Differential Equations Solutions 157

# Differential Equations Solutions 157 - ± by a pivot...

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Chapter 27 Solutions: Solving Sparse Linear Systems: Taking the Direct Approach CHALLENGE 27.1. (a) We notice that in Gauss elimination, we need only 5 row operations to zero elements in the lower triangle of the matrix, and the only row of the matrix that is changed is the last row. Since this row has no zeros, no new nonzeros can be produced. (b) Since P T P = I , we see that Ax = b ⇐⇒ PAx = Pb ⇐⇒ PAP T ( Px )= Pb , and this veriFes that the reordered system has the same solution as the original one. CHALLENGE 27.2. ±or part (b), the important observation is that if element k is the Frst nonzero in row ± , then we start the elimination on row
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Unformatted text preview: ± by a pivot operation with row k , after row k already has zeros in its Frst k − 1 positions. Therefore, an induction argument shows that no new nonzeros can be created before the Frst nonzero in a row. A similar argument works for the columns. Part (a) is a special case of this. CHALLENGE 27.3. The graph is shown in ±igure 27.1. The given matrix is a permutation of a band matrix with bandwidth 2, and Reverse Cuthill-McKee was able to determine this and produce an optimal ordering. The reorderings and number of nonzeros in the Cholesky factor (nz( L )) are 167...
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## This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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