Midterm 2010 with solution

# Midterm 2010 with solution - Applied Linear Algebra MATH...

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Applied Linear Algebra, MATH 270-2010 Fall Mid-term Quiz: Solutions (October 20, 2010.) 1. (25 points) Given the following matrix: A = - 1 3 2 0 1 1 2 - 2 0 1 - 3 - 2 1 - 3 - 2 0 1 1 0 0 0 0 0 0 (a) rank( A ) = 2; (b) null( A ) = 1; and the basis of ker( A ) = (1 , 1 , - 1) T . (c) Let b = (5 , 2 , c, d ) T . Find the values of c, d , such that the equation A x = b has a solution. The augmented matrix can reduced to the following echelon form: A # = - 1 3 2 5 0 1 1 2 2 - 2 0 c 1 - 3 - 2 d 1 - 3 - 2 5 0 1 1 2 0 0 0 2 + c 0 0 0 5 + d The consistency condition yields c = - 2 , d = - 5. 2. Given the following elements in the vector space ( M 2 , 3 ): A = 1 2 3 4 0 5 ; B = 2 4 7 10 1 13 ; C = 1 2 5 8 2 11 . (a) The usual basis ( B ) of the space ( M 2 , 3 ) is: E 1 = 1 0 0 0 0 0 ; E 2 = 0 1 0 0 0 0 ; E 3 = 0 0 1 0 0 0 ; E 4 = 0 0 0 1 0 0 ; E 5 = 0 0 0 0 1 0 ; E 6 = 0 0 0 0 0 1 . One can prove that they are linearly independent, and they span the vector space M 2 , 3 . The number of the vectors of the basis is 6.

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