Find all solutions to the following system of equations by row-reducing the coefficient matrix:
\[ \begin{array}{r l} \frac {1}{3} x _ {1} + 2 x _ {2} - 6 x _ {3} & = 0 \\ - 4 x _ {1} + 5 x _ {3} & = 0 \\ - 3 x _ {1} + 6 x _ {2} - 1 3 x _ {3} & = 0 \\ - \frac {7}{3} x _ {1} + 2 x _ {2} - \frac {8}{3} x _ {3} & = 0 \end{array} \]
Solution
The complete solution set is \[ \boxed{ (x_1,x_2,x_3) =t\left(\frac54,\frac{67}{24},1\right), \qquad t\in\mathbb C.} \]
1. Row-reduce the coefficient matrix.
The coefficient matrix is \[ A= \begin{pmatrix} 1/3&2&-6\\ -4&0&5\\ -3&6&-13\\ -7/3&2&-8/3 \end{pmatrix}. \] First replace \(R_1\) by \(3R_1\). Then apply \[ R_2\leftarrow R_2+4R_1, \qquad R_3\leftarrow R_3+3R_1, \qquad R_4\leftarrow R_4+\frac73R_1. \] This gives \[ \begin{pmatrix} 1&6&-18\\ 0&24&-67\\ 0&24&-67\\ 0&16&-134/3 \end{pmatrix}. \] Replace \(R_2\) by \(R_2/24\), then use \[ R_1\leftarrow R_1-6R_2, \qquad R_3\leftarrow R_3-24R_2, \qquad R_4\leftarrow R_4-16R_2. \] The row-reduced matrix is \[ \begin{pmatrix} 1&0&-5/4\\ 0&1&-67/24\\ 0&0&0\\ 0&0&0 \end{pmatrix}. \] Elementary row operations preserve the solution set of a homogeneous system [@HK71].
2. Read off the solutions.
The reduced equations are \[ x_1-\frac54x_3=0, \qquad x_2-\frac{67}{24}x_3=0. \] Thus \(x_3=t\) is free and \[ x_1=\frac54t, \qquad x_2=\frac{67}{24}t. \] Conversely every such triple satisfies the reduced system, hence also the original system. Therefore the displayed family is exhaustive.