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Unformatted text preview: Name: ID#: Solutions to Final Exam Math 20 Introduction to Linear Algebra and Multivariable Calculus 21 May 2004 Show all of your work. Full credit may not be given for an answer alone. You may use the backs of the pages or the extra pages for scratch work. Do not unstaple or remove pages. This is a noncalculator exam. Students who, for whatever reason, submit work not their own will ordinarily be required to withdraw from the College. —Handbook for Students 1 1 1. (20 Points) For this and the next three problems, let A =  1 1 1 2 2 1 1 1 . (a) Find the reduced row echelon form of A . Label each operation to receive partial credit in case of arithmetic mistakes. Solution. We have A =  1 1 1 2 2 1 1 1 ∼ 1 1 1 2 2 ← ( 1) · row 1 ← row 2 + ( 2) · row 1 ← row 3 + ( 1) · row 1 ∼ 1 1 1 1 ← 1 2 row 2 ← row 3 + ( 1) · row 2 ∼ 1 1 1 ← row 1 + ( 1) · row 2 (b) Find each of the following numbers. Explain your answers. (i) rank A Solution. The rank is the dimension of the column space, or the number of columns of the RREF which are standard basis vectors, or the number of leading entries in the RREF. This number is apparently 2. (ii) null A Solution. The rank plus the nullity is the number of columns in a matrix. In our case, this is 1. (iii) det A Solution. Since the nullity is positive, the matrix is not invert ible, so its determinant is zero. / 20 1 2 2 2. (15 Points) Remember, A =  1 1 1 2 2 1 1 1 . Let b = 1 2 1 . Find the parametric form of the general solution to the system of linear equations A x = b . Solution. We form the augmented matrix and find its RREF. A b =  1 1 1 1 2 2 2 1 1 1 1 ∼ 1 1 1 1 Therefore we can read off the general solution: x 3 is free, x 2 = 0, and x 1 = 1 + x 3 . Thus x 1 x 2 x 3 =  1 + x 3 x 3 =  1 + x 3 1 1 . / 15 2 3 3 3. (15 Points) Continuing to let A =  1 1 1 2 2 1 1 1 , let c = 1 2 2 . Find the parametric form of the general leastsquares solution to the system of linear equations A x = c . Solution. A leastsquares solution to A x = c is a solution to A T A x = A T c . We have A T A = 6 6 2 6 6 ; A T c =  7 1 7 . Using Gaussian Elimination, A T A A T c = 6 6 7 2 1 6 6 7 ∼ 1 1 7 6 1 1 2 ....
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 Spring '09
 MihaiBeligan
 Linear Algebra, Algebra, Multivariable Calculus, Row, Col C

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