[SOLVED] some simple inverse matrix help

• Mar 3rd 2010, 06:33 AM
llkkjj24
[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(Headbang)

Thanks!
• Mar 3rd 2010, 09:52 PM
Kasper
Quote:

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?
• Mar 4th 2010, 04:21 AM
HallsofIvy
Quote:

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}.)
Quote:

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?

Quote:

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(Headbang)

Thanks!
• Mar 4th 2010, 08:19 AM
llkkjj24
Quote:

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?
• Mar 4th 2010, 08:26 AM
llkkjj24
Quote:

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?(Giggle)
• Mar 4th 2010, 09:33 AM
HallsofIvy
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.
• Mar 4th 2010, 12:23 PM
llkkjj24
thanks, for no 2 i was given 2 matrices A and B

http://img718.imageshack.us/img718/5106/matrices1.jpg
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

http://img691.imageshack.us/img691/7769/matrices2.jpg
so since A . B = I , then A must be $B^{-1}$?
• Mar 4th 2010, 12:52 PM
HallsofIvy
Well, you should also show that BA= I but for square matrices if AB= I then BA= I also.