Math question

[Juice Box]

AB = BC

A, B, and C are Square matricies, and B is invertible. Solve for A.

Now, I know the answer is A = BCB^-1....but I'm not sure why. I looked through all the theorms and stuff I could find, but cannot find a logical reasoning as to why this is.

I don't need the reasoning for the homework...all i needed was the answer, which I have...I'm more looking for an explanation so I dont get raped come test-time

any ideas??

 

[ElMonoDelMar]

Is this MATH 225? I'd try to help you, but I sucked ass at that class.

 

[Juice Box]

Originally posted by: ElMonoDelMar
Is this MATH 225? I'd try to help you, but I sucked ass at that class.

haha yeah it is....its a really cool class, as I enjoy matricies in math....just a pain in the ass with the non-math, theory bullshit

 

[OOBradm]

Wouldnt A just be = to C?

since you have B * C * B^-1, would that not simplify to C? 3

EDIT: I dunno much about matrices...

 

[ElMonoDelMar]

Yeah, the math wasn't bad it's just all of the theory stuff.

 

[ElMonoDelMar]

Originally posted by: OOBradm
Wouldnt A just be = to C?

since you have B * C * B^-1, would that not simplify to C? 3

EDIT: I dunno much about matrices...

That's what I thought it would be. I thought a matrix times its inverse is just 1. That's probably why I did bad in that class.

 

[ts3433]

(erroneous post)

 

[chuckywang]

Originally posted by: Juice Box
AB = BC

A, B, and C are Square matricies, and B is invertible. Solve for A.

Now, I know the answer is A = BCB^-1....but I'm not sure why. I looked through all the theorms and stuff I could find, but cannot find a logical reasoning as to why this is.

I don't need the reasoning for the homework...all i needed was the answer, which I have...I'm more looking for an explanation so I dont get raped come test-time

any ideas??

B is invertible, so B^-1 exists. Multiply B^-1 on the RIGHT HAND SIDE of both sides of the equation.

You have ABB^-1 = BCB^-1. Since matrix multiplcation is associative, and BB^-1 = I (identity matrix), you have A*I = BCB^-1, or A=BCB^-1.

EDIT: I think the main point of this problem is showing that matrix multiplication is NOT commutative. Ie, AB does not equal BA in most cases.

 

[Juice Box]

Originally posted by: ElMonoDelMar

Originally posted by: OOBradm
Wouldnt A just be = to C?

since you have B * C * B^-1, would that not simplify to C? 3

EDIT: I dunno much about matrices...

That's what I thought it would be. I thought a matrix times its inverse is just 1. That's probably why I did bad in that class.

makes sense, but I believe it is B*C*B^-1....not just B*B^-1...which, you are correct, would just be 1....but the answer is BCB^-1...there has to be a reason for it!

 

[Syringer]

In matrices, ABC does not equal ACB or CBA, etc..

Basically you just need to get rid of the B in the AB part, and to do so you mutiply the right side by B^-1..and as we learned in algebra, what you do on one side you do to the other side as well, so the right side becomes BCB^-1

 

[Juice Box]

Originally posted by: chuckywang

Originally posted by: Juice Box
AB = BC

A, B, and C are Square matricies, and B is invertible. Solve for A.

Now, I know the answer is A = BCB^-1....but I'm not sure why. I looked through all the theorms and stuff I could find, but cannot find a logical reasoning as to why this is.

I don't need the reasoning for the homework...all i needed was the answer, which I have...I'm more looking for an explanation so I dont get raped come test-time

any ideas??

B is invertible, so B^-1 exists. Multiply B^-1 on the RIGHT HAND SIDE of both sides of the equation.

You have ABB^-1 = BCB^-1. Since matrix multiplcation is associative, and BB^-1 = I (identity matrix), you have A*I = BCB^-1, or A=BCB^-1.

EDIT: I think the main point of this problem is showing that matrix multiplication is NOT commutative. Ie, AB does not equal BA in most cases.

oh wow....that helps a lot! Chuckywang saves the day, yet again! Thanks man! =D

 

[ElMonoDelMar]

Hey, I'm starting to remember that stuff now.

 

[oboeguy]

Originally posted by: Juice Box

Originally posted by: chuckywang

Originally posted by: Juice Box
AB = BC

A, B, and C are Square matricies, and B is invertible. Solve for A.

Now, I know the answer is A = BCB^-1....but I'm not sure why. I looked through all the theorms and stuff I could find, but cannot find a logical reasoning as to why this is.

I don't need the reasoning for the homework...all i needed was the answer, which I have...I'm more looking for an explanation so I dont get raped come test-time

any ideas??

B is invertible, so B^-1 exists. Multiply B^-1 on the RIGHT HAND SIDE of both sides of the equation.

You have ABB^-1 = BCB^-1. Since matrix multiplcation is associative, and BB^-1 = I (identity matrix), you have A*I = BCB^-1, or A=BCB^-1.

EDIT: I think the main point of this problem is showing that matrix multiplication is NOT commutative. Ie, AB does not equal BA in most cases.

oh wow....that helps a lot! Chuckywang saves the day, yet again! Thanks man! =D

That's right, matrix multiplication does NOT commute. Chucky's solution is correct and to the point.