Matrix Algebra Inverse Matrix Question

[Lonely Lemon]

Homework Statement

If A is an nxn matrix such that A^3 = 0 (the zero matrix) then (I-A)^-1 = ...?

A. not invertible
B. I+A^2
C. I-A
D. I+A
E. I+A+A^2

Homework Equations

The Attempt at a Solution

I just don't know how to work out what the inverse of (I-A) is if I know A^3... how is this even helpful?? It's a multichoice question and apparently the answer is 'E'

 

[Hurkyl]

You don't need to work it out; you just need to check each one!

 

[Lonely Lemon]

If I do that, I get to option C That is (I-A)(I-A)^-1=(I-A)/(I-A)=I which seems correct, but the answer is apparently E. I+A+A^2

*EDIT: Meant I+A+A^2

 

[JonF]

By definition (I – A) (I – A)^-1 = I, so let C = (I – A)^-1

So C(I – A) = I thus C – CA = I so (C – CA)A^2 = A^2 thus CA^2 – CA^3 = A^2 and since we are given A^3 = 0, we get CA^2 = A^2 and since identity is unique for matrixes, C = I. so C = (I – A)^-1 = I. So (I – A)I = I. So I – A = I. So A = 0.

I could be off on this somewhere, but I’m pretty sure A=0 and your inverse is equal to I. So if this is true B, C,D and E should be solutions. Whereas E is trivial by plugging it in.

 

[Lonely Lemon]

Never mind! Click! Just figured out I should have been going (I-A)(I-A)=...

Thanks for the help

 

[JonF]

If I do that, I get to option C That is (I-A)(I-A)^-1=(I-A)/(I-A)=I

Here you seem to imply that (I-A)^-1 = 1/(I-A). What exactly does it mean to divide the number 1 by the difference of the identity matrix and matrix A? Can you divide a number by a matrix?

This illustrates how important it is to live by your definitions in math. How was A^-1 defined in your class, where A is a matrix? I bet the concept of division was never used.