2017-01-05