Find all solutions of
\[ \begin{array}{r l} 2 x _ {1} - 3 x _ {2} - 7 x _ {3} + 5 x _ {4} + 2 x _ {5} = & - 2 \\ x _ {1} - 2 x _ {2} - 4 x _ {3} + 3 x _ {4} + & x _ {5} = - 2 \\ 2 x _ {1} \qquad - 4 x _ {3} + 2 x _ {4} + & x _ {5} = 3 \\ x _ {1} - 5 x _ {2} - 7 x _ {3} + 6 x _ {4} + 2 x _ {5} = & - 7. \end{array} \]
Solution
All solutions are \[ (x_1,x_2,x_3,x_4,x_5) =(1+2s-t,\ 2-s+t,\ s,\ t,\ 1), \qquad s,t\in F. \]
1. Row reduction of the augmented matrix yields\[ \left[ \begin{array}{ccccc|c} 1&0&-2&1&0&1\\ 0&1&1&-1&0&2\\ 0&0&0&0&1&1\\ 0&0&0&0&0&0 \end{array} \right]. \]
Applying Gaussian elimination to the coefficient matrix together with the right-hand side gives the displayed reduced row-echelon augmented matrix.
2. The displayed family is exactly the solution set.
The reduced system is \[ x_1-2x_3+x_4=1, \qquad x_2+x_3-x_4=2, \qquad x_5=1. \] Set the free variables \(x_3=s\) and \(x_4=t\). Then \[ x_1=1+2s-t, \qquad x_2=2-s+t, \qquad x_5=1, \] which gives precisely the stated solutions.