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Applied Finite Mathematics HW Solutions 46

Applied Finite Mathematics HW Solutions 46 - 19 20 44...

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Unformatted text preview: 19. 20. 44 _ EXERCISES 3.4 68 " 612 24 y—9(Z§)——»12 =¢r y——12+E'_E' When we substitute y = 24/49 and z = 68/49 into the first equation, we get 24 68 43 272 11 m+2(&-§)—~4(E)=H3 = $——3—E+E—?. We present a sequence of elementary row operations that lead to a row echelon form for the augmented matrix of the given system. You may, however, use a difl’erent sequence of operations, resulting in a different row echelon form. The solution of the system, however, must be the same. 2 1 —4 13 2 1 —4 13 4 -1 2 _3 R2 —e -2R1 + R2 —r 0 —3 10 H29 R2 —) -—R3 6 2 —2 16 R3 —1 —3R1 +1133 0 -1 10 —23 R3 —> R2 2 1 -—4 13 2 1 —4 13 __., 0 1 -10 23 __., 0 1 —10 23 0 *3 10 —29 R343R2+R3 0 0 —20 40 Re—r—Ra/ZO 2 1 —4 13 R1 —+ 121/2 1 1/2 —2 13/2 __., 0 1 -10 23 __., 0 1 —10 23 0 0 1 —2 0 l} 1 —2 We n0w convert this augmented matrix to an equivalent system of equations, 1 13 y — 102 = 23, z = —2. When we substitute 2 = -—2 into the second equation, we obtain y—10(—2)=23 =:. y=23—20=3. ' When we substitute y = 3 and z = —2 into the first equation, we get 1 13 13 3 $+§(3)—'2(—2)—-E =}' {EH-Ewi—‘i-l. We present a sequence of elementary row operations that lead to a row echelon form for the augmented matrix of the given system. You may, however, use a difl‘erent sequence of ‘ ' operations, resulting in a difl‘erent row echelon form. The solution of the system, however, must be the same. 3‘ 2 —4 —15 R1—>¥R2+R1_ 1 1 _5 2 1 1 7 -—-—-+ 2 1 1 4 _2 2 —22 7 R2 —> —2R1 + R2 -2 Ream/2 2 _1 1 —1 Raee2Ri+Ra 1 1 —5 —-22 1 1 —5 F22 __., 0 —1 11 51 R2 —> —-R2 —» O 1 -11 —51 0 -3 11 43 0 —3 11 43 R, —> 3R2 + R3 1 1 —-5 —22 1 1 —-5 —22 __., O 1 —11 -51 ___, 0 1 —-11 —51 0 0 —22 —110 R3 _> —R3/22 0 0 1 5 We now convert this augmented matrix to an equivalent system of equations, .. __..L!L3¢..l /'—‘H. ...
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