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fall2010T2 - Do lo wn ad er ID 10 20 Te st ID 32 50 2 0 325...

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University of Toronto Department of Mathematics MAT223H1F Linear Algebra I Midterm Examination II November 25, 2010 H. Kim, F. Murnaghan, B. Rowe, S. Uppal Duration: 1 hour 20 minutes Last Name: Given Name: Student Number: Tutorial Code: No calculators or other aids are allowed. FOR MARKER USE ONLY Question Mark 1 /10 2 /10 3 /10 4 /10 5 /5 6 /5 TOTAL /50 1 of 9 est ID: 3250 Downloader ID: 10202 Test ID: 3250 Downloader ID: 10202 Oxdia @ http://www.oxdia.com This test is copyrighted by the uploader and/or course instructor. Downloader id is shown and also encrypted throughout the document. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. Downloader ID: 10202 Downloader ID: 10202 Test ID: 3250 Downloader ID: 10202 Downloader ID: 10202 Test ID: 3250 Downloader ID: 10202 Test ID: 3250 Test ID: 3250 Downloader ID: 10202 Test ID: 3250 Downloader ID: 10202 Test ID: 3250 Test ID: 3250 Test ID: 3250 Downloader ID: 10202 Downloader ID: 10202 1 / 9
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1. Consider the subspace W = s pan ( 1 0 0 - 1 , 1 - 1 0 0 , 0 1 0 1 ) of R 4 . (a) Use the Gram-Schmidt process to find an orthonormal basis for W . (b) Find the orthogonal projection of the vector 1 0 0 1 onto W . (a) Let v 1 = 1 0 0 - 1 , v 2 = 1 - 1 0 0 , and v 3 = 0 1 0 1 . Let u 1 = v 1 and let e 1 = u 1 || u 1 || = 2 2 1 0 0 - 1 . Let u 2 = v 2 - h u 1 , v 2 i h u 1 , u 1 i u 1 = 1 - 1 0 0 - 1 2 1 0 0 - 1 = 1 2 - 1 0 1 2 , and let e 2 = u 2 || u 2 || = 6 6 1 - 2 0 1 2 of 9 est ID: 3250 Downloader ID: 10202 Test ID: 3250 Downloader ID: 10202 Oxdia @ http://www.oxdia.com This test is copyrighted by the uploader and/or course instructor. Downloader id is shown and also encrypted throughout the document. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. Downloader ID: 10202 Test ID: 3250 Test ID: 3250 Downloader ID: 10202 Test ID: 3250 Test ID: 3250 Downloader ID: 10202 Test ID: 3250 Test ID: 3250 Test ID: 3250 Test ID: 3250 Test ID: 3250 2 / 9
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Let u 3 = v 3 - h u 2 , v 3 i h u 2 , u 2 i u 2 - h u 1 , v 3 i h u 1 , u 1 i u 1 = 1 3 2 2 0 2 , and let e 3 = u 3 || u 3 || = 3 3 1 1 0 1 . Then { e 1 , e 2 , e 3 } form an orthonormal basis of W .
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