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Mathematics 11 Online
OpenStudy (anonymous):

a) 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) b) 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.

OpenStudy (anonymous):

**For a), I don't understand what they mean by finding a formula...and thanks :)

OpenStudy (anonymous):

I'll give it a try. Just give me a minute.

OpenStudy (anonymous):

We are to assume that A and B are both nxn matrices?

OpenStudy (anonymous):

Yup!

OpenStudy (anonymous):

They are both nxn invertible matrices.

OpenStudy (anonymous):

Well I think I got the answer of the part a.

OpenStudy (anonymous):

By the theorem you wrote above, we can see that: \[A^{-1}(A+B)B^{-1}\] is an invertible matrix, since it's multiplication of three invertible matrices.

OpenStudy (anonymous):

Using properties of matrix multiplication, \[A^{-1}(A+B)B^{-1}=(A^{-1}A+A^{-1}B)B^{-1}=A^{-1}AB^{-1}+A^{-1}BB^{-1}=B^{-1}+A^{-1}=A^{-1}+B^{-1}\] Clearly A^-1+B^-1 is equal to an invertible matrix, and hence it's also an invertible matrix.

OpenStudy (anonymous):

Are you there meganchiu?

OpenStudy (anonymous):

yup im here

OpenStudy (anonymous):

Does the answer make sense to you?

OpenStudy (anonymous):

Would this have anything to do with it: Consider (A^-1(A+B)B^-1). (A^-1(A+B)B^-1)^-1 = (B^-1)^-1(A+B)^-1(A^-1)^-1 =B(A+B)^-1(A) <--- ** Since B and A are invertible and since A+B is invertible, then ** is invertible Does that have anything to do with it?

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