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Lecture 5: Inverse Matrices

Graduate Entrance Examination Mathematics study notes: Lecture 5: Inverse Matrices. Original formulas, diagrams, and examples are retained.

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5.1 Definition of Inverse Matrices

#####Definition: Inversematrix

description: A and B are n-order square matrices, E are n-order identity matrices; If AB=BA=E, then A is called an invertible matrix, and B is called the inverse of A, and the inverse is unique, denoted as $A^{-1}$

Explanation

  • The product of a matrix and its inverse matrix equals the identity matrix;

#####Definition: Thenecessaryandsufficientconditionfortheinversematrix

Description: A A sufficient and necessary condition for reversibility is that $|A|$ is not equal to 0;

Explanation

  • If the measure of the matrix is not equal to 0, then there is an inverse matrix;
  • Because $a^{-1}=\frac{1}{a}$ and the denominator cannot be zero, the necessary and sufficient condition for the inverse matrix is that the measure is not zero;

#####Definition: andopposedeachother

Description: first stage: $$A^{-1}A=E$$

Explanation

5.2 Properties and Important Formulas of Inverse Matrices

Properties: Properties of inverse matrices

  • 1. $(A^{-1})^{-1}=A$
  • 2. If k is not equal to 0, then $(kA)^{-1}=\frac{1}{k}A^{-1}$
  • When transposing matrices: $(kA)^T=kA^T$
  • When transposing matrices: $|kA|=k^n|A|$
  • 3. AB is also invertible, and $(AB)^{-1}=B^{-1}A^{-1}$
  • 4. $A^{\mathrm{T}}$ is also invertible, and $(A^{\mathrm{T}})^{-1}=(A^{-1})^{\mathrm{T}}$.
  • 5. $\left|A^{-1}\right|=|A|^{-1}$
  • Because: $|A^{-1}A|=|E|\rightarrow|E|=|A^{-1}||A|$ that is: $|A^{-1}|$ and $|A|$ are reciprocals of each other;

Note: $A+B$ need not be invertible, and in general $\left(A+B\right)^{-1}\neq A^{-1}+B^{-1}$.

  • $\left(A+B\right)^{T}=A^{T}+B^{T}$
  • $\vert A+B\vert$ is not equal to $\vert A\vert+\vert B\vert$

Supplement: Inverse matrix and block matrix conclusions

  • $$\left.\left(\begin{matrix}a&0\\0&b\end{matrix}\right.\right)^{-1}=\left(\begin{matrix}a&-1&0\\0&b^{-1}\end{matrix}\right)\left(\begin{matrix}A&0\\0&B\end{matrix}\right)^{-1}=\left(\begin{matrix}A&-1&0\\0&B^{-1}\end{matrix}\right)\\\left(\begin{matrix}0&a\\b&0\end{matrix}\right)^{-1}=\left(\begin{matrix}0&b^{-1}\\a^{-1}&0\end{matrix}\right)\left(\begin{matrix}0&A\\B&0\end{matrix}\right)^{-1}=\left(\begin{matrix}0&B^{-1}\\A^{-1}&0\end{matrix}\right)$$

5.3 Using the Definition Method to Find the Inverse of an Invertible Matricus

5.3.1 Method One: Definition Method

Method: Define and solve it, that is, find a matrix B such that AB=E, then A is invertible and $A^{-1}=B$

5.3.2 Method 2: Multiplication Method

Method: Divide A into the product of several invertible matrices. Since the product of two invertible matrices is still invertible, that is, if A=BC, where B、C are invertible, then A is invertible, and:

  • $$A^{-1}=(BC)^{-1}=C^{-1}B^{-1}$$