College Algebra (6th Edition)

Solution set: $\{( t-2$ , $t-2$ , $t )$ , $t\in \mathbb{R}\}$
Start with the augmented matrix. Use matrix row operations to get ls down the main diagonal from upper left to lower right, and 0s below the $1\mathrm{s}$ (row-echelon form). $\left[\begin{array}{lllll} 5 & -11 & 6 & | & 12\\ -1 & 3 & -2 & | & -4\\ 3 & -5 & 2 & | & 4 \end{array}\right]$ Swapping R1 and R2 brings a $-1$ into the first diagonal position, $\left[\begin{array}{lllll} -1 & 3 & -2 & | & -4\\ 5 & -11 & 6 & | & 12\\ 3 & -5 & 2 & | & 4 \end{array}\right]\left\{\begin{array}{l} \times(-1).\\ 5R1+R2\rightarrow R2.\\ 3R1+R3\rightarrow R3. \end{array}\right.$ $\left[\begin{array}{lllll} 1 & -3 & 2 & | & 4\\ 0 & 4 & -4 & | & -8\\ 0 & 4 & -4 & | & -8 \end{array}\right]\left\{\begin{array}{l} .\\ \div 4.\\ -R2+R3\rightarrow R3. \end{array}\right.$ $\left[\begin{array}{lllll} 1 & -3 & 2 & | & 4\\ 0 & 1 & -1 & | & -2\\ 0 & 0 & 0 & | & 0 \end{array}\right]$ Last row: all zeros, so the system is consistent, but there is no unique solution . We take z to be any number $t\in \mathbb{R}.$ Substituting into row 2: $y-t=-2\qquad/+t$ $y=t-2$ Substituting into row $1$: $x-3(t-2)+2t=4$ $x-t+6=4$ $x=t-2$ Solution set: $\{( t-2$ , $t-2$ , $t )$ , $t\in \mathbb{R}\}$