## Precalculus (6th Edition) Blitzer

$(-\frac{2t}{3}, \frac{t}{3}, 0, t)$
Step 1. Based on the system of equations, write an augmented matrix as $\begin{bmatrix} 1 & -1 & 0 & 1 & | & 0 \\ 1 & -4 & 1 & 2 & | & 0 \\ 3 & 0 & 1 & 2 & | & 0 \end{bmatrix} \begin{array} ..\\R1-R2\to R2\\-3R1+R3\to R3 \end{array}$ Step 2. Perform row operations given to the right of the matrix: $\begin{bmatrix} 1 & -1 & 0 & 1 & | & 0 \\ 0 & 3 & -1 & -1 & | & 0 \\ 0 & 3 & 1 & -1 & | & 0 \end{bmatrix} \begin{array} ..\\..\\R3-R2\to R3 \end{array}$ Step 3. Perform row operations given to the right of the matrix: $\begin{bmatrix} 1 & -1 & 0 & 1 & | & 0 \\ 0 & 3 & -1 & -1 & | & 0 \\ 0 & 0 & 2 & 0 & | & 0 \end{bmatrix} \begin{array} ..\\..\\.. \end{array}$ Step 4. The last row gives $y=0$. As the number of equations is less than the number of variables, the system is dependent. Let $z=t$; the middle row gives $3x-t=0$ and $x=\frac{t}{3}$ Step 5. The first row gives $w-\frac{t}{3}+t=0$ and $w=-\frac{2t}{3}$ Step 6. The solution set can be written as $(-\frac{2t}{3}, \frac{t}{3}, 0, t)$