means that A is invertible and B is its inverse. Hence and . Does that help at all?
Let A and B be n x n matrices such that AB=Isubn . Directly from the definition of linear independence, prove that the colums of B must be linearly independent.
Your answer should not involve row operations and any theorems on rank/invertibility.
This might get you thinking along the right lines. The idea is to suppose that the columns of B are not linearly independent and arrive at a contradiction.
Take A =
[a b c]
[d e f]
[g h i]
and B =
[1 2 0]
[2 4 1]
[3 6 0].
Then
AB =
[a + 2b + 3c; 2a + 4b + 6c; b]
[d + 2e + 3f; 2d + 4e + 6f; e]
[g + 2h + 3i; 2g + 4h + 6i; h]
But this is impossible because a + 2b + 3c = 1 and 2a + 4b + 6c = 0, but 2a + 4b + 6c = 0 implies a + 2b + 3c = 0. How can you generalize this argument?