Solve the equation by Jacobi's method : $5x + 2y +4z = 12$ ;$x+4y=2z=15$ ;$x+2y+5z=20$

Asked Apr 09 '16 at 14:19

Active Apr 09 '16 at 14:33

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Solve the equation by Jacobi's method : $$5x + 2y +7z = 30$$ $$x+4y+2z=15$$ $$x+2y+5z=20$$ I am getting the values as $$x=4.08,y=3.375,z=4.12.$$

My equations : $$x=\frac{1}{5} \{30-2y-7z\}$$ $$y=\frac{1}{4} \{15-x-2z\}$$ $$z=\frac{1}{5} \{20-x-2y\}$$

The values are given to be :$$x=1,y=2,z=3$$ (apologies for the previous errors)

edited Apr 09 '16 at 14:33

asked Apr 09 '16 at 14:19

PurpleShark

$1+4+2=7$ and $1+2+5=8$.... – Mastrem Apr 09 '16 at 14:24
sorry too much errors have to delete – PurpleShark Apr 09 '16 at 14:25
What are you using as the starting value? – Moo Apr 10 '16 at 06:57
itnitial values as zero – PurpleShark Apr 10 '16 at 07:27
With a starting value of $x_0 = (x, y, z) = (0, 0, 0)$, the Jacobi method converges to $x = (1, 2, 3)$ in 133 iterations (give or take depending on precision). – Moo Apr 10 '16 at 12:44
133 iterations? i am told to do only 3 the how am i supposed to do this – PurpleShark Apr 10 '16 at 13:31
Well, you pick $x_0 = (0, 0, 0)$ and use the iteration formula to calculate $x_1, x_2, x_3$ manually. It can be done by hand, but will not converge for a long while. – Moo Apr 11 '16 at 01:59

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