2004Spring20MidtermII-sol

# 2004Spring20MidtermII-sol - Name ID Solutions to Midterm...

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Name: ID#: Solutions to Midterm Examination II Math 20 Introduction to Linear Algebra and Multivariable Calculus 21 April 2004 Show all of your work. Full credit may not be given for an answer alone. You may use the backs of the pages or the extra pages for scratch work. Do not unstaple or remove pages. This is a non-calculator exam. Students who, for whatever reason, submit work not their own will ordinarily be required to withdraw from the College. —Handbook for Students In all situations where row operations are performed on a matrix, label each operation to receive partial credit in case of arithmetic mistakes.

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Problem Possible Points Number Points Earned 1 15 2 15 3 15 4 15 5 15 6 9 7 16 Total 100
1 1 1. (15 Points) Let W be the subspace spanned by - 1 2 2 , 2 - 4 - 4 , 1 1 - 3 , - 1 5 1 , - 2 7 3 Find a basis for W and the dimension of W . Solution. Notice that W = Col A , where A = - 1 2 1 - 1 - 2 2 - 4 1 5 7 2 - 4 - 3 1 3 . The reduced row echelon form of A is R = 1 - 2 0 2 3 0 0 1 1 1 0 0 0 0 0 . Since r 1 and r 3 form a basis Col R , we must have that a 1 and a 3 form a basis for Col A . Thus dim W = 2. / 15 1

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2 2 2. (15 Points) Find all numbers c such that 1 2 - 1 2 3 c 0 c - 15 is not invertible. Solution. By expanding along the bottom row, we see the determinant of this matrix is 1 2 - 1 2 3 c 0 c - 15 = - c 1 - 1 2 c - 15 1 2 2 3 = - c ( c + 2) - 15( - 1) = - c 2 - 2 c + 15 = - ( c - 3)( c + 5) .
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