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Unformatted text preview: Princeton University Department of Mathematics MAT 202 – Linear Algebra with Applications MIDTERM – Spring 2007 SOLUTIONS (1) Let A = 1 1 4 2 3 4 4 1 6 4 . (a) Find the reduced rowechelon form of A . (b) Determine all solutions of Ax = 0 . (c) Determine all solutions of Ax = b where the vector b is the sum of the first two columns of A . Solution: (a) We apply GaussJordan elimination: 2 4 1 1 4 2 3 4 4 1 6 4 3 5 → 2 4 1 1 4 3 2 4 1 6 4 3 5 → 2 4 1 1 4 1 6 4 3 2 4 3 5 → 2 4 1 1 4 1 6 4 16 16 3 5 → 2 4 1 1 4 1 6 4 1 1 3 5 → 2 4 1 3 1 2 1 1 3 5  {z } rref A (b) The relation given by the fourth column in the reduced rowechelon form of A above (namely, that the fourth column is 3 × (first column) 2 × (second column) + (third column)) shows that the solutions of Ax = 0 are the vectors of the form x = 2 6 6 4 3 2 1 1 3 7 7 5 t , t ∈ R . (a) Since b is the sum of the first two columns of A , the vector 2 6 6 4 1 1 3 7 7 5 is one solution of Ax = b . Any other solution differs from this one by a solution of Ax = 0 . Therefore, the solutions of Ax = b are the vectors of the form x = 2 6 6 4 1 1 3 7 7 5 + 2 6 6 4 3 2 1 1 3 7 7 5 t , t ∈ R . (2) Consider the matrix A = 1 2 1 1 2 1 5 . (a) Find a basis of the image of A . (b) Compute the coordinates of 3 1 5 in the basis you found in (a). (c) Find a basis for ( im A ) ⊥ , the orthogonal complement of the image of A . 2 Solution: (a) GaussJordan elimination gives: A = 2 4 1 2 1 1 2 1 5 3 5 → 2 4 1 2 1 1 1 1 3 5 → 2 4 1 2 1 1 3 5 = rref A Since the leading ones are on the first and second columns of rref A , we conclude that the first and second columns of...
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 Spring '08
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 Linear Algebra, Algebra, basis, IMA

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