Math416_HW4_sol

# Math416_HW4_sol - Math 416 Abstract Linear Algebra Fall...

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Math 416 - Abstract Linear Algebra Fall 2011, section E1 Homework 4 solutions Section 2.2 2.1.b. In matrix form, the equation is 1 - 2 - 1 2 - 3 1 3 - 5 0 1 0 5 x 1 x 2 x 3 = 1 6 7 9 . The corresponding vector equation is x 1 1 2 3 1 + x 2 - 2 - 3 - 5 0 + x 3 - 1 1 0 5 = 1 6 7 9 . We solve the system using row reduction: 1 - 2 - 1 1 2 - 3 1 6 3 - 5 0 7 1 0 5 9 1 - 2 - 1 1 0 1 3 4 0 1 3 4 0 2 6 8 1 - 2 - 1 1 0 1 3 4 0 0 0 0 0 0 0 0 1 0 5 9 0 1 3 4 0 0 0 0 0 0 0 0 Note that x 3 is a free variable. The solution set is 9 - 5 x 3 4 - 3 x 3 x 3 R 3 | x 3 R . 2.1.e. (2 pts) In matrix form, the equation is 1 2 - 1 3 2 4 - 1 6 0 1 0 2 x 1 x 2 x 3 x 4 = 2 5 3 . The corresponding vector equation is x 1 1 2 0 + x 2 2 4 1 + x 3 - 1 - 1 0 + x 4 3 6 2 = 2 5 3 . We solve the system using row reduction:

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Math416_HW4_sol - Math 416 Abstract Linear Algebra Fall...

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