Is There a Method for Canceling Matrices in Linear Algebra?

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Homework Help Overview

The discussion revolves around finding the inverse of matrices in the context of linear algebra, specifically through the equations AB=C and PA=LU. Participants are exploring the concept of "canceling" matrices and the implications of matrix multiplication order.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss their attempts to derive formulas for the inverse of matrices, questioning the validity of their cancellation methods. They explore the implications of matrix multiplication not being commutative and how this affects their reasoning.

Discussion Status

The discussion is active, with participants providing insights and clarifications regarding the placement of matrices in multiplication. Some guidance has been offered about the correct approach to finding inverses, and there is an ongoing exploration of the assumptions behind their methods.

Contextual Notes

Participants are navigating the complexities of matrix operations, particularly in relation to homework constraints and the definitions of matrix inverses. There are indications of differing interpretations of the cancellation process in matrix multiplication.

MurdocJensen
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Homework Statement



From AB=C find a formula for A^-1. Also, from PA=LU find A^-1

The Attempt at a Solution



My answers were (A^-1)=(C^-1)(B) and (L^-1)(U^-1)(P)=(A^-1).

The answers in the back of the book are (A^-1)=(B)(C^-1) and (A^-1)=(U^-1)(L^-1)(P)

Would I say that my canceling of matrices is wrong? Would I call what I am trying to do 'cancelling matrices'? What should I be considering when trying to carry out operations such as these?
 
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Why don't you show what steps you did to get those answers?
 
Good point.

First: AB = C --> (A^-1)(A)(B) = (A^-1)(C) = (B) --> (A^-1)=(C^-1)(B)

Second: PA = LU --> [want to cancel A on left side so I multiplied by A^-1 on both sides. assumed placement of A^-1 on left side didn't matter, as long as I could cancel] (P)(A^-1)(A)=(A^-1)(L)(U) --> [basically did the same thing as in the previous step, multiplying first by the inverse of the matrix closest to A^-1, which was L] (L^-1)(U^-1)(P)=(A^-1)

Let me know if I can make things clearer.
 
The placement of a matrix in matrix multiplication does matter, since matrix multiplication is not generally commutative.

For the first problem (and maybe you can get an idea for the second),

AB = C
==> A = CB-1
==> A-1 = (CB-1)-1 = (B-1)-1C-1
==> A-1 = BC-1

In the second step I multiplied on the right by B-1. In the third step I used the theorem that (AB)-1 = B-1C-1.
 
Okay. So...

PA=LU

A=P-1LU

A-1=(P-1LU)-1

A-1=U-1L-1P

...?

It seems like whatever placement I use to cancel something out (to the left or right of the term) I use that same placement (to the left or right of all the factors) on the other side of the equation.
 
That makes sense. If you take your term on the left, and your term on the right, and call them M and N for a second

We have that M=N

So if we want to cancel say a matrix C, we need to multiply one of these by C-1. MC-1=NC-1 just by directly replacing M with N. So the upshot is that whenever you multiply a matrix equation by another matrix, you have to make sure what you're multiplying by is either at the start or at the end of both expressions
 
MurdocJensen said:
Okay. So...

PA=LU

A=P-1LU

A-1=(P-1LU)-1

A-1=U-1L-1P
Which agrees with the answer in the book.
MurdocJensen said:
...?

It seems like whatever placement I use to cancel something out (to the left or right of the term) I use that same placement (to the left or right of all the factors) on the other side of the equation.
 

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