If `A=[(a+ib,c+id),(-c+id,a-ib)]` and `a^(2)+b^(2)+c^(2)+d^(2)=1`, then `A^(-1)` is equal to

Options

A. `[(a-ib,-c-id),(c-id,a+ib)]`
B. `[(a+ib,-c+id),(-c+id,a-ib)]`
C. `[(a-ib,-c-id),(-c-id,a+ib)]`
D. none of these

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} a + ib & c + id \\ -c + id & a - ib \end{pmatrix} \) given that \( a^2 + b^2 + c^2 + d^2 = 1 \), we will follow these steps:

### Step 1: Calculate the Determinant of Matrix A
The determinant of a 2x2 matrix \( \begin{pmatrix} p & q \\ r & s \end{pmatrix} \) is calculated as \( ps - qr \).

For our matrix \( A \):
\[ \text{det}(A) = (a + ib)(a - ib) - (c + id)(-c + id) \]

Calculating each part:
1. \( (a + ib)(a - ib) = a^2 + b^2 \)
2. \( (c + id)(-c + id) = -c^2 - d^2 \)

Thus, the determinant becomes:
\[ \text{det}(A) = (a^2 + b^2) - (-c^2 - d^2) = a^2 + b^2 + c^2 + d^2 \]

Given \( a^2 + b^2 + c^2 + d^2 = 1 \), we have:
\[ \text{det}(A) = 1 \]

### Step 2: Calculate the Adjoint of Matrix A
The adjoint of a 2x2 matrix \( \begin{pmatrix} p & q \\ r & s \end{pmatrix} \) is given by \( \begin{pmatrix} s & -q \\ -r & p \end{pmatrix} \).

For our matrix \( A \):
\[ \text{adj}(A) = \begin{pmatrix} a - ib & -(c + id) \\ -(-c + id) & a + ib \end{pmatrix} \]
This simplifies to:
\[ \text{adj}(A) = \begin{pmatrix} a - ib & -c - id \\ c - id & a + ib \end{pmatrix} \]

### Step 3: Calculate the Inverse of Matrix A
The inverse of a matrix is given by:
\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \]

Since we found that \( \text{det}(A) = 1 \), we have:
\[ A^{-1} = \text{adj}(A) \]
Thus:
\[ A^{-1} = \begin{pmatrix} a - ib & -c - id \\ c - id & a + ib \end{pmatrix} \]

### Final Result
The inverse of matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} a - ib & -c - id \\ c - id & a + ib \end{pmatrix} \]