## 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
The
$\Delta$
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
To understand this, observe that a change of 1 in
$z$
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
The solution is now
$10x=-5, \: 10y=18, \: 10z=17 \rightarrow x=-0.5, \: y=1.8, \: z=1.7$
.
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
The solution is now
$100x=-54, \: 100y=181, \: 100z=173 \rightarrow x=-0.54, \: y=1.81, \: z=1.73$
.
Continuing in this way gives the solution to 5 decimal place
$x=-0.55004, \: y=1.79216, \: z=1.73968$
.