Matrix Algebra

[cblackhand]

Help! I was attempting to help one of my co-workers with a matrix algebra problem. I looked at it, but it has been so long since I had that class that I can't remmeber the answer here. The problem is:

Find 2X2 matrices A,B,and C != 0 such that AC=BC but A!=B

Thanks in advance for any help.

[ClintPhoenix]

If I remember my matrix multiplication correctly(it's been a while,) this is pretty simple, one solution is:

10
00 = A

01
00 = B

01
01 = C

These would yield the same result,
01
00, but A!=B

I didn't mess it up, did I?

Clint.

[Tinker Grey]

how about

a=|1 2|
  |3 4|

b=|1 5|
  |3 7|

c=|0 1|
  |0 0|

a*c = |0 1|
      |0 3|

b*c = |0 1|
      |0 3|

Well, I see by preview that ClintPhoenix beat me. But here it is anyway.

[cblackhand]

Thanks. You guys are the best.

[jonfromdenver]

holy hell! i was just doing the same problem today for my math class! this person doesn't happen to go to metro denver does s/he?

[Punoqllads]

To have a nonzero matrix transform two other nonidentical nonzero matrices into the same matrix, it needs to be degenerate. That is, one or more of its rows are weighted sums of its other rows. With a 2x2 matrix, this is pretty easy. Just make the second row of C be a multiple of the first row.

In ClintPhoenix's examples, the multiplier is 1, while Tinker Grey's it's 0, but it would work equally for the matrices:

A = 2 0    B = 0 1    C = 1 2
    2 1        0 2        2 4

[Punoqllads]

Small error in my post:

Originally posted by Punoqllads
That is, one or more of its rows are weighted sums of its other rows

One or more of its rows are weighted sums of some of its other rows. So, for a 4x4 matrix, if rows 2 & 3 were multiples of each other, it would be degenerate.