Q. 13 - Precalculus Enhanced with Graphing Utilities

Question

solve each system of equations using matrices.

If the system has no solution, say that it is inconsistent.

$$\begin{gathered} 6x+3y=12 \\ 2x-y=-2 \end{gathered}$$

Step-by-Step Solution

Answer

The solution to the equation is $\left(\frac{1}{2},3\right)$.

Step 1. Given

$$\begin{gathered} 6x+3y=12 \\ 2x-y=-2 \end{gathered}$$

Step 2. Augmented matrix

$$\begin{gathered} 6x+3y=12 \\ 2x-y=-2 \\ Consider, \\ 6x+3y=12 \\ 2x+y=4 \\ therefore the augmented matrix will be \\ \left[\begin{array}{ccc}2& 1& 4\\ 2& -1& -2\end{array}\right] \end{gathered}$$

Step 3. Calculation using transformation

$$\begin{gathered} \left[\begin{array}{ccc}2& 1& 4\\ 2& -1& -2\end{array}\right] \\ {R}_1=\frac{r_1}{2} \\ \left[\begin{array}{ccc}2& 1& 4\\ 2& -1& -2\end{array}\right]\to \left[\begin{array}{ccc}1& \frac{1}{2}& 2\\ 2& -1& -2\end{array}\right] \\ Now R_2=r_2-2r_1 \\ \left[\begin{array}{ccc}1& \frac{1}{2}& 2\\ 2& -1& -2\end{array}\right]\to \left[\begin{array}{ccc}1& \frac{1}{2}& 2\\ 0& -2& -6\end{array}\right] \\ Now R_2=\frac{r_2}{-2} \\ \left[\begin{array}{ccc}1& \frac{1}{2}& 2\\ 0& -2& -6\end{array}\right]\to \left[\begin{array}{ccc}1& \frac{1}{2}& 2\\ 0& 1& 3\end{array}\right] \\ thus we get \\ x+\frac{1}{2}y=3 \\ y=3 \\ therefore (x,y)=\left(\frac{1}{2},3\right) \end{gathered}$$