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2002Exam2Solutions

# 2002Exam2Solutions - EXAM II SOLUTIONS Math 51 Spring 2002...

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EXAM II SOLUTIONS Math 51, Spring 2002. You have 2 hours. No notes, no books, no calculators. YOU MUST SHOW ALL WORK AND EXPLAIN ALL REASONING TO RECEIVE CREDIT Good luck! Name ID number 1. (/30 points) 2. (/30 points) 3. (/30 points) 4. (/30 points) 5. (/30 points) Bonus (/15 points) Total (/150 points) “On my honor, I have neither given nor received any aid on this examination. I have furthermore abided by all other aspects of the honor code with respect to this examination.” Signature: Circle your TA’s name: Tarn Adams (2 and 6) Mariel Saez (3 and 7) Yevgeniy Kovchegov (4 and 8) Heaseung Kwon (A02) Alex Meadows (A03) Circle your section meeting time: 11:00am 1:15pm 7pm 1

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1. (a) Without computing the inverse directly, and without row reducing, explain how you know that the matrix A given below is invertible. A = 3 0 2 0 2 4 0 0 1 Solution: We can compute the determinant of this matrix using minors: det A = 3det ± 2 4 0 1 - 0det ± 0 4 0 1 + 2det ± 0 2 0 0 = 6 6 = 0 Since the determinant of the matrix is not zero, we know that it must be invertible. (b) Find A - 1 . Solution: We row reduce ( A | I ): 3 0 2 0 2 4 0 0 1 1 0 0 0 1 0 0 0 1 3 0 0 0 2 0 0 0 1 1 0 - 2 0 1 - 4 0 0 1 (1 st ) - 2(3 rd ) (2 nd ) - 4(3 rd ) (3 rd ) 1 0 0 0 1 0 0 0 1 1 / 3 0 - 2 / 3 0 1 / 2 - 2 0 0 1 (1 st ) / 3 (2 nd ) / 2 (3 rd ) The end result is ( I | A - 1 ); so, A - 1 = 1 / 3 0 - 2 / 3 0 1 / 2 - 2 0 0 1 2
2. For each of the following, either (i) ﬁnd a matrix (or matrices) satisfying the given de- scriptions, or (ii) prove that such a matrix (or matrices) cannot exist. (a) The column space C ( M 1 ) contains the vectors 4 3 7 4 , 1 3 7 8 and the null space N ( M 1 ) contains the vectors 1 3 2 , 4 5 - 6 Solution: The given null space vectors have three components; since the null space is a subspace of the domain, we conclude that the domain is R 3 . Also, since the two null space vectors are independent, we conclude that the null space is at least two dimensional; similarly, we conclude that the column space must be at least two dimensional. However, the Rank-Nullity theorem says that the dimension of the null space plus the dimension of the column space must equal the dimension of the domain; which, with the given data, would be impossible (since 3 < 2 + 2).

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2002Exam2Solutions - EXAM II SOLUTIONS Math 51 Spring 2002...

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