Expansion Theorem for Determinants/Corollary

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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.

$\blacksquare$