Show that the matrix \(A=\dfrac12\begin{bmatrix}1+i & -1+i\\ 1+i & 1-i\end{bmatrix}\) is unitary and hence find \(A^{-1}\).

Answer: \(AA^{\theta}=I\), so \(A\) is unitary and \(A^{-1}=A^{\theta}=\dfrac12\begin{bmatrix}1-i & 1-i\\ -1-i & 1+i\end{bmatrix}\).

\(A^{\theta}=(\bar A)^T=\dfrac12\begin{bmatrix}1-i & 1-i\\ -1-i & 1+i\end{bmatrix}\).

\[AA^{\theta}=\frac14\begin{bmatrix}(1+i)(1-i)+(-1+i)(-1-i) & (1+i)(1-i)+(-1+i)(1+i)\\ (1+i)(1-i)+(1-i)(-1-i) & (1+i)(1-i)+(1-i)(1+i)\end{bmatrix}\]

Using \((1+i)(1-i)=2\), \((-1+i)(-1-i)=2\), \((-1+i)(1+i)=-2\), \((1-i)(-1-i)=-2\):

\[AA^{\theta}=\frac14\begin{bmatrix}4 & 0\\ 0 & 4\end{bmatrix}=I.\]

Hence \(A\) is unitary, and for a unitary matrix \(A^{-1}=A^{\theta}=\dfrac12\begin{bmatrix}1-i & 1-i\\ -1-i & 1+i\end{bmatrix}\).