- #1

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## Homework Statement

Let A be a square matrix.

If B is a square matrix satisfying AB=I

## Homework Equations

Proof that B=A^-1

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- Thread starter Ami
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- #1

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Let A be a square matrix.

If B is a square matrix satisfying AB=I

Proof that B=A^-1

- #2

radou

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

- #3

mjsd

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

- #4

radou

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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

- #5

AlephZero

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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

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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

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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

The main problem is to show that A is invertible

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

Last edited:

- #8

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isunique. 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

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

- #10

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

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

I still confused about it.

- #11

- #12

- #13

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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

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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

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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.

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