Applied Finite Mathematics HW Solutions 63

Applied Finite Mathematics HW Solutions 63 - EXERCISES 3.5...

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Unformatted text preview: EXERCISES 3.5 61 1 o 0 29 29 R1—1—29.R4+R1 1 o o c c 0 1 c 2 4 122442344122 0 1 o c 2 _’ 001—23—26 3342334+R3_' 0010—3 6 0 0 1 1 . o 0 c 1 1 The solution of the system is 3: = 0, y = 2, z = —3, w = 1. - 25. We present a sequence of elementary rowr 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 —3 1 4 1 2 —3 1 4 *1 1 4 —1 3 Rz—‘R1+R2 0 3 l U 7 1224—33 2 3 —4 *2 a4 114—1—2R,+R3 o —1 2 —4 w12 123—422 2 6 ‘3 —2 6 171-4—H2121+a.1 o 2 3 e4 —2 1 2 —3 1 4 R1—1—2RQ+R1 1 c 1 '—7 —2o _} 01—24 12 ' __, 01—24 12 o 3 1 o 7 age—334+R3 o o 7 —12 —29 o 2 3 -—4 —2 add—2122+R4 o o 7 H12 #26 me—me 0000s When we convert the last row to an equation, we obtain 0 = 3, an impossibility. The system has no solutions; it is inconsisteut. ' 26. 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, hex-(ever, use a difl'erent sequence of operations, resulting in a different row echelon form. The solution of the system, however, must be the same. 1 -2 e3 1 4 1 2 43 1 4 2 —3 1 —2 —5 Rg—>—2R1+R/z _} U *7 7 -4 ._13 Rz—tF-Rg 1 1 —4 6 1 33441214123 0 —1 —1 5 e3 123—ng 4 0 —6 5 0 34—1—4R1+s4 e —8 6 1' '—16 1 2 —3 1 4 R1—1~—2R2+R; 1 o —5 11 e2 0 1 1 —5 3 _} 01 1 —5 3 0 e7 7 ‘4 —13 R3—17R2+R3 c o 14 —39 8 0 —8 6 1 —16 34433244124 0 0 14 #39 8 Rx—>~R5+Rq 1 O —5 11 —-2 1 0 —5 11 —2 R145R3+R1 c 1 1 —5 3 _} 0 1 1 —5 3 Rg—ifiRad'Rg —’ c c 14 —39 3 aggro/14 c 0 1 —39/14 4/7 c 0 o c 0 o 0 0 c 0 1 c 0 —41/14 6/7 _) o 1 o —31/14 17/7 0 o 1 —39/14 4/7 c c o o 0 We now convert this augmented matrix to an equivalent system of equations, . -. . .. nu. .. a n -__‘___ I'.‘_._4L._---—-n.l...d:m :nnn-‘nl‘;n6!ll Erwififllunrnhihi‘md ...
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