==== One more example ====

Solve the following [[linear system]]:
\begin{align*} x+y+z&=2\\2x-z&=0\\x+3y+4z&=6\\3x+y&=2\end{align*}

== Solution ==

We remark that there are more equations than unknowns... but this isn't a problem! We proceed as usual:



\begin{align*} 
\def\go#1#2#3#4{\begin{bmatrix}#1\\#2\\#3\\#4\end{bmatrix}}
\def\ar#1{\\[6pt]\xrightarrow{#1}&}
&\go{1&1&1&2}{2&0&-1&0}{1&3&4&6}{3&1&0&2}
\ar{R2\to R2-2R1,\ R3\to R3-R1\text{ and }R4\to R4-3R1}
\go{1&1&1&2}{0&-2&-3&-4}{0&2&3&4}{0&-2&-3&-4}
\ar{R3\to R3+R2\text{ and }R4\to R4-R2}
\go{1&1&1&2}{0&1&1.5&2}{0&0&0&0}{0&0&0&0}
\ar{R1\to R1-R2}
\go{1&0&-0.5&0}{0&1&1.5&2}{0&0&0&0}{0&0&0&0}
\end{align*}

Now $z$ is a free variable, say $z=t$ where $t\in \mathbb{R}$, and from row 2 we get $y=2-1.5t$ and row 1 gives $x=0.5t$, so the solution is
\[ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\2\\0\end{bmatrix}+t\begin{bmatrix}0.5\\-1.5\\1\end{bmatrix},\quad t\in\mathbb{R}.\]

===== Observations about Gaussian elimination =====

{{page>gaussian elimination remarks}}