homework_6_solutions

# homework_6_solutions - MA 35100 HOMEWORK ASSIGNMENT#6...

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MA 35100 HOMEWORK ASSIGNMENT #6 SOLUTIONS Problem 1. pg. 133; prob. 21 In Exercises 21 through 25, ﬁnd the reduced row-echelon form for the given matrix A . Then ﬁnd a basis for the image of A and a basis for the kernel of A . A = 1 3 9 4 5 8 7 6 3 . Solution: First we ﬁnd the reduced row-echelon form for A . Subtract 4 times the ﬁrst row from the second, and 7 times the ﬁrst row from the third: 1 3 9 0 - 7 - 28 0 - 15 - 60 . Now divide the second row by - 7 and the third row by - 15: 1 3 9 0 1 4 0 1 4 FInally, subtract 3 times the second row from the ﬁrst, and subtract the second row from the third: rref( A ) = 1 0 - 3 0 1 4 0 0 0 The pivot columns of A are the ﬁrst and second columns. Using Theorem 3.3.5 on page 128 of the text, A basis for the image of A is 1 4 7 , 3 5 6 . Using the reduced row-echelon form for A , we see that an element in the kernel of A is necessarily in the form c 1 c 2 c 3 = 3 c 3 - 4 c 3 c 3 = c 3 3 - 4 1 We conclude that A basis for the kernel of A is 3 - 4 1 . Problem 2. pg. 133; prob. 22 1

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In Exercises 21 through 25, ﬁnd the reduced row-echelon form for the given matrix A . Then ﬁnd a basis for the image of A and a basis for the kernel of A . A = 2 4 8 4 5 1 7 9 3 . Solution: First we ﬁnd the reduced row-echelon form for A . Divide the ﬁrst row by 2: 1 2 4 4 5 1 7 9 3 Now subtract 4 times the ﬁrst row from the second, and subtract 7 times the ﬁrst row from the third: 1 2 4 0 - 3 - 15 0 - 5 - 25 Divide the second row by - 3, and the third row by - 5: 1 2 4 0 1 5 0 1 5 Finally, subtract twice the second row from the ﬁrst, and subtract the second row from the third: rref( A ) = 1 0 - 6 0 1 5 0 0 0 The pivot columns of A are the ﬁrst and second columns. Using Theorem 3.3.5 on page 128 of the text, A basis for the image of A is 2 4 7 , 4 5 9 . Using the reduced row-echelon form for A , we see that an element in the kernel of A is necessarily in the form c 1 c 2 c 3 = 6 c 3 - 5 c 3 c 3 = c 3 6 - 5 1 We conclude that A basis for the kernel of A is 6 - 5 1 . Problem 3.
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## This note was uploaded on 10/18/2010 for the course MATH 351 taught by Professor Egoins during the Spring '10 term at Purdue.

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homework_6_solutions - MA 35100 HOMEWORK ASSIGNMENT#6...

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