Let $\mathbf A = \sqbrk a_n$ be a square matrixoforder $n$.
That is, let:
$\quad \mathbf A = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$
Let $\lambda: \N_{> 0} \to \N_{> 0}$ be a permutation on $\N_{>0}$.
The determinant of $\mathbf A$ is defined as:
where:
The determinant of $\mathbf A$ is defined as follows:
For $n = 1$, the order $1$ determinant is defined as:
Thus the determinant of an order $1$ matrix is that element itself.
For $n > 1$, the determinantoforder $n$ is defined recursively as:
$\quad \ds \map \det {\mathbf A} := \begin {vmatrix}
a_{1 1} & a_{1 2} & a_{1 3} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & a_{2 3} & \cdots & a_{2 n} \\
a_{3 1} & a_{3 2} & a_{3 3} & \cdots & a_{3 n} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & a_{n 3} & \cdots & a_{n n} \\
\end {vmatrix} = a_{1 1} \begin {vmatrix}
a_{2 2} & a_{2 3} & \cdots & a_{2 n} \\
a_{3 2} & a_{3 3} & \cdots & a_{3 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n 2} & a_{n 3} & \cdots & a_{n n} \\
\end {vmatrix} - a_{1 2} \begin {vmatrix}
a_{2 1} & a_{2 3} & \cdots & a_{2 n} \\
a_{3 1} & a_{3 3} & \cdots & a_{3 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 3} & \cdots & a_{n n} \\
\end {vmatrix} + \cdots + \paren {-1}^{n + 1} a_{1 n} \begin {vmatrix}
a_{2 1} & a_{2 2} & \cdots & a_{2, n - 1} \\
a_{3 1} & a_{3 3} & \cdots & a_{3, n - 1} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 3} & \cdots & a_{n, n - 1} \\
\end {vmatrix}$
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 $\map \det A$ of $A$ is defined to be the determinant of any matrix of $A$ relative to some basis.
Let $\mathbf D$ be a determinant.
The rows of $\mathbf D$ are the lines of elements reading across the page.
Let $\mathbf D$ be a determinant.
The columns of $\mathbf D$ are the lines of elements reading across the page.
