## Related questions with answers

For two invertible $n \times n$

matrices $A$ and $B$,
determine the formulas stated below is necessarily true:

$A + B$ is invertible, and$(A + B)^{-1} = A^{-1} + B^{-1}$

Solutions

VerifiedCompute $(A+B)(A^{-1}+B^{-1})$ and $(A^{-1}+B^{-1})(A+B)$ to see if the results are $I$. If they are, then the inverse of $A+B$ is $(A+B)^{-1}=(A^{-1}+B^{-1})$.

Assume that $(A+B)^{-1}=A^{-1}+B^{-1}$ for the arbitrary invertible matrix $A,B$ for which $A+B$ is also invertible, then must be satisfied the equality $(A+B)(A^{-1}+B^{-1})=I_n$.

Let evaluate the product $(A+B)(A^{-1}+B^{-1})$ in the following

$\begin{align*}(A+B)(A^{-1}+B^{-1})&=(A+B)A^{-1}+(A+B)B^{-1}\\ &=AA^{-1}+BA^{-1}+AB^{-1}+BB^{-1}\\ &=I_n+BA^{-1}+AB^{-1}+I_n\\ &=2I_n+BA^{-1}+AB^{-1} \end{align*}$

First two equality are obtainted by using the distribution law of matrix product.

It is clear that equality $I_n=2I_n+BA^{-1}+AB^{-1}$ does not hold for the arbitrary invertible matrix $A,B$ for which $A+B$ is also invertible. As a counterexample we can use $A=B=I_n$, because in this case $A+B=2I_n$ is invertible matrix and $2I_n+BA^{-1}+AB^{-1}=4I_n\neq I_n$. We obtaint the contradiction with the assumption that $(A+B)^{-1}=A^{-1}+B^{-1}$ for the arbitrary invertible matrix $A,B$ for which $A+B$ is also invertible.

Therefore, the given formula is not correct!

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