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7.2 Solving a System WithAn LU-Factorization
2 1 0 −1 −3 1 7 −2 1 0 −3 −5 0 0 4 x1 x2 x3 = 12 17 5 We now let 7 −2 1 0 −3 −5 0 0 4 x1 x2 x3 = y1 y2 y3 (∗) and the above system becomes 1 0 0 2 1 0 −1 −3 1 y1 y2 y3 = 12 17 5 (∗∗) The system (∗∗) is easily solved for the vector y= [y1,y2,y3] by forward-substitution. From the first row we see that y1 = 12; from ...
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