Exercise 1632Linear algebra › DeterminantsL1Added 3 Nov 2026

The cofactor formula for the inverse

#cofactor#comatrix#inverse formula

Statement

Let $A$ be invertible. Define the cofactor $C_{ij}=(-1)^{i+j}\det M_{ij}$ , where $M_{ij}$ is the minor obtained by deleting row $i$ and column $j$ . Define the comatrix $\operatorname{com}(A)=(C_{ij})^T$ (transpose of the cofactor matrix). Show that
$A^{-1}=\frac{1}{\det A}\operatorname{com}(A).$

Hint

Show that $A\cdot\operatorname{com}(A)=\det(A)\cdot I$ , using the expansion of the determinant along a row.

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Solution

Entry $(i,j)$ of $A\cdot\operatorname{com}(A)$ .
$(A\cdot\operatorname{com}(A))_{ij}=\sum_k a_{ik}C_{jk}.$

Case $i=j$ . $\sum_k a_{ik}C_{ik}=\det A$ by expansion along row $i$ (each term is a cofactor times its entry).

Case $i\neq j$ . $\sum_k a_{ik}C_{jk}$ is the determinant of the matrix obtained from $A$ by replacing row $j$ with row $i$ . This matrix has two equal rows, so its determinant is zero.

Conclusion. $A\cdot\operatorname{com}(A)=\det(A)\cdot I$ , so $A^{-1}=\operatorname{com}(A)/\det A$ .

The cofactor formula for the inverse

Statement

Solution

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