detailed information: The input equation set is:
| | | | A | + | | 2 | B | + | | 3 | C | + | | 4 | D | + | | 5 | E | = | | 6 | | (1) |
| | 7 | A | + | | 8 | B | + | | 9 | C | + | | 10 | D | + | | 11 | E | = | | 12 | | (2) |
| | A | + | | 3 | B | + | | 5 | C | + | | 7 | D | + | | 9 | E | = | | 11 | | (3) |
| | 2 | A | + | | 4 | B | + | | 6 | C | + | | 8 | D | + | | 10 | E | = | | 12 | | (4) |
| | 5 | A | + | | 4 | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 0 | | (5) |
| Question solving process:
Multiply both sides of equation (1) by 7, the equation can be obtained: | | 7 | A | + | | 14 | B | + | | 21 | C | + | | 28 | D | + | | 35 | E | = | | 42 | (6) | , then subtract both sides of equation (6) from both sides of equation (2), the equations are reduced to:
| | | | A | + | | 2 | B | + | | 3 | C | + | | 4 | D | + | | 5 | E | = | | 6 | | (1) |
| -6 | B | | -12 | C | | -18 | D | | -24 | E | = | | -30 | | (2) |
| | A | + | | 3 | B | + | | 5 | C | + | | 7 | D | + | | 9 | E | = | | 11 | | (3) |
| | 2 | A | + | | 4 | B | + | | 6 | C | + | | 8 | D | + | | 10 | E | = | | 12 | | (4) |
| | 5 | A | + | | 4 | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 0 | | (5) |
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Subtract both sides of equation (1) from both sides of equation (3) ,the equations are reduced to:
| | | | A | + | | 2 | B | + | | 3 | C | + | | 4 | D | + | | 5 | E | = | | 6 | | (1) |
| -6 | B | | -12 | C | | -18 | D | | -24 | E | = | | -30 | | (2) |
| | | 2 | A | + | | 4 | B | + | | 6 | C | + | | 8 | D | + | | 10 | E | = | | 12 | | (4) |
| | 5 | A | + | | 4 | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 0 | | (5) |
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Multiply both sides of equation (1) by 2, the equation can be obtained: | | 2 | A | + | | 4 | B | + | | 6 | C | + | | 8 | D | + | | 10 | E | = | | 12 | (7) | , then subtract both sides of equation (7) from both sides of equation (4), the equations are reduced to:
| | | | A | + | | 2 | B | + | | 3 | C | + | | 4 | D | + | | 5 | E | = | | 6 | | (1) |
| -6 | B | | -12 | C | | -18 | D | | -24 | E | = | | -30 | | (2) |
| | | | 5 | A | + | | 4 | B | + | | 3 | C | + | | 2 | D | + | | E | = | | 0 | | (5) |
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Multiply both sides of equation (1) by 5, the equation can be obtained: | | 5 | A | + | | 10 | B | + | | 15 | C | + | | 20 | D | + | | 25 | E | = | | 30 | (8) | , then subtract both sides of equation (8) from both sides of equation (5), the equations are reduced to:
| | | | A | + | | 2 | B | + | | 3 | C | + | | 4 | D | + | | 5 | E | = | | 6 | | (1) |
| -6 | B | | -12 | C | | -18 | D | | -24 | E | = | | -30 | | (2) |
| | | -6 | B | | -12 | C | | -18 | D | | -24 | E | = | | -30 | | (5) |
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Divide the two sides of equation (2) by 6, the equation can be obtained: , then add the two sides of equation (9) to both sides of equation (3), the equations are reduced to:
| | | | A | + | | 2 | B | + | | 3 | C | + | | 4 | D | + | | 5 | E | = | | 6 | | (1) |
| -6 | B | | -12 | C | | -18 | D | | -24 | E | = | | -30 | | (2) |
| | | -6 | B | | -12 | C | | -18 | D | | -24 | E | = | | -30 | | (5) |
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Subtract both sides of equation (2) from both sides of equation (5) ,the equations are reduced to:
| | | | A | + | | 2 | B | + | | 3 | C | + | | 4 | D | + | | 5 | E | = | | 6 | | (1) |
| -6 | B | | -12 | C | | -18 | D | | -24 | E | = | | -30 | | (2) |
| | | |
Divide both sides of equation (2) by 3, get the equation: | -2 | B | | -4 | C | | -6 | D | | -8 | E | = | | -10 | (10) | , then add the two sides of equation (10) to both sides of equation (1), get the equation:
| | | | -6 | B | | -12 | C | | -18 | D | | -24 | E | = | | -30 | | (2) |
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The coefficient of the unknown number is reduced to 1, and the equations are reduced to:
Therefore, the solution of the equation set is:
Where: C, D, E 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》 |