Problems from Gaussian elimination method

Development:

The first thing is to re-write the system in matricial form: $$\left(\begin{array}{ccccc}1& A& 1& |& 1\\ A& 1& (A-1)& |& A\\ 1& 1& 1& |& A+1\end{array}\right)$$ We then use the Gauss method $$\left(\begin{array}{ccccc}1& A& 1& |& 1\\ A& 1& (A-1)& |& A\\ 1& 1& 1& |& A+1\end{array}\right)\to \{\begin{array}{c}r2-Ar1\\ r3-r1\end{array}\to \left(\begin{array}{ccccc}1& A& 1& |& 1\\ 0& 1-A^2& -1& |& 0\\ 0& 1-A& 0& |& A\end{array}\right)\to$$ $$\to (c3\leftrightarrow c2)\to \left(\begin{array}{ccccc}1& 1& A& |& 1\\ 0& -1& 1-A^2& |& 0\\ 0& 0& 1-A& |& A\end{array}\right)$$ We obtained: $(1-A)y=A$ *, then:

*See that, as we changed the columns, we obtain the solution for $y$ and not for $z$.

If $A=1$, we have an incompatible system.

Otherwise the system is compatible, and it has the following solution:

$y={\displaystyle \frac{A}{1-A}};z=(1+A)A;x={\displaystyle \frac{A^3-A^2-2A+1}{1-A}}$

Solution:

$A=1$

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