91Ó°ÊÓ

The given matrices areA and Bwhich is invertible and solve the given equation as follows:

By the definition of inverse, assume matrix C the inverse of $A^{−1}+B^{−1}$if

$C\left(A^{−1}+B^{−1}\right)=\left(A^{−1}+B^{−1}\right)C=I$

If a matrixC exists, then it is unique, hence it is the inverse.

That is, we have show that:

$A{\left(A+B\right)}^{−1}B\left(A^{−1}+B^{−1}\right)=\left(A^{−1}+B^{−1}\right)A{\left(A+B\right)}^{−1}B=I$

Now, solve as follows:

$\begin{array}{c}A{\left(A+B\right)}^{−1}B+A{\left(A+B\right)}^{−1}A=A{\left(A+B\right)}^{−1}\left(A+B\right)\\ =AI\\ =A\end{array}$

Thus, write as follows:

$A{\left(A+B\right)}^{−1}B=A−A{\left(A+B\right)}^{−1}A$

Similarly, also write as follows;

$A{\left(A+B\right)}^{−1}B=B−B{\left(A+B\right)}^{−1}B$

Now, further solve as follows:

$\begin{array}{c}\left(A^{−1}+B^{−1}\right)A{\left(A+B\right)}^{−1}B=A^{−1}\left(A{\left(A+B\right)}^{−1}B\right)+B^{−1}\left(A{\left(A+B\right)}^{−1}B\right)\\ =A^{−1}\left(A−A{\left(A+B\right)}^{−1}A\right)+B^{−1}\left(B−B{\left(A+B\right)}^{−1}B\right)\\ =I−{\left(A+B\right)}^{−1}A+I−{\left(A+B\right)}^{−1}B\\ =2I−{\left(A+B\right)}^{−1}\left(A+B\right)\\ =I\end{array}$

Then, similarly it gives $A{\left(A+B\right)}^{−1}B\left(A^{−1}+B^{−1}\right)=I$.

Thus, A + Bis invertible and${\left(A+B\right)}^{−1}=A^{−1}+B^{−1}$

Hence, the given equation is true.