Question #235891

Employ the Gauss-Seidel method, solve the system. 10𝑥 + 𝑦 + 𝑧 = 12 2𝑥 + 2𝑦 + 10𝑧 = 14 2𝑥 + 10𝑦 + 𝑧 = 13


1
Expert's answer
2021-09-13T17:26:18-0400

Rewrite


10x+y+z=1210x+y+z=12

2x+10y+z=132x+10y+z=13

2x+2y+10z=142x+2y+10z=14


xn+1=110(12ynzn)x_{n+1}=\dfrac{1}{10}(12-y_n-z_n)

yn+1=110(132xn+1zn)y_{n+1}=\dfrac{1}{10}(13-2x_{n+1}-z_n)

zn+1=110(142xn+12yn+1)z_{n+1}=\dfrac{1}{10}(14-2x_{n+1}-2y_{n+1})

Initial gauss (x,y,z)=(0.5,0.5,0.5)(x,y,z)=(0.5,0.5,0.5)

1st Approximation


x1=110(120.50.5)=1.1x_{1}=\dfrac{1}{10}(12-0.5-0.5)=1.1

y1=110(132(1.1)0.5)=1.03y_{1}=\dfrac{1}{10}(13-2(1.1)-0.5)=1.03

z1=110(142(1.1)2(1.03))=0.974z_{1}=\dfrac{1}{10}(14-2(1.1)-2(1.03))=0.974

2nd Approximation


x2=110(121.030.974)=0.9996x_{2}=\dfrac{1}{10}(12-1.03-0.974)=0.9996

y2=110(132(0.9996)0.974)=1.00268y_{2}=\dfrac{1}{10}(13-2(0.9996)-0.974)=1.00268

z2=110(142(0.9996)2(1.00268))=0.999544z_{2}=\dfrac{1}{10}(14-2(0.9996)-2(1.00268))=0.999544

3rd Approximation


x3=110(121.002680.999544)=0.9997776x_{3}=\dfrac{1}{10}(12-1.00268-0.999544)=0.9997776

y3=110(132(0.9997776)0.999544)=1.00009008y_{3}=\dfrac{1}{10}(13-2(0.9997776)-0.999544)=1.00009008

z3=110(142(0.9997776)2(1.00009008))=1.000026464z_{3}=\dfrac{1}{10}(14-2(0.9997776)-2(1.00009008))=1.000026464

Solution

x=1.000,y=1.000,z=1.000x=1.000, y=1.000, z=1.000

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