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MathHardAdv 2016 • Paper 1

Ordered Pairs of Exponents for Matrix Equation Involving Cube Roots of Unity

Consider z=−1+3i2z = \frac{-1 + \sqrt{3}i}{2}, where i=−1i = \sqrt{-1}, and r,s∈{1,2,3}r, s \in \{1, 2, 3\}. Let

P=(−z)rz2sz2szrP = \begin{array}{cc} (-z)^r & z^{2s} \\ z^{2s} & z^r \end{array}

and II be the identity matrix of order 2. Calculate the total number of ordered pairs (r,s)(r, s) for which P2=−IP^2 = -I.

Official Numerical Answer1

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Step-by-Step Solution

Given z=−1+3i2=ωz = \frac{-1 + \sqrt{3}i}{2} = \omega, where ω\omega is a non-real complex cube root of unity satisfying ω3=1\omega^3 = 1 and 1+ω+ω2=01 + \omega + \omega^2 = 0.

The matrix PP is given by:

P=[(−z)rz2sz2szr]=[(−ω)rω2sω2sωr]P = \begin{bmatrix} (-z)^r & z^{2s} \\ z^{2s} & z^r \end{bmatrix} = \begin{bmatrix} (-\omega)^r & \omega^{2s} \\ \omega^{2s} & \omega^r \end{bmatrix}

where r,s∈{1,2,3}r, s \in \{1, 2, 3\}.

Let a=(−ω)ra = (-\omega)^r, b=ω2sb = \omega^{2s}, and c=ωrc = \omega^r, so that P=[abbc]P = \begin{bmatrix} a & b \\ b & c \end{bmatrix}.

Computing P2P^2:

P2=[abbc][abbc]=[a2+b2b(a+c)b(a+c)b2+c2]P^2 = \begin{bmatrix} a & b \\ b & c \end{bmatrix} \begin{bmatrix} a & b \\ b & c \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & b(a+c) \\ b(a+c) & b^2 + c^2 \end{bmatrix}

We are given that P2=−I=[−100−1]P^2 = -I = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}.

Equating the entries of the matrices, we get:

  1. b(a+c)=0b(a+c) = 0
  2. a2+b2=−1a^2 + b^2 = -1
  3. b2+c2=−1b^2 + c^2 = -1

Since b=ω2s≠0b = \omega^{2s} \neq 0, from condition (1) we must have: a+c=0a + c = 0 (−ω)r+ωr=0(-\omega)^r + \omega^r = 0 ωr((−1)r+1)=0\omega^r \left( (-1)^r + 1 \right) = 0

Since ωr≠0\omega^r \neq 0, it follows that (−1)r+1=0  ⟹  (−1)r=−1(-1)^r + 1 = 0 \implies (-1)^r = -1. Therefore, rr must be an odd integer. Since r∈{1,2,3}r \in \{1, 2, 3\}, the possible values for rr are 11 and 33.

Case 1: r=1r = 1 For r=1r = 1, we have: a=−ω,c=ωa = -\omega, \quad c = \omega a2=ω2a^2 = \omega^2

Substituting a2a^2 into condition (2): a2+b2=−1  ⟹  ω2+(ω2s)2=−1a^2 + b^2 = -1 \implies \omega^2 + (\omega^{2s})^2 = -1 ω2+ω4s=−1\omega^2 + \omega^{4s} = -1 1+ω2+ω4s=01 + \omega^2 + \omega^{4s} = 0

Since 1+ω+ω2=01 + \omega + \omega^2 = 0, this requires: ω4s=ω\omega^{4s} = \omega

Using the identity ω3=1\omega^3 = 1, we have ω4s=ωs\omega^{4s} = \omega^s. Thus: ωs=ω  ⟹  s≡1(mod3)\omega^s = \omega \implies s \equiv 1 \pmod 3

Since s∈{1,2,3}s \in \{1, 2, 3\}, the only solution is s=1s = 1. Thus, (r,s)=(1,1)(r, s) = (1, 1) is a solution.

Case 2: r=3r = 3 For r=3r = 3, we have: a=(−ω)3=−1,c=ω3=1a = (-\omega)^3 = -1, \quad c = \omega^3 = 1 a2=1a^2 = 1

Substituting a2a^2 into condition (2): a2+b2=−1  ⟹  1+ω4s=−1a^2 + b^2 = -1 \implies 1 + \omega^{4s} = -1 ω4s=−2\omega^{4s} = -2

Since ∣ω4s∣=1|\omega^{4s}| = 1 for any integer ss, the equation ω4s=−2\omega^{4s} = -2 has no solutions.

Hence, the only valid ordered pair (r,s)(r, s) is (1,1)(1, 1).

The total number of ordered pairs (r,s)(r, s) is 11.

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