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2011 Final Exam

2011 Final Exam - MATH 304–505 May 6 2011 Final exam(with...

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Unformatted text preview: MATH 304–505 May 6, 2011 Final exam (with solutions) Problem 1 (15 pts.) Find a quadratic polynomial p ( x ) = ax 2 + bx + c such that p (1) = 2, p (2) = 5, and p (3) = 2 p ( − 2). Solution: p ( x ) = x 2 + 1. Problem 2 (20 pts.) Consider a linear operator L : R 3 → R 3 given by L ( x, y, z ) = ( x + 2 y − 3 z, 2 x + 5 y + z, 3 x + 8 y + 5 z ) . (i) Find the matrix of the operator L (relative to the standard basis). (ii) Find the dimensions of the range and the kernel of L . (iii) Find a basis for the kernel of L . Solution: The matrix is 1 2 − 3 2 5 1 3 8 5 . The dimensions of the range and the kernel are 2 and 1, respectively. A basis for the kernel consists of a single vector (17 , − 7 , 1). Problem 3 (20 pts.) Let V be a subspace of R 4 spanned by three vectors x 1 = (1 , 1 , 1 , 1), x 2 = (0 , 3 , 2 , 3), and x 3 = ( − 3 , 5 , 1 , 1). Find the distance from the point y = (0 , 12 , , 0) to the subspace V ....
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2011 Final Exam - MATH 304–505 May 6 2011 Final exam(with...

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