Basic knowledge of Linear Algebra -- concepts and relationships of common matrices

Transpose matrix

Symbol ：$A^T$
Concept ： Exchange of ranks
$A=\left(\begin{array}{ccc}1& 2& 99\\ 3& 4& 88\end{array}\right),A^T=\left(\begin{array}{cc}1& 3\\ 2& 4\\ 99& 88\end{array}\right)$

The remainder formula

Symbol ：$M_{ij}$
Concept ： For squares $A=(a_{ij}{)}_{n\times n}$, The matrix $A$ The elements of $a_{ij}$ In the second place $i$ Xing He $j$ After the column is crossed out , The rest of the elements are in the original order $n-1$ Determined by the order matrix determinant It's called the element $a_{ij}$ The Yu Zi form of , Write it down as $M_{ij}$.

Algebraic cofactor

Symbol ：$A_{ij}$
Concept ：$A_{ij}=(-1{)}^{i+j}{M}_{ij}$ It's called the element $a_{ij}$ The algebraic covalent of .
example ： matrix $A=\left(\begin{array}{ccc}2& 3& 1\\ 3& 4& 1\\ 3& 7& 2\end{array}\right)$
$M_{11}=\left|\begin{array}{cc}4& 1\\ 7& 2\end{array}\right|=1,A_{11}=(-1{)}^{1+1}{M}_{11}=1$
$M_{21}=\left|\begin{array}{cc}3& 1\\ 7& 2\end{array}\right|=-1,A_{11}=(-1{)}^{2+1}{M}_{11}=1$

Adjoint matrix

Symbol ：$A^{\ast }$
Concept ： For squares $A=(a_{ij}{)}_{n\times n}$, Every element $a_{ij}$ The algebraic covalent of $A_{ij}$ stay $a_{ij}$ The transpose matrix of the matrix composed of corresponding positions .
$A^{\ast }={\left(\begin{array}{cccc}{A}_{11}& A_{12}& \cdots & A_{1n}\\ A_{21}& A_{22}& \cdots & A_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ A_{n1}& A_{n2}& \cdots & A_{nn}\end{array}\right)}^T=\left(\begin{array}{cccc}{A}_{11}& A_{21}& \cdots & A_{n1}\\ A_{12}& A_{22}& \cdots & A_{n2}\\ ⋮& ⋮& \ddots & ⋮\\ A_{1n}& A_{2n}& \cdots & A_{nn}\end{array}\right)$
example ： matrix $A=\left(\begin{array}{cc}-3& 4\\ 6& 2\end{array}\right)$
$A^{\ast }={\left(\begin{array}{cc}{M}_{11}& -M_{12}\\ -M_{21}& M_{22}\end{array}\right)}^T={\left(\begin{array}{cc}2& -6\\ -4& -3\end{array}\right)}^T=\left(\begin{array}{cc}2& -4\\ -6& -3\end{array}\right)$

Inverse matrix

Symbol ：$A^{-1}$
Definition ：$A^{-1}=\frac{1}{|A|}{A}^{\ast }$
example ： matrix $A=\left(\begin{array}{cc}-3& 4\\ 6& 2\end{array}\right)$
According to the above calculation, we have obtained $A^{\ast }$,$|A|=-30$,$A^{-1}=\frac{1}{-30}\left(\begin{array}{cc}2& -4\\ -6& -3\end{array}\right)$
Check ：$A^{-1}A=\frac{1}{-30}\left(\begin{array}{cc}2& -4\\ -6& -3\end{array}\right)\left(\begin{array}{cc}-3& 4\\ 6& 2\end{array}\right)=\frac{1}{-30}\left(\begin{array}{cc}-30& 0\\ 0& -30\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$