## Precalculus (10th Edition)

Consistent Solution set: $\left\{\left(3,-\dfrac{8}{3},\dfrac{1}{9}\right)\right\}$
We are given the system of equations: $\begin{cases} x+4y-3z=-8\\ 3x-y+3z=12\\ x+y+6z=1 \end{cases}$ Write the augmented matrix: $\begin{bmatrix} 1&4&-3&|&-8\\3&-1&3&|&12\\1&1&6&|&1\end{bmatrix}$ Perform row operations to bring the matrix to the reduced row echelon form: $R_2=-3r_1+r_2$ $\begin{bmatrix} 1&4&-3&|&-8\\0&-13&12&|&36\\1&1&6&|&1\end{bmatrix}$ $R_3=-r_1+r_3$ $\begin{bmatrix}1&4&-3&|&-8\\0&-13&12&|&36\\0&-3&9&|&9\end{bmatrix}$ $R_2=-4r_3+r_2$ $\begin{bmatrix}1&4&-3&|&-8\\0&-1&-24&|&0\\0&-3&9&|&9\end{bmatrix}$ $R_2=-r_2$ $\begin{bmatrix}1&4&-3&|&-8\\0&1&24&|&0\\0&-3&9&|&9\end{bmatrix}$ $R_3=3r_2+r_3$ $\begin{bmatrix}1&4&-3&|&-8\\0&1&24&|&0\\0&0&81&|&9\end{bmatrix}$ $R_3=\dfrac{1}{81}r_3$ $\begin{bmatrix}1&4&-3&|&-8\\0&1&24&|&0\\0&0&1&|&\dfrac{1}{9}\end{bmatrix}$ Write the corresponding system of equations: $\begin{cases} x+4y-3z=-8\\ y+24z=0\\ z=\frac{1}{9} \end{cases}$ Solve the system by back substitution: $z=\frac{1}{9}$ $y+24z=0$ $y+24\left(\frac{1}{9}\right)=0$ $y=-\frac{8}{3}$ $x+4y-3z=-8$ $x+4\left(-\dfrac{8}{3}\right)-3\left(\dfrac{1}{9}\right)=-8$ $x-\dfrac{32}{3}-\dfrac{1}{3}=-8$ $x-11=-8$ $x=3$ The system is consistent. The solution is: $\left\{\left(3,-\dfrac{8}{3},\dfrac{1}{9}\right)\right\}$