# Linear algebra question **URL:** \ **Category:** Factual Questions\ **Created:** [January 8, 2003, 8:23pm UTC](https://boards.straightdope.com/t/linear-algebra-question/147361 "2003-01-08T20:23:36Z")\ **Posts on this page:** 5\ **Page:** 2
**Author:** ![Don\_Roberto](https://avatars.discourse-cdn.com/v4/letter/d/e47774/32.png) [@Don\_Roberto](https://boards.straightdope.com/u/Don_Roberto)\ **Post date:** [January 9, 2003, 7:25pm UTC](https://boards.straightdope.com/t/linear-algebra-question/147361/21 "2003-01-09T19:25:12Z")
> [@](#): > > But Shade said that [Bji] = (Bij)T This, by itself, is correct. But, Xj = [Bji]Yi is still simply X=BY in matrix notation, because the contracted indices are the i in the vector with the second subscript (the i) in the matrix. This implies that the rows of B are multiplied by the column Y. In tensor notation, X=B[sup]T[/sup]Y would be represented by Xj=BijYi. So, the upshot is that Bji and Bij are mutual transposes when they appear in the same expression; when they appear in different expressions they represent the same thing. What is important is the relative placements of indices which repeat (i.e., those which are contracted) in a single mulitiplicative expression. ---
**Author:** ![g8rguy](https://avatars.discourse-cdn.com/v4/letter/g/7cd45c/32.png) [@g8rguy](https://boards.straightdope.com/u/g8rguy)\ **Post date:** [January 9, 2003, 7:37pm UTC](https://boards.straightdope.com/t/linear-algebra-question/147361/22 "2003-01-09T19:37:11Z")
> [@](#): > > \*Originally posted by Sacroiliac \* > \*\*One final question and then I?ll drop this. In the first problem I posted > > 1. If [b[sup]ij[/sup]] = [a[sub]ij[/sub]][sup]-1[/sup] solve the n x n system > > y[sub]i[/sub] = a[sub]ij[/sub] x[sub]j[/sub] for x[sub]j[/sub] in terms of y[sub]i[/sub] > > In matrix notation this is just y = Ax, so x = A[sup]-1[/sup]y > > But the answer in index notation is x[sub]j[/sub] = [b[sub]ji[/sub]]y[sub]i[/sub] > > But Shade said that [b[sub]ji[/sub]] = (b[sub]ij[/sub])[sup]T[/sup] > > So given 1 wouldn?t this mean that x = [A[sup]-1[/sup]][sup]T[/sup]y – Vs. x = A[sup]-1[/sup]y? \*\* Short answer, no it wouldn’t. Long answer… Bear with me. **Don Roberto** is correct, but it may be more useful to you this way: The statement [b[sup]ij[/sup]] = [a[sub]ij[/sub]][sup]-1[/sup] means that b[sub]ij[/sub]a[sub]jk[/sub] = I[sub]ik[/sub], **I** being the identity matrix. Given that, if I take my original equation y[sub]i[/sub] = a[sub]ij[/sub]x[sub]j[/sub] and multiply on the left by b[sub]ki[/sub], I get b[sub]ki[/sub]y[sub]i[/sub] = b[sub]ki[/sub]a[sub]ij[/sub]x[sub]j[/sub] = I[sub]kj[/sub]x[sub]j[/sub] = x[sub]k[/sub] This is exactly the same thing as x[sub]j[/sub] = b[sub]ji[/sub]y[sub]i[/sub], where we’ve just replaced “j” with “k,” which we’re free to do, as already discussed. Oh, and a notational postscript. I’ve always seen **B** = **A** [sup]-1[/sup] expressed in this form via B[sub]ij[/sub] = (A[sup]-1[/sup])[sub]ij[/sub], which is I think much less confusing. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [January 9, 2003, 8:16pm UTC](https://boards.straightdope.com/t/linear-algebra-question/147361/23 "2003-01-09T20:16:58Z")
> [@](#): > > \*Originally posted by Sacroiliac \* > But the answer in index notation is x[sub]j[/sub] = [b[sub]ji[/sub]]y[sub]i[/sub] The answer could just as easily be x[sub]i[/sub] = [b[sub]ij[/sub]]y[sub]j[/sub] ---
**Author:** ![Sacroiliac](https://avatars.discourse-cdn.com/v4/letter/s/ecb155/32.png) [@Sacroiliac](https://boards.straightdope.com/u/Sacroiliac)\ **Post date:** [January 9, 2003, 8:22pm UTC](https://boards.straightdope.com/t/linear-algebra-question/147361/24 "2003-01-09T20:22:01Z")
Thank you gentlemen, that was very instructive. I think I’m beginning to understand this stuff a little. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [January 9, 2003, 8:25pm UTC](https://boards.straightdope.com/t/linear-algebra-question/147361/25 "2003-01-09T20:25:04Z")
Learning to switch between matrix notation and systems of equations is definitely one of the hardest parts of linear algebra. Stick with it; it’ll all make sense eventually. [Previous page](https://boards.straightdope.com/t/linear-algebra-question/147361.md?page=1)