Matrix Multiplication: A, B, and C - Exchange and Identity Solutions

[robertjford80]

Homework Statement

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

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.

 

[robertjford80]

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

 

[tiny-tim]

hi robertjford80!

Multiply A * B * C
…
Why not just exchange row 1 with row 3 and then you get an identity matrix …
…
Essentially it looks like they're exchanging column 3 with 1.

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

 

[robertjford80]

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

 

[tiny-tim]

i'm not sure what your strategy was

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

 

[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.

 

[tiny-tim]

multiply what?

(exchanging the rows or the columns was the multiplication)

(and i said columns for C !)

 

[robertjford80]

never mind, I give up.

 

[Mark44]

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.

 

[tiny-tim]

Multiplying AB on the right (postmultiplying) by C (= A) causes the 1st and 3rd rows of AB to be swapped

no, columns

 

[Ray Vickson]

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.

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

RGV

 

[Mark44]

I stand corrected.

 

[tiny-tim]

here's an interesting way of doing it …

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

then A_i,j = C_i,j = δ_i,-j, so (ABC)_i,j = …