detailed information: The input equation set is:
| | | | 4 | A | + | | 4 | B | + | | 4 | C | + | | 2 | D | + | | 2 | E | + | | F | = | | 0 | | (1) |
| | 16 | A | + | | 16 | B | + | | 24 | C | + | | 4 | D | + | | 6 | E | + | | F | = | | 0 | | (2) |
| | 16 | A | + | | 4 | B | + | | 8 | C | + | | 4 | D | + | | 2 | E | + | | F | = | | 0 | | (3) |
| | | | Question solving process:
Multiply both sides of equation (1) by 4, the equation can be obtained: | | 16 | A | + | | 16 | B | + | | 16 | C | + | | 8 | D | + | | 8 | E | + | | 4 | F | = | | 0 | (7) | , then subtract both sides of equation (7) from both sides of equation (2), the equations are reduced to:
| | | | 4 | A | + | | 4 | B | + | | 4 | C | + | | 2 | D | + | | 2 | E | + | | F | = | | 0 | | (1) |
| | | 16 | A | + | | 4 | B | + | | 8 | C | + | | 4 | D | + | | 2 | E | + | | F | = | | 0 | | (3) |
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Multiply both sides of equation (1) by 4, the equation can be obtained: | | 16 | A | + | | 16 | B | + | | 16 | C | + | | 8 | D | + | | 8 | E | + | | 4 | F | = | | 0 | (8) | , then subtract both sides of equation (8) from both sides of equation (3), the equations are reduced to:
| | | | 4 | A | + | | 4 | B | + | | 4 | C | + | | 2 | D | + | | 2 | E | + | | F | = | | 0 | | (1) |
| | -12 | B | | -8 | C | | -4 | D | | -6 | E | | -3 | F | = | | 0 | | (3) |
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交After the exchange of equation (2) and equation (3), the equation system becomes:
| | | | 4 | A | + | | 4 | B | + | | 4 | C | + | | 2 | D | + | | 2 | E | + | | F | = | | 0 | | (1) |
| -12 | B | | -8 | C | | -4 | D | | -6 | E | | -3 | F | = | | 0 | | (2) |
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Add both sides of equation (3) to both sides of equation (2), get the equation:
| | | | 4 | A | + | | 4 | B | + | | 4 | C | + | | 2 | D | + | | 2 | E | + | | F | = | | 0 | | (1) |
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Divide both sides of equation (3) by 2, get the equation:, then subtract both sides of equation (9) from both sides of equation (1), get the equation:
| | | | 4 | A | + | | 4 | B | + | | 4 | D | + | | 3 | E | + | | | 5 2 | F | = | | 0 | | (1) |
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Divide both sides of equation (2) by 3, get the equation: | -4 | B | | - | 8 3 | D | | - | 8 3 | E | | -2 | F | = | | 0 | (10) | , then add the two sides of equation (10) to both sides of equation (1), get the equation:
| | | | 4 | A | + | | | 4 3 | D | + | | | 1 3 | E | + | | | 1 2 | F | = | | 0 | | (1) |
| | | C | | - | 1 2 | D | | - | 1 4 | E | | - | 3 8 | F | = | | 0 | | (3) |
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The coefficient of the unknown number is reduced to 1, and the equations are reduced to:
| | | | A | + | | | 1 3 | D | + | | | 1 12 | E | + | | | 1 8 | F | = | | 0 | | (1) |
| | B | + | | | 2 3 | D | + | | | 2 3 | E | + | | | 1 2 | F | = | | 0 | | (2) |
| | C | | - | 1 2 | D | | - | 1 4 | E | | - | 3 8 | F | = | | 0 | | (3) |
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Therefore, the solution of the equation set is:
Where: D, E, F are arbitrary constants. 解方程组的详细方法请参阅:《多元一次方程组的解法》 |