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 =}' {EHEwi—‘il. 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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 Fall '11
 Dr.Cornell
 Math

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