# Matrix related to inverse

1. May 26, 2010

### songoku

1. The problem statement, all variables and given/known data
Let B = $$\left(\begin{array}{cc}2&4\\1&2\end{array}\right)$$ and A be a matrix satisfying A2 = 0, AB = 0

1. Calculate (I+A) (I-A), I = identity

2. Calculate (I+A)-1 (I+2A)-1B

2. Relevant equations

3. The attempt at a solution
1. Ans = I

2.
In the manual, it's written : (I+2A)-1 = 1/4 (I-2A)
I don't understand how 1/4 pops up...

Thanks

2. May 26, 2010

### Hurkyl

Staff Emeritus
What is the definition of the inverse of a matrix? What happens if you tried plugging (I-2A) into the definition?

Incidentally, it's worth making an analogy with the power series for 1/(1+2a).

3. May 26, 2010

### songoku

The inverse of a matrix A is a matrix A-1 such that AA-1=I

(I-2A) (I-2A)-1 = I...I still don't get it

power series of 1/(1+2a) = (1+2a)-1 = 1 - 2a - 8a2 - ...

What is the relation to the inverse?

Thanks

4. May 26, 2010

### Staff: Mentor

Should be an alternating series, and the coefficient on a2 is wrong.
1/(1 + 2a) can be written as (1 + 2a)-1.

(I + 2A)-1 has a similar expansion. That's the connection.
I also don't get what your solution manual shows for (I + 2A)-1; namely, the 1/4 factor. I think it's a typo.

5. May 26, 2010

### Hurkyl

Staff Emeritus
(I've changed the variable in the above to reduce confusion)

Sorry, I meant plugging in (I-2A) in for the "inverse of I+2A" -- i.e. B=I+2A and B-1=I-2A.

My mistake, though -- when you make the substitution, things do work out. The equation
(I+2A)-1 = (1/4) (I-2A)​
is, indeed, in error.

6. May 26, 2010

### songoku

yeah, it should be 4

a is variable and A is matrix. Can matrix also be expanded?

So, (I + 2A)-1 should be equal to (I-2A). Am I right?

Thanks

7. May 26, 2010

### Staff: Mentor

Yes, since (I + 2A)(I - 2A) = I. This means that I + 2A is the inverse of I - 2A.

8. May 26, 2010

### songoku

Thanks a lot Mark and Hurkyl

9. May 26, 2010

### Hurkyl

Staff Emeritus
Yes; you can do calculus with matrices! It's a little trickier, but useful. However, you don't have to know matrix analysis to use it as inspiration -- e.g. armed with the knowledge of geometric series, you can guess at a formula for the inverse of (I - A) is for any nilpotent matrix A... and once you have made the guess, it's easy to show directly that the formula does give the inverse.

10. May 26, 2010

### songoku

hm...haven't studied about it yet but I read about it a little on wiki. Thanks a lot hurkyl