"역행렬"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
14번째 줄: 14번째 줄:
 
:<math>\begin{array}{l}  \left( \begin{array}{ccc|ccc}  2 & -1 & 0 & 1 & 0 & 0 \\  -1 & 2 & -1 & 0 & 1 & 0 \\  0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\  -1 & 2 & -1 & 0 & 1 & 0 \\  0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\  0 & \frac{3}{2} & -1 & \frac{1}{2} & 1 & 0 \\  0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\  0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\  0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\  0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\  0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\  0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\  0 & 0 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\  0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\  0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & 0 & 0 & 1 & 1 & 1 \\  0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\  0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & 0 & 0 & 1 & 1 & 1 \\  0 & 1 & 0 & 1 & 2 & 2 \\  0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \end{array}</math>
 
:<math>\begin{array}{l}  \left( \begin{array}{ccc|ccc}  2 & -1 & 0 & 1 & 0 & 0 \\  -1 & 2 & -1 & 0 & 1 & 0 \\  0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\  -1 & 2 & -1 & 0 & 1 & 0 \\  0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\  0 & \frac{3}{2} & -1 & \frac{1}{2} & 1 & 0 \\  0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\  0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\  0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\  0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\  0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\  0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\  0 & 0 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & 1 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\  0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\  0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & 0 & 0 & 1 & 1 & 1 \\  0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\  0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \\  \left( \begin{array}{ccc|ccc}  1 & 0 & 0 & 1 & 1 & 1 \\  0 & 1 & 0 & 1 & 2 & 2 \\  0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \end{array}</math>
 
*  위의 결과로부터 주어진 행렬의 역행렬은 다음과 같음을 알 수 있다:<math>\left( \begin{array}{ccc}  1 & 1 & 1 \\  1 & 2 & 2 \\  1 & 2 & 3 \end{array} \right)</math>
 
*  위의 결과로부터 주어진 행렬의 역행렬은 다음과 같음을 알 수 있다:<math>\left( \begin{array}{ccc}  1 & 1 & 1 \\  1 & 2 & 2 \\  1 & 2 & 3 \end{array} \right)</math>
 
 
 
 
 
 
 
==역사==
 
 
 
 
 
* http://www.google.com/search?hl=en&tbs=tl:1&q=
 
* [[수학사 연표]]
 
  
 
 
 
 

2020년 11월 16일 (월) 08:29 판

개요

 

 

 

가우스-조단 소거법을 이용한 계산

  • 주어진 행렬은 다음과 같다\[\left( \begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{array} \right)\]
  • 가우스-조단 소거법 을 이용하기 위해, 다음과 같은 붙임행렬(augmented matrix)을 만든다\[\left( \begin{array}{ccc|ccc} 2 & -1 & 0 & 1 & 0 & 0 \\ -1 & 2 & -1 & 0 & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right)\]
  • 위의 행렬에 소거법을 적용하면, 다음의 행렬들을 얻는다

\[\begin{array}{l} \left( \begin{array}{ccc|ccc} 2 & -1 & 0 & 1 & 0 & 0 \\ -1 & 2 & -1 & 0 & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ -1 & 2 & -1 & 0 & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & \frac{3}{2} & -1 & \frac{1}{2} & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & 1 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \\ \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 & 2 & 3 \end{array} \right) \end{array}\]

  • 위의 결과로부터 주어진 행렬의 역행렬은 다음과 같음을 알 수 있다\[\left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} \right)\]

 

 

메모

 

 

 

관련된 항목들

 

 

수학용어번역

 

 

매스매티카 파일 및 계산 리소스

 

 

사전 형태의 자료

 

 

리뷰논문, 에세이, 강의노트