36_hwc1SolnsODDA

# 36_hwc1SolnsODDA - A Thus(column 2 of B t A t = 1 1 1 1 3 2...

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(c) Write the second row of AB as a linear combination of the rows of B . SOLUTION (a) By the first part of Theorem 13, or what is the same, V.I.F. 4 , the entry in question is given by the dot product of the second row of A with the thirst column of B . Thus, ( AB ) 2 , 3 = 1 3 1 2 · 1 2 1 1 = 10 (b) By the second part of Theorem 13, the answer is a = 1 , b = 0 , c = 0 , d = 2: These are the corresponding coefficients of the second column of the matrix B . (c) By the definition of the transpose, (row 2 of AB ) = (column 2 of ( AB ) t ) . ( * ) But by the second part of Theorem 14, ( AB ) t = B t A t and hence we have from this and ( ** ) that (row 2 of AB ) = (column 2 of B t A t ) . Now we are reduced to a question about columns, just like the one we handled in part (b) : By the second part of Theorem 13, (column 2 of B t A t ) is formed by taking the linear combination of the columns of B t with coefficients coming from the corresponding entries of the second column of A t – which are the corresponding entries of the second row of
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Unformatted text preview: A . Thus, (column 2 of B t A t ) = 1 1 1 1 + 3 2 2 + 1 1 + 2 3 2 1 . ( ** ) But the columns of B t are the rows of A , so by ( * ), the right hand side of ( ** ) expresses second row of AB as a linear combination of the rows of B . 5.5 Let A be a 3 × 3 matrix whose column representation is A = [ v 1 , v 2 , v 3 ]. Find an explicit numerical 3 × 3 matrix B so that AB = [ v 2 + v 3 , v 1 + v 2 , v 1 + v 2 ] . (You are looking for a single B that works no matter what v 1 , v 2 and v 3 happen to be. SOLUTION The given information expresses the ﬁrst column of AB as a linear combi-nation of the columns of A , namely v 2 + v 3 . By the second part of Theorem 13, the ﬁrst 21 /september/ 2005; 22:09 37...
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