Matrices proof

**Tutu** · January 18th 2013, 07:25 PM

If X = $P^{-1}AP and A^{3} = I, prove that X^{3} =I$

I did this
$X^{3} = (P^{-1}AP)(P^{-1}AP)(P^{-1}AP) = (P^{-3}A^{3}P^{3}) = (P^{-3+3}A^{3}) = A^{3} = I$

I used the law of indices, is it right to apply this for matrices?

Please help me, I am not sure if I did it right here!

Thank you!

**chiro** · January 18th 2013, 07:55 PM

Hey Tutu.

Hint: Group the P*P^(-1) = I (I is identity matrix) together to get a lot of cancellation and then use A^3 = I to get your final result.

**Tutu** · January 21st 2013, 03:53 AM

Thank you!
I thought I could not change the order within the multiplication, but so is it that I can do it in thie case? Why?
Is it
X^3 = (P^(-1)PA)(P^(-1)PA)(P^(-1)PA)
= P^(-3) (PA)^3
= I
Wait, I did this using laws of indices again, can you please explain how to do it, I'm afraid I do not understand..

Thank you ever so much!

**Deveno** · January 21st 2013, 06:08 AM

your proof is incorrect.

$X^3 = (P^{-1}AP)^3 = (P^{-1}AP)(P^{-1}AP)(P^{-1}AP)$

$= P^{-1}A(PP^{-1})A(PP^{-1})AP = P^{-1}AIAIAP = P^{-1}AAAP = P^{-1}A^3P$

NOT $P^{-3}A^3P^3$ .

and since $A^3 = I$ ,

$P^{-1}A^3P = P^{-1}IP = P^{-1}P = I$ .

no "switching around of order" is required.

**Tutu** · January 21st 2013, 06:56 AM

Thank you!
Does that mean that in matrices, PP^(-1) = P? I thought it equalled to I

**Tutu** · January 21st 2013, 06:57 AM

Sorry, ignore the previous post. I get it now!

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