Thread: [SOLVED] some simple inverse matrix help

1. [SOLVED] some simple inverse matrix help

1. IF matrices A, B and C are such that $A = B^{-1} C$ , Show that the inverse of $C^{-1}B$ is A

2. In this question i was given 2 matrices A and B . it tells me to work out AB which i know how to do, and then it says hence determine B.
how do i find the inverse of B from the product AB, im thinking i have to use I somewhere, do i make AB = I ?

i would appreciate if you can contribute your thinking process aswell. Im pretty good at all the matix calculations but when i see non standard questions like these which requres thinking, i just don't know where to start from and it really kills my confidence

Thanks!

2. Originally Posted by llkkjj24
1. IF matrices A, B and C are such that $A = {B^-1} C$ , Show that the inverse of $C^-1 B$ is A
A handy property of inverses (and transposes!) is the following:

$(AB)^{-1} = B^{-1}A^{-1}$, note the reverse order!

So if we have;

$A = B^{-1}C$ and we want to show that the inverse of $C^{-1}B$ is $A$, We just take the inverse of it.

$(C^{-1}B)^{-1} = B^{-1}[(C^{-1})^{-1}] = B^{-1}C = A$

Does this help?

3. Originally Posted by llkkjj24
1. IF matrices A, B and C are such that $A = {B^{-1}} C$ , Show that the inverse of $C^{-1} B$ is A
What is $A(C^{-1}B)$? What is $(C^{-1}B)A$?
(To get all of -1 as superscript put it in braces: {-1}.)
2. In this question i was given 2 matrices A and B . it tells me to work out AB which i know how to do, and then it says hence determine B.
how do i find the inverse of B from the product AB, im thinking i have to use I somewhere, do i make AB = I ?
What did you get for AB?

i would appreciate if you can contribute your thinking process aswell. Im pretty good at all the matix calculations but when i see non standard questions like these which requres thinking, i just don't know where to start from and it really kills my confidence

Thanks!

4. Originally Posted by Kasper
A handy property of inverses (and transposes!) is the following:

$(AB)^{-1} = B^{-1}A^{-1}$, note the reverse order!

So if we have;

$A = B^{-1}C$ and we want to show that the inverse of $C^{-1}B$ is $A$, We just take the inverse of it.

$(C^{-1}B)^{-1} = B^{-1}[(C^{-1})^{-1}] = B^{-1}C = A$

Does this help?
thanks, just wondering, is there a way to prove this using identities?

5. Originally Posted by HallsofIvy
What is $A(C^{-1}B)$? What is $(C^{-1}B)A$?
(To get all of -1 as superscript put it in braces: {-1}.)

What did you get for AB?
wow, working it out makes a huge difference, so i mutiplied A and B, all 3x3 matrices, and got back an identity matrix.

that means matrix A is the inverse of B right?

6. Where did you get that these are 3 by 3 matrices? You are not just looking at specific examples are you? And, no, I see nothing here to imply that A is the inverse of B.

You are told that $A= B^{-1}C$ and asked to show that A is the inverse of $C^{-1}B$. Two matrices, X and Y, are inverse to each other if and only if XY= YX= I, the identity matrix. To show that "A is the inverse of $C^{-1}B$", you need to show that $A(C^{-1}B)= I$ and that [tex](C^{-1}B)A= I[tex].

Of course, $A(C^{-1}B)= B^{-1}C(C^{-1}B)$ and $(C^{-1}B)A= (C^{-1}B)B^{-1}C$. Use the "associative law" to manipulate those.

For the second problem, you said you were able to find AB but you still have not said what you got.

7. thanks, for no 2 i was given 2 matrices A and B

I was told to work out AB and hence determince $B^{-1}$

I mulitplied A and B and got a matrix back. The identity matrix

so since A . B = I , then A must be $B^{-1}$?

8. Well, you should also show that BA= I but for square matrices if AB= I then BA= I also.