Intorduction to Linear Algebra(3) Determinants

Determinants

• introduction to determinants
• Properties of determinants

introduction to determinants

$det A=\sum^n_{j=1}(-1)^{i+j}a_{1j}A_{nj}$detA=∑j=1n​(−1)i+ja1j​Anj​
$det A=\sum^n_{i=1}(-1)^{i+j}a_{i1}A_{in}$detA=∑i=1n​(−1)i+jai1​Ain​
if $A$A is a triangular matrix the determinants is products of the entries on the main diagonal of $A$A

Properties of determinants

Row Operations:
Let $A$A be a square matrix.
a.If a multiple of one row of $A$A is added to another row to produce a matrix $B$B, then $det B=detA$detB=detA.
b.If two rows of $A$A are interchanged to produce $B$B, then $detB=-detA$detB=−detA.
c.If one row of $A$A is multipled by $k$k to produce $B$B, then $detB=k*detA$detB=k∗detA.
Theorem:
If $A$A is an $n\times n$n×n matrix, then $detA^T=detA$detAT=detA
If $A$A and $B$B are $n \times n$n×n matrix, then $det(AB)=detA \times detB$det(AB)=detA×detB
Linearity Properties of the determinant Function
$A(x)=[a_1 \cdots a_{i-1} \quad x \quad a_{i+1} \cdots a_{n}]$A(x)=[a1​⋯ai−1​xai+1​⋯an​]
$T(x)=det[a_1 \cdots a_{i-1} \quad x \quad a_{i+1} \cdots a_{n}]$T(x)=det[a1​⋯ai−1​xai+1​⋯an​]
$T(cx)=cT(x)$T(cx)=cT(x)
$T(a+b)=T(a)+T(b)$T(a+b)=T(a)+T(b)
Cramer’s Rule:
Let A be an ivertible $n\times n$n×n matrix. For any $b$b in $\mathbb R^n$Rn, the unique solution of $Ax=b$Ax=b is:
$x=\frac{detA_i(b)}{detA}$x=detAdetAi​(b)​
Thus, from $AA^{-1}=E$AA−1=E we could get $A^{-1}=\begin{bmatrix}\frac{detA_1(e_1)}{detA} & \cdots & \frac{detA_1{e_n}}{detA}\\ \vdots &\vdots & \vdots \\ \frac{detA_n{e_1}}{detA}&\cdots& \frac{detA_n{e_n}}{detA} \end{bmatrix}$A−1=⎣⎢⎡​detAdetA1​(e1​)​⋮detAdetAn​e1​​​⋯⋮⋯​detAdetA1​en​​⋮detAdetAn​en​​​⎦⎥⎤​
$A^{-1}=\frac{1}{detA}adjA$A−1=detA1​adjA

标签：Intorduction,AAA,Linear,nn,detA,frac,cdots,amp,Determinants
来源： https://blog.csdn.net/zbzhzhy/article/details/87900484