Find a row-reduced matrix which is row-equivalent to
\[ A = \left[ \begin{array}{c c c} i & - (1 + i) & 0 \\ 1 & - 2 & 1 \\ 1 & 2 i & - 1 \end{array} \right]. \]
Solution
A row-reduced matrix row-equivalent to \(A\) is \[ R=\begin{bmatrix} 1&0&i\\ 0&1&\dfrac{-1+i}{2}\\ 0&0&0 \end{bmatrix}. \]
1. Row reduction of \(A\) yields \(R\).
Applying elementary row operations to \[ A=\begin{bmatrix} i&-(1+i)&0\\ 1&-2&1\\ 1&2i&-1 \end{bmatrix} \] gives the reduced row-echelon form \[ \operatorname{rref}(A) =\begin{bmatrix} 1&0&i\\ 0&1&(-1+i)/2\\ 0&0&0 \end{bmatrix}. \] Each step in Gaussian elimination is an elementary row operation, so this matrix is row-equivalent to \(A\).
2. The displayed matrix is row-reduced.
Its two nonzero rows have leading entries \(1\) in columns \(1\) and \(2\), each pivot is the only nonzero entry in its column, the second pivot lies to the right of the first, and the zero row is last.