= Solution
An <invertible matrix> is a square matrix for which there is a matrix $A^{-1}$ with $AA^{-1}=A^{-1}A=I$. Write its assumed <diagonalization> as $A=P\Lambda P^{-1}$, where $P$ is invertible and $\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Then
$$
\det A=\det\Lambda=\prod_{j=1}^n\lambda_j.
$$
If the <determinant> is nonzero, every $\lambda_j$ is nonzero, so $\Lambda^{-1}=\operatorname{diag}(\lambda_j^{-1})$ exists and $A^{-1}=P\Lambda^{-1}P^{-1}$. If the <determinant> is zero, at least one <eigenvalue> is zero; its nonzero <eigenvector> is mapped to zero, which an invertible map cannot do. Hence
$$
\boxed{A\text{ is invertible}\quad\Longleftrightarrow\quad\det A\ne0.}
$$
The equivalence holds even without diagonalizability, but the eigenbasis proves the stated case directly.
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