 Solution

For a complex square matrix, define its determinant byThe cofactor is $C_{ij}=(-1{)}^{i+j}\det A^{(ij)}$, where $A^{(ij)}$ deletes row $i$ and column $j$. The adjugate is the transposed cofactor matrix, $\mathrm{adj}(A{)}_{ij}=C_{ji}$.

The $(i,j)$ entry of $\mathrm{adj}(A)A$ is ${\sum }_k{C}_{ki}{a}_{kj}$. For $j=i$, expansion along column $i$ gives $\det A$. For $j\ne i$, this same sum is the determinant of the matrix obtained by replacing column $i$ by column $j$: deleting the replaced column leaves all its cofactors unchanged. The new matrix has two equal columns, so its determinant is zero. ThusThis proof does not assume invertibility and therefore also covers singular matrices.