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Unformatted text preview: MATH 102 SOLUTIONS TO HW #5 Section 3.3, problem 2. In this case we are trying to solve the matrix equation: 1 2 D = 1 7 . This is obviously incompatible. Multiplying both sides by the row vector 1 2 , we have the normal equation: 5 D = 1 2 1 2 D = 1 2 1 7 = 15 . So D = 3 is the solution that minimizes the error. Thus, the vector in the column space closest to (1 , 7) T is (3 , 6) T . This is simply the projection of (1 , 7) T onto the line generated by (1 , 2) T . Section 3.3, problem 6. To solve this problem, we first compute the matrix for the projection onto the column space of A . It is also possible to solve this problem by simply computing the solution to the normal equations A T Ax = A T b , say by Gaussian elimination, and then computing the Ax = P b . The projection it computed via the formula P = A ( A T A ) 1 A T . To compute the middle term in this product we first derive: A T A = 6 8 8 18 . Therefore, by applying the simply formula for the inverse of a 2 2 matrix we have that: ( A T A ) 1 = 1 22 9 4 4 3 . A direct computation involving this last line shows that: P = A ( A T A ) 1 A T = 1 11 10 3 1 3 2 3 1 3 10 . Therefore, we directly have that: P b = 23 / 11 14 / 11 65 / 11 = p . The component of b orthogonal to C ( A ) is the vector: q = b p =  12 / 11 36 / 11 12 / 11 . 1 2 Notice that P q = 0 as should happen ( q is orthogonal to C ( A ) and thus in the kernel of P ). Thus q is in fact an element of N ( A T ), the orthogonal compliment of C ( A...
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This note was uploaded on 08/02/2011 for the course MATH 102 taught by Professor Szypowsk during the Fall '08 term at UCSD.
 Fall '08
 SZYPOWSK
 Math, Linear Algebra, Algebra

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