Expansion_Theorem_for_Determinants_v2

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$. Let $D = \map \det {\mathbf A}$ be the determinant of $\mathbf A$ $\displaystyle \map \det {\mathbf A} := \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} } = \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{n \map \lambda n}$ where the summation $\displaystyle \sum_\lambda$ goes over all the $n!$ permutations of $\set {1, 2, \ldots, n}$ $\map \sgn \lambda$ is the sign of the permutation $\lambda$. Let $a_{p q}$ be an element of $\mathbf A$. Let $A_{p q}$ be the cofactor of $a_{p q}$ in $D$. Then $(1): \quad \displaystyle \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$ $(2): \quad \displaystyle \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{k r} A_{k r}$ Thus the value of a determinant can be found either by multiplying all the elements in a row by their cofactors and adding up the products or multiplying all the elements in a column by their cofactors and adding up the products. The identity $\displaystyle D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$ is known as the '''expansion of $D$ in terms of row $r$''', while $\displaystyle D = \sum_{k \mathop = 1}^n a_{k r} A_{k r}$ is known as the '''expansion of $D$ in terms of column $r$'''.