Use the Jacobi method to find approximate solutions to x1 + 5æ2 – a3 = 5 21 +2 + 6r3 8 2x1 +2-x3 = 2 %3D starting with 0.5 as initial values of x , that is æi = x2 = *3 = 0.5 and iterating until the error is less than 1%. Round-off answer to 6 decimal places. Reminder: Arrange the system to be Diagonally Dominant before iteration. X1 =0.998935, x2 =0.998907, x3 =0.998519 none of the choices X =0.998611, x2 =0.999444, x3 =1.005556 X =0.999468, x2 =0.999454.x3 =0.999259 X1 1.003056, X2 1.001389, x3 1.000324

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Which of the following options is the correct answer? /Note: There is a none of the above option, but I believe it is only for show. So which of the four options with values is the correct answer?/

Use the Jacobi method to find approximate
solutions to
x1 + 5x2 – x3 = 5
21 +2 +6æz 8
2x1 +2-a3 2
starting with 0.5 as initial values of x , that is
x1 = x2 = *3 = 0.5 and iterating until the error
is less than 1%.
Round-off answer to 6 decimal places.
Reminder: Arrange the system to be Diagonally
Dominant before iteration.
X =0.998935, X2 0.998907, x3 =0.998519
none of the choices
X =0.998611, X2 =0.999444, x3 =1.005556
X =0.999468, xX =0.999454. x3 =0.999259
Ox1 =1.003056, x2 =1.001389, x3 =1.000324
Transcribed Image Text:Use the Jacobi method to find approximate solutions to x1 + 5x2 – x3 = 5 21 +2 +6æz 8 2x1 +2-a3 2 starting with 0.5 as initial values of x , that is x1 = x2 = *3 = 0.5 and iterating until the error is less than 1%. Round-off answer to 6 decimal places. Reminder: Arrange the system to be Diagonally Dominant before iteration. X =0.998935, X2 0.998907, x3 =0.998519 none of the choices X =0.998611, X2 =0.999444, x3 =1.005556 X =0.999468, xX =0.999454. x3 =0.999259 Ox1 =1.003056, x2 =1.001389, x3 =1.000324
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