past_B_sol - (b)The basis for the null space of A is ...

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THE CHINESE UNIVERSITY OF HONG KONG Department of Mathematics MAT2310 Linear Algebra and Applications (Fall 2006) Suggested Solutions For Past Examination Paper Note: This suggested solution is for reference only. 1. Solution: (a)The orthonormal basis for the column space of A is w 1 = 1 3 1 3 1 3 , w 2 = - 1 6 2 6 - 1 6 , w 3 = 1 2 0 - 1 2 . (b) Q = 1 3 - 1 6 1 2 1 3 2 6 0 1 3 - 1 6 - 1 2 ,R = 2 3 4 3 2 3 0 2 6 3 6 0 0 1 2 (c) (i)We obtain ˆx = - 1 3 0 . The associated least squares error is 0. (ii)HINT: Since A T b = 0 , we have ˆx = 0 2. Solution: (a)The best least squares is y = 2 5 x + 3 5 . (b)The best quadratic least squares is y = - 1 2 x 2 + 39 10 x - 49 10 . (c) k d k = 30 5 . 3. Solution: (a)The basis for the null space of A T is - 7 5 - 13 5 1 .
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Unformatted text preview: (b)The basis for the null space of A is     2-1 1     ,     3-2 1     . 4. Solution: HINT: Since Q T Q = I and det( Q ) = det( Q T ), we have (det( Q )) 2 = 1. 5. Solution: HINT: B T = A (( A T A )-1 ) T A T = A (( A T A ) T )-1 A T = B . 6. Solution: (a)HINT: The eigenvalues of A are 1 , 2 , 3, and the corresponding eigenvectors are   1-1 1   ,   1 1   ,   1 1   , respectively. Hence P =   1 1 0-1 0 1 1 1 1   . (b) C =   46 16-16-42 14 42-10 16 40  ...
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This note was uploaded on 05/18/2010 for the course MATHEMATIC MAT2310 taught by Professor Dr.jeffchak-fuwong during the Spring '06 term at CUHK.

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