Lecture 6: Adjugate Matrices
Graduate Entrance Examination Mathematics study notes: Lecture 6: Adjugate Matrices. Original formulas, diagrams, and examples are retained.
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6.1 Definition of adjugate matrices
6.1.1 Definition
#####Definition: AdjugateMatrix
Description: The adjugate matrix $A^$ is the transpose of the cofactor matrix of $A$: $$A^=\begin{bmatrix}A_{11}&A_{21}&\cdots&A_{n1}\\A_{12}&A_{22}&\cdots&A_{n2}\\\vdots&\vdots&&\vdots\\A_{1n}&A_{2n}&\cdots&A_{nn}\end{bmatrix}.$$
It satisfies $$AA^{}=A^{}A=\left|A\right|E.$$
Explanation
- Concept:
- The formation of the adjoint matrix comes from multiplying one row of dots by the product of one column;
- From $A^*=\left(\begin{matrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{matrix}\right)\rightarrow A=\left(\begin{matrix}a_{11}a_{12}\\a_{21}a_{22}\end{matrix}\right)$
- So: $AA^{*}=\left(\begin{matrix}|A|&0\\0&|A|\end{matrix}\right)=\left(\begin{matrix}1&0\\0&1\end{matrix}\right)|A|=E|A|$
- Calculation: Second order
- Find $A^*$ from matrix A
- Core:**Main switched, secondary code change#
- Example: $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\rightarrow A^*=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$
- Conclusion: At the second order, $(A^)^=A$
- Find $A^{-1}$ from $A^*$
- Example: $\frac{1}{|A|}A^{*}=\frac{1}{ad-bc}(\begin{matrix}a&-b\\-c&a\end{matrix})=A^{-1}$
- Calculation: Third order
- Find $A^*$ from matrix A:
- Honestly calculate the remaining equations of each algebraic term;
- Find $A^{-1}$ from $A^*$
- $A^{-1}=\frac{1}{A}A^{*}$
6.1.2 adjugate matrices and Invertible Matrices
Summary: Commutative matrices
- $$\begin{aligned}&A\cdot kE=kE\cdot A\\&AA^{-1}=A^{-1}A=E\\&AA^{}=A^{}A=|A|E\end{aligned}$$
Inference: $$A^{*}=|A|A^{-1}$$
- Meaning: obtained the relationship between the adjugate matrix and the invertible matrix;
Inference: $AA^{}A^{-1}=|A|EA^{-1}\rightarrow A^{}=|A|A^{-1}$
Conclusion: A method for finding reversibility:
- Method:
- When
Ais invertible, $A^{*}\text{ and }A^{-1}$ is only one nonzero multiple apart; - Meaning:
- Given adjunctions, think ofinvertible matrices
->That is, from the perspective of vectors, adjugate and invertible matrices are merely scaling->The properties of adjugate and invertible matrices are consistent;
6.2 Properties and Formulas of adjugate matrices
Concept: Summary of operations
- There are four operations: determinant, inverse, transpose, and adjugate;
- The following properties are the interactions, combinations, and exchanges of these properties;
Property 1: For any $n$-order square matrix $A$, there is an adjoint matrix A $^{}$, and the formula is: $$AA^{}=A^{}A=\left|A\right|E,\left|A^{}\right|=\left|A\right|^{n-1}$$
Property Two: When $|A|\neq0 A^{}=\left|A\right|A^{-1}, A^{-1}=\frac{1}{\left|A\right|}A^{}, A=\left|A\right|\left(A^{*}\right)^{-1}; $
- Premise: A has a measure of 0;
- First find the determinant, then the adjugate matrix, and finally the $\frac{1}{{|A|}}A^*$
Property 3: $(kA)(kA)^{*}=\left|kA\right|E$
Nature Four: $(A^T)^=(A^)^T$
Nature Five: $A^{-1}\left(A^{-1}\right)^{*}=\left|A^{-1}\right|E$
Nature Six: $A^\left(A^\right)^=\left|A^\right|E$
- Reason: $\left(A^\right)^=|A|^{n-2}A$
Nature 7:(Principle of Putting on and Doff) $\left(AB\right)^{}=B^{}A^{*}$
Supplement: About multiplication exchange
- $$\begin{aligned}&|kA|=k^{n}|A|\\&(kA)^{T}=kA^{T}\\&(kA)^{-1}=\frac{1}{k}A^{-1}\\&(kA)^{}=k^{n-1}A^{}\end{aligned}$$
Supplement: Summary - The auto-operation of the operation
- $$\begin{aligned}&|A^{-1}|=|A|^{-1}&&(A^{})^{-1}=(A^{-1})^{}\\&(A^{-1})^{T}=(A^{T})^{-1}&&|A^{}|=|A|^{n-1}\\&|A^{T}|=|A|^{T}\\&(A^{})^{T}=(A^{T})^{*}\end{aligned}$$
Supplement: Summary - Stacking Operations
- $$\begin{aligned}&||A||=|A|\\&(A^{T})^{T}=A\\&(A^{-1})^{-1}=A\\&(A^{})^{}=|A|^{n-2}A\end{aligned}$$
Additional note: Principles for putting on and taking off
- $$\begin{aligned}&|AB|=|B||A|\\&(AB)^{T}=B^{T}A^{T}\\&(AB)^{-1}=B^{-1}A^{-1}\\&(AB)^{}=B^{}A^{*}\end{aligned}$$
- Note: $\left(A+B\right)^{}\neq A^{}+B^{*}$
- Others: $\begin{aligned}&|A+B|\neq|A|+|B|\\&(A+B)^{-1}\neq A^{-1}+B^{-1}\\&(A+B)^{T}=A^{T}+B^{T}\end{aligned}$
6.3 Using Adjoint Matrices to Find the Inverse of an Invertible Matrice
#####Theorem: Findtheinverseoftheinvertiblematrixasitisadjoined
description: $$A^{-1}=\frac{1}{\left|A\right|}A^{*}=\frac{1}{\left|A\right|}\begin{bmatrix}A_{11}&A_{21}&\cdots&A_{n1}\\A_{12}&A_{22}&\cdots&A_{n2}\\\vdots&\vdots&&\vdots\\A_{1n}&A_{2n}&\cdots&A_{nn}\end{bmatrix}$$
Explanation
- Steps:
- Step 1: First, find the value of the determinant of the current matrix to see if it equals
0; if it equals zero, you cannot continue calculating; - Step 2: Find the adjugant of A;
- Step 3: Find $A^{-1}=\frac{1}{|A|}A^{*}$
- Note:
- $\text{ Note }A_{ij}\text{ Position and positive and negative signs }$
6.4 Methods for Finding Adjoint Matrices
Method One: Definition method. First find $A_{ij}$, then assemble into $A^*$
Method 2: Use the formula; If A is invertible, then $A^*=|A|A^{-1}$
- When encountering calculations related to adjoint matrices, first consider whether there is a formula, simplify it, perform some calculations on the formula, and then begin calculation;