Invertibility criterion via the determinant

($\Rightarrow$) Suppose $A$ is invertible, so there is a matrix $A^{-1}$ with $AA^{-1}=I$. Applying the multiplicativity theorem proved above, $\det(A)\det(A^{-1})=\det(AA^{-1})=\det(I)=1$. Since the product of the two numbers $\det(A)$ and $\det(A^{-1})$ equals $1$, neither of them can be $0$ — so $\det(A)\neq 0$.

($\Leftarrow$), worked concretely for $2\times 2$: suppose $A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ has $\det(A)=ad-bc\neq 0$. Define $\operatorname{adj}(A)=\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}$. Multiplying directly, $A\cdot\operatorname{adj}(A)=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}=\begin{pmatrix}ad-bc&-ab+ab\\ cd-cd&-cb+ad\end{pmatrix}=\begin{pmatrix}ad-bc&0\\ 0&ad-bc\end{pmatrix}=\det(A)\,I$, exactly the identity $A\cdot\operatorname{adj}(A)=\det(A)\,I$. The same computation in the other order gives $\operatorname{adj}(A)\cdot A=\det(A)\,I$ as well.

Since $\det(A)\neq 0$, divide both sides of $A\cdot\operatorname{adj}(A)=\det(A)I$ by the scalar $\det(A)$ to get $A\cdot\dfrac{\operatorname{adj}(A)}{\det(A)}=I$, and likewise from the other order $\dfrac{\operatorname{adj}(A)}{\det(A)}\cdot A=I$. This exhibits an explicit two-sided inverse $A^{-1}=\dfrac{1}{\det(A)}\operatorname{adj}(A)$, so $A$ is invertible.

For general $n$ (illustrated at $n=3$), the same argument works because the Laplace cofactor expansion gives $\sum_j a_{ij}C_{ij}=\det(A)$ (expanding along row $i$) while $\sum_j a_{ij}C_{kj}=0$ for $i\neq k$ (this sum is the cofactor expansion of a matrix with row $i$ copied into row $k$, which has two equal rows and hence determinant $0$ by the table above). Assembling these row-by-row identities into matrix form gives exactly $A\cdot\operatorname{adj}(A)=\det(A)I$ for every size $n$, so whenever $\det(A)\neq 0$ the same division argument as in the $2\times 2$ case produces $A^{-1}=\operatorname{adj}(A)/\det(A)$; conversely, if $\det(A)=0$, the first paragraph's contrapositive rules out invertibility entirely, since $\det(A)\det(A^{-1})=1$ would force $\det(A)\neq 0$. This proves the criterion in both directions.