A = [2, 5 ;-4, -10]

**question**: Does A**x** = **b** have a solution for every **b** in R2. If not determine conditions on **b** so that so that A**x** = **b** is consistent.

I'm just a little bit confused how to answer the question.

Here is what I have written:

Code:

-The columns of A are scalar multiples of each other => the set is linearly dependent. [5, 10] = (5/2)[2, -4]
-rref (A**x **= **b**) => x1 + (5/2)x2 = b1/2, 2b1 + b2 = 0
=> for A**x **= **b** to be consistent, 2b1 = -b2
=> no solution for **b **= [1, 1]

Have I answered this question correctly?