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anonymous
 5 years ago
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)
anonymous
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There 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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