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Find the first two iterations of the SOR method with ai = 1.1 for the following linear

Chapter 7, Problem 1

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QUESTION:

Find the first two iterations of the SOR method with \(\omega=1.1\) for the following linear systems, using \(\mathbf{x}^{(0)}=\mathbf{0}\):

a. \(\begin{array}{l}
3 x_{1}-x_{2}+x_{3}=1, \\
3 x_{1}+6 x_{2}+2 x_{3}=0, \\
3 x_{1}+3 x_{2}+7 x_{3}=4 .
\end{array}\)

b. \(\begin{aligned}
10 x_{1}-x_{2} & =9 \\
-x_{1}+10 x_{2}-2 x_{3} & =7 \\
-2 x_{2}+10 x_{3} & =6 .
\end{aligned}\)

c. \(\begin{aligned}
10 x_{1}+5 x_{2} & =6, \\
5 x_{1}+10 x_{2}-4 x_{3} & =25, \\
-4 x_{2}+8 x_{3}-x_{4} & =-11, \\
-x_{3}+5 x_{4} & =-11 .
\end{aligned}\)

d. \(\begin{aligned}
4 x_{1}+x_{2}+x_{3}+x_{5} & =6, \\
-x_{1}-3 x_{2}+x_{3}+x_{4} & =6, \\
2 x_{1}+x_{2}+5 x_{3}-x_{4}-x_{5} & =6, \\
-x_{1}-x_{2}-x_{3}+4 x_{4} & =6, \\
2 x_{2}-x_{3}+x_{4}+4 x_{5} & =6 .
\end{aligned}\)

Questions & Answers

QUESTION:

Find the first two iterations of the SOR method with \(\omega=1.1\) for the following linear systems, using \(\mathbf{x}^{(0)}=\mathbf{0}\):

a. \(\begin{array}{l}
3 x_{1}-x_{2}+x_{3}=1, \\
3 x_{1}+6 x_{2}+2 x_{3}=0, \\
3 x_{1}+3 x_{2}+7 x_{3}=4 .
\end{array}\)

b. \(\begin{aligned}
10 x_{1}-x_{2} & =9 \\
-x_{1}+10 x_{2}-2 x_{3} & =7 \\
-2 x_{2}+10 x_{3} & =6 .
\end{aligned}\)

c. \(\begin{aligned}
10 x_{1}+5 x_{2} & =6, \\
5 x_{1}+10 x_{2}-4 x_{3} & =25, \\
-4 x_{2}+8 x_{3}-x_{4} & =-11, \\
-x_{3}+5 x_{4} & =-11 .
\end{aligned}\)

d. \(\begin{aligned}
4 x_{1}+x_{2}+x_{3}+x_{5} & =6, \\
-x_{1}-3 x_{2}+x_{3}+x_{4} & =6, \\
2 x_{1}+x_{2}+5 x_{3}-x_{4}-x_{5} & =6, \\
-x_{1}-x_{2}-x_{3}+4 x_{4} & =6, \\
2 x_{2}-x_{3}+x_{4}+4 x_{5} & =6 .
\end{aligned}\)

ANSWER:

Step 1 of 30

Given: 

(a). \(\begin{array}{r} 3 x_{1}-x_{2}+x_{3}=1 \\ 3 x_{1}+6 x_{2}+2 x_{3}=0 \\ 3 x_{1}+3 x_{2}+7 x_{3}=4 \end{array}\)

(b). \(\begin{array}{r} 10 x_{1}-x_{2}=9 \\ -x_{1}+10 x_{2}-2 x_{3}=7 \\ -2 x_{2}+10 x_{3}=6 \end{array}\)

(c). \(\begin{aligned} 10 x_{1}+5 x_{2} & =6 \\ 5 x_{1}+10 x_{2}-4 x_{3} & =25 \\ -4 x_{2}+8 x_{3}-x_{4} & =-11 \\ -x_{3}+5 x_{4} & =-11 \end{aligned}\)
(d).  \(\begin{array}{r} 4 x_{1}+x_{2}+x_{3}+x_{5}=6 \\ -x_{1}-3 x_{2}+x_{3}+x_{4}=6 \\ 2 x_{1}+x_{2}+5 x_{3}-x_{4}-x_{5}=6 \\ -x_{1}-x_{2}-x_{3}+4 x_{4}=6 \\ 2 x_{2}-x_{3}+x_{4}+4 x_{5}=6 \end{array}\)

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