## The Relaxation Method of Solving Simultaneous Linear Equations

The method of relaxation is used to solve systems of linear equations by iteration. Given the system of equations\[3x+9y-2z=11\]

\[4x+2y+13z=24\]

\[11x-4y+3z=-8\]

Write the equations as

\[3x+9y-2z-11=0\]

\[4x+2y+13z-24=0\]

\[11x-4y+3z+8=0\]

and let

\[R_1=3x+9y-2z-11\]

, \[R_2=4x+2y+13z-24\]

and \[R_3=11x-4y+3z+8\]

Now construct the tableau.

\[\Delta x\] | \[\Delta y\] | \[\Delta z\] | \[\Delta R_1\] | \[\Delta R_2\] | \[\Delta R_3\] |

1 | 0 | 0 | 3 | 4 | 11 |

0 | 1 | 0 | 9 | 2 | -4 |

0 | 0 | 1 | -2 | 13 | 3 |

\[\Delta\]

symbols indicate an increment so increasing \[x\]

by 1 increases \[R_1\]

by 3.We take initial solution

\[x=y=z=0\]

then \[R_1=11, \: R_2=-24 \: R_3=8\]

. We aim for each \[R\]

to equal 0.If we make

\[R_1=0\]

we obtain the tableau below.\[x=0\] | \[y=0\] | \[z=0\] | \[R_1=-11\] | \[R_2=-24\] | \[R_3=8\] |

0 | 0 | 2 | -15 | 2 | 14 |

-1 | 0 | 0 | -18 | -2 | 3 |

0 | 2 | 0 | -0 | 2 | -5 |

x=-1 | y=2 | z=2 | 0 | 2 | -5 |

\[z\]

produces of -2 in \[R_1\]

.When

\[z\]

goes from 0 to 2, \[R_1\]

goes from -11 to -15.A change of 1 in \[z\]

produces a change of 13 in \[R_2 \]

so when \[z\]

changes from 0 to 2, \[R_2\]

changes from -24 to 2. When \[x\]

changes from 0 to -1, \[R_1\]

changes by to -3. Hence \[R_1\]

is now -18. When \[y\]

changes by 2, \[R_1\]

changes by 18 and is now 0. The other columns are similarly changed.In the next tableau, multiply the top row by 10 - in order to avoid decimals/fractions - to obtain

\[-10\] | \[20\] | \[20\] | \[0\] | \[20\] | \[-50\] |

5 | 0 | 0 | 15 | 40 | 5 |

0 | 0 | -3 | 21 | 1 | -4 |

0 | -2 | 0 | 3 | -3 | 4 |

x=-5 | y=18 | z=17 | 3 | -3 | 4 |

\[10x=-5, \: 10y=18, \: 10z=17 \rightarrow x=-0.5, \: y=1.8, \: z=1.7\]

.To obtain a second decimal place multiply the top row - containing the solutions - by 100.

\[-50\] | \[180\] | \[170\] | \[30\] | \[-30\] | \[40\] |

-4 | 0 | 0 | -18 | -46 | -4 |

0 | 0 | 4 | -10 | 6 | 8 |

0 | 1 | 0 | -1 | 8 | 4 |

0 | 0 | -1 | 1 | -5 | 1 |

-54 | 181 | 173 | 1 | -5 | 1 |

\[100x=-54, \: 100y=181, \: 100z=173 \rightarrow x=-0.54, \: y=1.81, \: z=1.73\]

.Continuing in this way gives the solution to 5 decimal place

\[x=-0.55004, \: y=1.79216, \: z=1.73968\]

.