How to Find the General Element of (AB)^T for (AB)^T [AB transposed]

In summary, the professor got the correct answer, but I think it would have been helpful if he had explained what was happening step-by-step.
  • #1
qwerty5
5
0

Homework Statement


Write the general element in terms of aij and bij for (AB)^T [AB transposed].


Homework Equations



(AB)^T = B^T*A^T; A=[aij]mxn; B=[bij]nxp

The Attempt at a Solution


n
AB= [sigma aik*bkj]mxp. Let this be equal to [xij]mxp
k=1
n
(AB)^T=[[sigma aik*bkj]mxp]^T
k=1

=[xji]pxm

n
=[sigma aki*bjk]mxp
k=1
n
so the general element xji=[sigma aki*bjk]
k=1


My teacher says this is wrong. Where did I go wrong?


------------------------
Alternate way I used to "check" my wrong answer:
n
(AB)^T=B^T*A^T=[bji]pxn[aji]nxm=[sigma bjk*aki]pxm=[sigma aki*bjk]pxm
k=1


We are using an differential equations/linear algebra textbook for engineers. It never discusses element-by-element proofs, and it leaves out many important differential equations topics, such as exact equations. I have a real diff eq book that my neighbor lent me, but I have to teach myself these types of problems through Wikipedia.
 
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  • #2
I'm really not certain what all that stuff that's not in LaTeX is, but consider this:

If [itex] A = \{a_{ij} \} [/itex] then you know that 1) [itex] A^T = \{ a_{ji} \} [/itex]
and you also know that 2) [itex] (AB)^T = B^T A^T [/itex].

Now if [itex] B = \{ b_{jk} \} [/itex] (where I've used the index "j" again since I know that j will iterate B in precisely the same manner as A for AB to make sense) you can write the matrix AB as [itex] AB = \displaystyle \left\{ a_{ij} b_{jk} \right\} [/itex]. Now let's say, that without being too rigorous, you were to apply the operations from 1) and 2) to this sum, what would you get?
 
  • #3
[bkj][aij] ?
 
  • #4
So I let A={a[tex]_{}ij[/tex]} and B={b[tex]_{}ij[/tex]}.

Then I know that A[tex]^{}T[/tex]={a[tex]_{}ji[/tex]} and B[tex]^{}T[/tex]={a[tex]_{}ji[/tex]}.

A typical element of the product B[tex]^{}T[/tex]A[tex]^{}T[/tex]={b[tex]_{}ji[/tex]a[tex]_{}ji[/tex]}.

However, B[tex]^{}T[/tex]A[tex]^{}T[/tex]={[tex]\Sigma[/tex][tex]^{}n[/tex][tex]_{}k=1[/tex]b[tex]_{}jk[/tex]a[tex]_{}kj[/tex]}.


Is this correct?
 
  • #5
So I let A={a[tex]_{}ij[/tex]} and B={b[tex]_{}ij[/tex]}.

Then I know that A[tex]^{}T[/tex]={a[tex]_{}ji[/tex]} and B[tex]^{}T[/tex]={a[tex]_{}ji[/tex]}.

A typical element of the product B[tex]^{}T[/tex]A[tex]^{}T[/tex]={b[tex]_{}ji[/tex]a[tex]_{}ji[/tex]}.

The sum would therefore be: B[tex]^{}T[/tex]A[tex]^{}T[/tex]={[tex]\Sigma[/tex]b[tex]_{}jk[/tex]a[tex]_{}ki[/tex]} from k=1 to n.

Is this correct?
 
  • #6
Max, I think we are both right.

What did the professor get?
 

Related to How to Find the General Element of (AB)^T for (AB)^T [AB transposed]

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