Exercises: Gaussian elimination

$\begin{array}{ccccc}3x& +& 4y& =& -1\\ 2x& +& 5y& =& -3\end{array}$

First, write down the extended coefficient matrix.

$\left(\begin{array}{ccc}3& 4& -1\\ 2& 5& -3\end{array}\right)$

Then divide the first line by 3, the second by 2.

$\stackrel{\stackrel{}{\mathrm{I}:3,\mathrm{I}\mathrm{I}:2}}{\to }\left(\begin{array}{ccc}1& \frac{4}{3}& -\frac{1}{3}\\ 1& \frac{5}{2}& -\frac{3}{2}\end{array}\right)$

Now subtract the first line from the second line.

$\stackrel{\stackrel{}{\mathrm{I}\mathrm{I}-\mathrm{I}}}{\to }\left(\begin{array}{ccc}1& \frac{4}{3}& -\frac{1}{3}\\ 0& \frac{7}{6}& -\frac{7}{6}\end{array}\right)$

Then divide the second line by $\frac{7}{6}$.

$\stackrel{\stackrel{}{\mathrm{I}\mathrm{I}:\frac{7}{6}}}{\to }\left(\begin{array}{ccc}1& \frac{4}{3}& -\frac{1}{3}\\ 0& 1& -1\end{array}\right)$

Then subtract from the first line $\frac{4}{3}$ times the second line.

$\stackrel{\stackrel{}{\mathrm{I}-\frac{4}{3}\cdot \mathrm{I}\mathrm{I}}}{\to }\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& -1\end{array}\right)$

Now you can read the solution in the right column:

$\Rightarrow x=1;y=-1$

$\begin{array}{ccrcc}3x& -& 4y& =& -26\\ 2x& +& 3y& =& 28\end{array}$

First, write down the extended coefficient matrix.

$\left(\begin{array}{ccc}3& -4& -26\\ 2& 3& 28\end{array}\right)$

Then divide the first line by 3 and the second by 2.

$\stackrel{\stackrel{}{\mathrm{I}:3,\mathrm{I}\mathrm{I}:2}}{\to }\left(\begin{array}{ccc}1& -\frac{4}{3}& -\frac{26}{3}\\ 1& \frac{3}{2}& 14\end{array}\right)$

Now subtract the first line from the second line.

$\stackrel{\stackrel{}{\mathrm{I}\mathrm{I}-\mathrm{I}}}{\to }\left(\begin{array}{ccc}1& -\frac{4}{3}& -\frac{26}{3}\\ 0& \frac{17}{6}& \frac{68}{3}\end{array}\right)$

Then divide the second line by $\frac{17}{6}$.

$\stackrel{\stackrel{}{\mathrm{I}\mathrm{I}:\frac{17}{6}}}{\to }\left(\begin{array}{ccc}1& -\frac{4}{3}& -\frac{26}{3}\\ 0& 1& 8\end{array}\right)$

Then subtract from the first line $-\frac{4}{3}$ times the second line.

$\stackrel{\stackrel{}{\mathrm{I}-(-\frac{4}{3})\cdot \mathrm{I}\mathrm{I}}}{\to }\left(\begin{array}{ccc}1& 0& 2\\ 0& 1& 8\end{array}\right)$

Now you can read the solution in the right column:

$\Rightarrow x=2;y=8$

$\begin{array}{rcrcrcr}x& +& 2y& -& z& =& 2\\ x& +& y& +& 2z& =& 9\\ 2x& +& 3y& -& 3z& =& -1\end{array}$

First, write down the extended coefficient matrix.

$\left(\begin{array}{ccr}1& 2& -1\\ 1& 1& 2\\ 2& 3& -3\end{array}|\begin{array}{r}2\\ 9\\ -1\end{array}\right)\stackrel{\mathrm{I}\mathrm{I}-\mathrm{I}}{\underset{\mathrm{I}\mathrm{I}\mathrm{I}-2\cdot \mathrm{I}}{⟶}}\left(\begin{array}{crr}1& 2& -1\\ 0& -1& 3\\ 0& -1& -1\end{array}|\begin{array}{r}2\\ 7\\ -5\end{array}\right)\stackrel{\mathrm{I}\mathrm{I}\mathrm{I}-\mathrm{I}\mathrm{I}}{⟶}\left(\begin{array}{crr}1& 2& -1\\ 0& -1& 3\\ 0& 0& -4\end{array}|\begin{array}{r}2\\ 7\\ -12\end{array}\right)$

$\stackrel{\mathrm{I}\mathrm{I}:(-1)}{\underset{\mathrm{I}\mathrm{I}\mathrm{I}:(-4)}{⟶}}\left(\begin{array}{ccc}1& 2& -1\\ 0& 1& -3\\ 0& 0& 1\end{array}|\begin{array}{c}2\\ -7\\ 3\end{array}\right)$

The third line implies:

$z=3$

Plug this into the second line.

$y-9=-7$

$\Rightarrow y=2$

Plug $y$ and $z$ into the first line.

$x+4-3=2$

$x=1$

$\Rightarrow x=1;y=2;z=3$