## Precalculus (6th Edition) Blitzer

$(7,-4,6)$
Step 1. List the system of equations in matrix form: $\begin{bmatrix} 1 & -1 & 1 & 17 \\ -4 & 1 & 5 & -2 \\ 2 & 3 & 1 & 8 \end{bmatrix}$ $\begin{array}{lcl} .. \\ 4R1+R2\to R2\\-2R1+R3\to R3 \end{array}$ Step 2. Perform the operations given on the right side of the matrix: $\begin{bmatrix} 1 & -1 & 1 & 17 \\ 0 & -3 & 9 & 66 \\ 0 & 5 & -1 & -26 \end{bmatrix}$ $\begin{array}{lcl} .. \\ -R2/3\to R2\\.. \end{array}$ Step 3. Perform the operations given on the right side of the matrix: $\begin{bmatrix} 1 & -1 & 1 & 17 \\ 0 & 1 & -3 & -22 \\ 0 & 5 & -1 & -26 \end{bmatrix}$ $\begin{array}{lcl} .. \\ ..\\-5R2+R3\to R3 \end{array}$ Step 4. Perform the operations given on the right side of the matrix: $\begin{bmatrix} 1 & -1 & 1 & 17 \\ 0 & 1 & -3 & -22 \\ 0 & 0 & 14 & 84 \end{bmatrix}$ $\begin{array}{lcl} .. \\ ..\\.. \end{array}$ Step 5. The last equation given is $14z=84$. Thus $z=6$ Step 6. Back-substitute and solve for $x,y$. We have $y-3z=-22$. Thus $y=-4$ and $x=7$ Step 7. The solution to the system is $(7,-4,6)$