linear system (2)

Raoh · #1 $Old$ November 2nd, 2009, 07:40 AM

hi

i need little help

$\left\{\begin{matrix} x+z+t=0\\ ax+y+(a-1)z+t=0\\ x-y+2z+at=0\\ 2x+ay+z+2t=0 \end{matrix}\right.$
thanks.

HallsofIvy · #2 $Old$ November 3rd, 2009, 06:53 AM

Quote:

Originally Posted by Raoh $View Post$

hi

i need little help

$\left\{\begin{matrix} x+z+t=0\\ ax+y+(a-1)z+t=0\\ x-y+2z+at=0\\ 2x+ay+z+2t=0 \end{matrix}\right.$
thanks.

To do what? Solve the system of equations? Find the value of a such that these equations have a non-trival solution? That will happen if and only if the determinant is 0.

Raoh · #3 $Old$ November 6th, 2009, 12:26 PM

I arrived here,what should i do next ?
$\left\{\begin{matrix} x+z+t=0\\ y-z+t(1-a)=0\\ -t(1-a)=0 \end{matrix}\right.$
thanks.

Drexel28 · #4 $Old$ November 6th, 2009, 12:32 PM

Quote:

Originally Posted by Raoh $View Post$

I arrived here,what should i do next ?
$\left\{\begin{matrix}x+z+t=0\\ y-z+t(1-a)=0\\ -t(1-a)=0\end{matrix}\right.$
thanks.

What? What's a variable and what's a constant?

Raoh · #5 $Old$ November 6th, 2009, 12:43 PM

$t$ and $a$ are parameters.