Inverse of a sum of matrices

1. Aug 30, 2010

CentreShifter

Inverse of a sum of matrices [solved]

The problem is relatively simple. Given the equation:

$$(I+2A)^{-1}= \begin{bmatrix} -1 & 2 \\ 4 & 5 \end{bmatrix}$$

Find A.

My problem seems to be that I'm distributing the inverse on the LHS incorrectly. My real question then is, is the following correct?

$$(I+2A)^{-1}=I^{-1}+\frac{1}{2}A^{-1}$$

Last edited: Aug 30, 2010
2. Aug 30, 2010

Staff: Mentor

No. There is no distributive property for exponents, which is what you seem to be doing.

Since
$$(I+2A)^{-1}= \begin{bmatrix} -1 & 2 \\ 4 & 5 \end{bmatrix}$$

The matrix on the right is clearly invertible, so why don't you take the inverse of both sides?

3. Aug 30, 2010

Hurkyl

Staff Emeritus
Well, exponents do distribute over products in some fashion, and there is the binomial theorem and its generalizations. But the first is not relevant and the second a long and torturous path for this problem.

4. Aug 30, 2010

Staff: Mentor

I should have been more clear - that exponents don't distribute over a sum; in other words, that (A + B)n $\neq$ An + Bn.

5. Aug 30, 2010

CentreShifter

Beautiful.

Inverting both sides did the trick.

Inverse of the RHS is $$\begin{bmatrix}\frac{-5}{13} & \frac{2}{13} \\ \frac{4}{13} & \frac{1}{13} \end{bmatrix}$$. I'll call this $$B^{-1}$$

So then I'm left with
\begin{align}I+2A&=B^{-1} \\ 2A&=B^{-1}-I \\ A&=\frac{B^{-1}-I}{2}=\begin{bmatrix} \frac{-9}{13} & \frac{1}{13} \\ \frac{2}{13} & \frac{-6}{13}\end{bmatrix} \end{align}

Plugging this into the original LHS yields the correct result. Thanks.