Let A be a square matrix.
If B is a square matrix satisfying AB=I
Proof that B=A^-1
Or, you have to show that, for a regular matrix A, and matrices B, C (where C is the inverse of A, the existence of which you have the right to assume after working out the hint in post #2), the cancellation law holds.you can show this by showing that the inverse is unique.. ie. if AB=AC=I then B=C
Can you please show me how I can prove this???I imagine that the point of this exercise is to show that the inverse is unique. You know that AB= I. Can you use that to prove that BA= I?
Suppose AC= CA= I. Can you prove that B= C (hint, if AB= I = AC, multiply on both sides, on the left, by C.)
If you need to show that A is invertible, the upper attempt is invalid, since you can't multiply an equation with something you don't even know exists (A^-1).I'm sorry.I need to slove this problem without using the determinants.
My attempt at the solution is:_
First: If A is invertible:-
By multiplying by A^-1 on both sides on the left:_
Second: I need to show now, that A is invertible, to complete the solution.
This is the part which I'm confused about.