Show that the matrix \(A=\frac{1}{\sqrt3}\begin{bmatrix}1 & 1+i\\ 1-i & -1\end{bmatrix}\) is unitary and hence find \(A^{-1}\).

Given: \(A=\frac{1}{\sqrt3}\begin{bmatrix}1 & 1+i\\ 1-i & -1\end{bmatrix}\). A is unitary if \(AA^{\theta}=I\), where \(A^{\theta}=(\bar A)^{T}\).

Step 1: \(\bar A=\frac{1}{\sqrt3}\begin{bmatrix}1 & 1-i\\ 1+i & -1\end{bmatrix}\), so \(A^{\theta}=(\bar A)^T=\frac{1}{\sqrt3}\begin{bmatrix}1 & 1+i\\ 1-i & -1\end{bmatrix}=A\).

Step 2: \(AA^{\theta}=\frac13\begin{bmatrix}1 & 1+i\\ 1-i & -1\end{bmatrix}\begin{bmatrix}1 & 1+i\\ 1-i & -1\end{bmatrix}=\frac13\begin{bmatrix}1+(1+i)(1-i) & (1+i)-(1+i)\\ (1-i)-(1-i) & (1-i)(1+i)+1\end{bmatrix}=\frac13\begin{bmatrix}3 & 0\\ 0 & 3\end{bmatrix}=I\).

Since \(AA^{\theta}=I\), A is unitary.

Step 3: For a unitary matrix \(A^{-1}=A^{\theta}\). Hence \(A^{-1}=A^{\theta}=\frac{1}{\sqrt3}\begin{bmatrix}1 & 1+i\\ 1-i & -1\end{bmatrix}\) (i.e. \(A^{-1}=A\)).