mat211rev - 1.2 6 The system is already in row reduced form...

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§ 1.2 6. The system is already in row reduced form with leading variables x 1 , x 3 , x 4 and free variables x 2 , x 5 . Let s = x 2 and t = x 5 . The set of solutions is 3 + 7 s - t s 2 + 2 t 1 - t t : s, t R . 7. The reduced row echelon form of the system is fl fl fl fl fl fl fl fl x 1 +2 x 2 = 0 x 3 = 0 x 4 = 0 x 5 = 0 fl fl fl fl fl fl fl fl . We have leading variables x 1 , x 3 , x 4 , x 5 and free variable x 2 . If t = x 2 then every solution is - 2 t t 0 0 0 with t R . 8. The rref of the system is fl fl fl fl x 2 - x 5 = 0 x 4 +2 x 5 = 0 fl fl fl fl . The leading variables are x 2 , x 4 and the free variables are x 1 , x 3 , x 5 . We have x 2 = x 5 x 4 = 2 x 5 Set t = x 5 , then x 2 x 4 x 5 = t 2 t t for t R where of course x 1 , x 3 can be equal to anything.
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