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?

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- Nov 28th 2005, 05:18 AMprime_66simple 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 AMCaptainBlackQuote:

Originally Posted by**prime_66**

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

$\displaystyle -4y\ +\ 5z\ =\ 4$

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 AMprime_66
thanx, I thought z had to be equal to just t. but now i understand :)