A square matrix a satisfies a^2=i-a , where i is the identity matrix. If a^n =5a-3i, find the value of n.
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Answer:
n = 5
Step-by-step explanation:
The numbers are small enough that we don't need to do anything fancy.
a² = i - a
a³ = a ( i - a ) = a - a² = a - ( i - a ) = 2a - i
a⁴ = a ( 2a - i ) = 2a² - a = 2 ( i - a ) - a = 2i - 3a
a⁵ = a ( 2i - 3a ) = 2a - 3a² = 2a - 3 ( i - a ) = 5a - 3i
So n = 5.
If there were larger numbers involved, you'd start the same way, notice the pattern that aⁿ = ( -1 )ⁿ ( Fₙ₋₁ i - Fₙ a ), where Fₙ is the n-th Fibonacci number, prove that the pattern does indeed hold, and take it from there.
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