Stitz-Zeager_College_Algebra_e-book

1 5 2 1 16 2 1 3 1 3 12 39 83 matrix arithmetic 481

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Unformatted text preview: − (E 3) −z = −6 Finally, we apply the second move from Theorem 8.1 one last time and multiply E 3 by −1 to satisfy the conditions of Definition 8.3 for the variable z . 5 (E 1) x − y + z = (E 2) y − 1 z = −3 2 (E 3) −z = −6 5 (E 1) x − y + z = 1 (E 2) y − 2 z = −3 −− − − − − −→ −−−−−−− (E 3) z= 6 Replace E 3 with −1E 3 Now we proceed to substitute. Plugging in z = 6 into E 2 gives y − 3 = −3 so that y = 0. With y = 0 and z = 6, E 1 becomes x − 0 + 6 = 5, or x = −1. Our solution is (−1, 0, 6). We leave it to the reader to check that substituting the respective values for x, y , and z into the original system results in three identities. Since we have found a solution, the system is consistent; since there are no free variables, it is independent. 2. Proceeding as we did in 1, our first step is to get an equation with x in the E 1 position with 1 1 as its coefficient. Since there is no easy fix, we multiply E 1 by 2 . (E 1) 2x + 3y − z = 1 (E 2) 10x − z...
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