Show that the following matrix is a unitary matrix. $$ \left(\begin{array}{ll} (1+i \sqrt{3}) / 4 & \frac{\sqrt{3}}{2 \sqrt{2}}(1+i) \\ \frac{-\sqrt{3}}{2 \sqrt{2}}(1+i) & (\sqrt{3}+i) / 4 \end{array}\right) $$

A complex conjugate is formed by changing the sign of the imaginary part of a complex number. For example, if we have a complex number like \(1 + i\sqrt{3}\), its complex conjugate will be \(1 - i\sqrt{3}\). This transformation is essential when working with complex numbers in matrices.

To better understand, let's consider one element from our matrix \(A\): \(\frac{(1+i\sqrt{3})}{4}\). When finding the complex conjugate, we will change \(i\sqrt{3}\) to \(-i\sqrt{3}\), giving us \(\frac{(1-i\sqrt{3})}{4}\).
Complex conjugates play a crucial role in ensuring mathematical consistency in operations such as finding the conjugate transpose (also known as the Hermitian transpose) of a matrix, which is key in proving that a matrix is unitary.

Matrix multiplication is a fundamental operation in linear algebra. It involves the dot product of rows from the first matrix with columns of the second matrix.
Consider the matrices \(A\) and \(A^{\dagger}\) from our example. To multiply these, each element in the resulting matrix is found by multiplying rows of \(A\) with columns of \(A^{\dagger}\).

For instance, the top-left element of the product matrix is calculated as:

\[\left(\frac{(1+i\sqrt{3})}{4} \times \frac{(1-i\sqrt{3})}{4} + \frac{\sqrt{3}}{2\sqrt{2}}(1+i) \times \frac{\sqrt{3}}{2\sqrt{2}}(1-i) \right) \]

Each multiplication step involves combining like terms, ensuring all imaginary parts and real parts are correctly managed.
Matrix multiplication confirms whether \(A \cdot A^{\dagger} = I\), where \(I\) is the identity matrix, which proves the unitarity of the matrix \(A\).

An identity matrix is a special kind of square matrix where all the elements on the main diagonal are 1, and all other elements are 0. The identity matrix is denoted as \(I\), and it has the property \(A \cdot I = I \cdot A = A\) for any matrix \(A\) of compatible dimensions.

For a 2x2 matrix, the identity matrix looks like this:
\[I = \left( \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right) \]
The identity matrix acts like the number 1 in matrix multiplication.

In our exercise, proving that \(A \cdot A^{\dagger} = I\) involves showing that the product of matrix \(A\) and its conjugate transpose yields the identity matrix:
\(\left( \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right)\).
This confirms that matrix \(A\) is indeed a unitary matrix, possessing the unique property \(A^{-1} = A^{\dagger}\).