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## Homework Statement

Given an nxn matrix, if a b exists so Ax=b has no solutions, can A be one-to-one?

## Homework Equations

I understand that as a linear transformation, you need things such as (to be one-to-one as a linear trans)

1. n pivots

2. Only the trivial solution exists to Ax=0

Ax=b having no solutions is a bit of an oddball for me.

## The Attempt at a Solution

If I set up an augmented matrix such as [I | b] (Identity matrix | some vector b)

By removing one pivot from the Ident. matrix I can see that I've created an Ax=b where b has no solution. This is akin to a linearly independent set (which contains the zero vector). in Rn space.

It technically doesn't span Rn space nor Rm space fully... I think? As it will have a row of zeroes and a column of zeroes.

But does this truly affect its ability to be one-to-one?

I don't see how it does........

For instance, let's say you have a 3x3 ID matrix A, but one pivot is missing.

If that's put into an augmented matrix[A|b], if a row with zeroes has a number to the right of it, it means there's no solution for this b. So doesn't this mean the situation is just n/a?