Question

True / false with explaination. If Ax=b has atleast one solution for b in Rn then A is an invertible matrix

Answer 1

Can you create a noninvertible matrix A, such that Ax = b?

Answer 2

No, atleast I think so, since if you have Ax=b then there exists x=A^1b. Do I have that right?

Answer 3

\[\begin{pmatrix}1&2\\2&4\end{pmatrix}\binom{5}{2}=\binom{9}{17}\]

Answer 4

Well that's what I originally thought. So is there not any way for those two statements to be logically equivalent?

Answer 5

Well, one counterexample is all that is needed to prove a statement false.

If my setup is valid, then I have a noninvertible matrix A, that has x and b solutions from it.

Answer 6

granted b = (9 18) .... but thats immaterial :)

Answer 7

So then how is it that all the statements in the invertible matrix theorem are logically equivalent? or am I miss understanding the situation?

Answer 8

looks like you are refering to "g"

Answer 9

has at least one solution for "each" b in R; looks like a might have misread your post

Answer 10

the columns of my matrix do not for a basis for R^2

Answer 11

if "a" is true; then the rest of the list, "b to t" is true.

Answer 12

my matrix is essentially a line in R^2, therefore its not going to have a solution for all (x,y) in R^2

Answer 13

So I know that (g) is true but I'm not sure how to go about proving that A is invertible as a result.

Answer 14

I know the columns must be linearly independent. How do I generally show that? ( I tend to over complicate things which causes me to overlook things)

Answer 15

row echelon form is an equivalent setup; if you can row reduce it to the Identity matrix; they are independant

Answer 16

if a matrix contains a row or column of zeros, the determinant is zero; if it has a row or column of zeros, it is dependant

Answer 17

but how to prove that: Ax = b has a solution for every b in R^n hmmm

Answer 18

i wonder if assuming A is invertible and Ax=b does NOT have a solution for every b in R^n, and showing it a contradiction might be easy

Answer 19

Let A be an nxn invertible matrix; therefore it has an inverse

spose that Ax = b has no solution for some b in R^n

Ax = b

A^-1A x = A^-1 b

x = A^-1 b

since x is a solution in R^n, we have a contradiction