Determinant

• categories: Linear algebra, Definition

Definition
The determinant of a square Matrix $A\in {\mathbb{R}}^{n\times n}$, denoted $\det (A)$ or $|A|$, is a scalar value that encodes certain properties of the matrix, such as whether it is invertible. Formally, for $A=[a_{ij}]$, the determinant is defined recursively:

• For $n=1$, $\det (A)=a_{11}$.
• For $n>1$,
  $\det (A)={\sum }_{j=1}^n(-1{)}^{1+j}{a}_{1j}\det (A_{1j}),$
  where $A_{1j}$ is the $(n-1)\times (n-1)$ submatrix obtained by removing the first row and $j$-th column of $A$.

Intuition
The determinant measures the “scaling factor” of the transformation represented by the matrix. A matrix with a determinant of $0$ represents a transformation that collapses space (e.g., maps a 2D plane to a line or a point), meaning the matrix is not invertible.

Key Properties

1. Invertibility: $A$ is invertible if and only if $\det (A)\ne 0$.
2. Multiplicative Property: $\det (AB)=\det (A)\det (B)$.
3. Transposition: $\det (A^T)=\det (A)$.
4. Row Operations:
   • Swapping two rows changes the sign of the determinant.
   • Multiplying a row by a scalar $c$ multiplies the determinant by $c$.
   • Adding a scalar multiple of one row to another does not change the determinant.
5. Diagonal Matrices: For a diagonal or triangular matrix, $\det (A)={\prod }_{i=1}^n{a}_{ii}$.

Applications

• Matrix Inversion: $\det (A)$ is used in the computation of the inverse via the adjugate formula: $A^{-1}=\frac{\text{adj}(A)}{\det (A)}$.
• Eigenvalues and eigenvectors: The determinant of $A-\lambda I$ (characteristic polynomial) determines the eigenvalues of $A$.
• Volume Scaling: $\det (A)$ gives the scaling factor of the volume of a parallelepiped defined by the column vectors of $A$.
• System of Linear Equation: A system $Ax=b$ has a unique solution if $\det (A)\ne 0$.

Examples

1. For $A=\left[\begin{array}{cc}3& 8\\ 4& 6\end{array}\right]$,
   $\det (A)=(3)(6)-(8)(4)=-14.$
2. For $A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 4\\ 5& 6& 0\end{array}\right]$,
   $\det (A)=1\cdot \det \left[\begin{array}{cc}1& 4\\ 6& 0\end{array}\right]-2\cdot \det \left[\begin{array}{cc}0& 4\\ 5& 0\end{array}\right]+3\cdot \det \left[\begin{array}{cc}0& 1\\ 5& 6\end{array}\right],$
   leading to $\det (A)=1(-24)-2(-20)+3(-5)=-24+40-15=1$.