1. ## matrix

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.

2. $AB = I_n$ means that A is invertible and B is its inverse. Hence $\det{A} \ne 0$ and $\det{B} \ne 0$. Does that help at all?

3. sorry, i have not learnt det yet. Do you have the other way to prove this? thanks

4. 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?