# Thread: Does Ax = b have a solution for every b in R2. If not determine conditions on b ...

1. ## Does Ax = b have a solution for every b in R2. If not determine conditions on b ...

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:

Code:
-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]
Have I answered this question correctly?

2. ## Re: Does Ax = b have a solution for every b in R2. If not determine conditions on b

Hey StudentMCCS.

Here is some output from Octave for your matrix:

>> A
A =

2 5
-4 10

>> inv(A)
ans =

0.250000 -0.125000
0.100000 0.050000

>> det(A)
ans = 40
>> rref(A)
ans =

1 0
0 1

All of this (although you only need one of them) says that the matrix is of full rank so you will never ever get any inconsistent situations for solving x = A^(-1)*b since a unique solution exists for all b in R^2.

The only time you get a chance of inconsistent solutions is when your matrix is not full-rank (i.e. has linearly dependent vectors in the rows of the matrix) and if your matrix is square, the easiest way to check this computationally is to check det(A). If you want to find out what rows are linearly independent, do a row reduction and see what rows go to zero and what rows don't and the number of non-zero rows in rref(A) give the number of linearly independent rows and you can get your rank and your nullity (nullity(A) = dim(space) - rank(A)).

When you have nullity, then and only then do have a chance of inconsistency and this can be determined by doing the normal reduction with a variable matrix and seeing when you get the 0 = 1 situation and those situations are the cases where you get inconsistencies).

3. ## Re: Does Ax = b have a solution for every b in R2. If not determine conditions on b

supposing that what you meant is:

$A = \begin{bmatrix}2&5\\-4&-10 \end{bmatrix}$

we can try to solve: Ax = b.

the easiest way is to use an augmented matrix and row-reduce it. that gives us:

$A' = \begin{bmatrix}2&5&|&b_1\\-4&-10&|&b_2 \end{bmatrix}$

adding 2 times row 1 to row 2 gives us:

$\begin{bmatrix}2&5&|&b_1\\0&0&|&2b_1 + b_2 \end{bmatrix}$

so for the equation Ax = b to HAVE a solution, we must have: $2b_1 + b_2 = 0$.

equivalently $b_2 = -2b_1$.

well this obviously isn't true for EVERY b = (b1,b2), but it is true for SOME b: namely all scalar multiples of the vector (1,-2).