## College Algebra 7th Edition

Dependent. The solution set is $\{(2-3t\ ,\ 3-5t,\ t) \ |\ \ t\in \mathbb{R}\}$
Start with the augmented matrix, row reduce to reduced row echelon form (Gauss-Jordan.) $\left[\begin{array}{llll} 2 & -3 & -9 & -5\\ 1 & 0 & 3 & 2\\ -3 & 1 & -4 & -3 \end{array}\right]\rightarrow\left(\begin{array}{l} R_{1}\leftrightarrow R_{2}.\\ .\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 0 & 3 & 2\\ 2 & -3 & -9 & -5\\ -3 & 1 & -4 & -3 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ -2R_{1}.\\ +3R_{1}. \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 0 & 3 & 2\\ 0 & -3 & -15 & -9\\ 0 & 1 & 5 & 3 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ \div(-3).\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 0 & 3 & 2\\ 0 & 1 & 5 & 3\\ 0 & 1 & 5 & 3 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ .\\ -R_{2}. \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 0 & 3 & 2\\ 0 & 1 & 5 & 3\\ 0 & 0 & 0 & 0 \end{array}\right]$ The last row represents 0=0, which is always true. The system is consistent (has solutions) and dependent. Let $z=t\in \mathbb{R}$, an arbitrary number, Row 2 $\Rightarrow y=3-5t$ Row $1 \Rightarrow x=2-3t$ The solution set is $\{(2-3t\ ,\ 3-5t,\ t) \ |\ \ t\in \mathbb{R}\}$