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Unformatted text preview: Hour Exam I – October 15, 2010 1. Find the inverse of the matrix A = 1 1 1 2 1 2 . Solution: For this you just need to compute the reduced row echelon form of ( A I ): 1 1 1 0 0 1 2 0 1 0 1 2 0 0 1 ∼ 1 2 0 1 0 1 1 1 0 0 1 2 0 0 1 ∼ 1 0 2 0 1 0 0 1 1 1 0 0 0 0 1 1 0 1 ∼ 1 0 0 2 1 2 0 1 0 2 1 0 0 1 1 1 . So A 1 =  2 1 2 2 1 1 1 . 2. Find the solution to x 1 + 2 x 2 + 3 x 3 = 0 2 x 1 + 4 x 2 + 7 x 3 + 2 x 4 = 1 in the form ~x = ~v + s~v 1 + t~v 2 , where s and t are arbitrary real numbers. Solution: Written in augmented matrix form this is 1 2 3 0  2 4 7 2  1 ¶ ∼ 1 2 3 0  0 0 1 2  1 ¶ ∼ 1 2 0 6   3 0 0 1 2  1 ¶ That means x 1 x 2 x 3 x 4 =  3 2 s + 6 t s 1 2 t t =  3 1 + s  2 1 + t 6 2 1 2 3. Suppose that A is an invertible n × n matrix, B...
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This note was uploaded on 02/06/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.
 Fall '08
 lee

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