We need to make the number in the upper left corner a one. There are several ways to do this. One way would be to use the elementary row operation \(\frac{1}{6}{R_{\,1}}\). However, this would put fractions into the other two entries in the first row.

We’re not going to be able to avoid fractions after this step and the above idea would do what we need but it would lead to two fractions. Note however that if we interchange the two rows we get,

\[\left[ {\begin{array}{rr|r}6&{ - 5}&8\\{ - 12}&2&0\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} \leftrightarrow {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}{ - 12}&2&0\\6&{ - 5}&8\end{array}} \right]\]

We could now do the elementary row operation \( - \frac{1}{{12}}{R_{\,1}}\) and we’ll only end up with one fraction in the first row instead of two so let’s do that.

\[\left[ {\begin{array}{rr|r}{ - 12}&2&0\\6&{ - 5}&8\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{ - \frac{1}{{12}}{R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&{ - \frac{1}{6}}&0\\6&{ - 5}&8\end{array}} \right]\]

Note that as this step has shown there are several different paths to do these problems. Some will result in “messier” intermediate steps, but the solution we get in the end will be the same regardless of the path we chose to follow in the solution process.

Show Step 3

Next, we need to convert the 6 below the 1 into a zero and we can do that with the following elementary row operation.

\[\left[ {\begin{array}{rr|r}1&{ - \frac{1}{6}}&0\\6&{ - 5}&8\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} - 6{R_{\,1}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&{ - \frac{1}{6}}&0\\0&{ - 4}&8\end{array}} \right]\] Show Step 4

The next step is to turn the number at the bottom of the second column (-4 in this case) into a one. The following elementary row operation will do that for us.

\[\left[ {\begin{array}{rr|r}1&{ - \frac{1}{6}}&0\\0&{ - 4}&8\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{ - \frac{1}{4}{R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&{ - \frac{1}{6}}&0\\0&1&{ - 2}\end{array}} \right]\] Show Step 5

Finally, we need to convert the number above the one we got in Step 4 into a zero. To do that we can use the following elementary row operation.

\[\left[ {\begin{array}{rr|r}1&{ - \frac{1}{6}}&0\\0&1&{ - 2}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} + \frac{1}{6}{R_{\,2}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&0&{ - \frac{1}{3}}\\0&1&{ - 2}\end{array}} \right]\] Show Step 6

From the final augmented matrix we found in Step 5 we get the solution to the system is : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{1}{3},\,\,y = - 2}}\).