Section 7.3 : Augmented Matrices
In this section we need to take a look at the third method for solving systems of equations. For systems of two equations it is probably a little more complicated than the methods we looked at in the first section. However, for systems with more equations it is probably easier than using the method we saw in the previous section.
Before we get into the method we first need to get some definitions out of the way.
An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.
Let’s take a look at an example. Here is the system of equations that we looked at in the previous section.
\[\begin{align*}x - 2y + 3z & = 7\\ 2x + y + z & = 4\\ - 3x + 2y - 2z & = - 10\end{align*}\]Here is the augmented matrix for this system.
\[\left[ {\begin{array}{rrr|r}1&{ - 2}&3&7\\2&1&1&4\\{ - 3}&2&{ - 2}&{ - 10}\end{array}} \right]\]The first row consists of all the constants from the first equation with the coefficient of the \(x\) in the first column, the coefficient of the \(y\) in the second column, the coefficient of the \(z\) in the third column and the constant in the final column. The second row is the constants from the second equation with the same placement and likewise for the third row. The dashed line represents where the equal sign was in the original system of equations and is not always included. This is mostly dependent on the instructor and/or textbook being used.
Next, we need to discuss elementary row operations. There are three of them and we will give both the notation used for each one as well as an example using the augmented matrix given above.
- Interchange Two Rows. With this operation we will interchange all the entries in row \(i\) and row \(j\). The notation we’ll use here is \({R_i} \leftrightarrow {R_j}\). Here is an example.
\[\left[ {\begin{array}{rrr|r}1&{ - 2}&3&7\\2&1&1&4\\{ - 3}&2&{ - 2}&{ - 10}\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}{ - 3}&2&{ - 2}&{ - 10}\\2&1&1&4\\1&{ - 2}&3&7\end{array}} \right]\]
So, we do exactly what the operation says. Every entry in the third row moves up to the first row and every entry in the first row moves down to the third row. Make sure that you move all the entries. One of the more common mistakes is to forget to move one or more entries.
- Multiply a Row by a Constant. In this operation we will multiply row \(i\) by a constant \(c\) and the notation will use here is \(c{R_i}\). Note that we can also divide a row by a constant using the notation \(\frac{1}{c}{R_i}\). Here is an example.
\[\left[ {\begin{array}{rrr|r}1&{ - 2}&3&7\\2&1&1&4\\{ - 3}&2&{ - 2}&{ - 10}\end{array}} \right]\begin{array}{*{20}{c}}{ - 4{R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&3&7\\2&1&1&4\\{12}&{ - 8}&8&{40}\end{array}} \right]\]
So, when we say we will multiply a row by a constant this really means that we will multiply every entry in that row by the constant. Watch out for signs in this operation and make sure that you multiply every entry.
- Add a Multiple of a Row to Another Row. In this operation we will replace row \(i\) with the sum of row \(i\) and a constant, \(c\), times row \(j\). The notation we’ll use for this operation is \({R_i} + c{R_j} \to {R_i}\). To perform this operation we will take an entry from row \(i\) and add to it \(c\) times the corresponding entry from row \(j\) and put the result back into row \(i\). Here is an example of this operation.
\[\left[ {\begin{array}{rrr|r}1&{ - 2}&3&7\\2&1&1&4\\{ - 3}&2&{ - 2}&{ - 10}\end{array}} \right]\begin{array}{*{20}{c}}{{R_3} - 4{R_1} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1&{ - 2}&3&7\\2&1&1&4\\{ - 7}&{10}&{ - 14}&{ - 38}\end{array}} \right]\]
Let’s go through the individual computation to make sure you followed this.
\[\begin{align*} - 3 - 4\left( 1 \right) & = - 7\hspace{0.25in}\\ 2 - 4\left( { - 2} \right) & = 10\\ - 2 - 4\left( 3 \right) & = - 14\\ - 10 - 4\left( 7 \right) & = - 38\end{align*}\]Be very careful with signs here. We will be doing these computations in our head for the most part and it is very easy to get signs mixed up and add one in that doesn’t belong or lose one that should be there.
