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Unformatted text preview: Linear Algebra Explore Assignment 1 I. Exercise on Linear Systems (a). A = [1 1 1 1  0 ] [ 1 2 2 2   4] [0 3 2 1  10] [2 0 0 1  4 ] (b). B = [1 0 0 0  4 ] [ 0 1 0 0   6 ] [ 0 0 1 0   10] [0 0 0 1  12 ] (c). Yes, x4 =12, x3 = 10, x2 = 6, and x1 = 4. (d). A = [1 1 1 1  0 ] [ 2 2 2 2   4] [0 3 2 1  10] [2 0 0 1  4 ] (e). ). C = [1 0 0 .5  0] [0 1 0 .4  0] [0 0 1 1.1  0] [0 0 0 0  2] (f). No, because the last row is all zeros and has inconsistency equal 2. (g). B is in the reduced echelon form and C isn’t. C has rows 1 through 3 equaling to zero while the last row is 2. This means that B has a unique solution while C does not have a solution. I I. Finding Polynomials – Cardiac Hill (a). A = [0 0 1  5] [1 1 1  3] [9 3 1  2] B = [1 0 0  .5 ] [0 1 0  2.5] [0 0 1  5 ] a = .5 b = 2.5 c = 5 (b). (c). (0,5), (1,3), (3,2), and (2,10). (0^3)a + (0^2)b + (0)c + 1d = 5 (1^3)a + (1^2)b + (1)c + 1d = 3 (3^3)a + (3^2)b + (3)c + 1d = 2 (2^3)a + (2^2)b + (2)c + 1d = 10 A = [0 0 0 1  5 ] [1 1 1 1  3 ] [27...
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This note was uploaded on 01/27/2010 for the course MATH 280 taught by Professor Dwang during the Fall '09 term at Pittsburgh.
 Fall '09
 Dwang
 Algebra

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