**Re: MATH1231/1241/1251 SOS Thread**
I'm stuck on a similar one

How do I find this vector???

You essentially have two conditions:

x -(1/4)y + 0z = 0

and

0x -(3/4)y + z = 0

(assuming your answer was right, I didn't check it).

So the vector (x,y,z) is in the column space of A iff it satisfies those two conditions.

You can turn those conditions into a matrix and get it into row-echelon form (actually in this case, it already is in row-echelon form).

Then do the usual procedure of setting a non-leading column's variable to a free parameter (in this case, that variable is z, so set z = lambda say), then use back substitution as usual to get x and y in terms of lambda.

This will mean you'll end up with x, y, z in terms of lambda, which means you can get a vector

**b** as desired. For the sake of example, if you ended up with x = 2lambda, y = -lambda, z = lambda, we'd have (x,y,z) = (2lambda, -lambda, lambda) = lambda (2,-1,1), and thus a vector we could choose for

**b** would be (2,-1,1).