If
\[ A = \left[ \begin{array}{c c c} 3 & - 1 & 2 \\ 2 & 1 & 1 \\ 1 & - 3 & 0 \end{array} \right] \]
find all solutions of AX = 0 by row-reducing A.
Solution
Row-reduce \(A\): \[ \begin{bmatrix} 3&-1&2\\ 2&1&1\\ 1&-3&0 \end{bmatrix} \sim \begin{bmatrix} 1&-3&0\\ 0&7&1\\ 0&8&2 \end{bmatrix} \sim \begin{bmatrix} 1&-3&0\\ 0&7&1\\ 0&0&6 \end{bmatrix}. \]
1. The matrix has a pivot in every column.
The displayed echelon form has three nonzero pivots; equivalently its reduced row-echelon form is \(I_3\).
2. The only solution of \(AX=0\) is\[ X=\begin{bmatrix}0\\0\\0\end{bmatrix}. \]
Row operations preserve the solution set of a homogeneous system. Since the reduced system is \(I_3X=0\), all three coordinates vanish.