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September 18, 2014

September 18, 2014

Posted by **David** on Tuesday, February 9, 2010 at 5:37pm.

To prove A is invertible, this is what I did although I don't that it's right:

A^3 - 2I = 0

A(A^2 - 2A^-1) = 0 because A * A^-1 = I

so (A^2 - 2A^-1) must = A^-1

(A^2 - 2A^-1) = A^-1

so (1/3)A^2 = A^-1 thus proving A is invertible. However I don't know that this is correct, and I don't know how to prove that (A - I) is also invertible. I know I need to factor the equation, but the 2I makes it difficult.

- College Math : Linear Algebra -
**David**, Thursday, February 11, 2010 at 5:50pmWell I guess no one can help. :/

Oh well, I'll figure it out

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