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HE1 - Hour Exam I 1 Find the inverse of the matrix 1 2 2 0...

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Hour Exam I – October 15, 2010 1. Find the inverse of the matrix A = 0 1 - 1 1 0 2 0 - 1 2 . Solution: For this you just need to compute the reduced row echelon form of ( A I ): 0 1 - 1 1 0 0 1 0 2 0 1 0 0 - 1 2 0 0 1 1 0 2 0 1 0 0 1 - 1 1 0 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 0 1 0 0 1 1 0 1 . So A - 1 = - 2 1 - 2 2 0 1 1 0 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 0 + 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 | 0 2 4 7 2 | 1 1 2 3 0 | 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 0 1 0 + s - 2 1 0 0 + t 6 0 - 2 1
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2 3. Suppose that A is an invertible n × n matrix, B is an n × r matrix, ~x is an n-component vector, and ~ y is an r-component vector. If ( A B ) ~x ~ y = ~ b, find a formula for ~x in terms of A - 1 , B , ~ b and ~ y .
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