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Unformatted text preview: MATH 205, SPRING, 2011, 4:00 EXAM # 2 SOLUTIONS (1) Find bases for the rowspace, columnspace, and nullspace of the matrix A given by A = 1 2 0 3 0 0 1 2 1 2 0 8 Solution: First, we have to rowreduce the matrix: A = 1 2 0 3 0 0 1 2 1 2 0 8 ∼ 1 2 0 3 0 0 1 2 0 0 0 5 ∼ 1 2 0 0 0 0 1 0 0 0 0 1 , so the matrix has rank 3. You can then nd the rowspace, since the rows of the reduced rowechelon form are automatically a basis of the rowspace: Rowspacebasis = { (1 , 2 , , 0) , (0 , , 1 , 0) , (0 , , , 1) } . Now, the nullspace is the space of solutions of AX = 0 , which is the same as the solutions for the reduced rowechelon form. Since the second column is the only one without a leading one, you set x 2 = 1 , and the equations given by the reduced matrix determines the other components: Nullspacebasis =  2 1 . The columnspace is a bit trickier. You have to go back to the original matrix A , and nd 3 linearly independent columns. Why 3? Because the dimension of the columnspace is the same as the dimension of the rowspace, which is the rank of the matrix, 3. Note that taking the rst two columns is a bad idea, since they are linearly dependent, so you take the rst, third, and fourth columns (those which end up having leading ones) Columnspacebasis = 1 1 , 1 , 3 2 8 (2) Is the set of functions e 2 x , sin( x ) linearly independent as functions on R ? If so, explain why, if not, explain why not. Solution: This set is linearly independent. There are several ways to prove this. 1 2 MATH 205, SPRING, 2011, 4:00 EXAM # 2 SOLUTIONS 1: Use the Wronskian. In this case, the Wronskian determinant w ( e 2 x , sin( x ) ) is w ( e 2 x , sin( x ) ) = e 2 x sin( x ) 2 e 2 x cos( x ) = e 2 x cos( x ) 2 e 2 x sin( x ) . The set of functions is linearly independent if this Wronskian is not 0 (as a function), and since at x = 0 , w ( e 2 x , sin( x ) ) x =0 = e 2 x cos( x ) 2 e 2 x sin( x ) x =0 = 1 , the functions have to be linearly independent....
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 Spring '08
 zhang
 Math, Linear Algebra, basis

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