Question

For two unimodular complex numbers \(z_1\) and \(z_2\), \(\begin{bmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{bmatrix}^{-1} \begin{bmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{bmatrix}^{-1}\) is equal to

• A. \(\begin{bmatrix} z_1 & z_2 \\ z_1 & z_2 \end{bmatrix}\)
• B. \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
• C. \(\begin{bmatrix} 1/2 & 0 \\ 0 & 1/2 \end{bmatrix}\)
• D. None of these

Hint:

Use \(U^{-1} = \frac{1}{\det U} \text{adj}(U)\) and properties of unitary matrices.

Answer 1

The Correct Option is C

We are given two matrices where the entries are related to unimodular complex numbers \( z_1 \) and \( z_2 \). Unimodular complex numbers have a modulus of 1. Our task is to find the product of the inverses of these two matrices and determine which option it equates to.

The matrices we need to consider are: 

\[\begin{bmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{bmatrix}^{-1}\]

\[\begin{bmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{bmatrix}^{-1}\]

First, we need to compute the inverse of each matrix.

Matrix Inversion:

1. Inverse of a 2x2 matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is given by: \(\frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\).

Compute the inverse of the first matrix:

Matrix:

\[\begin{bmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{bmatrix}\]

1. Determinant: \( z_1\bar{z}_1 + z_2\bar{z}_2 \)
2. Since \( z_1 \) and \( z_2 \) are unimodular: \( |z_1|^2 = 1 \) and \( |z_2|^2 = 1 \).
3. Therefore, the determinant simplifies to \( 1 + 1 = 2 \).
4. Inverse is then given by: \(\frac{1}{2} \begin{bmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{bmatrix}\).

Compute the inverse of the second matrix:

Matrix:

\[\begin{bmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{bmatrix}\]

1. Determinant: \(\bar{z}_1 z_1 + z_2 (-\bar{z}_2)\)
2. Again by unimodular property: \(|z_1|^2 = 1\) and \(|z_2|^2 = 1\).
3. Therefore, the determinant is \(1 + 1 = 2\).
4. Inverse is then: \(\frac{1}{2} \begin{bmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{bmatrix}\).

Determine the product of inverses:

Product:

\(\left( \frac{1}{2}\begin{bmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{bmatrix} \right) \left(\frac{1}{2} \begin{bmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{bmatrix}\right)\)

1. Calculate product: \(\frac{1}{4} \begin{bmatrix} \bar{z}_1 z_1 + z_2 \bar{z}_2 & \bar{z}_1 z_2 + -z_2 \bar{z}_1 \\ \bar{z}_2 z_1 - \bar{z}_2 z_1 & \bar{z}_2 z_2 + z_1 \bar{z}_1 \end{bmatrix}\).
2. Notice that \(|z_1|^2 = 1\), \(|z_2|^2 = 1\), and \(0 + 0\) for off-diagonals.
3. This results in: \(\frac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).

Thus, the matrix simplifies to option \(\begin{bmatrix} 1/2 & 0 \\ 0 & 1/2 \end{bmatrix}\).