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Unformatted text preview: MATH321 HOMEWORK SOLUTIONS HOMEWORK #7 Section 3.3: 1, 2(c), 3(b), 4(a), 5, 6, 7(b), 10 Section 3.4: 1, 2(f), 3, 4(b), 5, 9, 10, 12 Krzysztof Galicki Problem 3.3.1 (See Answers to Selected Exercises ). Problem 3.3.2(c) We have 1 2 1 2 1 1 1 2 1 3 3 1 2 1 1 1 1 1 1 1 . Hence, any solution of the homogeneous system can be written as ( x 1 , x 2 , x 3 ) = ( t, t, t ) = t ( 1 , 1 , 1) . The space of solutions is one dimensional (a line in R 3 ) and a basis for this line is { ( 1 , 1 , 1) } . Problem 3.3.3(b) We have 1 1 1 4 1 2 1 3 1 1 1 3 2 1 1 1 1 1 1 2 / 3 1 1 / 3 1 1 / 3 1 2 / 3 2 / 3 1 / 3 . Hence, any solution of the homogeneous system can be written as ( x 1 , x 2 , x 3 ) = (2 / 3 + t/ 3 , 1 / 3 + 2 t/ 3 , t ) = t (1 / 3 , 2 / 3 , 1) + (2 / 3 , 1 / 3 , 0) . The space of solutions is one dimensional (a line in R 3 ) but it is not a line through the origin (not a vector subspace). Note that t (1 / 3 , 2 / 3 , 1) is a general solution of the associated homogeneous system and (2 / 3 , 1 / 3 , 0) is a solution of the inhomogeneous system. Problem 3.3.4(a) We have A = 1 3 2 5 . One can easily compute the inverse using any method A 1 = 5 3 2 1 . Hence, the solution is given by x 1 x 2 = A 1 4 3 = 5 3 2 1 4 3 = 11 5 . Problem 3.3.5 Take any homogeneous n n linear system with the last equation the same as first equation, for example. Problem 3.3.6 Since T ( a, b, c ) = ( a + b, 2 a c ) the preimage of the point (1 , 11) consists of all vectors in R 3 such that a + b = 1 , 2 a c = 11 ....
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 Winter '07
 Liu
 Math, Linear Algebra, Algebra

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