# Linear Algebra and Canceling

1. Sep 13, 2010

### MurdocJensen

1. The problem statement, all variables and given/known data

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

3. 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?

2. Sep 14, 2010

### Office_Shredder

Staff Emeritus
Why don't you show what steps you did to get those answers?

3. Sep 14, 2010

### MurdocJensen

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.

4. Sep 14, 2010

### Staff: Mentor

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.

5. Sep 14, 2010

### MurdocJensen

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.

6. Sep 14, 2010

### Office_Shredder

Staff Emeritus
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

7. Sep 14, 2010

### Staff: Mentor

Which agrees with the answer in the book.