Lemma 10.15.6. Let $R$ be a ring. Let $n \geq m$. Let $A = (a_{ij})$ be an $n \times m$ matrix with coefficients in $R$, written in block form as

\[ A = \left( \begin{matrix} A_1
\\ A_2
\end{matrix} \right) \]

where $A_1$ has size $m \times m$. Let $B$ be the adjugate (transpose of cofactor) matrix to $A_1$. Then

\[ AB = \left( \begin{matrix} f 1_{m \times m}
\\ C
\end{matrix} \right) \]

where $f = \det (A_1)$ and $c_{ij}$ is (up to sign) the determinant of the $m \times m$ minor of $A$ corresponding to the rows $1, \ldots , \hat j, \ldots , m, i$.

**Proof.**
Since the adjugate has the property $A_1B = B A_1 = f$ the first block of the expression for $AB$ is correct. Note that

\[ c_{ij} = \sum \nolimits _ k a_{ik}b_{kj} = \sum (-1)^{j + k}a_{ik} \det (A_1^{jk}) \]

where $A_1^{ij}$ means $A_1$ with the $j$th row and $k$th column removed. This last expression is the row expansion of the determinant of the matrix in the statement of the lemma.
$\square$

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