## Elementary Linear Algebra 7th Edition

$$x_1=0, \quad x_2=0 , \quad x_3=4, \quad x_4=-1, \quad x_5=2.$$
The augmented matrix is given by $$\left[ \begin {array}{cccccc} 1&5&3&0&0&14\\ 0&4&2& 5&0&3\\ 0&0&3&8&6&16\\ 2&4&0&0&-2&0 \\ 2&0&-1&0&0&0\end {array} \right] .$$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[ \begin {array}{cccccc} 1&0&0&0&0&2\\ 0&1&0&0 &0&0\\ 0&0&1&0&0&4\\ 0&0&0&1&0&-1 \\ 0&0&0&0&1&2\end {array} \right] .$$ From which we get the solution $$x_1=0, \quad x_2=0 , \quad x_3=4, \quad x_4=-1, \quad x_5=2.$$