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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.

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

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Start Solution

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

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

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