# Invertible matrix

1. Mar 17, 2007

### Ami

1. The problem statement, all variables and given/known data
Let A be a square matrix.
If B is a square matrix satisfying AB=I

2. Relevant equations
Proof that B=A^-1

3. The attempt at a solution

2. Mar 17, 2007

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

3. Mar 18, 2007

### mjsd

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

4. Mar 18, 2007

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.

5. Mar 18, 2007

### AlephZero

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.

6. Mar 18, 2007

### HallsofIvy

Staff Emeritus
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.)

7. Mar 21, 2007

### Ami

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: Mar 21, 2007
8. Mar 21, 2007

### Ami

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

9. Mar 21, 2007

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

10. Mar 21, 2007

### Ami

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

11. Mar 21, 2007

No. You show us exactly which part you're confused about.

12. Mar 21, 2007

### AlephZero

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

13. Mar 22, 2007

### Ami

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.

14. Mar 22, 2007

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).

15. Mar 22, 2007

### Ami

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.