# Invertible matrix

## Homework Statement

Let A be a square matrix.
If B is a square matrix satisfying AB=I

## Homework Equations

Proof that B=A^-1

## The Attempt at a Solution

Homework Helper
Well, you can first use the relation det(AB) = det(A)det(B). What does it tell you?

mjsd
Homework Helper
you can show this by showing that the inverse is unique.. ie. if AB=AC=I then B=C

Homework Helper
you can show this by showing that the inverse is unique.. ie. if AB=AC=I then B=C

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.

AlephZero
Homework Helper
You seem to be making this more complicated than it is. You don't know anything about A and B except that A has an inverse. So the only thing that you can use is the definition of what the inverse is:

A^-1.A = A.A^-1 = I

So just use that definition, to get from the equation AB = I to the equation B = A^-1.

Hint: the associative law for multiplication says (XY)Z = X(YZ) for any matrices X Y and Z.

HallsofIvy
Homework Helper
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.)

Thanks so much to all of for replying

The main problem is to show that A is invertible

than I can show That [B=A^-1] easily

Last edited:
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.)

Can you please show me how I can prove this???

Homework Helper
Actually, the problem is already solved for you. Just re-read the replies.

Actually, the problem is already solved for you. Just re-read the replies.
Thanks.
But can you show me the solution more clearly ,please???

Homework Helper
Thanks.
But can you show me the solution more clearly ,please???

No. You show us exactly which part you're confused about. AlephZero
Homework Helper
Thanks so much to all of for replying

The main problem is to show that A is invertible

than I can show That [B=A^-1] easily

It A is not invertible, then det(A) = 0

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:_
(A^-1)(AB)=(A^-1)I
IB=A^-1
B=A^-1

Second: I need to show now, that A is invertible, to complete the solution.
This is the part which I'm confused about.

please,I need the solution very quickly.And sorry for the disturbance.

Homework Helper
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:_
(A^-1)(AB)=(A^-1)I
IB=A^-1
B=A^-1

Second: I need to show now, that A is invertible, to complete the solution.
This is the part which I'm confused about.

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).

Yes.but it can be valid if we show that A is invertible.

There was another part of this problem,but it was about BA=I,The reference book solved it like the way I've written above.
But It was easier:_

To show that A is invertible: We can show that AX=0, have only the trivial solution:_
So by multypling AX=0 by B on the left:_
B(AX)=B0
(BA)X=0 And, since BA=I
IX=0
X=0 And this is the trivial solution
So A is invertible.

But in our case,The problem is that we have [AB=I],not [BA=I],so we can't multiply by B, and derive that easily.