If the product of two nxn matrices A and B is invertible, how can you prove A and B must also be invertible?

Question

The determinant answers require you to prove that det(AB) = det(A) * det(B) which seems to me like the same thing as (AB)^-1 = B^-1*A^-1...

Answer 1

if AB is invertible, then det(AB) is not equal to zero. Therefore since det(AB)=det(A)*det(B), neither det(A) nor det(B) can be zero, hence both A and B are invertible.

EDIT Here's a second proof. Suppose B is not invertible. Then there exist two distinct vectors x and y such that Bx=By. This implies ABx=ABy, so then AB would not be invertible. Hence B is invertible. Next, A=(AB)*(B^(-1)), the product of invertible matrices is invertible, so A is invertible.

Answer 2

Let C = A B. Since C is a square invertible matrix, we must have det(C) <>0 (det = determinant).

We know det(C) = det(A) det(B). Since det(C) <> 0, it follows det(A) <>0 and det(C) <> 0. We also know a square matrix is invertible if, and only if, its determinant is different from zero. So, A and B are invertible.

Answer 3

Where are you stuck? I'm not going to simply do your homework for you. However, I'll take a guess ...

Try it from the other angle: what property of a matrix makes it *not* invertible? If AB is *not* invertible, what does that tell you about A and/or B?

Does that lead you to a solution? If not, write again.

Answer 4

ab is invertle then
(ab).c = c.(ab)=1
If space is associated then
1. a.(b.c)=1
2. (c.a).b=1
now it is sufficient proof that (b.c).a=1 and b.(c.a)=1
from 1. (b.c).a.(b.c)=(b.c)*1=(b.c) this means that (b.c).a=1 (a*1=1*a=a)
therefore (b.c)=a^(-1)
And from 2. (c.a).b.(c.a)=1*(c.a)=(c.a)
This means that (c.a).b=1 then (c.a)=b^(-1)
(1 is unique)