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Let A and B be two non-singular square matrices such that A⁶ = I and A * B² = B * A. If K is the least positive integer m (with B not equal to I) such that B^m = I, find the value of K / 9.
- K/9 = 1 (so K = 9)
- K/9 = 2 (so K = 18)
- K/9 = 3 (so K = 27)
- K/9 = 7 (so K = 63)
Correct answer: K/9 = 7 (so K = 63)
Solution
From AB² = BA we get A * B² * A^(-1) = B. This means conjugation by A maps B² to B. Applying repeatedly: Aⁿ * B * A^(-n) = B^(2ⁿ). Setting n = 6 and using A⁶ = I gives B^(2⁶) = B⁶⁴ = B. So B⁶³ = I. Since B is not I and 63 is the least such value (63 = 7 * 9), K = 63 and K/9 = 7.
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