Unformatted text preview: EXERCISES 3.5 '59 —22 . 1 1 5
51 . Rz —! ~R2 —* 0 1 ——11
43 0 —3 11 —51 1 1 —5
—1 0 —I 11
' 43 (Jr311 —22) R; —¥—R2+Rl
R3—13R2+R3 1 0 6 29 1 0 6 29 R1 —? "633 + 31
—r 0 1 —11 —51 —* U 1 11 —51 Rg —¥ 11R3 +R2
0 U —22 "110 R3 —> "Ra/22 U 0 1 5 The solution of the system is a: = —1, y = 4, z = 5. . 21. We present a sequence of elementary row operations that lead to the reduced row‘echelon term for the augmented matrix of the given system. You might use a diﬂerent sequence of _ operations, but since the reduced row echelon form is unique, you must arrive at the same
augmented matrix. _ 5 "4 2 —3 R1 —¥ ""233 +31 1 —2 4
2 ——1 3 6 31—1
2 2*13
_ 1——2—4
—1 0711 ——7
6 R2 —> —3R1+R2
2 R3 —> —2R1 + R3 —'i' 1 2 —4
27 Rz—+——ZR;3+R2 —* 0 1 —11 7 R1 —* 2R2 + RI
—5
0 3 11 16 0 3 11 16 Re, —¥ "3R2 +R3 ' 1 0 ——26 "17 1 0 F26 n17 Rl—r2ﬁR3+R1
_.. 0 1 _11 —5  —+ 0 1 411 5)Rg_,1133+32
‘ 0 0 44 31 123—4123/44 0 0 1 31/44
1 0 0 29/22
_.. 0 1 0 11/4 
0 0 1 31/44 The solutiouof the system is m 2 29/22, 1; = 11/4, 2: = 31/44. 22. We present a sequence of elementary row operations that lead to the reduced row echelon
form for the augmented matrix of the given system. You might use a different sequence of
operations, but since the reduced row echelon form is unique, you must arrive at the same augmented matrix.
1 2‘
5 Rz—v—gﬁl+R2 —1 0 1 0 4
0 4? 2 1 —1
' 4 3 ~2
6 1 1 "4 R3 —’ ""331 + Rs 2 0 —1
41 0 1 0
0 0 4 1134~21Rz+s3 1)Rl—1——Rz+Rl n3 2 0 ——1
4 —+ 0 1 0
9 Ra—iRa/4 o o 1 The solution of the system is m : 3/8, y = 4, z = 9/4. ——L!L=..J ...
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 Fall '11
 Dr.Cornell
 Math

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