# simple gauss elimination prob.

• Nov 28th 2005, 06:18 AM
prime_66
simple gauss elimination prob.
For some reason I can't see how:
x+3y-z = 1
-4y + 5z = 4
can become:
x = 2-11t
y =-1+5t
z=4t

Can someone explain the step/steps between?
• Nov 28th 2005, 10:11 AM
CaptainBlack
Quote:

Originally Posted by prime_66
For some reason I can't see how:
x+3y-z = 1
-4y + 5z = 4
can become:
x = 2-11t
y =-1+5t
z=4t

Can someone explain the step/steps between?

$x\ +\ 3y\ -\ z\ =\ 1$
$-4y\ +\ 5z\ =\ 4$

is an under-determined system, so there will be
a free parameter $t$ in any solution. So we choose
$t$so that it maximises the convenience of solving
the system in terms of it.

The constant on the RHS of the second equation is $4$, as is the
coefficient of $y$ on the LHS. So setting $z\ =\ 4t$ allows
us to divide through by $4$, and solve for $y$ in terms of $t$.

Having found $y$ and $z$ in terms of $t$
we just substitute these into the first equation to solve for $x$
in terms of $t$.

RonL
• Nov 28th 2005, 10:15 AM
prime_66
thanx, I thought z had to be equal to just t. but now i understand :)