I have this linear system:
$$
\left\{
\begin{array}{r}
x+2y=4 \\
y-z=0 \\
x+2z=4
\end{array}
\right.
$$
This Linear system can be converted to echelon form as follows:
-p1+p3
$$ \left[ \begin{array}{ccc|c} 1&2&0&4\\ 0&1&-1&0\\ 0&-2&2&0 \end{array} \right] $$
2p2+p3
$$ \left[ \begin{array}{ccc|c} 1&2&0&4\\ 0&1&-1&0\\ 0&0&0&0 \end{array} \right] $$
The solution set here is $\{(4-2z,z,z) | z \in\Bbb R \}$
My question is:
I really don't understand why "$(4-2z,z,z)$" is the solution set?
What is the meaning of two single $z$ in this solution set?
Why not $x+2y=4$, $y-z=0$ is the solution set?
1 Answer
$\begingroup$The last tableau exhibits the equations
$$x+2y=4$$ and $$y-z=0$$
$z$ can be chosen arbitarily.
It follows $y=z$ and $x=4-2y=4-2z$, so the general solution is $(4-2z/z/z)$
In general, if you have an echelon form which has not diagonal shape, the variables right from the diagonal part can be chosen arbitarily. The rest of the variables can be calculated easily depending on the chosen variables. There are infinite many solutions in this case because there are infinite many possible choices.
$\endgroup$ 4