It is very important that you can do this operation as this operation is the one that we will be using more than the other two combined.
Okay, so how do we use augmented matrices and row operations to solve systems? Let’s start with a system of two equations and two unknowns.
\[\begin{align*}ax + by & = p\\ cx + dy & = q\end{align*}\]We first write down the augmented matrix for this system,
\[\left[ {\begin{array}{rr|r}a&b&p\\c&d&q\end{array}} \right]\]and use elementary row operations to convert it into the following augmented matrix.
\[\left[ {\begin{array}{rr|r}1&0&h\\0&1&k\end{array}} \right]\]Once we have the augmented matrix in this form we are done. The solution to the system will be \(x = h\) and \(y = k\).
This method is called Gauss-Jordan Elimination.
- \(\begin{align*}3x - 2y & = 14\\ x + 3y & = 1\end{align*}\)
- \(\begin{align*} - 2x + y & = - 3\\ x - 4y & = - 2\end{align*}\)
- \(\begin{align*}3x - 6y & = - 9\\ - 2x - 2y & = 12\end{align*}\)
The first step here is to write down the augmented matrix for this system.
\[\require{color}\left[ {\begin{array}{rr|r} {\color{Red} 3}&{ - 2}&{14}\\1&3&1\end{array}} \right]\]To convert it into the final form we will start in the upper left corner and work in a counter-clockwise direction until the first two columns appear as they should be.
So, the first step is to make the red three in the augmented matrix above into a 1. We can use any of the row operations that we’d like to. We should always try to minimize the work as much as possible however.
So, since there is a one in the first column already it just isn’t in the correct row let’s use the first row operation and interchange the two rows.
\[\require{color}\left[ {\begin{array}{rr|r}3&{ - 2}&{14}\\1&3&1\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&3&1\\{\color{Red} 3}&{ - 2}&{14}\end{array}} \right]\]The next step is to get a zero below the 1 that we just got in the upper left hand corner. This means that we need to change the red three into a zero. This will almost always require us to use third row operation. If we add -3 times row 1 onto row 2 we can convert that 3 into a 0. Here is that operation.
\[\require{color}\left[ {\begin{array}{rr|r}1&3&1\\3&{ - 2}&{14}\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} - 3{R_1} \to {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&3&1\\0&{\color{Red} - 11}&{11}\end{array}} \right]\]Next, we need to get a 1 into the lower right corner of the first two columns. This means changing the red -11 into a 1. This is usually accomplished with the second row operation. If we divide the second row by -11 we will get the 1 in that spot that we need.
\[\require{color}\left[ {\begin{array}{rr|r}1&3&1\\0&{ - 11}&{11}\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{{11}}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{\color{Red} 3}&1\\0&1&{ - 1}\end{array}} \right]\]Okay, we’re almost done. The final step is to turn the red three into a zero. Again, this almost always requires the third row operation. Here is the operation for this final step.
\[\require{color}\left[ {\begin{array}{rr|r}1&3&1\\0&1&{ - 1}\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} - 3{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&0&4\\0&1&{ - 1}\end{array}} \right]\]We have the augmented matrix in the required form and so we’re done. The solution to this system is \(x = 4\) and \(y = - 1\).
b \(\begin{align*} - 2x + y & = - 3\\ x - 4y & = - 2\end{align*}\) Show Solution
In this part we won’t put in as much explanation for each step. We will mark the next number that we need to change in red as we did in the previous part.
We’ll first write down the augmented matrix and then get started with the row operations.
\[\require{color}\left[ {\begin{array}{rr|r}{\color{Red} - 2}&1&{ - 3}\\1&{ - 4}&{ - 2}\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{ - 4}&{ - 2}\\{\color{Red} - 2}&1&{ - 3}\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} + 2{R_1} \to {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{ - 4}&{ - 2}\\0&{\color{Red} - 7}&{ - 7}\end{array}} \right]\]Before proceeding with the next step let’s notice that in the second matrix we had one’s in both spots that we needed them. However, the only way to change the -2 into a zero that we had to have as well was to also change the 1 in the lower right corner as well. This is okay. Sometimes it will happen and trying to keep both ones will only cause problems.
