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



Write the equations as



and let  ,   and  
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

The    symbols indicate an increment so increasing    by 1 increases    by 3.
We take initial solution    then  . We aim for each    to equal 0.
If we make    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

To understand this, observe that a change of 1 in    produces of -2 in  .
When    goes from 0 to 2,    goes from -11 to -15.A change of 1 in    produces a change of 13 in    so when    changes from 0 to 2,    changes from -24 to 2. When    changes from 0 to -1,    changes by to -3. Hence    is now -18. When    changes by 2,    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

The solution is now  .
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

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