Finding the Inverse of a Sum of Matrices

[CentreShifter]

Inverse of a sum of matrices [solved]

The problem is relatively simple. Given the equation:

(I+2A)^{-1}= \begin{bmatrix} <br /> -1 &amp; 2 \\ <br /> 4 &amp; 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}

 

[Mark44]

The problem is relatively simple. Given the equation:

(I+2A)^{-1}= \begin{bmatrix} <br /> -1 &amp; 2 \\ <br /> 4 &amp; 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}

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

Since
(I+2A)^{-1}= \begin{bmatrix} <br /> -1 &amp; 2 \\ <br /> 4 &amp; 5 \end{bmatrix}

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

 

[Hurkyl]

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.

 

[Mark44]

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

 

[CentreShifter]

Beautiful.

Inverting both sides did the trick.

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

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

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