Homework Help: The exchange matrix

1. May 7, 2012

robertjford80

1. The problem statement, all variables and given/known data
Multiply the following matrices

Multiply A * B * C

A =
0 0 1
0 1 0
1 0 0

B =
1 2 3
4 5 6
7 8 9

C =
0 0 1
0 1 0
1 0 0

3. The attempt at a solution

Why not just exchange row 1 with row 3 and then you get an identity matrix and the answer will be

7 8 9
4 5 6
1 2 3?

The book says the answer is

9 8 7
6 5 4
3 2 1

Essentially it looks like they're exchanging column 3 with 1.

2. May 7, 2012

robertjford80

Wait a second, maybe my answer and the book's answer is the same. If so, I'd like to know.

3. May 7, 2012

tiny-tim

hi robertjford80!
do you mean A*C = I ?

yes, but that doesn't help because A*B*C ≠ A*C*B​

put A (or C) on the left, it exchanges row 1 with row 3

put A (or C) on the right, it exchanges column 1 with column 3

4. May 7, 2012

robertjford80

I don't understand, TT, are you saying my strategy was illegal? And if so, why?

5. May 7, 2012

tiny-tim

i'm not sure what your strategy was

if i'm understanding it correctly, you've only calculated A*B

6. May 7, 2012

robertjford80

Well in that case, why not exchange rows 1 and 3 for A and B, multiply, then exchange rows 1 and 3 again for B and C and multiply. In that case B would be the same in the beginning as in the end.

7. May 7, 2012

tiny-tim

multiply what?

(exchanging the rows or the columns was the multiplication)

(and i said columns for C !)

8. May 7, 2012

robertjford80

never mind, I give up.

9. May 7, 2012

Staff: Mentor

Multiplying B on the left (premultiplying) by A causes the 1st and 3rd rows of B to be swapped.

Multiplying AB on the right (postmultiplying) by C (= A) causes the 1st and 3rd rows of AB to be swapped, taking you right back to B.

The net result is that ABC = B.

10. May 7, 2012

tiny-tim

no, columns

11. May 7, 2012

Ray Vickson

No. Multiplying by A on the left swaps rows; multiplying by C on the right swaps columns.

RGV

12. May 7, 2012

Staff: Mentor

I stand corrected.

13. May 7, 2012

tiny-tim

here's an interesting way of doing it …

number the rows and columns -1, 0, 1 (instead of 1, 2, 3) …

then Ai,j = Ci,j = δi,-j, so (ABC)i,j = …