Problems from Linear systems of three equations with three unknowns

Development:

First of all, the third equation is placed at the top, and the variable $x$ is eliminated from the second equation. $$\{\begin{array}{rcl}x+y-z& =& 1\\ 2x+2y+2z& =& 2\\ 2x-2y+2z& =& 3\end{array}$$ $$E{2}^{\prime }=E2-2\cdot E1$$ $$\{\begin{array}{rcl}x+y-z& =& 1\\ 4z& =& 0\\ 2x-2y+2z& =& 3\end{array}$$

It is possible to observe,also, that the variable y has been eliminated from the second equation. So $z=0$ and we have a system of two equations and two variables, which it is possible to solve easily by replacement: $$\begin{gathered} \{\begin{array}{rcl}x+y& =& 1\\ 2x-2y& =& 3\end{array} \\ 2(1-y)-2y=3\Rightarrow 2-4y=3\Rightarrow y=-{\displaystyle \frac{1}{4}} \\ x={\displaystyle \frac{5}{4}} \end{gathered}$$

And so, $$x={\displaystyle \frac{5}{4}};y=-{\displaystyle \frac{1}{4}};z=0$$

The change of sign in the third equation shows that equations 1 and 3 are proportional, that is, they contain the same information. $$\{\begin{array}{rcl}2x+2y+2z& =& 2\\ 2x-2y+2z& =& 3\\ x+y\boxed{\text{+}}z& =& 1\end{array}$$ $$\{\begin{array}{rcl}2x-2y+2z& =& 3\\ x+y+z& =& 1\end{array}$$ $$E{1}^{\prime }=E1-2\cdot E2$$ $$\{\begin{array}{rcl}z& =& 1\\ x+y+z& =& 1\end{array}$$ $$x+y+1=1\Rightarrow y=-x$$

Thus we have more variables than equations, so it will not be possible to find a simple solution. The degree of freedom that there is results in the solution being a straight line (intersection of two planes).

Solution:

$x={\displaystyle \frac{5}{4}};y=-{\displaystyle \frac{1}{4}};z=0$

$y=-x;z=1$

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