## Elementary Linear Algebra 7th Edition

The system has a unique solution, that is, $$x_1=1, \quad x_2=-1, \quad x_3=2.$$
Given \begin{aligned} x_{1}-2 x_{2}+3 x_{3} &=9 \\ -x_{1}+3 x_{2}-x_{3} &=-6 \\ 2 x_{1}-5 x_{2}+5 x_{3} &=17 \end{aligned}. The augmented matrix is given by $$\left[\begin{array}{rrrr} {1} & {-2} & {3} & {9} \\ {-1} & {3} & {-1} & {-6} \\ {2} & {-5} & {5} & {17} \end{array}\right].$$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&-1 \\ 0&0&1&2\end {array} \right].$$ The system has a unique solution, that is, $$x_1=1, \quad x_2=-1, \quad x_3=2.$$