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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 Fall '11
 Dr.Cornell
 Math

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