Define the determinant recursively by $\det[a]=a$ and, for $n>1$, $\det A=\sum_{j=1}^n(-1)^{1+j}a_{1j}\det A_{\widehat1,\widehat j}$. The empty determinant is $1$, making cofactors of $1\times1$ matrices consistent. For $2\times2$ matrices this becomes $ad-bc$. Cofactor expansion is useful for sparse symbolic examples but inefficient for large dense matrices.
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