sample_final_2_answ

Sample_final_2_answ - 1 a If the rank is 4 and there are 4 linear equations then there is a leading 1 in each row of the coecient matrix and hence

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1. a) If the rank is 4 and there are 4 linear equations, then there is a leading 1 in each row of the coeFcient matrix and hence the system is consistent for all ± b . The number of parameters in the general solution is 5-4=1. b) ±v = 1( x 2 + 1) + 2( x 2 +2 x ) + 3( x + 1) = 3 x 2 +7 x + 4. c) ±x · ± y = 1( - 2) + ( - 1)(1) + 2(1) = - 1, × ± y =( - 3 , - 5 , - 1). d) There are many, many possible choices. e) proj ±a (perp ( )) = · ( - · ± ± 2 ) ± ± 2 = · - · ± ± 2 ± ± 2 ± ± 2 = · - · ± ± 2 =0 2. Consider the system of linear equations x 1 x 2 + - x 4 =2 - x 1 - 2 x 2 + x 3 +4 x 4 = - 4 2 x 1 x 2 x 3 x 4 a) 1 2 0 - 1 - 1 - 2 1 4 2 4 2 4 1 2 0 - 1 2 0 0 1 3 - 2 0 0 2 6 - 4 b) x 1 x 2 x 3 x 4 = 2 0 - 2 0 + - 2 1 0 0 s + 1 0 - 3 1 t. c) ±rom our work in b) we have A = 1 0 0 - 1 1 0 0 0 1 1 0 0 0 1 0 2 0 1 1 0 0 0 1 0 21 1 2 0 - 1 0 0 1 3 0 0 0 0 .
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This note was uploaded on 04/30/2010 for the course MATH 136 taught by Professor All during the Winter '08 term at Waterloo.

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Sample_final_2_answ - 1 a If the rank is 4 and there are 4 linear equations then there is a leading 1 in each row of the coecient matrix and hence

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