2010sol1 - M340L First Midterm Exam Solutions, February 18,...

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Unformatted text preview: M340L First Midterm Exam Solutions, February 18, 2010 1.Let A = 1 2 5 2 1 7 6 3 3 12 6 . a) Compute A rref . A rref = 1 3 4 1 1- 2 . b) Find all solutions to A x = 0. The equations become z 1 =- 3 x 3- 4 x 4 , x 2 =- x 3 + 2 x 4 , with x 3 = x 3 and x 4 = x 4 . In other words x = x 3 - 3- 1 1 + x 4 - 4 2 1 , where x 3 and x 4 are arbitrary. c) Find all solutions to A x = 1- 1 . This is similar, only with an extra term on the right-hand side. Row- reducing [ A | b ] gives 1 3 4 | - 1 1 1- 2 | 1 | . This means z 1 =- 3 x 3- 4 x 4- 1, x 2 =- x 3 + 2 x 4 + 1, with x 3 = x 3 and x 4 = x 4 . In other words x = x 3 - 3- 1 1 + x 4 - 4 2 1 + - 1 1 , where x 3 and x 4 are arbitrary. d) Find all solutions to A x = 1- 1 1 . Row-reducing [ A | b ] gives 1 3 4 | - 1 1 1- 2 | 1 | 1 , so there are no so- lutions....
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2010sol1 - M340L First Midterm Exam Solutions, February 18,...

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