By the definition of the transpose, and what we have just seen, we have:(rowkofAB) = (columnkof (AB)t)= (columnkofBtA)=(columniofBt)ifj=i0ifj=i=(rowiofB)ifj=i0ifj=i.That is,(rowiofAB) = (rowiofB)andall other rows ofABare zero.which tells us what the rows ofABare.SECOND SOLUTION FOR (a)We can also start by directly working out what therows ofABare. By the first part of Theorem 13, or what is the same,V.I.F. 4,(AB)k,= (rowkofA)·(columnofB).(*)By the given information,(rowkofA) =eiifk=i0ifk=i.(**)Combining (*) and (**), and observing thatei·(columnofB) =Bi,, we see that(AB)k,=Bi,ifk=i0ifk=i.This says that(rowiofAB) = (rowiofB)andall other rows ofABare zero.We can now use the transpose to handle the part about the columns, using what we
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