Q. 15 - Precalculus Enhanced with Graphing Utilities

Question

solve each system of equations using matrices.

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

$$\begin{gathered} x+2y+4z=-3 \\ 2x+7y+15z=-12 \\ 4x+7y+13z=-10 \end{gathered}$$

Step-by-Step Solution

Answer

The solution to the given equations is $\left(1,-2,0\right).$

Step 1. Given

$$\begin{gathered} x+2y+4z=-3 \\ 2x+7y+15z=-12 \\ 4x+7y+13z=-10 \end{gathered}$$

Step 2. Augmented matrix

$\left[\begin{array}{cccc}1& 2& 4& -3\\ 2& 7& 15& -12\\ 4& 7& 13& -10\end{array}\right]$

Step 3. Transformation

Transform the matrix to get it into echelon form

$$\begin{gathered} R_2=r_2-2r_1 \\ \left[\begin{array}{cccc}1& 2& 4& -3\\ 2-2\left(1\right)& 7-2\left(2\right)& 15-2\left(4\right)& -12-2(-3)\\ 4& 7& 13& -10\end{array}\right]\to \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 3& 7& -6\\ 4& 7& 13& -10\end{array}\right] \\ {R}_3=r_3-4r_1 \\ \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 3& 7& -6\\ 4& 7& 13& -10\end{array}\right]\to \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 3& 7& -6\\ 0& -1& -3& 2\end{array}\right] \\ {R}_2=\frac{r_2}{3} \\ \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 3& 7& -6\\ 0& -1& -3& 2\end{array}\right]\to \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 1& \frac{7}{3}& -2\\ 0& -1& -3& 2\end{array}\right] \\ {R}_3=r_3+r_2 \\ \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 1& \frac{7}{3}& -2\\ 0& -1& -3& 2\end{array}\right]\to \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 1& \frac{7}{3}& -2\\ 0& 0& -\frac{2}{3}& 0\end{array}\right] \\ {R}_3=-\frac{3}{2}{r}_3 \\ \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 1& \frac{7}{3}& -2\\ 0& 0& -\frac{2}{3}& 0\end{array}\right]\to \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 1& \frac{7}{3}& -2\\ 0& 0& 1& 0\end{array}\right] \\ Thus the equation becomes \\ x+2y+4z=-3 \\ y+\frac{7}{3}z=-2 \\ z=0 \\ Thus solving we get \\ y+0=-2 \\ y=--2 \\ x+2(-2)+4\left(0\right)=-3 \\ x=1 \end{gathered}$$