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Section 7-4 : More on the Augmented Matrix

8. For the following system of equations convert the system into an augmented matrix and use the augmented matrix techniques to determine the solution to the system or to determine if the system is inconsistent or dependent.

\[\begin{align*}3x - 9z & = 33\\ 7x - 4y - z & = - 15\\ 4x + 6y + 5z & = - 6\end{align*}\]

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Start Solution

The first step is to write down the augmented matrix for the system of equations.

\[\left[ {\begin{array}{rrr|r}3&0&{ - 9}&{33}\\7&{ - 4}&{ - 1}&{ - 15}\\4&6&5&{ - 6}\end{array}} \right]\]

Note the zero in the second column of the first row. Recall that the second column corresponds to the coefficients of the \(y\)’s in each equation and because there is no \(y\) in the first equation that coefficient must be zero.

Show Step 2

We need to make the number in the upper left corner a one. We can easily do that with the following elementary row operation.

\[\left[ {\begin{array}{rrr|r}3&0&{ - 9}&{33}\\7&{ - 4}&{ - 1}&{ - 15}\\4&6&5&{ - 6}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{3}{R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\7&{ - 4}&{ - 1}&{ - 15}\\4&6&5&{ - 6}\end{array}} \right]\] Show Step 3

Next, we need to convert the 7 and the 4 below the 1 in the first column into zeroes and we can do that with the following elementary row operations.

\[\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\7&{ - 4}&{ - 1}&{ - 15}\\4&6&5&{ - 6}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} - 7{R_{\,1}} \to {R_{\,2}}}\\{{R_{\,3}} - 4{R_{\,1}} \to {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&{ - 4}&{20}&{ - 92}\\0&6&{17}&{ - 50}\end{array}} \right]\] Show Step 4

We now need to turn the -4 in the second row into a one and that can be done with the following elementary row operation.

\[\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&{ - 4}&{20}&{ - 92}\\0&6&{17}&{ - 50}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{ - \frac{1}{4}{R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&6&{17}&{ - 50}\end{array}} \right]\] Show Step 5

Next, we need to turn the 6 in the third row into a zero. The following elementary row operation will do that for us.

\[\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&6&{17}&{ - 50}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,3}} - 6{R_{\,2}} \to {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&0&{47}&{ - 188}\end{array}} \right]\] Show Step 6

Okay, we need to turn the 47 in the third row into a one and we can do that as follows,

\[\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&0&{47}&{ - 188}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{{47}}{R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&0&1&{ - 4}\end{array}} \right]\] Show Step 7

Next, we need to turn the -5 and the -3 in the third column into zeroes. The following elementary row operations will do that for us.

\[\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&0&1&{ - 4}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} + 3{R_{\,3}} \to {R_{\,1}}}\\{{R_{\,2}} + 5{R_{\,3}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&0&{ - 1}\\0&1&0&3\\0&0&1&{ - 4}\end{array}} \right]\] Show Step 8

Normally we would have another step to do. We would need to turn the number in the first row and second column into a zero. However, in this case there is already a zero there and so there is no work to do in this step.

The final form of the augmented matrix is then,

\[\left[ {\begin{array}{rrr|r}1&0&0&{ - 1}\\0&1&0&3\\0&0&1&{ - 4}\end{array}} \right]\]

As this step has shown we occasionally will get a number “for free”. In other words, the work we put into an intermediate step will give us not only the number we were looking for in that step but will also put in a number that we need in a later step. Or, as in this case, the number we needed was actually there from the start.

Show Step 9

From the final augmented matrix we found in Step 8 we get the solution to the system is : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 1,\,\,y = 3,\,\,z = - 4}}\).