Question

if A^4=0, then I-A is ivertible, (I-A)^-1=I+A+A^2+A^3. T or F, prove it

Answer 1

Let me write it clearly then you can continue with him or her:

\[(I-A)^{-1}=I+A+A^2+A^3\]

Answer 2

:)

Answer 3

Oh, so finally you fell down here.. :P

Answer 4

i think it is T

Answer 5

but can not prove it

Answer 6

How you can say that??

Because of your positive and optimistic nature towards life?? :P

Answer 7

hmm is this a given or need to prove ?
(I-A)^-1=I+A+A^2+A^3

Answer 8

The question is asking whether RHS is equal to LHS or not..

Answer 9

We have to solve for \((1-A^{-1})\) and get the right hand side thing. Or you can do reverse also.. :)

Answer 10

But how can we do this, I don't know still.. :P

Answer 11

Oh, -1 is outside brackets..

Answer 12

it is quite equal to I^4-A^4

Answer 13

how ?
(I-A)^-1 = (I-A^-1)

Answer 14

I said -1 is outside brackets.. :P

Do you wear glasses, ikram?? :P

Answer 15

:))
two guys...

Answer 16

:3
nvm

Answer 17

Never mind from my side too, I was just kidding.. :)

Answer 18

@wonnguyen1995 and ur 3 bhahaha

Answer 19

May be, we need experts' advice here.. He he he.. :)

@ganeshie8

Answer 20

Which topic are you studying @wonnguyen1995

Answer 21

Apart from this I mean..

Have you studied of Binomial Theorems??

Answer 22

not yet

Answer 23

This topic is under Matrices And Determinants, right??

Answer 24

of course

Answer 25

but my major just need basic for this

Answer 26

You are learning Matrices And Determinants and Binomial has not been taught to you yet??

Answer 27

i have not studied determinant and BO

Answer 28

I have a strong feeling that Binomial expansion has a significance here.. :)

Answer 29

you know any guy who could solve it here?

Answer 30

@hartnn

Answer 31

Let see the expansion of \((1-x)^{-1}\)..

\[(1-x)^{-1} = 1 + x + x^2 + x^3 + x^4 + x^5 + ........\]

Answer 32

This is matching to your right hand side.. Miracle..!!! :P

Answer 33

If you can replace x with A, and you know \(A^4 = 0\), so every term afer \(A^4\) will be 0 I think..

Answer 34

So, you will remain upto exponent of 3 only..

\[(1-A)^{-1} = 1 + A + A^2 + A^3\]

Answer 35

\[I^4-A^4 = (I-A)(I^3+I^2*A+I*A^2+A^3))\]

Answer 36

Really??

Answer 37

Wait, let me see it:

\[I^4 - A^4 = (I-A)(I+A)(I^2 + A^2) \implies \color{green}{I^4 - A^4 = (I-A)(I + A + A^2 + A^3)}\]

Answer 38

Yeah you went really great..

Appreciable..

Answer 39

:)

Answer 40

Now as \(A^4 = 0\), and divide both sides by \((I-A)\) as it is invertible so you can divide..

Just fantastic approach.. :)

Answer 41

Got?? Or still in doubt??

Answer 42

:) you are enthusiastic

Answer 43

Me ?? :P