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Applied Finite Mathematics HW Solutions 50

# Applied Finite Mathematics HW Solutions 50 - 48 EXERCISES...

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Unformatted text preview: 48 EXERCISES 3.4 1 2 .43 1 4 1 2 43 1 4 __} 0 1 42 4 12 __} 0 1 —2 4 12 0 3 1 0 7 R3—1——3R2+Ra 0 0 7 —12 _29 0 2 3 —4 —2 mau2R2+Ri 0 0 7 412 —26 34—3423er 1 2 ——3 1 4 0 1 —2 4 12 ' 0 0 7 -—12 ~29 0 0 0 0 3 When we convert the last row to an equation, we obtain 0 = 3, an impossibility. The system has no solutions; it is inconsistent. 26. We present a sequence of elementary rowr operations that lead to a. row echelon form for the augmented matrix of the given system. You may, however, use a diﬁ'erent sequence of operations, resulting in a different row echelon form. The solution of the system, however, must he the same. 1 2 —3 1 4 1 2 —3 1 4 2 —3 1 —2 _5 R2 _. H2171 + R2 __’ 0 -—7 7 —4 ~13 R2 —» "Hg 1 1 —4 6 1 R3—>—R1+R3 0 -—1 —1 5 —3 Ra—»R2 4 0 —ﬁ 5 '0 R4 -—) ~4R1+ 34 0 H8 6 1 —16 ——> 1 2 ﬁ3 1 4 1 2 —3 . 1 4 _} 0 1 1 —5 3 _} 01 1 —5 3 0 -7 7 —-4 ~13 R347R2+R3 0 0 14 ~39 8 0 —8 e 1 "—16 R4—»8R2+R4 0 0 14 —39 8 B4—>~R.3+R4 1 2 —3 1 4 1 2 -—3 1 4 __} 0 l 1 m5 3 _’ 0 1 1 —5 3 0 0 14 "39 8 123—».83/14 0 0 1 ~39/14 4f? 0 0 0 0 0 0 0 0 0 0 We nowr convert this augmented matrix to an equivalent system of equations, m+2y-—3z+w=4, y+zﬂ5w=3, z__3_?.,,,_£ 14 “7' When we solve the last equation for 2:, we get 3911: 4. “171+? When we substitute this into the second equation, we ﬁnd 391:: 4 39m 4_31w 1‘? y+(1—4+§)-5w:3 3} y—3+5IU—"i4——7?—1—4+ 7. Finally, substituting y = 31w/14 +1777, and z z 3910/14 +4}? into the ﬁrst equationgives 31w 17 39111 4 _ _ 3110 34 11711: 2.. _§_ ﬂ m+2("ff+'7')"3(1_4+7)+m_4 =" “4‘7“?“ 14 +7 ”77+ 14- Thus, there is an inﬁnity of solutions, F 7’ 14 7’ 14 7 ....t.__4|. _ rt._:___ "-4.. .131 l__!4_L_ n- "--“a- n...nl._..__.l....a:.... L. ...\_s A- :n a.“ L. 4&d1..—ML:L:-lm1 .r “\I ...
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