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b) The coefficients of the following system are taken from GF(2). Solve it using Gaus- sian elimination. X1 + x2 + x4 = 0 X1 + X3 + X4 = 1 X2 + X5 = 0 X1 + x2 + x3 + X5 = 0 X1 + x3 = 0

b) The coefficients of the following system are taken from GF(2). Solve it using Gaus- sian elimination. X1 + x2 + x4 = 0 X1 + X3 + X4 = 1 X2 + X5 = 0 X1 + x2 + x3 + X5 = 0 X1 + x3 = 0

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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b) The coefficients of the following system are taken from GF(2). Solve it using Gaus- sian elimination. X1 + X2 + X4 = 0 X1 + X3 + X4 =1 X2 + X5 = 0 X1 + X2 + X3 + x5 = 0 X1 + X3 = 0 (b) Gaussian elimination in compressed form should look somewhat like the following. 1 1 0 1 0|0 1 0 1 1 01 0 1 0 0 10 1 1 10 10 \1 0 1 0 0|0 1 1 0 1 0|0 0 1 1 0 0 1 0 1 0 0 10 0 0 1 1 1 0 0 1 1 1 0|0 add row 1 to rows 2,4,5 1 1 0 1 0 0 0 1 10 0|1 0 0 1 0 1 1 0 0 1 1 10 0 0 0 1 0 1 add row 2 to rows 3,5 1 1 0 10|0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 1 0|1 0 0 0 1 0 1 row 4 + row 3 + row 4 1 1 0 1 0 0 0 1 100 0 0 1 0 1 |1 0 0 0 1 0 1 0 0 0 0 0|0 row 5 + row 5 + row 4 1 1 0 1 0|0 0 1 0 0 10 0 0 10 1|1 0 0 0 10 1 0 0 0 0 0 0 row 2 + row 2 + row 3 1 0 0 1 1|0 0 1 0 0 10 0 0 1 0 11 0 0 0 1 0|1 0 0 0 0 0|0 row 1 + row 1 + row 2 0 11 0 1 0 0 10 0 0 10 11 0 0 0 1 0 |1 0 0 0 0 00 1 row 1 + row 1 + row 4 This shows xs is free. As GF(2) only has two elements (0 and 1), the solutions are X5 = 0, X4 = 1, X3 = 1, X2 = 0, X1 = 1 and X5 = 1, X4 = 1, X3 = 0, X2 = 1, X1 = 0.
Transcribed Image Text:b) The coefficients of the following system are taken from GF(2). Solve it using Gaus- sian elimination. X1 + X2 + X4 = 0 X1 + X3 + X4 =1 X2 + X5 = 0 X1 + X2 + X3 + x5 = 0 X1 + X3 = 0 (b) Gaussian elimination in compressed form should look somewhat like the following. 1 1 0 1 0|0 1 0 1 1 01 0 1 0 0 10 1 1 10 10 \1 0 1 0 0|0 1 1 0 1 0|0 0 1 1 0 0 1 0 1 0 0 10 0 0 1 1 1 0 0 1 1 1 0|0 add row 1 to rows 2,4,5 1 1 0 1 0 0 0 1 10 0|1 0 0 1 0 1 1 0 0 1 1 10 0 0 0 1 0 1 add row 2 to rows 3,5 1 1 0 10|0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 1 0|1 0 0 0 1 0 1 row 4 + row 3 + row 4 1 1 0 1 0 0 0 1 100 0 0 1 0 1 |1 0 0 0 1 0 1 0 0 0 0 0|0 row 5 + row 5 + row 4 1 1 0 1 0|0 0 1 0 0 10 0 0 10 1|1 0 0 0 10 1 0 0 0 0 0 0 row 2 + row 2 + row 3 1 0 0 1 1|0 0 1 0 0 10 0 0 1 0 11 0 0 0 1 0|1 0 0 0 0 0|0 row 1 + row 1 + row 2 0 11 0 1 0 0 10 0 0 10 11 0 0 0 1 0 |1 0 0 0 0 00 1 row 1 + row 1 + row 4 This shows xs is free. As GF(2) only has two elements (0 and 1), the solutions are X5 = 0, X4 = 1, X3 = 1, X2 = 0, X1 = 1 and X5 = 1, X4 = 1, X3 = 0, X2 = 1, X1 = 0.
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