HW12-solutions

# HW12-solutions - huang(dh34953 HW12 gilbert(54160 This...

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huang (dh34953) – HW12 – gilbert – (54160) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points If the vectors in an orthogonal set of non- zero vectors are normalized, then some of the new vectors may not be orthogonal. True or False? 1. TRUE 2. FALSE correct Explanation: Consider the orthogonal set { u 1 , u 2 , ..., u p } in R n . Because they are orthogonal, u j · u k = 0 for all j negationslash = k . So if v j is the result of normalizing u j , then v j = 1 bardbl u j bardbl u j , and v j · v k = parenleftbigg 1 bardbl u j bardbl u j parenrightbigg · parenleftbigg 1 bardbl u k bardbl u k parenrightbigg = 1 bardbl u j bardbl 1 bardbl u k bardbl ( u j · u k ) = 0 for j negationslash = k . But this means that v j , v k also are orthogonal. Consequently, the statement is FALSE . 002 10.0points Determine the vector z in R 3 such that y z is the projection of y in Span( u ) when y = 11 5 2 , u = 1 1 2 . 1. z = 6 2 4 2. z = 6 2 4 3. z = 9 3 6 correct 4. z = 9 3 6 Explanation: By definition, y = proj u y + z = y · u u · u u + z . But y · u u · u u = parenleftbigg (11)(1) + (5)(1) + ( 2)(2) 1 + 1 + 4 parenrightbigg 1 1 2 = 2 1 1 2 , so after rearrangement, we see that z = y proj u y = 11 5 2 2 2 4 . Consequently, z = 9 3 6 . 003 10.0points Compute the distance from y to the line through u and the origin when y = 1 0 3 , u = 4 0 4 .

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huang (dh34953) – HW12 – gilbert – (54160) 2 1. 5 2 2. 10 3. 3 2 4. 4 2 5. 2 2 correct Explanation: The shortest distance from y to u will be the distance between y and the orthogonal projection of y onto u . Hence the shortest distance between y and the line through u and the origin is dist( y , proj u y ) = bardbl y proj u y bardbl . So in this case, proj u y = y · u u · u u = ( 4) + (0) + (12) (16) + (0) + (16) u = 1 4 u = 1 0 1 . Now we calculate y proj u y = 1 0 3 1 0 1 = 2 1 0 1 . Finally, bardbl y proj u y bardbl = radicalBig (2) 2 + (0) 2 + (2) 2 = 4 + 0 + 4 = 8 = 2 2 Consequently, bardbl y proj u y bardbl = 2 2 . 004 10.0points Use the fact that A = 2 2 1 1 1 1 3 3 5 3 3 5 0 0 1 0 0 1 to determine an orthogonal basis for Col( A ). 1. 4 2 6 , 1 1 5 2. 4 2 6 , 1 1 1 3. 2 1 3 , 3 0 2 correct 4. 1 1 5 , 1 4 1 Explanation: The pivot columns of A provide a basis for Col( A ). But by row reduction, A 3 3 5 0 0 1 0 0 1 3 3 0 0 0 1 0 0 0 , so the pivot columns of A are a 1 = 2 1 3 , a 3 = 1 1 5 .
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