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### Section 7.4 : More on the Augmented Matrix

1. 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*}x - 7y & = - 11\\ 5x + 2y & = - 18\end{align*}

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Start Solution

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

$\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 11}\\5&2&{ - 18}\end{array}} \right]$ Show Step 2

We need to make the number in the upper left corner a one. In this case it already is and so there really isnâ€™t anything to do in this step for this particular problem.

Show Step 3

Next, we need to convert the 5 below the 1 into a zero and we can do that with the following elementary row operation.

$\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 11}\\5&2&{ - 18}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} - 5{R_{\,1}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 11}\\0&{37}&{37}\end{array}} \right]$ Show Step 4

The next step is to turn the number at the bottom of the second column (37 in this case) into a one. The following elementary row operation will do that for us.

$\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 11}\\0&{37}&{37}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{{37}}{R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 11}\\0&1&1\end{array}} \right]$ 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.

$\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 11}\\0&1&1\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} + 7{R_{\,2}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&0&{ - 4}\\0&1&1\end{array}} \right]$ Show Step 6

From the final augmented matrix we found in Step 5 we get the solution to the system is : $$\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 4,\,\,y = 1}}$$.