Solution_4

# Solution_4 - THE CHINESE UNIVERSITY OF HONG KONG Department...

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THE CHINESE UNIVERSITY OF HONG KONG Department of Mathematics MAT2310 Linear Algebra and Applications (Fall 2007) Suggested Solutions For Homework 4 1. Let B be the space of all the solutions x such that A x = 0 . Since the rank of the matrix A is n - r , the dimension of B must be n - ( n - r ) = r . By assumption, x 1 , x 2 , ··· , x r be linearly independent vectors satisfying A x = 0 , it is a base for B . Hence any x B can be expressed as a linear combination of x 1 , x 2 , ··· , x r . 2. (a) A = 1 - 1 3 5 - 4 - 4 7 - 6 2 (-5) times the ﬁrst row was added to the second row, (-7) times the ﬁrst row was added to the third row, we obtain 1 - 1 3 0 1 - 19 0 1 - 19 (1) times the second row was added to the ﬁrst row, (-1) times the second row was added to the third row, we obtain 1 0 - 16 0 1 - 19 0 0 0 A basis for the row space is { (1 , 0 , - 16) , (0 , 1 , - 19) } , a basis for the null space is 16 19 1 . The transpose of A is 1 5 7 - 1 - 4 - 6 3 - 4 2 Using elementary row operations, we obtain 1 0 2 0 1 1 0 0 0 Hence a basis for the column space is 1 0 2 , 0 1 1 .

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2 (b) Similar to (a), performing the elementary row operations, we obtain 1 0 1 - 2 / 7 0 1 1 4 / 7 0 0 0 0 Hence a basis for the row space is { (1 , 0 , 1 , - 2 / 7) , (0 , 1 , 1 , 4 / 7) } , a basis for the null space is { ( - 1 , - 1 , 1 , 0) T , (2 / 7 , - 4 / 7 , 0 , 1) T } . The transpose of A is 1 2 - 1 4 1 3 5 3 2 2 0 2 Using the elementary row operations, we obtain
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## This note was uploaded on 05/18/2010 for the course MATHEMATIC MAT2310 taught by Professor Dr.jeffchak-fuwong during the Spring '06 term at CUHK.

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Solution_4 - THE CHINESE UNIVERSITY OF HONG KONG Department...

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