# Thread: Consider 2 homogeneous systems

1. ## Consider 2 homogeneous systems

Find all possible solutions to W1 and W2.
Find the set of all vectors which are solutions to both systems and why it is a vector space.

2. The augmented matrix for the first is

$\displaystyle \begin{bmatrix} 1 && 2 && 1 && 0 \\ 2 && 4 && 2 && 0 \\ 1 && 2 && 1 && 0 \\ \end{bmatrix}$

In reduced tow eschelon form you get...

$\displaystyle \begin{bmatrix} 1 && 2 && 1 && 0 \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 \\ \end{bmatrix}$

so we have two "free" variables we get the solution

$\displaystyle x=-2s-t$
$\displaystyle y= s$
$\displaystyle z=t$

$\displaystyle v_=\begin{bmatrix} -2s-t \\ s \\ t \end{bmatrix}= s\begin{bmatrix} -2 \\ 1 \\ 0 \\ \end{bmatrix}+ t\begin{bmatrix} -1 \\ 0 \\ 1 \\ \end{bmatrix}$

The last two vectors span the solution space of the first matrix.

Try this proceedure with the second matrix and compare the solution spaces.

I hope this helps

Good luck.

3. thanks man.
I got the following solution for second one.
So is S[-2，1，0] the solution for both systems?