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I am trying to get a better grasp on invertibility of matrices. My current mental model is basically that, a Matrix is invertible if every value of b can be mapped back to a unique x. Do I have the right idea here?

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Yes, "invertible" means exactly the same thing as an invertible function. Using your own words (emphasis mine): Every value of $b$ (this requirement is basically surjectivity) can be mapped back to a unique $x$ (this is injectivity). – Srivatsan Sep 23 '11 at 0:00
Yes you do, more specifically a linear transformation $T$ can be invertible if it maps $T : V\rightarrow W$, where $V$ and $W$ have the same dimension and is an isomorphism. To be invertible it simply means that we have some $S$ such that $S\circ T = T\circ S$ is the identity transformation. Of course the matrix $[S\circ T] = [T]\times[S]$ – Deven Ware Sep 23 '11 at 0:01
awesome thanks for the comments everyone! – weezybizzle Sep 23 '11 at 0:24
But be careful,suppose you have $T = \pmatrix{1 &0\\0 &1\\0&0}$ and $b=Tx$ then you can again obtain $x$ from $b$. And $T$ is not invertible. Hence, you have to pick your wording slightly more rigorously. – user13838 Sep 23 '11 at 0:26
@Deven Ware: I might be wrong, but are you sure that the matrix representation of $[S \circ T]$ is $T \times S$, not the other way around? – Gerben Sep 23 '11 at 1:42

As remarked in the comments, your idea is correct.

Specifically, a linear transformation $T: V \to W$ is invertible iff (that is, if and only if) there is a linear transformation $S: W \to V$ such that $T\circ S: V \to V$ and $S \circ T: W \to W$ are both the identity linear transformation.

In matrix notation, if $[T]$ is an $n \times m$-matrix, we need a $m \times n$-matrix $[S]$ such that $[T][S]$ and $[S][T]$ are identity matrices (this relates to the above by $[T][S] = [T\circ S]$). It follows by e.g. Rank-Nullity that necessarily $n = m$.

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