detailed information: The input equation set is:
 | | | | A | + | | 2 | B | | -1 | C | | -2 | D | + | | E | = | | 0 | | (1) |
| | -1 | A | + | | B | + | | C | + | | 6 | D | + | | 7 | E | = | | 0 | | (3) |
| -2 | A | + | | B | | -1 | C | + | | 5 | D | | -3 | E | = | | 0 | | (4) |
| | Question solving process:
Multiply both sides of equation (1) by 2, the equation can be obtained: | | 2 | A | + | | 4 | B | | -2 | C | | -4 | D | + | | 2 | E | = | | 0 | (6) | , then subtract both sides of equation (6) from both sides of equation (2), the equations are reduced to:
 | | | | A | + | | 2 | B | | -1 | C | | -2 | D | + | | E | = | | 0 | | (1) |
| | -1 | A | + | | B | + | | C | + | | 6 | D | + | | 7 | E | = | | 0 | | (3) |
| -2 | A | + | | B | | -1 | C | + | | 5 | D | | -3 | E | = | | 0 | | (4) |
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Add both sides of equation (1) to both sides of equation (3) ,the equations are reduced to:
 | | | | A | + | | 2 | B | | -1 | C | | -2 | D | + | | E | = | | 0 | | (1) |
| | | -2 | A | + | | B | | -1 | C | + | | 5 | D | | -3 | E | = | | 0 | | (4) |
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Multiply both sides of equation (1) by 2, the equation can be obtained: | | 2 | A | + | | 4 | B | | -2 | C | | -4 | D | + | | 2 | E | = | | 0 | (7) | , then add the two sides of equation (7) to both sides of equation (4), the equations are reduced to:
 | | | | A | + | | 2 | B | | -1 | C | | -2 | D | + | | E | = | | 0 | | (1) |
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Add both sides of equation (2) to both sides of equation (3) ,the equations are reduced to:
 | | | | A | + | | 2 | B | | -1 | C | | -2 | D | + | | E | = | | 0 | | (1) |
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Multiply both sides of equation (2) by 5 Divide the two sides of equation (2) by 3, the equation can be obtained: | -5 | B | + | | | 10 3 | C | | - | 5 3 | D | | - | 20 3 | E | = | | 0 | (8) | , then add the two sides of equation (8) to both sides of equation (4), the equations are reduced to:
 | | | | A | + | | 2 | B | | -1 | C | | -2 | D | + | | E | = | | 0 | | (1) |
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Divide the two sides of equation (3) by 6, the equation can be obtained: , then subtract both sides of equation (9) from both sides of equation (4), the equations are reduced to:
 | | | | A | + | | 2 | B | | -1 | C | | -2 | D | + | | E | = | | 0 | | (1) |
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Multiply both sides of equation (4) by 18 Divide both sides of equation (4) by 7, get the equation:, then add the two sides of equation (10) to both sides of equation (3), get the equation:
 | | | | A | + | | 2 | B | | -1 | C | | -2 | D | + | | E | = | | 0 | | (1) |
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Multiply both sides of equation (4) by 6 Divide both sides of equation (4) by 7, get the equation:, then subtract both sides of equation (11) from both sides of equation (2), get the equation:
 | | | | A | + | | 2 | B | | -1 | C | | -2 | D | + | | E | = | | 0 | | (1) |
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Multiply both sides of equation (4) by 12 Divide both sides of equation (4) by 7, get the equation:, then subtract both sides of equation (12) from both sides of equation (1), get the equation:
 | | | | A | + | | 2 | B | | -1 | C | + | | | 107 7 | E | = | | 0 | | (1) |
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Subtract both sides of equation (3) from both sides of equation (2), get the equation:
 | | | | A | + | | 2 | B | | -1 | C | + | | | 107 7 | E | = | | 0 | | (1) |
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Divide both sides of equation (3) by 2, get the equation:, then add the two sides of equation (13) to both sides of equation (1), get the equation:
Multiply both sides of equation (2) by 2 Divide both sides of equation (2) by 3, get the equation:, then add the two sides of equation (14) to both sides of equation (1), get the equation:
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: 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》 |