Stitz-Zeager_College_Algebra_e-book

From our discussion at the beginning of the section

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Unformatted text preview: −−−−−−− 4 −4 − − − − − − − − → 0 −7 Replace R3 with −2R1 + R3 2 3 −4 10 0 −1 −2 2 Now we repeat the above process for the variable y which means we need to get the leading entry in R2 to be 1. 1 2 −1 4 1 2 −1 4 1 Replace R2 with − 7 R2 4 0 −7 4 −4 − − − − − − − → 0 1 −7 4 −−−−−−− 7 0 −1 −2 2 0 −1 −2 2 To guarantee the leading 1 in R3 is to the right of the leading 1 in R2, we get a 0 in the second column of R3. 1 2 −1 4 1 2 −1 4 Replace R3 with R2 + R3 4 0 1 −4 4 −− − − − − − −→ 0 1 −4 −−−−−−−− 7 7 7 7 18 18 0 −1 −2 2 0 0 −7 7 Finally, we get the leading entry in R3 to be 1. 1 2 −1 4 1 7 Replace R3 with − 18 R3 4 4 1 −7 − − − − − − −→ 0 −− − − − − − − 0 7 0 0 0 − 18 18 7 7 2 −1 4 4 1 −4 7 7 0 1 −1 Decoding from the matrix gives a system in triangular form 1 2 −1 4 4 x + 2y − z = Decode from the matrix 4 4 4 4 0 1 −7 y − 7z = −−−−−−−→ −−−−−−− 7 7 0 0 1 −1 z = −1 4 4 We get z = −1, y = 4 z + 7 = 4 (−1) + 7 = 0 and x = −2y + z + 4 = −2(0) + (−1) + 4 = 3 for a 7 7 final answer of (3, 0, −1). We l...
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