If A + B is also invertible, then show that A^-1 + B^-1 is also invertible by finding a formula for it. Hint: Consider A^-1(A+B)B^-1 and use Theorem 1.39. Theorem 1.39 If A and B are invertible nxn matrices, then AB is invertible and (AB)^−1 = (B^-1)(A^-1)

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If A + B is also invertible, then show that A^-1 + B^-1 is also invertible by finding a formula for it. Hint: Consider A^-1(A+B)B^-1 and use Theorem 1.39. Theorem 1.39 If A and B are invertible nxn matrices, then AB is invertible and (AB)^−1 = (B^-1)(A^-1)

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I think : A^-1 + B^-1 = A^-1(A+B)B^-1 (If you expand RHS you should see this). Furthermore using Th1.39 : (A^-1(A+B)B^-1)^-1 simplifies to B(A+B)^-1A ... Eq1 and since we are given (A+B) is invertible then Eq1 is invertible and thus the hint expression is invertible and thus (A^-1 + B^-1) is invertible
Is there supposed to be a particular formula? Also: Generalize the previous result: If cA + dB is invertible, for real numbers c and d then show that dA−1 + cB−1 is also invertible by finding a formula for it. Cite any theorems or definitions used.
There is a general formula according to this second question and deriving it is very similar to the answer to the pervious question. you need to write dA−1 + cB−1 as an expression that has cA + dB as part of it and then invert that expression using the theorem. Onece you simplify that I would assume you've answered the question sufficiently. So state (dA−1 + cB−1)^-1 = (A^-1(cA+dB)B^-1)^-1 and simplify RHS. =).

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