a \(\begin{align*}3x + y - 2z & = 2\\ x - 2y + z & = 3\\ 2x - y - 3z & = 3\end{align*}\)

Show Solution
Let’s first write down the augmented matrix for this system.

\[\require{color}\left[ {\begin{array}{rrr|r} {\color{Red} 3}&1&{ - 2}&2\\1&{ - 2}&1&3\\2&{ - 1}&{ - 3}&3\end{array}} \right]\]

As with the previous examples we will mark the number(s) that we want to change in a given step in red. The first step here is to get a 1 in the upper left hand corner and again, we have many ways to do this. In this case we’ll notice that if we interchange the first and second row we can get a 1 in that spot with relatively little work.

\[\require{color}\left[ {\begin{array}{rrr|r}{\color{Red} 3}&1&{ - 2}&2\\1&{ - 2}&1&3\\2&{ - 1}&{ - 3}&3\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_2}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\{\color{Red} 3}&1&{ - 2}&2\\{\color{Red} 2}&{ - 1}&{ - 3}&3\end{array}} \right]\]

The next step is to get the two numbers below this 1 to be 0’s. Note as well that this will almost always require the third row operation to do. Also, we can do both of these in one step as follows.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\{\color{Red} 3}&1&{ - 2}&2\\{\color{Red} 2}&{ - 1}&{ - 3}&3\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} - 3{R_1} \to {R_2}}\\{{R_3} - 2{R_1} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&{\color{Red} 7}&{ - 5}&{ - 7}\\0&3&{ - 5}&{ - 3}\end{array}} \right]\]

Next, we want to turn the 7 into a 1. We can do this by dividing the second row by 7.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&{\color{Red} 7}&{ - 5}&{ - 7}\\0&3&{ - 5}&{ - 3}\end{array}} \right]\begin{array}{*{20}{c}}{\frac{1}{7}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&1&{ - \frac{5}{7}}&{ - 1}\\0&{\color{Red} 3}&{ - 5}&{ - 3}\end{array}} \right]\]

So, we got a fraction showing up here. That will happen on occasion so don’t get all that excited about it. The next step is to change the 3 below this new 1 into a 0. Note that we aren’t going to bother with the -2 above it quite yet. Sometimes it is just as easy to turn this into a 0 in the same step. In this case however, it’s probably just as easy to do it later as we’ll see.

So, using the third row operation we get,

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&1&{ - \frac{5}{7}}&{ - 1}\\0&{\color{Red} 3}&{ - 5}&{ - 3}\end{array}} \right]\begin{array}{*{20}{c}}{{R_3} - 3{R_2} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&1&{ - \frac{5}{7}}&{ - 1}\\0&0&{\color{Red} - \frac{{20}}{7}}&0\end{array}} \right]\]

Next, we need to get the number in the bottom right corner into a 1. We can do that with the second row operation.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 2}&1&3\\0&1&{ - \frac{5}{7}}&{ - 1}\\0&0&{\color{Red} - \frac{{20}}{7}}&0\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{7}{{20}}{R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&{\color{Red} 1}&3\\0&1&{\color{Red} - \frac{5}{7}}&{ - 1}\\0&0&1&0\end{array}} \right]\]

Now, we need zeroes above this new 1. So, using the third row operation twice as follows will do what we need done.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 2}&{\color{Red} 1}&3\\0&1&{\color{Red} - \frac{5}{7}}&{ - 1}\\0&0&1&0\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} + \frac{5}{7}{R_3} \to {R_2}}\\{{R_1} - {R_3} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{\color{Red} - 2}&0&3\\0&1&0&{ - 1}\\0&0&1&0\end{array}} \right]\]

Notice that in this case the final column didn’t change in this step. That was only because the final entry in that column was zero. In general, this won’t happen.

