Stitz-Zeager_College_Algebra_e-book

# Using entry notation we have r2c 3 a21 b13 a22 b23 a23

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Unformatted text preview: − − − − − −→ (E 3) 0= 6 The last equation, 0 = 6, is a contradiction so the system has no solution. According to Theorem 8.1, since this system has no solutions, neither does the original, thus we have an inconsistent system. 3. For our last system, we begin by multiplying E 1 by (E 1) 3x1 + x2 + x4 = 6 (E 2) 2x1 + x2 − x3 = 4 (E 3) x2 − 3x3 2x4 = 0 1 3 to get a coeﬃcient of 1 on x1 . 1 (E 1) x1 + 3 x2 + 1 x4 = 2 3 (E 2) 2x1 + x2 − x3 = 4 −−−−−−− −−−−−−→ (E 3) x2 − 3x3 − 2x4 = 0 Replace E 1 with 1 E1 3 Next we eliminate x1 from E 2 (E 1) x1 + 1 x2 + 1 x4 = 2 3 3 (E 2) 2x1 + x2 − x3 = 4 (E 3) x2 − 3x3 − 2x4 = 0 1 1 (E 1) x1 + 3 x2 + 3 x4 = 2 Replace E 2 2 (E 2) 1 x2 − x3 − 3 x4 = 0 −− − − −→ −−−−− 3 with −2E 1 + E 2 (E 3) x2 − 3x3 − 2x4 = 0 We switch E 2 and E 3 to get a coeﬃcient of 1 for x2 . 1 (E 1) x1 + 1 x2 + 3 x4 = 2 3 (E 2) 1 x2 − x3 − 2 x4 = 0 3 3 (E 3) x2 − 3x3 − 2x4 = 0 Finally, we eliminate x2 in E 3. 1 1 (E 1) x1 + 3 x2 + 3 x4 = 2 Switch E...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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