## Precalculus (10th Edition)

Consistent Solution set: $\left\{\left(-3,2,1\right)\right\}$
We are given the system of equations: $\begin{cases} 2x+y=-4\\ -2y+4z=0\\ 3x-2z=-11 \end{cases}$ Write the augmented matrix: $\begin{bmatrix} 2&1&0&|&-4\\0&-2&4&|&0\\3&0&-2&|&-11\end{bmatrix}$ Perform row operations to bring the matrix to the reduced row echelon form: $R_3=-r_1+r_3$ $\begin{bmatrix} 2&1&0&|&-4\\0&-2&4&|&0\\1&-1&-2&|&-7\end{bmatrix}$ Interchange $R_1$ and $R_3$ $\begin{bmatrix}1&-1&-2&|&-7\\0&-2&4&|&0\\2&1&0&|&-4\end{bmatrix}$ $R_3=-2r_1+r_3$ $\begin{bmatrix}1&-1&-2&|&-7\\0&-2&4&|&0\\0&3&4&|&10\end{bmatrix}$ $R_2=-\dfrac{1}{2}r_2$ $\begin{bmatrix}1&-1&-2&|&-7\\0&1&-2&|&0\\0&3&4&|&10\end{bmatrix}$ $R_1=-r_2+r_3$ $\begin{bmatrix}1&0&-4&|&-7\\0&1&-2&|&0\\0&3&4&|&10\end{bmatrix}$ $R_3=-3r_2+r_1$ $\begin{bmatrix}1&0&-4&|&-7\\0&1&-2&|&0\\0&0&10&|&10\end{bmatrix}$ $R_3=\dfrac{1}{10}r_3$ $\begin{bmatrix}1&0&-4&|&-7\\0&1&-2&|&0\\0&0&1&|&1\end{bmatrix}$ $R_1=4r_3+r_1$ $\begin{bmatrix}1&0&0&|&-3\\0&1&-2&|&0\\0&0&1&|&1\end{bmatrix}$ $R_2=2r_3+r_2$ $\begin{bmatrix}1&0&0&|&-3\\0&1&0&|&2\\0&0&1&|&1\end{bmatrix}$ Write the corresponding system of equations: $\begin{cases} 1x+0y+0z=-3\\ 0x+1y+0z=2\\ 0x+0y+1z=1 \end{cases}$ $\begin{cases} x=-3\\ y=2\\ z=1 \end{cases}$ The system is consistent. The solution set is: $\left\{\left(-3,2,1\right)\right\}$