Expansion Theorem for Determinants/Corollary

Corollary to Expansion Theorem for Determinants
Let $\mathbf A = \left[{a}\right]_{n}$ be a square matrix of order $n$.

Let $D = \det \left({\mathbf A}\right)$ be the determinant of $\mathbf A$.

Let $a_{pq}$ be an element of $\mathbf A$.

Let $A_{pq}$ be the cofactor of $a_{pq}$ in $D$. Let $\delta_{rs}$ be the Kronecker delta.

Then:


 * $(1): \quad \displaystyle \forall r \in \left[{1 \,.\,.\, n}\right]: \sum_{k \mathop = 1}^n a_{rk} A_{sk} = \delta_{rs} D$
 * $(2): \quad \displaystyle \forall r \in \left[{1 \,.\,.\, n}\right]: \sum_{k \mathop = 1}^n a_{kr} A_{ks} = \delta_{rs} D$

That is, if you multiply each element of a row or column by the cofactor of another row or column, the sum of those products is zero.

Proof
Let $D\,'$ be the determinant obtained by replacing row $s$ with row $r$.

Let $a'_{ij}$ be an element of $D\,'$.

Then:


 * $a'_{ij} = \begin{cases}

a_{ij} & : i \ne s \\ a_{rj} & : i = s \end{cases}$

Let the cofactor of $a'_{ij}$ in $D\,'$ be denoted by $A'_{ij}$.

Then:


 * $\forall k \in \left[{1 \,.\,.\, n}\right]: A'_{sk} = A_{sk}$

So by the Expansion Theorem for Determinants:


 * $\displaystyle D\,' = \sum_{k \mathop = 1}^n a'_{sk} A'_{sk}$

But the $r$th and $s$th rows are identical.

So by Square Matrix with Duplicate Rows has Zero Determinant:
 * $D\,' = 0$

Hence the result.

The result for columns follows from Determinant of Transpose.