Thread: linear system

hi
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 and .
thanks.

Originally Posted by Raoh

hi
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 and .
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

okay,so for every value and ,the system has a unique solution,
and .
is it okay ?
thanks.

Yes, that is what I get.

by the way,is my Gauss method correct,did anyone check it

I hadn't before but I just did. Yes, that is correct.