Could someone show me the proofs of the following properties of square invertible matrices.?

Question

Let A and B be square invertible matrices. Then,

(a) (A^-1)^-1 = A;
(b) t = transpose. (At)^-1 = (A^-1)t;
(c) AB is invertible, and (AB)^-1 = B^-1A^-1

Answer 1

Remember the definition of inverses: A(A^-1) = I

a)
A(A^-1) = I
A(A^-1)[(A^-1)^-1] = (A^-1)^-1
A * I = (A^-1)^-1
A = (A^-1)^-1

b) Start with A(A^-1) = I and take the transpose of both sides:
(A(A^-1))^t = I^t
(A^-1)^t * A^t = I
(A^-1)^t * A^t * (A^t)^(-1)= I * (A^t)^(-1)
(A^-1)^t * I = (A^t)^(-1)
(A^-1)^t = (A^t)^(-1)

c) Start with AB = AB and multiply both sides by the inverse of A:
(A^-1)(AB) = (A^-1)(AB)
(A^-1)(AB) = [(A^-1)(A)]B
(A^-1)(AB) = B
(B^-1)(A^-1)(AB) = (B^-1)B
(B^-1)(A^-1)(AB) = I
(B^-1)(A^-1)(AB)(AB)^-1 = I*(AB)^-1
(B^-1)(A^-1) = (AB)^-1

Answer 2

a) The inverse of A^-1 is (A^-1)^-1. So, [A^-1][(A^-1)^-1] = Identity.
Left multiply both sides by A. A[A^-1][(A^-1)^-1] = A, and so since A[A^-1] = Identity, [(A^-1)^-1]=A.

b) This hinges on the property (AB)t = (Bt)(At) and also that
(Identity)t = Identity. So, Identity = [AA^-1]t = (A^-1)t(At). Hence,
Idnetity = (A^-1)t(At). Right multiply by (At)^-1 and thus,
(At)^-1 = (A^-1)t.

c) (AB)(AB)^-1 = Identity
Left multiply by A^-1.
B(AB)^-1 = A^-1
Left multiply by B^-1.
(AB)^-1 = B^-1A^-1

Hope this helps

Answer 3

I think that would all be easier to read if you put parentheses around each of the (-1)s. That said:

For (c), multiply the RHS by AB and see what happens.
For (b), multiply the RHS by At and see what happens.
For (a), multiply the RHS by A-inverse and see what happens.