Invertible matrix

  • Thread starter Ami
  • Start date
  • #1
Ami
14
0

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

 

Answers and Replies

  • #2
radou
Homework Helper
3,115
6
Well, you can first use the relation det(AB) = det(A)det(B). What does it tell you?
 
  • #3
mjsd
Homework Helper
726
3
you can show this by showing that the inverse is unique.. ie. if AB=AC=I then B=C
 
  • #4
radou
Homework Helper
3,115
6
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.
 
  • #5
AlephZero
Science Advisor
Homework Helper
6,994
292
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
HallsofIvy
Science Advisor
Homework Helper
41,833
962
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
Ami
14
0
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:
  • #8
Ami
14
0
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???
 
  • #9
radou
Homework Helper
3,115
6
Actually, the problem is already solved for you. Just re-read the replies.
 
  • #10
Ami
14
0
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.
 
  • #11
radou
Homework Helper
3,115
6
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. :smile:
 
  • #12
AlephZero
Science Advisor
Homework Helper
6,994
292
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.
 
  • #13
Ami
14
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.
 
  • #14
radou
Homework Helper
3,115
6
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).
 
  • #15
Ami
14
0
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.
 

Related Threads on Invertible matrix

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
5
Views
8K
Top