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College Algebra, 2013a

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8.1 Systems of Linear Equations: Gaussian Elimination 559<br />

Our next step is to get the coefficient of y in E2 equal to 1. To that end, we have<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

(E1) x + 3 2 y − 1 2 z = 1<br />

2<br />

(E2) −15y +4z = −3<br />

(E3) −15y +4z = 3<br />

Replace E2 with− 1<br />

15 E2<br />

−−−−−−−−−−−−−−−→<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

(E1) x + 3 2 y − 1 2 z = 1 2<br />

(E2) y − 4<br />

15 z = 1 5<br />

(E3) −15y +4z = 3<br />

Finally, we rid E3 ofy.<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

(E1) x + 3 2 y − 1 2 z = 1 2<br />

(E2) y − 4<br />

15 z = 1 5<br />

(E3) −15y +4z = 3<br />

Replace E3 with15E2+E3<br />

−−−−−−−−−−−−−−−−−→<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

(E1) x − y + z = 5<br />

(E2) y − 1 2 z = −3<br />

(E3) 0 = 6<br />

The last equation, 0 = 6, is a contradiction so the system has no solution. According to<br />

Theorem 8.1, since this system has no solutions, neither does the original, thus we have an<br />

inconsistent system.<br />

3. For our last system, we begin by multiplying E1 by 1 3 to get a coefficient of 1 on x 1.<br />

⎧<br />

⎨<br />

⎩<br />

(E1) 3x 1 + x 2 + x 4 = 6<br />

(E2) 2x 1 + x 2 − x 3 = 4<br />

(E3) x 2 − 3x 3 − 2x 4 = 0<br />

Replace E1 with 1 3 E1<br />

−−−−−−−−−−−−−→<br />

⎧<br />

⎨<br />

⎩<br />

(E1) x 1 + 1 3 x 2 + 1 3 x 4 = 2<br />

(E2) 2x 1 + x 2 − x 3 = 4<br />

(E3) x 2 − 3x 3 − 2x 4 = 0<br />

Next we eliminate x 1 from E2<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

(E1) x 1 + 1 3 x 2 + 1 3 x 4 = 2<br />

(E2) 2x 1 + x 2 − x 3 = 4<br />

(E3) x 2 − 3x 3 − 2x 4 = 0<br />

Replace E2<br />

−−−−−−−−−−→<br />

with −2E1+E2<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

(E1) x 1 + 1 3 x 2 + 1 3 x 4 = 2<br />

(E2)<br />

1<br />

3 x 2 − x 3 − 2 3 x 4 = 0<br />

(E3) x 2 − 3x 3 − 2x 4 = 0<br />

We switch E2 andE3 togetacoefficientof1forx 2 .<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

(E1) x 1 + 1 3 x 2 + 1 3 x 4 = 2<br />

(E2)<br />

1<br />

3 x 2 − x 3 − 2 3 x 4 = 0<br />

(E3) x 2 − 3x 3 − 2x 4 = 0<br />

Finally, we eliminate x 2 in E3.<br />

Switch E2 and E3<br />

−−−−−−−−−−−→<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

(E1) x 1 + 1 3 x 2 + 1 3 x 4 = 2<br />

(E2) x 2 − 3x 3 − 2x 4 = 0<br />

(E3)<br />

1<br />

3 x 2 − x 3 − 2 3 x 4 = 0

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