hw7 - Solutions of Selected Problems in HW7 5.2.4 Let y =...

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Solutions of Selected Problems in HW7 5.2.4 Let y = ( y 1 ,y 2 ,y 3 ,y 4 ) T S . Then < x 1 , y > = < x 1 , y > = 0 or " 1 0 - 2 1 0 1 3 - 2 # y 1 y 2 y 3 y 4 = " 0 0 # . Solving this equation gives the general solution ( y 1 ,y 2 ,y 3 ,y 4 ) T = (2 α - β, - 3 α +2 β,α,β ) . Indeed y = α 2 - 3 1 0 + β - 1 2 0 1 . Therefore (2 , - 3 , 1 , 0) T and ( - 1 , 2 , 0 , 1) T form a basis for S . 5.2.14 (a) It is enough to show an arbitrary vector x N ( B ) is always in N ( C ). x N ( B ) B x = 0 AB x = A 0 = 0 C x = 0 x N ( C ) 5.3.3 (b) 1 1 3 - 1 3 1 1 2 4 x 1 x 2 x 3 = - 2 0 8 This system is inconsistent. The corresponding normal equation is 1 1 3 - 1 3 1 1 2 4 T 1 1 3 - 1 3 1 1 2 4 x 1 x 2 x 3 = 1 1 3 - 1 3 1 1 2 4

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This note was uploaded on 07/17/2008 for the course MATH 571 taught by Professor Kim during the Spring '08 term at Ohio State.

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hw7 - Solutions of Selected Problems in HW7 5.2.4 Let y =...

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