To solve the problem, we need to calculate \( A^9 \) where \( A = \begin{pmatrix} \omega & -\omega \\ -\omega & \omega \end{pmatrix} \) and \( \omega \) is a complex cube root of unity. We also need to show that \( A^9 = 2^k B \) for some integer \( k \), where \( B = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \).
### Step 1: Understand the properties of \( \omega \)
Since \( \omega \) is a cube root of unity, we have:
\[
\omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0
\]
### Step 2: Calculate \( A^2 \)
We will first calculate \( A^2 \):
\[
A^2 = A \cdot A = \begin{pmatrix} \omega & -\omega \\ -\omega & \omega \end{pmatrix} \begin{pmatrix} \omega & -\omega \\ -\omega & \omega \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( \omega \cdot \omega + (-\omega)(-\omega) = \omega^2 + \omega^2 = 2\omega^2 \)
- First row, second column: \( \omega \cdot (-\omega) + (-\omega)\omega = -\omega^2 - \omega^2 = -2\omega^2 \)
- Second row, first column: \( -\omega \cdot \omega + \omega(-\omega) = -\omega^2 - \omega^2 = -2\omega^2 \)
- Second row, second column: \( -\omega \cdot (-\omega) + \omega \cdot \omega = \omega^2 + \omega^2 = 2\omega^2 \)
Thus, we have:
\[
A^2 = \begin{pmatrix} 2\omega^2 & -2\omega^2 \\ -2\omega^2 & 2\omega^2 \end{pmatrix}
\]
### Step 3: Calculate \( A^3 \)
Next, we calculate \( A^3 = A^2 \cdot A \):
\[
A^3 = \begin{pmatrix} 2\omega^2 & -2\omega^2 \\ -2\omega^2 & 2\omega^2 \end{pmatrix} \begin{pmatrix} \omega & -\omega \\ -\omega & \omega \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 2\omega^2 \cdot \omega + (-2\omega^2)(-\omega) = 2\omega^3 + 2\omega^3 = 4 \)
- First row, second column: \( 2\omega^2 \cdot (-\omega) + (-2\omega^2)\omega = -2\omega^3 - 2\omega^3 = -4 \)
- Second row, first column: \( -2\omega^2 \cdot \omega + 2\omega^2(-\omega) = -2\omega^3 - 2\omega^3 = -4 \)
- Second row, second column: \( -2\omega^2 \cdot (-\omega) + 2\omega^2 \cdot \omega = 2\omega^3 + 2\omega^3 = 4 \)
Thus, we have:
\[
A^3 = \begin{pmatrix} 4 & -4 \\ -4 & 4 \end{pmatrix}
\]
### Step 4: Calculate \( A^6 \)
Next, we calculate \( A^6 = (A^3)^2 \):
\[
A^6 = \begin{pmatrix} 4 & -4 \\ -4 & 4 \end{pmatrix} \begin{pmatrix} 4 & -4 \\ -4 & 4 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 4 \cdot 4 + (-4)(-4) = 16 + 16 = 32 \)
- First row, second column: \( 4 \cdot (-4) + (-4) \cdot 4 = -16 - 16 = -32 \)
- Second row, first column: \( -4 \cdot 4 + 4 \cdot (-4) = -16 - 16 = -32 \)
- Second row, second column: \( -4 \cdot (-4) + 4 \cdot 4 = 16 + 16 = 32 \)
Thus, we have:
\[
A^6 = \begin{pmatrix} 32 & -32 \\ -32 & 32 \end{pmatrix}
\]
### Step 5: Calculate \( A^9 \)
Now, we calculate \( A^9 = A^6 \cdot A^3 \):
\[
A^9 = \begin{pmatrix} 32 & -32 \\ -32 & 32 \end{pmatrix} \begin{pmatrix} 4 & -4 \\ -4 & 4 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 32 \cdot 4 + (-32)(-4) = 128 + 128 = 256 \)
- First row, second column: \( 32 \cdot (-4) + (-32) \cdot 4 = -128 - 128 = -256 \)
- Second row, first column: \( -32 \cdot 4 + 32 \cdot (-4) = -128 - 128 = -256 \)
- Second row, second column: \( -32 \cdot (-4) + 32 \cdot 4 = 128 + 128 = 256 \)
Thus, we have:
\[
A^9 = \begin{pmatrix} 256 & -256 \\ -256 & 256 \end{pmatrix}
\]
### Step 6: Relate \( A^9 \) to \( B \)
We know that:
\[
B = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}
\]
We can express \( A^9 \) as:
\[
A^9 = 256 \cdot \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = 256B
\]
### Step 7: Determine \( k \)
From the equation \( A^9 = 2^k B \), we can see that:
\[
256 = 2^8
\]
Thus, \( k = 8 \).
### Final Answer
\[
k = 8
\]
