Applied Finite Mathematics HW Solutions 61

Applied Finite Mathematics HW Solutions 61 - EXERCISES...

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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 (Jr-311 —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 4-1 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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