Paul's Online Notes
Paul's Online Notes
Home / Algebra / Systems of Equations / More on the Augmented Matrix
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 7.4 : More on the Augmented Matrix

3. 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.

\[\begin{align*}3x + 9y & = - 6\\ - 4x - 12y & = 8\end{align*}\]

Show All Steps Hide All Steps

Start Solution

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

\[\left[ {\begin{array}{rr|r}3&9&{ - 6}\\{ - 4}&{ - 12}&8\end{array}} \right]\] Show Step 2

We need to make the number in the upper left corner a one. In this case we can quickly do that by dividing the top row by 3.

\[\left[ {\begin{array}{rr|r}3&9&{ - 6}\\{ - 4}&{ - 12}&8\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{3}{R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&3&{ - 2}\\{ - 4}&{ - 12}&8\end{array}} \right]\] Show Step 3

Next, we need to convert the -4 below the 1 into a zero and we can do that with the following elementary row operation.

\[\left[ {\begin{array}{rr|r}1&3&{ - 2}\\{ - 4}&{ - 12}&8\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} + 4{R_{\,1}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&3&{ - 2}\\0&0&0\end{array}} \right]\] Show Step 4

The minute we see the bottom row of all zeroes we know that the system if dependent. We can convert the top row into an equation and solve for \(x\) as follows,

\[x + 3y = - 2\hspace{0.25in} \to \hspace{0.25in}x = - 3y - 2\]

From this we can write the solution as,

\[\begin{array}{l}{x = - 3t - 2}\\{y = t}\end{array}\hspace{0.25in}t{\mbox{ is any number}}\]