We need to make the number in the upper left corner a one. Much like with the previous problems (i.e. solving systems with two variables) we can quickly do it with the elementary row operation \(\frac{1}{2}{R_{\,1}}\) but that will put fractions into the augmented matrix and they would probably be around for quite a few steps and it would be really nice to avoid them for as long as possible when the augmented matrix starts getting this size.

So, let’s start with the following elementary row operation.

\[\left[ {\begin{array}{rrr|r}2&5&2&{ - 38}\\3&{ - 2}&4&{17}\\{ - 6}&1&{ - 7}&{ - 12}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} - {R_{\,2}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}{ - 1}&7&{ - 2}&{ - 55}\\3&{ - 2}&4&{17}\\{ - 6}&1&{ - 7}&{ - 12}\end{array}} \right]\]

With this operation we got a negative one in the spot where we needed a plus one, but we can easily fix that with the next elementary row operation.

\[\left[ {\begin{array}{rrr|r}{ - 1}&7&{ - 2}&{ - 55}\\3&{ - 2}&4&{17}\\{ - 6}&1&{ - 7}&{ - 12}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{rrr|r}{ - {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{r}}1&{ - 7}&2&{55}\\3&{ - 2}&4&{17}\\{ - 6}&1&{ - 7}&{ - 12}\end{array}} \right]\]

Now, a quick note before we really jump into the rest of this problem. Using augmented matrices to solve systems with three variables can be a very tedious process and there are a great number of possible paths to take in the solution process so your solution may well vary from this solution depending on the path you took. The final answers however will the same regardless of the path we take provided we did all the arithmetic correctly.

Show Step 3

Next, we need to convert the 3 and the -6 below the 1 in the first column into zeroes and we can do that with the following elementary row operations.

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\3&{ - 2}&4&{17}\\{ - 6}&1&{ - 7}&{ - 12}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{rrr|r}{{R_{\,2}} - 3{R_{\,1}} \to {R_{\,2}}}\\{{R_{\,3}} + 6{R_{\,1}} \to {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{r}}1&{ - 7}&2&{55}\\0&{19}&{ - 2}&{ - 148}\\0&{ - 41}&5&{318}\end{array}} \right]\] Show Step 4

We now need to turn the 19 in the second row into a one and it seems like the only easy way to do that is the following elementary row operation.

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&{19}&{ - 2}&{ - 148}\\0&{ - 41}&5&{318}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{{19}}{R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&{ - 41}&5&{318}\end{array}} \right]\]

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. With augmented matrices for systems with three variables fractions will almost inevitably show up and they will often be “messy” when they do.

This is just something we’ll need to deal with when solving these systems. We try to avoid them for as long as possible but except it when they show up and continue with the solution process.

Show Step 5

Next, we need to turn the -41 in the third row into a zero. The following elementary row operation will do that for us.

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&{ - 41}&5&{318}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,3}} + 41{R_{\,2}} \to {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&0&{\frac{{13}}{{19}}}&{ - \frac{{26}}{{19}}}\end{array}} \right]\]

Again, we had to put more fraction into the augmented matrix. This is just a fact of life with these types of problems. However, as we’ll see in the next step they do often disappear as well.

Show Step 6

Okay, we need to turn the \(\frac{{13}}{{19}}\) in the third row into a one and we can do that as follows,

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&0&{\frac{{13}}{{19}}}&{ - \frac{{26}}{{19}}}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{{19}}{{13}}{R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&0&1&{ - 2}\end{array}} \right]\] Show Step 7

Next, we need to turn the \( - \frac{2}{{19}}\) and the 2 in the third column into zeroes. The following elementary row operations will do that for us.

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&0&1&{ - 2}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} - 2{R_{\,3}} \to {R_{\,1}}}\\{{R_{\,2}} + \frac{2}{{19}}{R_{\,3}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&{ - 7}&0&{59}\\0&1&0&{ - 8}\\0&0&1&{ - 2}\end{array}} \right]\]

Note that the fractions are now completely gone! This won’t always happen but it also will happen fairly regularly that fractions get introduced in intermediate steps and then go away in later steps.

Show Step 8

For the final operation we need to turn the -7 in the second column into a zero and we can do that as follows,

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&0&{59}\\0&1&0&{ - 8}\\0&0&1&{ - 2}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} + 7{R_{\,2}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&0&3\\0&1&0&{ - 8}\\0&0&1&{ - 2}\end{array}} \right]\] Show Step 9

From the final augmented matrix we found in Step 8 we get the solution to the system is : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 3,\,\,y = - 8,\,\,z = - 2}}\).