## Elementary Linear Algebra 7th Edition

$$x_1=4, \quad x_2=-5, \quad x_3=2.$$
Given \begin{aligned} x_{1}- x_{2}+4 x_{3} &=17 \\ x_{1}+ 3x_{2} &=-11 \\ - 6x_{2}+5 x_{3} &=40 \end{aligned}. The augmented matrix is given by $$\left[ \begin {array}{cccc} 1&-1&-4&17\\ 1&3&0&-11 \\ 0&-6&5&40\end {array} \right] .$$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[ \begin {array}{cccc} 1&0&0&4 \\ 0&1&0&-5\\ 0&0&1&2\end {array} \right] .$$ From which the solution is $$x_1=4, \quad x_2=-5, \quad x_3=2.$$