quartz/content/Obsidian Vault/determinants.md

#math/linear_algebra

theorem on product of determinants

$det(AB)=(detA)(detB)$ WARNING det(A+B)\neq detA-detB

theorem for determinant of invertible matrix

det(A^{-1})=\frac{1}{detA}

characteristic equation

\operatorname{det}(A-\lambda I)=0

ways of computing

!

Definition (Determinant of A ) ! ! For a 1 \times 1 matrix A=[a], the determinant of A, \operatorname{denoted} by \operatorname{det} A, is defined to be

\operatorname{det} A=a

For n \geq 2, the determinant of an n \times n matrix A=\left[a_{i j}\right] is the sum

\begin{aligned}
\operatorname{det} A &=a_{11} \operatorname{det} A_{11}-a_{12} \operatorname{det} A_{12}+\ldots+(-1)^{n+1} a_{1 n} \operatorname{det} A_{1 n} \\
&=\sum_{i=1}^{n}(-1)^{i+1} a_{1 i} \operatorname{det} A_{1 i}
\end{aligned}

Sometimes we use absolute value brackets to denote the determinant, i.e. we sometimes write |A| to \operatorname{denote} \operatorname{det} A.