Khác biệt giữa các bản “Phép khử Gauss-Jordan”

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== Ví dụ:Giải hệ phương trình sau bằng phương pháp khử Gauss-Jordan ==
2''x''<sub>1</sub> + 3''x''<sub>2</sub> + ''x''<sub>3</sub> = 11
2x1 + 3x2 +x3 = 11
 
-''x''<sub>1</sub> + 2''x''<sub>2</sub> - ''x''<sub>3<sub/> = 0
-x1 + 2x2 -x3 = 0
 
3''x''<sub>1</sub> + 2''x''<sub>3</sub> = 9
3x1 + 2x3 =9
 
Bước1Bước 1: Hệ phương trình trên tương đương với
 
3''x''<sub>1</sub> + 2''x''<sub>3</sub> = 9
3x1 + 2x3=9
 
''x''<sub>1</sub> – 2''x''<sub>2</sub> + ''x''<sub>3</sub> = 0
x1 - 2x2 +x3 = 0
 
2''x''<sub>1</sub> + 3''x''<sub>2</sub> + ''x<sub>3</sub> = 11
2x1 + 3x2 +x3 = 11
 
Bước 1:
 
3''x''<sub>1</sub> + 0 + 2''x''<sub>3</sub> = 9
3x1 + 0 +2x3 =9
 
2''x''<sub>2</sub> – ''x''<sub>3</sub>/3 = 3
2x2 -x3/3 = 3
 
3''x''<sub>2</sub> – ''x''<sub>3</sub>//3 = 5
3x2 -x3//3 = 5
 
''h''<sub>1</sub> = ''h''<sub>1</sub>; ''h''<sub>2</sub> = ''h''<sub>2</sub> + ''h''<sub>1</sub>/3; 3 = ''h''<sub>3</sub> – 2''h''<sub>1</sub>/3;
h1=h1 h2=h2+h1/3; 3=h3-2*h1/3;
 
Bước 2:
 
3x13''x''<sub>1</sub> + 0 +2x3 2''x''<sub>3</sub> = 9
 
3''x''<sub>2</sub> – ''x''<sub>3</sub>/3 = 5
3x2 - x3/3 = 5
 
2''x''<sub>2</sub> – ''x''<sub>3</sub>/3 = 3
2x2 - x3/3 = 3
 
''h''<sub>1</sub> = ''h''<sub>1</sub>; ''h''<sub>2</sub> = ''h''<sub>3</sub>; ''h''<sub>3</sub> = ''h''<sub>2</sub>
h1=h1; h2=h3; h3=h2
 
3x13''x''<sub>1</sub> + 0 + 2x32''x''<sub>3</sub> = 9
 
3''x''<sub>2</sub> – ''x''<sub>3</sub>/3 = 5
3x2 - x3/3 = 5
 
-x3–''x''<sub>3</sub>/9 = -1–1/3
 
''h''<sub>1</sub> = ''h''<sub>1</sub>; ''h''<sub>2</sub> = ''h''<sub>2</sub>; ''h''<sub>3</sub> = ''h''<sub>3</sub> – 2''h''<sub>2</sub>/3
h1=h1; h2=h2; h3=h3-2*h2/3
 
Bước 3:
Vậy
 
3x13''x''<sub>1</sub> + 0 + 0 = 3
 
3x23''x''<sub>2</sub> - 0 = 6
 
-x3–x<sub>3</sub>/9 =-1 –1/3
 
''x''<sub>1</sub> = 1; ''x''<sub>2</sub> = 2; ''x''<sub>3</sub> = 3
x1=1; x2=2; x3=3
 
''h''<sub>1</sub> = ''h''<sub>1</sub> – 2''h''<sub>3</sub>/(–1/9); ''h''<sub>2</sub> = ''h''<sub>2</sub> – (1/3)''h''<sub>3</sub>/(–1/9); ''h''<sub>3</sub> = ''h''<sub>3</sub>/(–1/9);
h1=h1-2*h3/(-1/9); h2=h2-(1/3)*h3/(-1/9); h3=h3/(-1/9);