The final step is then to make the -2 above the 1 in the second column into a zero. This can easily be done with the third row operation.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{\color{Red} - 2}&0&3\\0&1&0&{ - 1}\\0&0&1&0\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 2{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&0&0&1\\0&1&0&{ - 1}\\0&0&1&0\end{array}} \right]\]

So, we have the augmented matrix in the final form and the solution will be,

\[x = 1,\,\,\,y = - 1,\,\,\,z = 0\]

This can be verified by plugging these into all three equations and making sure that they are all satisfied.

b \(\begin{align*}3x + y - 2z & = - 7\\ 2x + 2y + z & = 9\\ - x - y + 3z & = 6\end{align*}\)

Show Solution
Again, the first step is to write down the augmented matrix.

\[\require{color}\left[ {\begin{array}{rrr|r}{\color{Red} 3}&1&{ - 2}&{ - 7}\\2&2&1&9\\{ - 1}&{ - 1}&3&6\end{array}} \right]\]

We can’t get a 1 in the upper left corner simply by interchanging rows this time. We could interchange the first and last row, but that would also require another operation to turn the -1 into a 1. While this isn’t difficult it’s two operations. Note that we could use the third row operation to get a 1 in that spot as follows.

\[\require{color}\left[ {\begin{array}{rrr|r}{\color{Red} 3}&1&{ - 2}&{ - 7}\\2&2&1&9\\{ - 1}&{ - 1}&3&6\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} - {R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\{\color{Red} 2}&2&1&9\\{\color{Red} - 1}&{ - 1}&3&6\end{array}} \right]\]

Now, we can use the third row operation to turn the two red numbers into zeroes.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\{\color{Red} 2}&2&1&9\\{\color{Red} - 1}&{ - 1}&3&6\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} - 2{R_1} \to {R_2}}\\{{R_3} + {R_1} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&{\color{Red} 4}&7&{41}\\0&{ - 2}&0&{ - 10}\end{array}} \right]\]

The next step is to get a 1 in the spot occupied by the red 4. We could do that by dividing the whole row by 4, but that would put in a couple of somewhat unpleasant fractions. So, instead of doing that we are going to interchange the second and third row. The reason for this will be apparent soon enough.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&{\color{Red} 4}&7&{41}\\0&{ - 2}&0&{ - 10}\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} \leftrightarrow {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&{\color{Red} - 2}&0&{ - 10}\\0&4&7&{41}\end{array}} \right]\]

Now, if we divide the second row by -2 we get the 1 in that spot that we want.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&{\color{Red} - 2}&0&{ - 10}\\0&4&7&{41}\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{2}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&1&0&5\\0&{\color{Red} 4}&7&{41}\end{array}} \right]\]

Before moving onto the next step let’s think notice a couple of things here. First, we managed to avoid fractions, which is always a good thing, and second this row is now done. We would have eventually needed a zero in that third spot and we’ve got it there for free. Not only that, but it won’t change in any of the later operations. This doesn’t always happen, but if it does that will make our life easier.

Now, let’s use the third row operation to change the red 4 into a zero.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&1&0&5\\0&{\color{Red} 4}&7&{41}\end{array}} \right]\begin{array}{*{20}{c}}{{R_3} - 4{R_2} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&1&0&5\\0&0&{\color{Red} 7}&{21}\end{array}} \right]\]

We now can divide the third row by 7 to get that the number in the lower right corner into a one.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{ - 3}&{ - 16}\\0&1&0&5\\0&0&{\color{Red} 7}&{21}\end{array}} \right]\begin{array}{*{20}{c}}{\frac{1}{7}{R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 1}&{\color{Red} - 3}&{ - 16}\\0&1&0&5\\0&0&1&3\end{array}} \right]\]

Next, we can use the third row operation to get the -3 changed into a zero.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{ - 1}&{\color{Red} - 3}&{ - 16}\\0&1&0&5\\0&0&1&3\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 3{R_3} \to {R_{\kern 1pt} }}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{\color{Red} - 1}&0&{ - 7}\\0&1&0&5\\0&0&1&3\end{array}} \right]\]

The final step is to then make the -1 into a 0 using the third row operation again.

\[\require{color}\left[ {\begin{array}{rrr|r}1&{\color{Red} - 1}&0&{ - 7}\\0&1&0&5\\0&0&1&3\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + {R_2} \to {R_{\kern 1pt} }}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&0&0&{ - 2}\\0&1&0&5\\0&0&1&3\end{array}} \right]\]

The solution to this system is then,

\[x = - 2,\,\,\,y = 5,\,\,\,z = 3\]