Let \(A\in\mathbb{F}^{\,n\times n}\) be a square matrix. If the \(i^{\mathrm{th}}\) row and \(j^{\mathrm{th}}\) column is removed from \(A\) the resulting matrix \(B_{ij} \in \mathbb{F}^{\,(n-1)\times (n-1)}\) is called the \((i,j)\)-sub-matrix of \(A.\) The \((i,j)\)-minor of \(A\) is the determinant \(M_{ij}\in\mathbb{F}\) of the \((i,j)\)-sub-matrix \(B_{ij}.\) The \((i,j)\)-co-factor of \(A,\) \(C_{ij}\in\mathbb{F},\) is the \((i,j)\)-minor \(M_{ij}\) scaled by \((-1)^{i+j}.\)

\[M_{ij} = \lvert B_{ij}\rvert\] \[C_{ij} = (-1)^{i+j}M_{ij}\]

The determinant, \(\mathrm{det}(A)\) or \(\lvert A\rvert,\) of \(A\) is defined recursively. For \(n=1,\) \(\lvert A \rvert = a_{11}.\) For \(n \geq 2\) it is defined with the Laplace expansion of \(A\) along any fixed row \(i \in \{1,\ldots,n\}:\)

\[\lvert A \rvert = \sum_{j=1}^{n} a_{ij}C_{ij}\]

Every choice of row \(i\) gives the same value.

For the simple case \(n=2,\) \(\lvert A \rvert = a_{11}a_{22}-a_{12}a_{21}.\)

The determinant of a real, square matrix \(A\in\mathbb{R}^{\,n\times n}\) is the signed volume of the parallelepiped formed by taking each row of \(A\) as a vector in \(\mathbb{R}^n.\) The volume itself is \(\lvert \mathrm{det}(A) \rvert\) and the sign records whether the rows have the same orientation as the standard basis.

The determinant of square matrices \(A,B\in\mathbb{F}^{\,n\times n}\) satisfies the following properties:

  • Determinant of transpose: \(\lvert A \rvert\,= \lvert A^\intercal \rvert\)
  • Determinant of scalar multiplication: \(\lvert \lambda A \rvert\,= \lambda^n\lvert A \rvert\,\,\forall\,\lambda\in\mathbb{F}\)
  • Determinant of matrix multiplication: \(\lvert AB \rvert\,= \lvert A \rvert \lvert B \rvert\)
  • Determinant of matrix inverse: \(\lvert A^{-1} \rvert\,=\frac{1}{\lvert A \rvert}\) if \(A\) is invertible
  • Invertibility: \(\lvert A \rvert \,\neq 0 \Leftrightarrow A\) is invertible
  • Full rank: \(\lvert A \rvert \,\neq 0 \Leftrightarrow A\) has rank \(n\)

Let \(A\in\mathbb{F}^{\,n\times n}\) be an invertible square matrix, so that \(\lvert A\rvert \neq 0,\) and let \(C\in\mathbb{F}^{\,n\times n}\) be its co-factor matrix whose \((i,j)^{\mathrm{th}}\) element is \(C_{ij}.\) The matrix inverse \(A^{-1}\) can be calculated from the determinant and co-factors of \(A:\)

\[A^{-1}=\frac{C^\intercal}{\lvert A\rvert}\]

where the \((i,j)^{\mathrm{th}}\) element of \(A^{-1}\) is \(\frac{C_{ji}}{\lvert A\rvert}.\)