Determinant and rank

Equivalence chain.
$$\det A\neq0 \iff A \text{ is invertible} \iff \ker A=\{0\} \iff \operatorname{rank} A=n.$$

Proof of each step.
- $\det A\neq 0\iff A$ invertible: standard result (the inverse is given by the cofactor formula, which involves division by $\det A$).
- $A$ invertible $\iff\ker A=\{0\}$: if invertible, $Ax=0\Rightarrow x=0$; if $\ker A=\{0\}$, then $A$ is injective, hence bijective (finite dimension), so invertible.
- $\ker A=\{0\}\iff\operatorname{rank} A=n$: by rank-nullity, $\operatorname{rank} A=n-\dim\ker A$, so $\operatorname{rank} A=n\iff\dim\ker A=0\iff\ker A=\{0\}$.

Conclusion. $\det A\neq0$ iff $\operatorname{rank} A=n$ (i.e. $A$ has full rank).