Some linear maps can be undone: there is a second map that returns every vector to where it started. Others cannot, because they send different vectors to the same place and the information is gone.

Converting a price to another currency can be undone by converting back. Rounding it to the nearest whole unit cannot, because several prices round to the same figure.

This map sends $\mathbf{u}$ and $\mathbf{v}$ to the very same place, so no map could send that place back to both. Every vector lands somewhere on the dashed line, and one whole direction is gone — which is why squareness alone is not enough for an inverse.

A square matrix $A$ is invertible when some matrix undoes it in both directions. That matrix is written $A^{-1}$. Only square matrices are candidates here, and squareness is not enough: a matrix whose columns fail to reach the whole space has no inverse.

• $A$ is invertible exactly when $A\mathbf{x}=\mathbf{b}$ has one solution for every $\mathbf{b}$.
• If $A$ is invertible then that solution is $\mathbf{x}=A^{-1}\mathbf{b}$.

Suppose one matrix undoes $A$ from the left and another undoes it from the right. Sandwich $A$ between them and read the product two ways: grouping to the left uses the first assumption, grouping to the right uses the second, and the two readings must agree.

Composing three maps in a row gives the same result however the pairs are bracketed.

Read the left grouping first, using $AC=I$ and then the fact that $I$ changes nothing.

Read the other grouping, using $BA=I$ in the same way.

The two readings are of one and the same matrix, so the results agree and there is only one inverse.

$AA^{-1}=I$ and $A^{-1}A=I$

One equation alone is not the definition.

$A\mathbf{x}=\mathbf{b}$ has one solution for every $\mathbf{b}$

That single condition is what invertibility amounts to.

A non-zero vector lands on the zero vector, where $\mathbf{0}$ already goes.

Applied on the left of both sides, since order matters.