If `A=[[(-1+isqrt(3))/(2i),(-1-isqrt(3))/(2i)],[(1+isqrt(3))/(2i),(1-isqrt(3))/(2i)]]`, `i = sqrt(-1) and f (x) = x^(2) + 2, `
then `f(A)` equals to

Options

A. `[[1,0],[0,1]]`
B. `((3-isqrt(3))/2)[[1,0],[0,1]]`
C. `((5-isqrt(3))/2)[[1,0],[0,1]]`
D. `(2+isqrt(3))[[1,0],[0,1]]`

Explanation

To solve the problem, we need to find \( f(A) \) where \( A \) is given as:

\[ A = \begin{pmatrix} \frac{-1 + i\sqrt{3}}{2i} & \frac{-1 - i\sqrt{3}}{2i} \\ \frac{1 + i\sqrt{3}}{2i} & \frac{1 - i\sqrt{3}}{2i} \end{pmatrix} \]

and \( f(x) = x^2 + 2 \).

### Step 1: Identify the values of \( \omega \) and \( \omega^2 \)

We recognize that the elements of matrix \( A \) can be expressed in terms of the cube roots of unity. Let:

\[ \omega = \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad \omega^2 = \frac{-1 - i\sqrt{3}}{2} \]

These correspond to the cube roots of unity, where \( \omega^3 = 1 \) and \( \omega + \omega^2 + 1 = 0 \).

### Step 2: Rewrite matrix \( A \)

Using the values of \( \omega \) and \( \omega^2 \), we can rewrite matrix \( A \):

\[ A = \begin{pmatrix} -\frac{i}{2} \omega & -\frac{i}{2} \omega^2 \\ \frac{i}{2} \omega^2 & \frac{i}{2} \omega \end{pmatrix} \]

### Step 3: Calculate \( A^2 \)

To find \( f(A) = A^2 + 2I \), we first need to compute \( A^2 \):

\[ A^2 = A \cdot A = \begin{pmatrix} -\frac{i}{2} \omega & -\frac{i}{2} \omega^2 \\ \frac{i}{2} \omega^2 & \frac{i}{2} \omega \end{pmatrix} \cdot \begin{pmatrix} -\frac{i}{2} \omega & -\frac{i}{2} \omega^2 \\ \frac{i}{2} \omega^2 & \frac{i}{2} \omega \end{pmatrix} \]

Calculating the elements of \( A^2 \):

1. First row, first column:
\[ \left(-\frac{i}{2} \omega\right)\left(-\frac{i}{2} \omega\right) + \left(-\frac{i}{2} \omega^2\right)\left(\frac{i}{2} \omega^2\right) = \frac{-1}{4} \omega^2 + \frac{-1}{4} \omega^4 = \frac{-1}{4}(\omega^2 + \omega) = \frac{-1}{4}(-1) = \frac{1}{4} \]

2. First row, second column:
\[ \left(-\frac{i}{2} \omega\right)\left(-\frac{i}{2} \omega^2\right) + \left(-\frac{i}{2} \omega^2\right)\left(\frac{i}{2} \omega\right) = \frac{-1}{4} \omega^3 + \frac{-1}{4} \omega^3 = \frac{-1}{2} = 0 \]

3. Second row, first column:
\[ \left(\frac{i}{2} \omega^2\right)\left(-\frac{i}{2} \omega\right) + \left(\frac{i}{2} \omega\right)\left(\frac{i}{2} \omega^2\right) = \frac{-1}{4} \omega^3 + \frac{-1}{4} \omega^3 = \frac{-1}{2} = 0 \]

4. Second row, second column:
\[ \left(\frac{i}{2} \omega^2\right)\left(-\frac{i}{2} \omega^2\right) + \left(\frac{i}{2} \omega\right)\left(\frac{i}{2} \omega\right) = \frac{-1}{4} \omega^4 + \frac{-1}{4} \omega^2 = \frac{-1}{4}(-1) = \frac{1}{4} \]

Thus, we have:

\[ A^2 = \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & \frac{1}{4} \end{pmatrix} \]

### Step 4: Calculate \( f(A) \)

Now we can calculate \( f(A) \):

\[ f(A) = A^2 + 2I = \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & \frac{1}{4} \end{pmatrix} + 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{4} + 2 & 0 \\ 0 & \frac{1}{4} + 2 \end{pmatrix} \]

Calculating the final values:

\[ f(A) = \begin{pmatrix} \frac{1}{4} + \frac{8}{4} & 0 \\ 0 & \frac{1}{4} + \frac{8}{4} \end{pmatrix} = \begin{pmatrix} \frac{9}{4} & 0 \\ 0 & \frac{9}{4} \end{pmatrix} \]

### Final Answer

Thus, the final result is:

\[ f(A) = \frac{9}{4} I = \frac{9}{4} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]