## Precalculus (6th Edition) Blitzer

$(2,1,-4)$
Step 1. Based on the given system of equations, we can set up an augmented matrix and perform the row operations: $\begin{bmatrix} 1 & -2 & 1 & | & -4 \\ 2 & 2 & -1 & | & 10 \\ 4 & -1 & 2 & | & -1 \end{bmatrix} \begin{array} ..\\R2-2R1\to R2\\R3-4R1\to R3 \end{array}$ $\begin{bmatrix} 1 & -2 & 1 & | & -4 \\ 0 & 6 & -3 & | & 18 \\ 0 & 7 & -2 & | & 15 \end{bmatrix} \begin{array} ..\\R2/3\to R2\\.. \end{array}$ $\begin{bmatrix} 1 & -2 & 1 & | & -4 \\ 0 & 2 & -1 & | & 6 \\ 0 & 7 & -2 & | & 15 \end{bmatrix} \begin{array} ..\\..\\2R3-7R2\to R3 \end{array}$ $\begin{bmatrix} 1 & -2 & 1 & | & -4 \\ 0 & 2 & -1 & | & 6 \\ 0 & 0 & 3 & | & -12 \end{bmatrix} \begin{array} ..\\..\\.. \end{array}$ Step 2. The last row gives $3z=-12$; thus $z=-4$ Step 3. Using back-substitution, we have $2y-z=6$ and $2y=6+z=2$; thus $y=1$ Step 4. Using back-substitution, we have $x-2y+z=-4$ and $x=-4+2+4=2$ Step 5. The solution set for the system is $(2,1,-4)$