Do Exercise 1, but with
\[ A = \left[ \begin{array}{c c c} 2 & 0 & i \\ 1 & - 3 & - i \\ i & 1 & 1 \end{array} \right]. \]
Solution
The matrix is invertible, so its reduced row echelon form is \(R=I_3\). One may take \(P=A^{-1}\), namely \[ P= \begin{bmatrix} \frac13&\frac1{30}-\frac{i}{10}&\frac1{10}-\frac{3i}{10}\\ 0&-\frac3{10}-\frac{i}{10}&\frac1{10}-\frac{3i}{10}\\ -\frac{i}{3}&\frac15+\frac{i}{15}&\frac35+\frac{i}{5} \end{bmatrix}. \] Direct multiplication gives \(PA=I_3\), hence \(R=PA\). Since \(P\) has inverse \(A\), it is invertible as required.