A = [2, 5 ;-4, -10]
question: Does Ax = b have a solution for every b in R2. If not determine conditions on b so that so that Ax = b is consistent.
I'm just a little bit confused how to answer the question.
Here is what I have written:
Have I answered this question correctly?
-The columns of A are scalar multiples of each other => the set is linearly dependent. [5, 10] = (5/2)[2, -4]
-rref (Ax = b) => x1 + (5/2)x2 = b1/2, 2b1 + b2 = 0
=> for Ax = b to be consistent, 2b1 = -b2
=> no solution for b = [1, 1]