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Math question

J

Juice Box

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

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 😛
 
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...
 
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...
Click to expand...

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.

 
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?? 😛
Click to expand...

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.
 
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...
Click to expand...

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.
Click to expand...

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!
 
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
 
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?? 😛
Click to expand...

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.
Click to expand...

oh wow....that helps a lot! Chuckywang saves the day, yet again! 😛 Thanks man! =D
 
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?? 😛
Click to expand...

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.
Click to expand...

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

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