Let z = (-1 + √3i)/2, where i = √-1, and r, s in {1, 2, 3}. Let P = [[(-z)^r, z^(2s)], [z^(2s), z^r]

Q: Wait, why do we reject the negative root here? Or rather, why does $r=3$ fail?

That is a fantastic question! When we tested $r=3$, we arrived at the equation $\omega^{4s} = -2$. Remember that $\omega$ lies on the unit circle, meaning its modulus is $|\omega| = 1$. Consequently, any integer power of $\omega$ must also have a modulus of $1$. Since $|-2| = 2$, it is physically and mathematically impossible for $\omega^{4s}$ to equal $-2$. Always check the magnitude when you are stuck with complex equations!

Q: Why can't we just use energy conservation? Wait, wrong subject... why can't we just calculate $P^2$ and equate it to $-I$ directly without checking off-diagonals?

You absolutely can! However, in JEE Advanced, time is your most precious resource. By looking at the off-diagonal elements first, we immediately restricted $r$ to be an odd number. This saved us from testing all nine combinations of $(r, s)$. Strategic observation is the hallmark of a top-tier problem solver.

Q: Is there a trick to handle $(-\omega)^r$?

Yes! The trick is to treat the negative sign as $(-1)^r$. So, $(-\omega)^r = (-1)^r \cdot \omega^r$. If $r$ is even, $(-1)^r$ becomes $1$, and you are left with $\omega^r$. If $r$ is odd, $(-1)^r$ becomes $-1$, and you are left with $-\omega^r$. This parity check is a powerful tool for simplifying expressions involving roots of unity.

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

Sigma Percentile · Accepted Answer

### The Elegance of the Cube Root of Unity

Welcome, student. Today, we are not just solving a matrix problem; we are embarking on a journey through the symmetry of complex numbers.

When you first look at $z = \frac{-1 + \sqrt{3}i}{2}$, I want you to see past the fraction. I want you to see the geometry. This is $\omega$, the non-real cube root of unity.

It sits on the unit circle at an angle of $120^\circ$. It is a number that dances in cycles of three. Whenever you see $\omega$, remember its two greatest gifts:

$$\omega^3 = 1 \quad 	ext{and} \quad 1 + \omega + \omega^2 = 0$$

These are not just equations; they are the keys that will unlock this entire problem.

### Decoding the Matrix Structure

We are given the matrix $P = \begin{bmatrix} (-\omega)^r & \omega^{2s} \ \omega^{2s} & \omega^r \end{bmatrix}$. Our goal is to find the number of pairs $(r, s)$ such that $P^2 = -I$.

Let us perform the matrix multiplication $P \cdot P$. As we multiply the rows by the columns, we get:

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

Look at the off-diagonal elements. They are identical. For $P^2$ to equal $-I$, which is $\begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$, these off-diagonal elements must be zero. This is our first major breakthrough.

### The Constraint Hunt

We set the off-diagonal term to zero: $\omega^{2s}((-\omega)^r + \omega^r) = 0$. Since $\omega^{2s}$ is never zero, we must have $(-\omega)^r + \omega^r = 0$.

This is where the parity of $r$ becomes the hero of our story. If $r$ is even, $(-\omega)^r$ becomes $\omega^r$, and the sum becomes $2\omega^r$, which is not zero.

But if $r$ is odd, $(-\omega)^r$ becomes $-\omega^r$, and the terms cancel out perfectly! Thus, $r$ must be an odd number. Given $r \in \{1, 2, 3\}$, our candidates are $r=1$ and $r=3$.

### The Final Verdict

Now, we test our candidates against the diagonal condition: $\omega^{2r} + \omega^{4s} = -1$.

**Case 1: $r=1$**
The equation becomes $\omega^2 + \omega^{4s} = -1$. Rearranging, we get $\omega^{4s} = -1 - \omega^2$.

Using our identity $1 + \omega + \omega^2 = 0$, we know that $-1 - \omega^2 = \omega$. So, $\omega^{4s} = \omega$. This implies $4s \equiv 1 \pmod 3$. Testing $s \in \{1, 2, 3\}$, we find that $s=1$ works perfectly ($4 \equiv 1 \pmod 3$).

**Case 2: $r=3$**
The equation becomes $\omega^6 + \omega^{4s} = -1$. Since $\omega^3 = 1$, $\omega^6 = 1$.

This leads to $1 + \omega^{4s} = -1$, or $\omega^{4s} = -2$. As we discussed, the magnitude of any power of $\omega$ is $1$, so this is impossible.

We have systematically dismantled the problem. Only one pair, $(1, 1)$, survives the scrutiny. The **answer is 1**. You have mastered the symmetry, the algebra, and the logic. Well done.

Animated Solution for Mathematics - Matrices and Determinants: Let $z = \frac{- 1 + 3 i}{2}$ , where $i = - 1$ , and $r, s \in {1, 2, 3}$ . Let $P = [(- z)^{r} z^{2 s} z^{2 s} z^{r}]$ and I be the identity matrix of order 2. Then the total number of ordered pairs $(r, s)$ for which $P^{2} = - I$ is

Enter Numerical Value:

Visualized Solution

Identifying the Complex Number $z$

Key Properties of $ω$

Expressing Matrix $P$ in terms of $ω$

Setting up the Matrix Equation $P^{2} = - I$

Computing the Matrix Product $P^{2}$

Formulating the System of Equations

Analyzing the Off-Diagonal Constraint

Determining the Allowed Values of $r$

Testing Case 1: $r = 1$

Solving for $s$ when $r = 1$

Testing Case 2: $r = 3$

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

The Elegance of the Cube Root of Unity

Decoding the Matrix Structure

The Constraint Hunt

The Final Verdict

Similar Questions

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Enter Numerical Value:

Visualized Solution

Identifying the Complex Number .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Key Properties of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω

Expressing Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω

Setting up the Matrix Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2=−I

Computing the Matrix Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2

Formulating the System of Equations

Analyzing the Off-Diagonal Constraint

Determining the Allowed Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Testing Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=1

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s when .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=1

Testing Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=3