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Unformatted text preview: Solutions to the first midterm exam (October 10) Problem 1. Writing the augmented matrix of the system and reducing it, we get 1 1  2 2 1 2  5 1 1 a 2  a + 2 → 1 1  2 1  1 1 a 2 − 1  a → 1 1  2 1  1 a 2 − 1  a − 1 . If a 2 − 1 negationslash = 0, we can find z = ( a − 1) / ( a 2 − 1) = 1 / ( a + 1), y = 1 and x = 2 − z = 2 − 1 / ( a + 1) = (2 a + 1) / ( a + 1), so there is a unique solution. If a = 1 then the last equation reads 0 = 0 and we get infinitely many solutions z = t , y = 1, x = 2 − t , where t is any number. If a = − 1 then the last equation reads 0 = − 2 and there are no solutions. Answer: a) there are no solutions when a = − 1; b) there is a unique solution x = (2 a + 1) / ( a + 1), y = 1, z = 1 / ( a + 1) when a negationslash = ± 1; c) there are infinitely many solutions x = 2 − t , y = 1, z = t , where t can be any number, when a = 1....
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 Fall '08
 Conger
 Linear Algebra, Algebra, Elementary algebra, Linear map

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