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 \(\frac{1}{7}{R_{\,1}}\). However, this would put fractions into the other two entries in the first row and it might be nice to avoid them.

So, instead let’s do the following elementary row operation.

\[\left[ {\begin{array}{rr|r}7&{ - 8}&{ - 12}\\{ - 4}&2&3\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} + 2{R_{\,2}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}{ - 1}&{ - 4}&{ - 6}\\{ - 4}&2&3\end{array}} \right]\]

Now, this isn’t quite what we want since the number in the upper left is a minus one and not a positive one. However, we can easily fix that by multiplying the first row by -1.

\[\left[ {\begin{array}{rr|r}{ - 1}&{ - 4}&{ - 6}\\{ - 4}&2&3\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{ - {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&4&6\\{ - 4}&2&3\end{array}} \right]\]

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 -4 below the 1 into a zero and we can do that with the following elementary row operation.

\[\left[ {\begin{array}{rr|r}1&4&6\\{ - 4}&2&3\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} + 4{R_{\,1}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&4&6\\0&{18}&{27}\end{array}} \right]\] Show Step 4

The next step is to turn the number at the bottom of the second column (18 in this case) into a one. The following elementary row operation will do that for us.

\[\left[ {\begin{array}{rr|r}1&4&6\\0&{18}&{27}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{{18}}{R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{r}}1&4&6\\0&1&{\frac{{27}}{{18}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&4&6\\0&1&{\frac{3}{2}}\end{array}} \right]\]

In the first step we chose to avoid the step that put fractions into the augmented matrix, but sometimes, as in this step, they can’t be avoided.

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.

\[\left[ {\begin{array}{rr|r}1&4&6\\0&1&{\frac{3}{2}}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} - 4{R_{\,2}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&0&0\\0&1&{\frac{3}{2}}\end{array}} \right]\] Show Step 6

From the final augmented matrix we found in Step 5 we get the solution to the system is : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 0,\,\,y = \frac{3}{2}}}\).