Question

For any 2 matrices A and B of suitable orders, (AB)^T = 

1. B^T A^T

2. A^T B^T

3. A B^T

4. B A^T​

Answer 1

Step-by-step explanation:

Write the matrices A and B as A=[aij] and B=[bij], meaning that their (i,j)-th entries are aij and bij, respectively.

Let C=AB=[cij], where cij=∑nk=1aikbkj, the standard multiplication definition.

We want (AB)T=CT=[cji]. That is the element in position j,i is ∑nk=1aikbkj. For instance, if i=2,j=3, then the element in 2,3 of C is that sum, but the element in position 3,2 of the transpose is that sum.

I need to get the same value for the element in position 3,2 of the right side.

The transpose matrices are BT=[bji],AT=[aji]. They are size p×n and n×m. That is, they switch rows and columns.

Let D=BTAT=[dji]. I write the indices backwards because if I want the element in position 3,2, that is, i=2,j=3 just like on the other side.

So I need the summation for dji. But I get as dji=∑nk=1bjkaki, which does not match

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