Question

Let $z = \frac{-1+\sqrt{3}i}{2}$, where $i = \sqrt{-1}$, and r, s $\in$ {1, 2, 3}. Let $P = \begin{bmatrix} (-z)^r & z^{2s} \\ z^{2s} & z^r \end{bmatrix}$ and I be the identity matrix of order 2. Then the total number of ordered pairs (r, s) for which $P^2 = -I$ is ______.

Answer 1

1

Answer 2

The given complex number is $z = \frac{-1+\sqrt{3}i}{2}$. This is the complex cube root of unity, commonly denoted as $\omega$. So, $z = \omega$. The properties of $\omega$ are $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.

The given matrix is $P = \begin{bmatrix} (-\omega)^r & \omega^{2s} \\ \omega^{2s} & \omega^r \end{bmatrix}$, where r, s $\in$ {1, 2, 3}. We are looking for the ordered pairs (r, s) such that $P^2 = -I$, where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Let's compute $P^2$: $P^2 = \begin{bmatrix} (-\omega)^r & \omega^{2s} \\ \omega^{2s} & \omega^r \end{bmatrix} \begin{bmatrix} (-\omega)^r & \omega^{2s} \\ \omega^{2s} & \omega^r \end{bmatrix} = \begin{bmatrix} ((-\omega)^r)^2 + (\omega^{2s})^2 & (-\omega)^r \omega^{2s} + \omega^{2s} \omega^r \\ \omega^{2s} (-\omega)^r + \omega^r \omega^{2s} & (\omega^{2s})^2 + (\omega^r)^2 \end{bmatrix}$

Since $(-\omega)^r = (-1)^r \omega^r$, we have $((-\omega)^r)^2 = ((-1)^r \omega^r)^2 = (-1)^{2r} \omega^{2r} = 1 \cdot \omega^{2r} = \omega^{2r}$. Also, $(\omega^{2s})^2 = \omega^{4s}$.

The diagonal elements of $P^2$ are $\omega^{2r} + \omega^{4s}$. The off-diagonal elements are $(-\omega)^r \omega^{2s} + \omega^{2s} \omega^r = (-1)^r \omega^r \omega^{2s} + \omega^{r+2s} = (-1)^r \omega^{r+2s} + \omega^{r+2s} = \omega^{r+2s}((-1)^r + 1)$.

So, $P^2 = \begin{bmatrix} \omega^{2r} + \omega^{4s} & \omega^{r+2s}((-1)^r + 1) \\ \omega^{r+2s}((-1)^r + 1) & \omega^{2r} + \omega^{4s} \end{bmatrix}$.

We are given $P^2 = -I = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$. Comparing the elements, we get two equations:

1. $\omega^{2r} + \omega^{4s} = -1$
2. $\omega^{r+2s}((-1)^r + 1) = 0$

Let's analyze the second equation. Since $\omega^k \neq 0$ for any integer $k$, $\omega^{r+2s} \neq 0$. Thus, we must have $(-1)^r + 1 = 0$, which implies $(-1)^r = -1$. This condition holds if and only if r is an odd integer. The possible values for r are {1, 2, 3}. Among these, the odd values are 1 and 3. So, r must be 1 or 3.

Now let's analyze the first equation: $\omega^{2r} + \omega^{4s} = -1$. We know that $1 + \omega + \omega^2 = 0$, so $\omega + \omega^2 = -1$. The possible values for $\omega^k$ are $1, \omega, \omega^2$ (since $\omega^3=1$, $\omega^k = \omega^{k \pmod 3}$). The equation $\omega^{2r} + \omega^{4s} = -1$ implies that the two terms $\omega^{2r}$ and $\omega^{4s}$ must be $\omega$ and $\omega^2$ in some order. That is, $\{\omega^{2r}, \omega^{4s}\} = \{\omega, \omega^2\}$.

We check the possible values of r (1 and 3) and s (1, 2, 3).

Case 1: r = 1 $2r = 2(1) = 2$. So $\omega^{2r} = \omega^2$. The first equation becomes $\omega^2 + \omega^{4s} = -1$. Since $\omega^2 + \omega = -1$, this requires $\omega^{4s} = \omega$. This implies $4s \equiv 1 \pmod 3$. Let's test s $\in$ {1, 2, 3}:

• If s=1, $4s = 4$. $4 \equiv 1 \pmod 3$. So $\omega^4 = \omega$. This works.
• If s=2, $4s = 8$. $8 \equiv 2 \pmod 3$. So $\omega^8 = \omega^2$. This does not work ($\omega^2 + \omega^2 = 2\omega^2 \neq -1$).
• If s=3, $4s = 12$. $12 \equiv 0 \pmod 3$. So $\omega^{12} = \omega^0 = 1$. This does not work ($\omega^2 + 1 = -\omega \neq -1$). So, for r=1, the only possible value for s is 1. This gives the ordered pair (1, 1).

Let's verify the pair (1, 1). r=1 is odd, so the second equation is satisfied. For r=1, s=1, $\omega^{2r} = \omega^2$ and $\omega^{4s} = \omega^4 = \omega$. $\omega^{2r} + \omega^{4s} = \omega^2 + \omega = -1$. The first equation is satisfied. Thus, (1, 1) is a valid pair.

Case 2: r = 3 $2r = 2(3) = 6$. So $\omega^{2r} = \omega^6 = (\omega^3)^2 = 1^2 = 1$. The first equation becomes $1 + \omega^{4s} = -1$. This implies $\omega^{4s} = -2$. The possible values for $\omega^{4s}$ are $1, \omega, \omega^2$. None of these is equal to -2. So, there are no values of s for r=3 that satisfy the first equation.

Therefore, the only ordered pair (r, s) from {1, 2, 3} that satisfies the conditions is (1, 1). The total number of such ordered pairs is 1.

The final answer is $\boxed{1}$.

Explanation of the solution:

1. Identify $z$ as the complex cube root of unity $\omega$.

2. Calculate $P^2$ in terms of $\omega$.

3. Equate $P^2$ to $-I$ to obtain a system of equations.

4. Solve the off-diagonal equation: $\omega^{r+2s}((-1)^r + 1) = 0$. Since $\omega^k \neq 0$, this implies $(-1)^r + 1 = 0$, so $(-1)^r = -1$. This means r must be odd. Given r $\in \{1, 2, 3\}$, r must be 1 or 3.

5. Solve the diagonal equation: $\omega^{2r} + \omega^{4s} = -1$. Using the property $\omega + \omega^2 = -1$, this equation is satisfied if and only if $\{\omega^{2r}, \omega^{4s}\} = \{\omega, \omega^2\}$.

6. Check the possible values of r (1 and 3) and s (1, 2, 3).

   • If r=1, $\omega^{2r} = \omega^2$. We need $\omega^{4s} = \omega$. This requires $4s \equiv 1 \pmod 3$. For s $\in \{1, 2, 3\}$, only s=1 satisfies this. This gives the pair (1, 1).
   • If r=3, $\omega^{2r} = \omega^6 = 1$. We need $1 + \omega^{4s} = -1$, which means $\omega^{4s} = -2$. No value of $\omega^k$ is -2, so there are no solutions for r=3.

7. The only valid ordered pair is (1, 1).

8. The total number of ordered pairs is 1.

The final answer is $\boxed{1}$.