x= 10 7 t, 3 9 y= 1 5 + t, 3 9 z = t.

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Note: Although the solution looks slightly different than the solution obtained previously, it is equivalent in the sense that it describes precisely the same solution set.

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Another popular method, LU decomposition, is useful, particularly if you have many systems, all having the same coefficient matrix. The idea behind the method is simple. If A is a square matrix, it may be possible to factor A = LU where L is lower triangular with 1s on the main diagonal and U is upper triangular. The system Ax = b then reads (LU)x = b, which can be written L(Ux) = b. If we let y = Ux, we can solve Ly = b for y. Once we have determined y, we solve Ux = y for x. Even though the solution of a system by LU decomposition involves solving two systems of equations, each involves a triangular matrix so the computation is efficient. Thus, there are two steps to solving a system of equations by LU decomposition: factorization and back substitution. The corresponding Mathematica commands are LUDecomposition and LUBackSubstitution.

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LUDecomposition[matrix] finds the LU decomposition of matrix. LUBackSubstitution[data, b]uses the output of LUDecomposition[matrix] to solve the system matrix.x = b.

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The output of LUDecomposition consists of three parts: (1) the matrices L and U packed as a single matrix, (2) a permutation vector, and (3) the L condition number of the matrix. The output of LUDecomposition, data, is fed into LUBackSubstitution to solve the system. The permutation vector rearranges the rows in order to ensure a maximum degree of numerical stability. The condition number will be of no concern to us in this chapter. LUDecomposition and LUBackSubstitution cannot be used on systems that possess an infiite number of solutions.

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Linear Algebra

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EXAMPLE 19 To solve the system 2 x + y + z = 7, x 4 y + 3z = 2, 3 x + 2 y + 2 z = 13 using LU decomposition, we must first obtain the matrix factorization of the coefficient matrix.

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2 1 1 a = 1 4 3 ; 3 2 2

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b ={7,2,13};

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data = LUDecomposition[a]

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{1, 4,3},{2,9, 5}, 3, 14, 7 ,{2,1,3},1 9 9

LUBackSubstitution[data, b] {1, 2, 3}

The next two examples illustrate the structure of data.

EXAMPLE 20

2 3 4 m = 4 11 14 ; 6 29 43 {lu, p, cond} = LUDecomposition[m]

{{{2,

3, 4}, {2, 5, 6}, {3, 4, 7}}, {1, 2, 3}, 1}

In this example, no rearrangement of the rows was performed because the permutation vector, p, is 2, 3) . The first part of LUDecomposition[m] is given in a packed format. Since LU is known to be

1 0 0 x x x x x x 1 0 0

of the form x 1 0 0 x x , only nine entries (represented by x) need be specified. The first part of LUDecomposition[m] specifies these nine numbers as a single matrix.

lu //MatrixForm 2 3 4 2 5 6 3 4 7

The numbers, although combined into one matrix, are in their correct positions. l = 2 1 0

3 4 1

EXAMPLE 21

1 0 0

u = 0 5 6

0 0 7

2 3 4

2 1 1 m = 1 4 3 ; 3 2 2 {lu, p, cond} = LUDecomposition[m]

{1, 4,3},{2,9, 5}, 3, 14, 7 ,{2,1,3},1 9 9

lu // MatrixForm 1 4 3 2 9 5 3 14 7 9 9

Linear Algebra

If we proceed as in the previous example, we would be tempted to say that 1 0 0 l = 2 1 0 3 14 1 9 and 1 4 3 u = 0 9 5 0 0 7 9

However, multiplying l by u does not give back the original matrix: l.u //MatrixForm 1 4 3 2 1 1 3 2 2 The permutation vector, p = {2, 1, 3}, indicates that the rows of the matrix have been interchanged. Indeed rows 1 and 2 have been switched. If we permute the rows of l, we should get back our original matrix upon multiplication by u. 2 1 0 1 4 3 l = 1 0 0 ; u = 0 9 5 ; 3 14 1 0 0 7 9 9 l.u // MatrixForm 2 1 1 1 4 3 3 2 2