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A^3=0. Prove that (I-A)^3= I+A+A^2. A=Matrix
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Use the binomial theorem, which works without problems here because IA = AI = A
and check your identity, because I'm not sure it's right.
(which is a polite way of saying it's false.)
I dont think we've covered binomial theorem. ah wait. i wrote it wrong
A^3=0 (I-A)^-1= I+A+A^2
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(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
Ahh ... you don't need the binomial theorem.
Multiply both sides by I-A and show that the right-hand side equals the left-hand side.
wow. i had the step (I-A)(I+a+A^2) with me I just didnt know where it came from. Thanks, you're a life saver
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