# simple gauss elimination prob.

• Nov 28th 2005, 05:18 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, 09:11 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?

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

is an under-determined system, so there will be
a free parameter \$\displaystyle t\$ in any solution. So we choose
\$\displaystyle 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 \$\displaystyle 4\$, as is the
coefficient of \$\displaystyle y\$ on the LHS. So setting \$\displaystyle z\ =\ 4t\$ allows
us to divide through by \$\displaystyle 4\$, and solve for \$\displaystyle y\$ in terms of \$\displaystyle t\$.

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

RonL
• Nov 28th 2005, 09:15 AM
prime_66
thanx, I thought z had to be equal to just t. but now i understand :)