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9.8 Perform the same calculations as in Example 9.5, but for the tridiagonal system: 0.8 -0.4 XI 41 -0.4 0.8 -0.4 x2 25 -0.4 0.8 X3 105

9.8 Perform the same calculations as in Example 9.5, but for the tridiagonal system: 0.8 -0.4 XI 41 -0.4 0.8 -0.4 x2 25 -0.4 0.8 X3 105

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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9.8 Perform the same calculations as in Example 9.5, but for the tridiagonal system: 0.8 -0.4 X1 41 -0.4 0.8 -0.4 X2 25 -0.4 0.8 X3 105
Transcribed Image Text:9.8 Perform the same calculations as in Example 9.5, but for the tridiagonal system: 0.8 -0.4 X1 41 -0.4 0.8 -0.4 X2 25 -0.4 0.8 X3 105
EXAMPLE 9.5 Solution of a Tridiagonal System Problem Statement. Solve the following tridiagonal system: [2.04 -1 -1 2.04 40.8 -1 0.8 -1 2.04 -1 X3 0.8 -1 2.04 X4 200.8 Solution. As with Gauss elimination, the first step involves transforming the matrix to upper triangular form. This is done by multiplying the first equation by the factor ez/fi and subtracting the result from the second equation. This creates a zero in place of ez and trans- forms the other coefficients to new values, -1 e2 f2 = fr -g1 = 2.04 fi (-1) = 1.550 2.04 e2 r2 = r2 --ri = 0.8 – -1 (40.8) = 20.8 2.04 fi Notice that g, is unmodified because the element above it in the first row is zero. After performing a similar calculation for the third and fourth rows, the system is trans- formed to the upper triangular form Г2.04 -1 40.8 1.550 -1 X2 20.8 1.395 -1 X3 14.221 1.323. X4 210.996 Now back substitution can be applied to generate the final solution: 210.996 r4 X4 = f4 159.480 1.323 14.221 – (-1)159.480 r3 - 83X4 X3 = f3 = 124.538 1.395 20.800 – (-1)124.538 r2 - 82X3 f2 X2 = = 93.778 1.550 40,800 – - (-1)93.778 ri - g1x2 fi = 65.970 2.040
Transcribed Image Text:EXAMPLE 9.5 Solution of a Tridiagonal System Problem Statement. Solve the following tridiagonal system: [2.04 -1 -1 2.04 40.8 -1 0.8 -1 2.04 -1 X3 0.8 -1 2.04 X4 200.8 Solution. As with Gauss elimination, the first step involves transforming the matrix to upper triangular form. This is done by multiplying the first equation by the factor ez/fi and subtracting the result from the second equation. This creates a zero in place of ez and trans- forms the other coefficients to new values, -1 e2 f2 = fr -g1 = 2.04 fi (-1) = 1.550 2.04 e2 r2 = r2 --ri = 0.8 – -1 (40.8) = 20.8 2.04 fi Notice that g, is unmodified because the element above it in the first row is zero. After performing a similar calculation for the third and fourth rows, the system is trans- formed to the upper triangular form Г2.04 -1 40.8 1.550 -1 X2 20.8 1.395 -1 X3 14.221 1.323. X4 210.996 Now back substitution can be applied to generate the final solution: 210.996 r4 X4 = f4 159.480 1.323 14.221 – (-1)159.480 r3 - 83X4 X3 = f3 = 124.538 1.395 20.800 – (-1)124.538 r2 - 82X3 f2 X2 = = 93.778 1.550 40,800 – - (-1)93.778 ri - g1x2 fi = 65.970 2.040
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