# linear system

• Oct 31st 2009, 08:50 AM
Raoh
linear system
hi(Happy)
i have the following system.
x+2y+3z+4t+5u=6
x+3y+5z+7t+4u=13
x+4y+7z+10t+3u=20
x+2y+4z+6t+8u=13
with Gauss method,i ended up here,
x+2y+3z+4t+5u=6
y+2z+3t-u=7
z+2t+3u=7
i don't know how to work with the parameters \$\displaystyle t\$ and \$\displaystyle u\$.
thanks.
• Nov 1st 2009, 01:48 AM
CaptainBlack
Quote:

Originally Posted by Raoh
hi(Happy)
i have the following system.
x+2y+3z+4t+5u=6
x+3y+5z+7t+4u=13
x+4y+7z+10t+3u=20
x+2y+4z+6t+8u=13
with Gauss method,i ended up here,
x+2y+3z+4t+5u=6
y+2z+3t-u=7
z+2t+3u=7
i don't know how to work with the parameters \$\displaystyle t\$ and \$\displaystyle u\$.
thanks.

Use t and u as free parameters and solve for x, y and z in terms of them.

Start with z, from the last equation z=7-2t-3u, now do y and then x.

CB
• Nov 1st 2009, 02:37 AM
Raoh
okay,so for every value \$\displaystyle u \$and \$\displaystyle t\$,the system has a unique solution,
\$\displaystyle x = -1-10u, y= -7 +t + 7u\$ and \$\displaystyle z = 7 -2t -3u\$.
is it okay ?
thanks.
• Nov 1st 2009, 02:53 AM
HallsofIvy
Yes, that is what I get.
• Nov 1st 2009, 02:57 AM
Raoh
by the way,is my Gauss method correct,did anyone check it(Happy)
• Nov 1st 2009, 04:31 AM
HallsofIvy
I hadn't before but I just did. Yes, that is correct.