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108s1 2007 exam solutions

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MATH 108-07S1 Final Exam - Solutions Q1. [20 marks] (a) [8 marks] The augmented matrix for the system is 1 2 −1 1 1 3 2 k 3 6 −3 k The row operatio ns R 2 := R 2 − R 1 , R 3 := R 3 − 3R 1 lead to 1 2 −1 1 0 1 3 k − 1 0 0 0 k − 3 So for the system to be consistent, we need k = 3. When k = 3 the matrix is 1 2 −1 1 0 1 3 2 0 0 0 0 Solve by back substitution with z = t to obtain y + 3t = 2, i.e. y = 2 − 3t x + 2(2 − 3t) − t = 1, i.e. x = −3 + 7t i.e. (x,y,z) = (−3 + 7t, 2 − 3t, t). Geometrically, the system describes three planes which intersect in a line. (b) [4 marks] (i) (I − A)(I + A 2 ) = I − A + A 2 − A 2 2 = I − A + A 2 + A 2 = I and similarly (I + A 2 )(I − A) = I So I − A is invertible and (I − A) −1 = I + A 2 . (ii) If A is invertible, A −1 (A 2 ) = −A −1 A i.e. (A −1 A)A = −I I A = −I A = −I 1

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