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### Invertible Matrices

```
Date: 12/07/2001 at 09:17:44
From: Julie
Subject: Invertible matrices

Here is my problem: show that if A and B are nxn invertible matrices,
then A^(-1) = (A+B)^(-1) + (A+AB^(-1)A)^(-1); assume that the
expressions in the parenthesis on the right are invertible.

I've tried substitution and multiplying by inverses (repeatedly).
I don't know where to go from here. Can you help?
```

```
Date: 12/08/2001 at 20:03:15
From: Doctor Pete
Subject: Re: Invertible matrices

Hi, and thanks for writing to Dr. Math.

To make the work a little more compact, I will use the following
notation for the inverse of a matrix: We'll write

M' = M^(-1).

Next, it might be useful to do a little preliminary work, which we
might call a lemma:  If C = AB, then what is C' = (AB)'? Well, we have

B = A'AB = A'C,
and
A = ABB' = CB',
so
C = AB = (CB')(A'C) = CB'A'C
or
I = B'A'C,
or
C' = B'A'.

Therefore, for any two nxn matrices A, B, we have

[Lemma.]   (AB)' = B'A'.

Next, we want to find (A+B)'.  We have

(A+B)'(A+B) = (A+B)'A + (A+B)'B = I,
so
(A+B)'A = I - (A+B)'B.

Hence

[1]     (A+B)' = (A+B)'AA'
[2]            = (I - (A+B)'B)A'
[3]            = A' - (A+B)'BA'
[4]            = A' - (A+B)'(AB')'
[5]            = A' - ((AB')(A+B))'
[6]            = A' - (AB'A + ABB')'
[7]            = A' - (AB'A + A)'.

If we then add (AB'A + A)' to both sides, we find

(AB'A + A)' + (A+B)' = A',

which is equivalent to what was to be shown. Notice that I used the
Lemma twice, once in step 4 and once in step 5.  A little tricky but
really quite elegant.

- Doctor Pete, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Linear Algebra
High School Linear Algebra

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