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Determinant of Matrix

Let $\mathbf A = \left[{a}\right]_n$ be a square matrix of order $n$.

That is, let:

$\mathbf A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}$

Let $\lambda: \N_{> 0} \to \N_{> 0}$ be a permutation on $\N_{> 0}$.

Then the determinant of $\mathbf A$ is defined as:

$\displaystyle \det \left({\mathbf A}\right) := \sum_{\lambda} \left({\operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n a_{k \lambda \left({k}\right)}}\right) = \sum_{\lambda} \operatorname{sgn} \left({\lambda}\right) a_{1 \lambda \left({1}\right)} a_{2 \lambda \left({2}\right)} \cdots a_{n \lambda \left({n}\right)}$


the summation $\displaystyle \sum_\lambda$ goes over all the $n!$ permutations of $\left\{{1, 2, \ldots, n}\right\}$.
$\operatorname{sgn} \left({\lambda}\right)$ is the sign of the permutation $\lambda$.

Determinant of Linear Operator

Let $V$ be a finite-dimensional vector space over a field $K$.

Let $A: V \to V$ be a linear operator of $V$.

The determinant $\det \left({A}\right)$ of $A$ is defined to be the determinant of any matrix of $A$ relative to some basis.

Also see