I have the following system of equation $(S)$, where $m$ is a parameter in $\mathbb{R}$ $$ (S) \begin{cases} (m-1)x+my+z=1\\ mx+2y+3z=3\\ (m+1)x+my+(m-1)z=m-1 \end{cases} $$ I've tried solving it using the Gauss pivot method but it is very long and takes a lot of calculation. Is there any other simpler way to find the solution ?


2 Answers 2


Solving a $3 \times 3$ just takes a certain amount of work. Here subtracting the first from the third gives $$2x+(m-2)z=m-2$$ and adding the first and third gives $$2mx + 2my + mz=m$$ If $m \neq 0$ you can divide it out. If $m=0$ you will be an equation short.


$$ (S) \begin{cases} (m-1)x+my+z=1\\ mx+2y+3z=3\\ (m+1)x+my+(m-1)z=m-1 \end{cases} $$

No need for heavy calculus. Simple inspection obviously gives $$\begin{cases} x=0 \\ y=0 \\ z=1 \end{cases}$$


The above result is true any value of $m$.

But if $\quad\left|\left|\begin{matrix} m-1 & m & 1 \\ m & 2 & 3 \\ m+1 & m & m-1 \end{matrix}\right|\right| =m^2(4-m)=0\quad$ ,

that is if $m=0$ or $m=4$ , the three equations are not linearly independent.

Then they are an infinity of solutions.


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