35_hwc1SolnsODDA

35_hwc1SolnsODDA - 1 4.17 Let u be the unit vector u = √...

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Unformatted text preview: 1 4.17 Let u be the unit vector u = √ 6 given by u. −2 1 −1 and x = 1 1 . Compute x and x⊥ , where the direction is 1 SOLUTION We use the formula x = (x · u)u to compute x . Since x·u= this gives −2 −1 1 . x= 3 −1 Now, since x = x + x⊥ , we have x⊥ = x − x , which gives −2 3 −2 1 1 1 − −1 1 = 1 3 + 1 = 1 4 . x⊥ = 3 3 3 1 3 −1 −1 2 −2 + 1 − 1 −2 √ =√ , 6 6 Section 5 1 2 5.1 Let A = 0 1 2 0 1 2 3 2 and let v = 1 3 1 2 . Use a single dot product to compute the second entry of Av. 1 SOLUTION By the second part of Theorem 12, or what is the same, V.I.F. 2, the entry in question is given by the dot product of the second row of A with the ector v, so the 2 1 0 · 2 = 4. (Remember that by convention, we write all vectors as column answer is 2 1 vectors, so when we “pull out” the second row of A, we write it vertically.) 1 1 1 2 3 2 0 1 2 1 2 2 1 2 and B = 0 3 5.3 Consider the matrices A = 1 0 0 2 1 2 . 1 1 (a) Compute (AB )2,3 without computing the whole matrix product AB . (b) Write the second column of AB as a linear combination of 1 a, b, c and d so that the third column of AB equals a 1 + b 1 21/september/2005; 22:09 the columns of A. That is, ﬁnd numbers 2 0 1 3 +c 1 +d 2 . 2 2 2 36 ...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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