Why is the matrix $(A^{-1}+B^{-1})$ not equal to $(A+B)^{-1}$?

[alingy1]

Show that if A, B and A+B are invertible matrices with the same size, then
$$A(A^{-1}+B^{-1})B(A+B)^{-1}=I$$

What does the result in the first part tell you about the matrix $$(A^{-1}+B^{-1})$$?

I get the first part. Help me with the second part. My book says that the matrix $$(A^{-1}+B^{-1})$$ is not equal to $$(A+B)^{-1}$$
How did they mathematically prove that?

 

[ehild]

My book says that the matrix $$(A^{-1}+B^{-1})$$ is not equal to $$(A+B)^{-1}$$
How did they mathematically prove that?

Multiply $(A^{-1}+B^{-1})$ with $(A+B)$. Do you get $I$?

ehild

 

[STEMucator]

Show that if A, B and A+B are invertible matrices with the same size, then
$$A(A^{-1}+B^{-1})B(A+B)^{-1}=I$$

What does the result in the first part tell you about the matrix $$(A^{-1}+B^{-1})$$?

I get the first part. Help me with the second part. My book says that the matrix $$(A^{-1}+B^{-1})$$ is not equal to $$(A+B)^{-1}$$
How did they mathematically prove that?

Think about it, is adding two matrices together and then taking the inverse of the resulting matrix the same as taking the inverse of the two matrices individually and summing the result? If you try this for some easy 2x2 cases you will see it does not hold.

 

[HallsofIvy]

Does it surprise you? For a and b numbers, \frac{1}{a}+ \frac{1}{b} is generally NOT equal to \frac{1}{a+ b}.

 

[AlephZero]

Hmmm ... the statement that $(A^{-1} + B^{-1}) \ne (A+B)^{-1}$ should be rather "obvious" for the reasons given in the other posts, but I don't quite see the why the result of the first part should make you think of it.