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38_hwc1SolnsODDA - By the denition of the transpose and...

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By the definition of the transpose, and what we have just seen, we have: (row k of AB ) = (column k of ( AB ) t ) = (column k of B t A ) = (column i of B t ) if j = i 0 if j = i = (row i of B ) if j = i 0 if j = i . That is, (row i of AB ) = (row i of B ) and all other rows of AB are zero . which tells us what the rows of AB are. SECOND SOLUTION FOR (a) We can also start by directly working out what the rows of AB are. By the first part of Theorem 13, or what is the same, V.I.F. 4 , ( AB ) k, = (row k of A ) · (column of B ) . ( * ) By the given information, (row k of A ) = e i if k = i 0 if k = i . ( ** ) Combining ( * ) and ( ** ), and observing that e i · (column of B ) = B i, , we see that ( AB ) k, = B i, if k = i 0 if k = i . This says that (row i of AB ) = (row i of B ) and all other rows of AB are zero . We can now use the transpose to handle the part about the columns, using what we
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