If A+B matrix and A and B are invertible then prove :

Question

(A^-1 + B^-1 )^-1 = A(A+B)^-1 B =B(A+B)^-1 A

Answer 1

If you may appeal to the result that (MN)^-1 = N^-1M^-1 for any invertible matrices M and N of equal size, then the statement you are trying to prove is an immediate consequence:

A(A+B)^-1B

= (B^-1(A+B)A^-1)^-1

= (B^-1AA^-1 + B^-1BA^-1)^-1

= (B^-1I + IA^-1)^-1

= (B^-1 + A^-1)^-1.

If you need to first prove that (MN)^-1 = N^-1M^-1, then just multiply MN by N^-1M^-1 and see what you get!

Answer 2

I'll show that (A^-1 + B^-1)*A(A+B)^-1B is the identity matrix I

I'll multiply on the left by B^-1B (you'll see why in a moment):

B^-1B*(A^-1 + B^-1)*A(A+B)^-1B equals the expression we started with (i.e., (A^-1 + B^-1)*A(A+B)^-1B), because B^-1B is the identity matrix I.

Distributing the A into the left-hand set of parentheses gives you (I boldfaced it just so you'd be able to see which instance of A I'm talking about):

B^-1B*(A^-1 + B^-1)*A(A + B)^-1B = B^-1B*(A^-1A + B^-1A)*(A + B)^-1B

= B^-1B*(I + B^-1A)*(A + B)^-1B

= B^-1(B + BB^-1A)*(A + B)^-1B (I distributed the first B into the left-hand parens)

= B^-1(B + A)*(A + B)^-1B

= B^-1(A + B)*(A + B)^-1B (because matrix addition is commutative)

= B^-1IB

= I

In similar fashion, you can show that multiplying A(A+B)^-1B and A^-1 + B^-1 in the other order also gives you the identity. Working with B(A+B)^-1A is very similar

Answer 3

Let C=A^-1+B^-1, and D=A(A+B)^-1B. So, to show that D=C^-1, we need to show that CD=DC=I, the identity matrix.

CD=(A^-1+B^-1)A(A+B)^-1B=(I+B^-1A)(A+B)^-1B. At this point we need a little trick. Since B^{^-1}B=I

(I+B^-1A)(A+B)^-1B=I(I+B^-1A)(A+B)^-1B=B^-1B(I+B^-1A)(A+B)^-1B=B^-1(B+A)(A+B)^-1B=B^-1B=I.

Showing that CD=I, is similar and is showing CE=EC=I, where E=B(A+B)^-1A.