# Matrix Algebra **URL:** \ **Category:** Factual Questions\ **Created:** [October 1, 2001, 7:01pm UTC](https://boards.straightdope.com/t/matrix-algebra/84806 "2001-10-01T19:01:57Z")\ **Posts on this page:** 7\ **Page:** 1
**Author:** ![cblackhand](https://avatars.discourse-cdn.com/v4/letter/c/ecc23a/32.png) [@cblackhand](https://boards.straightdope.com/u/cblackhand)\ **Post date:** [October 1, 2001, 7:01pm UTC](https://boards.straightdope.com/t/matrix-algebra/84806/1 "2001-10-01T19:01:57Z")
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. ---
**Author:** ![ClintPhoenix](https://avatars.discourse-cdn.com/v4/letter/c/b9bd4f/32.png) [@ClintPhoenix](https://boards.straightdope.com/u/ClintPhoenix)\ **Post date:** [October 1, 2001, 7:16pm UTC](https://boards.straightdope.com/t/matrix-algebra/84806/2 "2001-10-01T19:16:29Z")
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. ---
**Author:** ![Tinker\_Grey](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tinker_grey/32/8136_2.png) [@Tinker\_Grey](https://boards.straightdope.com/u/Tinker_Grey)\ **Post date:** [October 1, 2001, 7:17pm UTC](https://boards.straightdope.com/t/matrix-algebra/84806/3 "2001-10-01T19:17:50Z")
how about ```auto 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. ---
**Author:** ![cblackhand](https://avatars.discourse-cdn.com/v4/letter/c/ecc23a/32.png) [@cblackhand](https://boards.straightdope.com/u/cblackhand)\ **Post date:** [October 1, 2001, 7:25pm UTC](https://boards.straightdope.com/t/matrix-algebra/84806/4 "2001-10-01T19:25:42Z")
Thanks. You guys are the best. ---
**Author:** ![jonfromdenver](https://avatars.discourse-cdn.com/v4/letter/j/82dd89/32.png) [@jonfromdenver](https://boards.straightdope.com/u/jonfromdenver)\ **Post date:** [October 1, 2001, 7:30pm UTC](https://boards.straightdope.com/t/matrix-algebra/84806/5 "2001-10-01T19:30:18Z")
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? ---
**Author:** ![Punoqllads](https://avatars.discourse-cdn.com/v4/letter/p/d2c977/32.png) [@Punoqllads](https://boards.straightdope.com/u/Punoqllads)\ **Post date:** [October 1, 2001, 7:33pm UTC](https://boards.straightdope.com/t/matrix-algebra/84806/6 "2001-10-01T19:33:07Z")
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: ```auto A = 2 0 B = 0 1 C = 1 2 2 1 0 2 2 4 ``` ---
**Author:** ![Punoqllads](https://avatars.discourse-cdn.com/v4/letter/p/d2c977/32.png) [@Punoqllads](https://boards.straightdope.com/u/Punoqllads)\ **Post date:** [October 1, 2001, 7:37pm UTC](https://boards.straightdope.com/t/matrix-algebra/84806/7 "2001-10-01T19:37:31Z")
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.