# Matrix proofs

1. Apr 20, 2013

### colt

1. The problem statement, all variables and given/known data

First Question: Be A and B square matrix. Show that if A and B are invertible matrix, then: (A + B)^(-1) = A^(-1) * [I + B*A^(-1)]^(-1)

3. The attempt at a solution

First Queston: (A+B)^(-1) = [A^(-1) + A^(-1)*B*A^(-1)]^(-1) (I distributed the a^(-1) outside the []). Then I don't know what else I can do

1. The problem statement, all variables and given/known data

Second Question: Be A and P matrix of nxn dimension with P being invertible. So det (P^(-1)*A*P) = det(A)

3. The attempt at a solution

I would like to transform the (P^(-1)*A*P) into (I*A), but since matrix multiplication ain't commutative, I don't know how to get there

2. Apr 21, 2013

### jbunniii

You can verify this by showing that
$$(A+B)(A^{-1}[I + BA^{-1}]^{-1}) = I$$
What properties of determinants do you know about?

3. Apr 21, 2013

### colt

I was only able to develop it until here:

$A(A^{-1}[I + BA^{-1}]^{-1}) + B(A^{-1}[I + BA^{-1}]^{-1}) =$

$AA^{-1}[I + BA^{-1}]^{-}) + BA^{-1}[I + BA^{-1}]^{-1} =$

$I[I + BA^{-1}]^{-1} + BA^{-1}[I + BA^{-1}]^{-1} =$

$[I + BA^{-1}]^{-1} + B[A^{-1} + A^{-1}BA^{-1}]^{-1} =$

I know that de determinant of A^(-1) = 1/ det A, that det A = det A transpose, that the det A with a constant K that multiplies one of its lines is equal to K that multiplies det A and so that if this constant multiplies all lines. then det A = K^n * det A

4. Apr 21, 2013

### Dick

I think the problem is asking you to assume that the wrong things are invertible. A and B invertible doesn't show A+B is invertible. You need to assume that everything is invertible. Then for the first use that C^(-1)D^(-1)=(DC)^(-1). Figure out what to put D and C equal to.

5. Apr 21, 2013

### jbunniii

Now try using the distributive property.

6. Apr 21, 2013

### jbunniii

Another important property: $\det(AB) = \det(A)\det(B)$. Do you know this one? If so, the second problem should be easy.

7. Apr 22, 2013

### colt

How? The identity is irrelevant, so only the $a^{-1}$ is left, and I already distributed it without seeing any meaningful result

Ah yes, had forgotten this one.

Didn't really understand. Can you try to explain it once more?

8. Apr 22, 2013

### Dick

What I mean is take the inverse of both sides of (A + B)^(-1) = A^(-1) * [I + B*A^(-1)]^(-1). What you first did, distributing the A^(-1) inside of the [] is quite wrong. There's a ^(-1) on the bracket.

9. Apr 22, 2013

### jbunniii

Both terms have a common factor on the right, namely $(I + BA^{-1})^{-1}$:

$$I(I+BA^{-1})^{-1} + BA^{-1}(I+BA^{-1})^{-1}$$

So factor out that common term and see what you get.

10. Apr 24, 2013