If `a^(r) = (cos 2 r pi + I sin 2 r pi )^(1//9)` , then the value of `|{:(a_(1), a_(2), a_(3)),(a_(4), a_(5), a^(6)),(a_(7) , a_(8), a_(9)):}|` is

Options

A. 1
B. `-1`
C. 0
D. 2

Answer

To solve the problem given, we start with the expression for \( a^r \):

\[ a^r = \left( \cos(2r\pi) + i \sin(2r\pi) \right)^{\frac{1}{9}} \]

### Step 1: Convert to Exponential Form
Using Euler's formula, we can express the right-hand side in exponential form:

\[ a^r = \left( e^{i \cdot 2r\pi} \right)^{\frac{1}{9}} = e^{i \cdot \frac{2r\pi}{9}} \]

### Step 2: Find Values of \( a \)
Now we can express \( a \) in terms of \( r \):

\[ a = e^{i \cdot \frac{2\pi}{9}} \quad \text{for } r = 1 \]

To find the other values of \( a \), we can consider the different integer values of \( r \) from 0 to 8 (since we want \( a_1, a_2, \ldots, a_9 \)):

\[ \begin{align*} a_1 & = e^{i \cdot \frac{2\pi \cdot 0}{9}} = e^{0} = 1 \\ a_2 & = e^{i \cdot \frac{2\pi \cdot 1}{9}} \\ a_3 & = e^{i \cdot \frac{2\pi \cdot 2}{9}} \\ a_4 & = e^{i \cdot \frac{2\pi \cdot 3}{9}} \\ a_5 & = e^{i \cdot \frac{2\pi \cdot 4}{9}} \\ a_6 & = e^{i \cdot \frac{2\pi \cdot 5}{9}} \\ a_7 & = e^{i \cdot \frac{2\pi \cdot 6}{9}} \\ a_8 & = e^{i \cdot \frac{2\pi \cdot 7}{9}} \\ a_9 & = e^{i \cdot \frac{2\pi \cdot 8}{9}} \\ \end{align*} \]

### Step 3: Construct the Matrix
Now we can construct the matrix \( A \):

\[ A = \begin{pmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{pmatrix} \]

### Step 4: Calculate the Determinant
To find the value of \( |A| \), we can use the property of determinants of matrices formed by roots of unity. The determinant of a matrix formed by the \( n \)-th roots of unity is given by:

\[ |A| = \prod_{j=0}^{n-1} \left( e^{i \cdot \frac{2\pi j}{n}} \right) = n \]

In our case, \( n = 9 \):

\[ |A| = 9 \]

### Final Answer
Thus, the value of \( |A| \) is:

\[ \boxed{9} \]