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Applied Finite Mathematics HW Solutions 40

Applied Finite Mathematics HW Solutions 40 - 38 EXERCISES...

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Unformatted text preview: 38 . ' _ ' EXERCISES 3.4 (k) If we convert the augmented matrix back to a system of equations, the result is x—y+3w=4. z+4w=6. The system has an infinity of solutions. They can he expressed in the form a: = 4 + y —— 3w, 2 = 6 — 410, where y and w are arbitrary. EXERCISES 3.4 6. (a) This augmented matrix is in row echelon form.- (b) This augmented matrix is not in row echelon form. The 4 in the last column should be a 1. (c) This augmented matrix is not in row echelon form. The row of zeros should be at the hottom of the matrix. (d) This augmented matrix is not in row echelon form. The —1 in the (2, 2) position should be a 1. (e) This augmented matrix is in row echelon form. (f) This augmented matrix is not in row echelon form. There cannot be a 1 in the (3,2) position; it should be a zero. 7. We present a sequence of elementary row operations that lead to a row echelon iiorm for the augmented matrix of the given system. You may, however, use a different sequence of operations, resulting in a different row echelon form; The solution of the System, however, _ must be the sarne. ' ' 1 2 —3 13 2 1 —4 —13 3 —2 1 —44) R3 —'» —3R2+R3 13 l 2 —3 13 R2 —> —2R, + R2 —) 0 —3 2. —5 R3 —> —3R1 'i" R3 . 0 —8 10 1 2 —3 13 1 2 H3 13 ~---! 0 —3 2 —13 R2 —> R3 —) 0 l 4 "5 0 l 4 45 R3 —) R2 0 -3 2 -13 Ra —) 3R2 + R3 1 2 -3 13 1 2 —3 13 -—1 D 1 4 —5 "-4 0 1 4 "5 0 0 14 #28 R3 —) 3.3/14 0 0 1 *2 We now convert this augmented matrix to an equivalent system of equations, :1: +23; — 32': 13, y + 43 = —5, z = —2. When we substitute 2 = —2 into the second equation, we obtain y+4(—-2) == —5 => y: —5+8=3. When we substitute 3; = 3 and z = —2 into the first equation, we get :1:+2(3) _3(_2) = 13 => m213-—12=1. '1'”... “Mun-:4 1...... km M.“ 3.. comm-Ham mm. mun-:n-‘ni In“: In: flu I Iniumih: nFMnni-lnhn Rmimfm-a Further mmvinmim in nuri nr in fin". is. Rfridlv nmhihiied. ...
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