Linear Prog COP4355_Part_57

Linear Prog COP4355_Part_57 - Appendix A: Review of...

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232 Appendix A: Review of Mathematical Tools A.6 (continued) (j) -1 64 - 1 4 10 10 = -1 6 1 1 10 10    (using the shortcut method for 2 x 2’s) (k) -1 23 - 4 3 55 = 1 2 1 4 - (l) Check: 2 3 -1 2 3/2 5 1 0 0 0 4 2 1 -1/2 -2 = 0 1 0 1 0 -1 -2 3/2 4 0 0 1 0 I 3 -1 1 0 0 4 0 0 -1 1 3/2 -1/2 1/2 0 0 0 0 -3/2 -1/2 -1/2 0 1 -5/4 1/2 -3/8 0 1/2 1/4 0 1/4 II III -1/2 3/8 1 -2 3/2 5 1 -1/2 -2 4
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Solutions Manual and Workbook 233 A.6 (continued) (m) Check: 1 1 0 0 1 0 1 0 0 1 0 0 1 -1 0 = 0 1 0 0 2 -1 2 -2 -1 0 0 1    DETERMINANTS A.7 (a) 4 =+4 (b) (c) 5 -3 = 10 - (-12) = 22 4 2 1 1 0 1 0 0 1 0 0 0 1 0 0 2 -1 0 0 1 1 1 0 1 0 0 0 -1 0 -1 1 0 0 2 -1 0 0 1 1 0 0 0 1 0 0 1 9 1 -1 0 0 0 -1 -2 2 1 1 0 0 0 1 0 0 1 0 1 -1 0 0 0 1 2 -2 -1 4 2 = (4)(1) - (2)(3) = 4 - 6 = -2 3 1
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234 Appendix A: Review of Mathematical Tools A.7 (continued) (d) 3 2 = 18 - 18 = 0 9 6 (e) (f) 2+4 2 1 1 0 2 1 1 1 2 1 4 = 0 + (-1) (4) 0 1 2 + 0 + 0 0 1 2 0 1 1 0 1 1 0 0 = 4(0 + 2 + 0 -1 -4 + 0) = 4(-3) = -12 A.8 (a) [] [ ] 2 -7 3 7 (2)(7) - (-7)(3) 14 + 21 35 5 Y = = = = = - -14 -14 -14 -14 2 [ ] 6 -1 5 6 -1 2 1 0 2 4 3 3 = (6)(1)(4) + (-1)(0)(3) + (5)(2)(3) - (5)(1)(3) + (6)(0)(3) + (-1)(2)(4) = 24 + 0 + 30 - 15 + 0 + 8 = 54 - 7 = 47 [ ] -7 6 7 2 (-7)(2) - (6)(7) -14 - 42 -56 X = = = = = 4 2 6 (2)(2) - (6)(3) 4 - 18 -14 3 2
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This note was uploaded on 11/13/2011 for the course COP 4355 taught by Professor Koslov during the Spring '10 term at University of Florida.

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Linear Prog COP4355_Part_57 - Appendix A: Review of...

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