Invertible matrix

[Ami]

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

 

[radou]

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

 

[mjsd]

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

 

[radou]

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]

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]

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.)

 

[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

 

[Ami]

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?

 

[radou]

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

 

[Ami]

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

Thanks.
But can you show me the solution more clearly ,please?
I still confused about it.

 

[radou]

Thanks.
But can you show me the solution more clearly ,please?
I still confused about it.

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

 

[AlephZero]

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

That leads to a contradiction - see Radou's first hint.

 

[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.

 

[radou]

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).

 

[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.