Question

Linear Algebra: Let A be a 4x3 matrix and suppose that the vectors z1=(1,1,2)^T and (1,0,-1)^T form a basis for N(A). If the b = a1 + 2a2 + a3 (where b is a vector and a1, a2, a3 are column vectors of A), find all solutions of the system Ax = b.

Answer 1

z2 = (1,0,-1)^T *correction

Answer 2

So first of all we know that Az1 = Az2 = 0. Hence for any solution x = y of
Ax = b, then x = y + c1.z1 + c2.z2 is also a solution for any scalars c1 and c2.

Answer 3

Now the question is what is such a vector y? Given the form of b, you have been given a serious clue by the question. Look carefully at the form of b and figure out what the entries of the column vector y would have te be to get
Ay = b = a1 + 2a2 + a3

Answer 4

If you're really stuck, ask yourself this: if a1, a2 and a3 are the column of A, then what, for instance is the result of
A(1,0,0)^T? And what about A(0,1,0)^T?

Answer 5

i'm sorry if this sounds really dumb, but could you please explain how were you able to say Az1 = Az2 = 0 from the fact that {z1,z2} is a basis for N(A)?

Answer 6

read the definition of null space

Answer 7

Please tell me if this is a correct thought process:

{z1, z2} is a basis for N(A).
Therefore any linear combination of z1 and z2 will be z : Az = 0
Therefore if I let scalar c1=1 and c2=0, c1z1 will be z : Az=0
The same can be said for z2 if c1=0 and c2=1.
Therefore Az1 = Az2 = 0

Answer 8

{z1, z2} is a basis for N(A). clearly z1 and z2 are in the nullspace thus Az1=0 and Az2=0

Answer 9

Lol i really have tendencies of making everything complicated. Thank you.

Answer 10

Thank you all, I understand the solution now. My professor already gave the class a solution, but I couldn't make sense of how the reminiscent method of solving non-homogeneous ODE's could appear here. You all gave me another way of looking at this problem, which makes more sense to me. Thanks.