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Unformatted text preview: THE CHINESE UNIVERSITY OF HONG KONG Department of Mathematics MAT2310B Linear Algebra and Applications (Fall 2007) Midterm Examination Solutions 1. (a) First we perform the elementary row operation to reduce the matrix A . • ( 35 / 2) times the first row and add to the second row • ( 3) times the first row and add to the third row, we obtain 2 1 2 0 31 / 2 39 1 5 so m 21 = 35 / 2 ,m 31 = 3. • (2 / 31) times the second row and add to the third row, we obtain 2 1 2 0 31 / 2 39 233 / 31 so m 32 = 2 / 31. Hence U = 2 1 2 0 31 / 2 39 233 / 31 L = 1 0 0 35 / 2 1 0 3 2 / 31 1 We have gain the decomposition A = LU (b) Now solving A x = b is equivalent to solving LU x = b . We can solve LU x = b by the following two steps. First, let U x = z and then solve L z = b by forward substitution. We obtain z 1 = 1 z 2 = 4 (35 / 2) z 1 = 27 / 2 z 3 = 3 (3) z 1 + (2 / 31) z 2 = 27 / 31 Next we solve U x = z and get x 3 = ( 31 / 233) × ( 27 / 31) = 27 / 233 x 2 = 2 31 (( 27 / 2) + 39 x 3 ) = 135 / 233 x 1 = 1 2 (1 + x 2 2 x 3 ) = 22 / 233 Thus the solution to the given linear system is x = 22 / 233 135 / 233 27 / 233 . 2 2. (a) Let A be 1 1 α 1 α 1 α 1 1 The system has an unique solution if and only if det( A ) 6 = 0 But det( A ) = α + α + α α 3 1 1 = α 3 + 3 α 2 which equals to ( α 1) 2 ( α + 2) Hence the system has an unique solution if and only if α 6 = 1 and α 6 = 2 (b) The augmented matrix of the given system is...
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This note was uploaded on 05/18/2010 for the course MATHEMATIC MAT2310 taught by Professor Dr.jeffchakfuwong during the Spring '06 term at CUHK.
 Spring '06
 Dr.JeffChakFuWONG
 Linear Algebra, Algebra

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