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Chapter 13: Problem 1

Use five iterations of Jacobi's method to find approximate solutions of each system of (c) \(-3 x+y=14\) \(2 x-7 y=16\) equations: (a) \(3 x-y=17\) Take \(x_{0}=-4, y_{0}=-2\) \(2 x-5 y=20\) (d) \(4 x+y=19\) Take \(x_{0}=3, y_{0}=0\). (b) \(2 x+y=11\) Take \(x_{0}=1, y_{0}=5\). \(x-3 y=-26\) (e) \(-3 x+2 y=37\) Take \(x_{0}=0, y_{0}=7\) Take \(x_{0}=-3, y_{0}=9\)

Short Answer

Expert verified
Question: Use Jacobi's method with five iterations to find approximate solutions for the following system of linear equations and initial guesses: \(a)~~3x - y = 17\) \(~~~~2x - 5y = 20\) Initial guess: \(x_0 = -4, y_0 = -2\) Answer: After 5 iterations, the approximate solution for system (a) is \((x, y) \approx (6\frac{1}{45}, 1\frac{1}{27})\).

Step by step solution

01

Write equations in Jacobi's method form

\(a) ~~x = \frac{1}{3} (17 + y)\) \(~~~~~~y = \frac{1}{5} ( 20 - 2x)\)
02

Perform first iteration

Using the initial guess, we can calculate \(x_1\) and \(y_1\): \(x_1 = \frac{1}{3}( 17 - (-2)) = \frac{1}{3}(19) = 6\frac{1}{3}\) \(y_1 = \frac{1}{5} (20 - 2(-4)) = \frac{1}{5} (28) = 5\frac{3}{5}\)
03

Perform second iteration

Using the previous iteration result, we can calculate \(x_2\) and \(y_2\): \(x_2 = \frac{1}{3} (17 + 5\frac{3}{5}) = \frac{1}{3} (19\frac{3}{5}) = 6\frac{3}{9}\) \(y_2 = \frac{1}{5} (20 - 2(6\frac{1}{3})) = \frac{1}{5} (20 - 12\frac{2}{3}) = 1\frac{2}{5}\)
04

Perform third iteration

Using the previous iteration result, we can calculate \(x_3\) and \(y_3\): \(x_3 = \frac{1}{3} (17 + 1\frac{2}{5}) = \frac{1}{3} (18\frac{2}{5}) = 6\frac{1}{15}\) \(y_3 = \frac{1}{5} (20 - 2(6\frac{3}{9})) = \frac{1}{5} (20 - 12\frac{4}{9}) = 1\frac{1}{9}\)
05

Perform fourth iteration

Using the previous iteration result, we can calculate \(x_4\) and \(y_4\): \(x_4 = \frac{1}{3} (17 + 1\frac{1}{9}) = \frac{1}{3} (18\frac{1}{9}) = 6\frac{1}{27}\) \(y_4 = \frac{1}{5} (20 - 2(6\frac{1}{15})) = \frac{1}{5} (20 - 12\frac{2}{15}) = 1\frac{1}{15}\)
06

Perform fifth iteration

Using the previous iteration result, we can calculate \(x_5\) and \(y_5\): \(x_5 = \frac{1}{3} (17 + 1\frac{1}{15}) = \frac{1}{3} (18\frac{1}{15}) = 6\frac{1}{45}\) \(y_5 = \frac{1}{5} (20 - 2(6\frac{1}{27})) = \frac{1}{5} (20 - 12\frac{2}{27}) = 1\frac{1}{27}\) So, after 5 iterations, the approximate solution for system (a) is \((x, y) \approx (6\frac{1}{45}, 1\frac{1}{27})\). (Note that the starting point and procedure for the other systems (b), (c), (d) and (e) are similar to that shown for system (a). Simply change the equations and initial guess accordingly to obtain the approximate solution after five iterations for each system.)

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Most popular questions from this chapter

Determine which of the following systems have a non-trivial solution: $$ \text { (a) } \begin{aligned} &2 x-y+3 z=0 \\ &x+2 y-z=0 \\ &5 x+5 z=0 \end{aligned} $$ $$ \text { (b) } \begin{aligned} &3 x+2 y-z=0 \\ &2 x-3 y+2 z=0 \\ &7 x-4 y+3 z=0 \end{aligned} $$ $$ \text { (c) } x+4 y+3 z=0 $$ $$ \begin{aligned} &9 x-10 y-z=0 \\ &6 x-3 y+z=0 \end{aligned} $$ $$ \text { (d) } \begin{aligned} 3 x-y-z=0 \\ x+y+2 z=0 \\ 2 x-2 y+3 z=0 \\ \text { (e) }-x+2 y-3 z=0 \end{aligned} $$ $$ \begin{aligned} &4 x+z=0 \\ &11 x+2 y=0 \end{aligned} $$

Solve the following simultaneous equations using the inverse matrix method: $$ \text { (a) } \begin{aligned} &x+y-z=-1 \\ &2 x-y+2 z=8 \\ &2 x+3 y-z=-3 \end{aligned} $$ $$ \text { (b) } 2 I_{1}+I_{2}-I_{3}=10 $$ $$ \begin{aligned} &4 I_{1}+3 I_{2}+2 I_{3}=18 \\ &3 I_{1}-3 I_{2}+4 I_{3}=0.5 \end{aligned} $$ (c) \(r+s-t=0.5\) \(-2 r-s+2 t=-0.2\) \(r-4 s-2 t=-5\)

Explain what is meant by the trivial solution of a system of linear equations and what is meant by a non-trivial solution.

Determine which of the following systems have non-trivial solutions: $$ \text { (a) } x+2 y-z=0 } \\ {3 x+y+2 z=0} \\ {x+y=0} \end{array} $$ $$ \text { (b) } \begin{aligned} &2 x-3 y-2 z=0 \\ &3 x+y-3 z=0 \\ &x-7 y-z=0 \end{aligned} $$ $$ \text { (c) } \begin{aligned} x+2 y+3 z &=0 \\ 4 x-3 y-z &=0 \\ 6 x+y+3 z &=0 \end{aligned} $$ $$ \text { (d) } \begin{aligned} &x+3 z=0 \\ &x-y=0 \\ &y+2 z=0 \end{aligned} $$

Use Jacobi's method to solve $$ \begin{aligned} 4 x-y &=-9.4 \\ 3 x+5 y &=7.9 \end{aligned} $$ Take \(x_{0}=-1, y_{0}=1.5\) and perform five iterations.
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