Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.4
Solution: We have $$A\rightarrow\left[\begin{array}{ccc}
1 & -2 & 1\\
i & -(1+i) & 0\\
1 & 2i & -1
\end{array}\right]\rightarrow
\left[\begin{array}{ccc}
1 & -2 & 1\\
0 & -1+i & -i\\
0 & 2+2i & -2
\end{array}\right]\rightarrow
\left[\begin{array}{ccc}
1 & -2 & 1\\
0 & 1 & \frac{1-i}2\\
0 &2+ 2i & -2
\end{array}\right]
$$ $$\rightarrow
\left[\begin{array}{ccc}
1 & -2 & 1\\
0 & 1 & \frac{i-1}2\\
0 & 0 & 0
\end{array}\right]\rightarrow
\left[\begin{array}{ccc}
1 & 0 & i\\
0 & 1 & \frac{i-1}2\\
0 & 0 & 0
\end{array}\right].
$$
