Let \[ A = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 5 & -3 \\ 0 & 0 & 0 \end{pmatrix}. \] Show that \(\ker A = \operatorname{span}\{(1,3,5)\}\).
Solution
Row-reducing \(A\) (already nearly echelon) gives the system \(x-2y+z=0\) and \(5y-3z=0\). Thus \(y=\frac{3}{5}z\) and \(x=2y-z=\frac{1}{5}z\). Taking \(z=5\) yields the basis vector \((1,3,5)\). The kernel is one-dimensional, so \(\ker A = \operatorname{span}\{(1,3,5)\}\).