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Unformatted text preview: M340L Second Midterm Exam Solutions, March 7, 2003 1. Consider the 4 5 matrix A = 1 2 1 4 1 1 1 1 2 6 2 1 11 4 4 7 1 a) What is the rank of A ? [Note: the row reduction of A can be done using integers only. If you start getting fractions, you probably made a mistake] After row reduction, A becomes 1 2 1 1 1 2 1 1 . There are 3 pivots, so the rank is three. b) Find a basis for the column space of A . The pivots are in the 1st, 2nd and 4th columns, so the 1st, 2nd and 4th columns of the ORIGINAL matrix form a basis for Col ( A ): 1 2 , 2 1 6 4 , 1 1 1 7 c) Find a basis for the null space of A . 2 1 1 , 1 2 1 1 2. Consider the linear transformation T : R 2 R 2 defined by T parenleftbigg...
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 Spring '08
 PAVLOVIC
 Linear Algebra, Algebra, Fractions, Integers

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