Concept of Invertible Matrices

[war485]

Homework Statement

If A is an invertible square matrix, then Ax = b is consistent for each b in R^n

2. The attempt at a solution

If A multiplied by A inverse is identity, then it would always be consistent. So I thought , if A is just randomly multiplied by some x, then it will still be consistent right? I can't seem to find anything wrong with the statement above.

 

[cse63146]

I think Ax = b holds true regardless of A being invertible or not

 

[Dick]

I think Ax = b holds true regardless of A being invertible or not

Ax=b is not exactly 'true' if there no solutions x to the equation given a b. If A is invertible then A^(-1)Ax=Ix=x=A^(-1)b. So given any b, x=A^(-1)b. Yes, it's consistent.

 

[HallsofIvy]

I think Ax = b holds true regardless of A being invertible or not

I have no idea what you mean by this. What is "Ax= b" that it could "hold true"? Certainly if A is a matrix and x a vector with as many components as A has columns, then there exist a vector b such that Ax= b. Is that what you meant?

If A is invertible, then given any such vector b, there exist a vector x such that Ax= b.

If A is not invertible then, given b, there may not exist such an x or there may exist an infinite number of the them.

 

[war485]

Thanks for the help guys. That little algebra helped me see it properly Dick.