Let’s finish the problem.
\[\require{color}\left[ {\begin{array}{rr|r}1&{ - 4}&{ - 2}\\0&{\color{Red} - 7}&{ - 7}\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{7}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{\color{Red} - 4}&{ - 2}\\0&1&1\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 4{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&0&2\\0&1&1\end{array}} \right]\]The solution to this system is then \(x = 2\) and \(y = 1\).
c \(\begin{align*}3x - 6y & = - 9\\ - 2x - 2y & = 12\end{align*}\) Show Solution
Let’s first write down the augmented matrix for this system.
\[\require{color}\left[ {\begin{array}{rr|r}{\color{Red} 3}&{ - 6}&{ - 9}\\{ - 2}&{ - 2}&{12}\end{array}} \right]\]Now, in this case there isn’t a 1 in the first column and so we can’t just interchange two rows as the first step. However, notice that since all the entries in the first row have 3 as a factor we can divide the first row by 3 which will get a 1 in that spot and we won’t put any fractions into the problem.
Here is the work for this system.
\[\require{color}\left[ {\begin{array}{rr|r}{\color{Red} 3}&{ - 6}&{ - 9}\\{ - 2}&{ - 2}&{12}\end{array}} \right]\begin{array}{*{20}{c}}{\frac{1}{3}{R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{ - 2}&{ - 3}\\{\color{Red} - 2}&{ - 2}&{12}\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} + 2{R_1} \to {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{ - 2}&{ - 3}\\0&{\color{Red} - 6}&6\end{array}} \right]\] \[\require{color}\left[ {\begin{array}{rr|r}1&{ - 2}&{ - 3}\\0&{\color{Red} - 6}&6\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{6}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&{\color{Red} - 2}&{ - 3}\\0&1&{ - 1}\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 2{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1&0&{ - 5}\\0&1&{ - 1}\end{array}} \right]\]The solution to this system is \(x = - 5\) and \(y = - 1\).
It is important to note that the path we took to get the augmented matrices in this example into the final form is not the only path that we could have used. There are many different paths that we could have gone down. All the paths would have arrived at the same final augmented matrix however so we should always choose the path that we feel is the easiest path. Note as well that different people may well feel that different paths are easier and so may well solve the systems differently. They will get the same solution however.
For two equations and two unknowns this process is probably a little more complicated than just the straight forward solution process we used in the first section of this chapter. This process does start becoming useful when we start looking at larger systems. So, let’s take a look at a couple of systems with three equations in them.
In this case the process is basically identical except that there’s going to be more to do. As with two equations we will first set up the augmented matrix and then use row operations to put it into the form,
\[\left[ {\begin{array}{rrr|r}1&0&0&p\\0&1&0&q\\0&0&1&r\end{array}} \right]\]Once the augmented matrix is in this form the solution is \(x = p\), \(y = q\) and \(z = r\). As with the two equations case there really isn’t any set path to take in getting the augmented matrix into this form. The usual path is to get the 1’s in the correct places and 0’s below them. Once this is done we then try to get zeroes above the 1’s.
Let’s work a couple of examples to see how this works.
- \(\begin{align*}3x + y - 2z & = 2\\ x - 2y + z & = 3\\ 2x - y - 3z & = 3\end{align*}\)
- \(\begin{align*}3x + y - 2z & = - 7\\ 2x + 2y + z & = 9\\ - x - y + 3z & = 6\end{align*}\)
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,
\[x = - 2,\,\,\,y = 5,\,\,\,z = 3\]Using Gauss-Jordan elimination to solve a system of three equations can be a lot of work, but it is often no more work than solving directly and is many cases less work. If we were to do a system of four equations (which we aren’t going to do) at that point Gauss-Jordan elimination would be less work in all likelihood that if we solved directly.
Also, as we saw in the final example worked in this section, there really is no one set path to take through these problems. Each system is different and may require a different path and set of operations to make. Also, the path that one person finds to be the easiest may not by the path that another person finds to be the easiest. Regardless of the path however, the final answer will be the same.