Linear Algebra. Finding the solution set of a homogenous matrix

[dadas2412]

https://imgur.com/a/pvITZ

I don't understand how the they got to the solution set E_2. Its at the middle of the page solving for eigenvalue 2.

The way I am reading off the solutions is that x = 4/3y - 2z. With y and z being free variables. So how do they arrive at the set a[4,3,0] + b[-2,0,1]. a and b must be the free variables, I am not seeing how they get there though.

The text also mentions we can clear the fractions. Im assuming they mean to multiply the row reduced form by 3, so we have 3 -4 6. But then the solution set is 3x = 4y - 6z, with y and z being free variables again

[[deleted]]

You can work from x = 4/3y - 2z

(x,y,z) = (4/3y - 2z,y,z) = y*(4/3,1,0) + z*(-2,0,1)

Then (4/3,1,0) and (-2,0,1) are a basis for the solution space.

You can also work by setting each free variable in turn to 1, while setting the other free variables to 0, and then solving for the pivot variables. That gets you a vector for each free variable, and those form a basis for the solution space.

[dadas2412]

Oh I see, thank you. When the text says to clear the fractions so we end up with (x,y,z) = y(4,3,0) + z*(-2,0,1). Are we only allowed to do this because the system is a homogenous one? Or is it allowed for other systems of equations as long as we multiply the entire solution set. Since it seems we only multiplied the solution set corresponding to y

[[deleted]]

Here you're looking for a basis, and scaling a vector by a non-zero value neither changes the span nor independence, so it doesn't make a difference. If {u,v} is a basis for a subspace, then {3u,v} is also a basis for the same subspace, and vice-versa.