By Isac G.

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**Sample text**

3 Gaussian Elimination 27 done is to reverse the order of the two equations. The solution can be seen to be x1 = 1, x2 = 1. To illustrate the effect of rounding error we consider the following problem. Problem Solve the following systems rounding the results of all arithmetic operations to three significant figures. 661. 44). 89. 0140. 0200. 020. 863. In this case the solution obtained by using arithmetic which retains three significant figures is nothing like the true solution. 46). 530 to three significant figures.

9) using 3x4 − 2 + 6 + 6 = 2, to give x4 = −2 23 . 8) to obtain 2x3 − 10 + 9 = 5 and conclude that x3 = 3. 6) we have x2 = 10 and x1 = 0 to complete the solution. To summarise, we have x1 = 0, x2 = 10, x3 = 3, x4 = −2 23 , x5 = 1, x6 = 3, x7 = −2, x8 = 5 and x9 = 1. Discussion The system of equations we have just solved is an example of an upper triangular system. If we were to write the system in the matrix and vector form Ax = b, A would be an upper triangular matrix, that is a matrix with zero entries below the diagonal.

X1 + 2x2 + 3x3 + x4 = 8 2x1 + 4x2 − x3 + x4 = 1 x1 + 3x2 − x3 + 2x4 = 1 −3x1 − x2 − 3x3 + 2x4 = −7. Use the lu command in the form [L U] =lu(A), which returns an upper triangular U and a (possibly) permuted lower triangular L so that Ax = LUb, where b is the column vector of right-sides. Having established matrices L and U we have LUx = b. Writing Ly = b, find y by using the left-division operator. We now have Ux = y. Find x by using the left-division operator. Check the result by comparing Ax and b.

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