The Relaxation Method of Solving Simultaneous Linear 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
\[x=-0.55004, \: y=1.79216, \: z=1.73968\]
. 
