Show that the following matrix is a unitary matrix.

$\left(\begin{array}{cc}\frac{1+i\sqrt{3}}{4}& \frac{\sqrt{3}\left(1+i\right)}{2\sqrt{2}}\\ \frac{-\sqrt{3}\left(1+i\right)}{2\sqrt{2}}& \frac{\left(\sqrt{3}+i\right)}{4}\end{array}\right)$

The given matrix is:

A unitary matrix is one whose conjugate transpose equals its inverse. Real orthogonal matrices have a complex analogue in the form of unitary matrices.

If the Matrix A satisfies the condition:

$\mathit{A}{\mathit{A}}^{\mathbf{†}}\mathbf{=}{\mathit{A}}^{\mathbf{†}}\mathit{A}\mathbf{=}\mathit{I}$, then Matrix Ais a unitary matrix.

The Hermitian conjugate of the given matrix A is given below:

$A^{†}=\left[\begin{array}{cc}\frac{\left(1-\sqrt{3}\right)}{4}& -\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\\ \frac{\sqrt{3}}{2\sqrt{2}}(1-i)& \frac{(\sqrt{3}-i)}{4}\end{array}\right]$

The products are:

role="math" localid="1664265399341" $$\begin{gathered} A{A}^{†}=\left[\begin{array}{cc}\frac{(1-i\sqrt{3})}{4}& -\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\\ \frac{\sqrt{3}}{2\sqrt{2}}(1-i)& \frac{(\sqrt{3}-i)}{4}\end{array}\right]\left[\begin{array}{cc}\frac{{\displaystyle (1-i\sqrt{3})}}{{\displaystyle 4}}& -\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2\sqrt{2}}}(1-i)\\ \frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2\sqrt{2}}}(1-i)& \frac{{\displaystyle (\sqrt{3}-i)}}{{\displaystyle 4}}\end{array}\right] \\ =\left[\begin{array}{cc}\left(\frac{(1-i\sqrt{3})}{4}\times \frac{(1-i\sqrt{3})}{4}\right)+\left(-\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\times -\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\right)& \left(\frac{(1-i\sqrt{3})}{4}\times \frac{\sqrt{3}}{2\sqrt{2}}(1-i)\right)+\left(\frac{(1-i\sqrt{3})}{4}\times \frac{(\sqrt{3}-i)}{4}\right)\\ \left(\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\times \frac{(1-i\sqrt{3})}{4}\right)+\left(\frac{(\sqrt{3}-i)}{4}\times -\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\right)& \left(\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\times \frac{\sqrt{3}}{2\sqrt{2}}(1-i)\right)+\left(\frac{(\sqrt{3}-i)}{4}\times \frac{(\sqrt{3}-i)}{4}\right)\end{array}\right] \\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

$A{A}^{†}=I$

And

$$\begin{gathered} A^{†}A=\left[\begin{array}{cc}\frac{(1-i\sqrt{3})}{4}& \frac{\sqrt{3}}{2\sqrt{2}}(1-i)\\ -\frac{\sqrt{3}}{2\sqrt{2}}(1-i)& \frac{(\sqrt{3}-i)}{4}\end{array}\right]\left[\begin{array}{cc}\frac{(1-i\sqrt{3})}{4}& -\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\\ \frac{\sqrt{3}}{2\sqrt{2}}(1-i)& \frac{(\sqrt{3}-i)}{4}\end{array}\right] \\ =\left[\begin{array}{cc}\left(\frac{(1-i\sqrt{3})}{4}\times \frac{(1-i\sqrt{3})}{4}\right)+\left(-\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\times -\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\right)& \left(\frac{(1-i\sqrt{3})}{4}\times \frac{\sqrt{3}}{2\sqrt{2}}(1-i)\right)+\left(\frac{(1-i\sqrt{3})}{4}\times \frac{(\sqrt{3}-i)}{4}\right)\\ \left(\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\times \frac{(1-i\sqrt{3})}{4}\right)+\left(\frac{(\sqrt{3}-i)}{4}\times -\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\right)& \left(\frac{\sqrt{3}}{2\sqrt{2}}(1-i)\times \frac{\sqrt{3}}{2\sqrt{2}}(1-i)\right)+\left(\frac{(\sqrt{3}-i)}{4}\times \frac{(\sqrt{3}-i)}{4}\right)\end{array}\right] \\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ {A}^{†}A=I \end{gathered}$$

Which means matrix is a unitary matrix.