detailed information: The input equation set is:
| | | | -1 | A | + | | 5 | B | + | | 6 | C | | -1 | D | = | | 0 | | (2) |
| -2 | A | + | | 4 | B | + | | 3 | C | | -1 | D | = | | 0 | | (3) |
| | Question solving process:
Divide the two sides of equation (1) by 4, the equation can be obtained: | | A | | - | 5 4 | B | + | | | 1 2 | C | | - | 1 4 | D | = | | 0 | (5) | , then add the two sides of equation (5) to both sides of equation (2), the equations are reduced to:
| | | | | | 15 4 | B | + | | | 13 2 | C | | - | 5 4 | D | = | | 0 | | (2) |
| -2 | A | + | | 4 | B | + | | 3 | C | | -1 | D | = | | 0 | | (3) |
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Divide the two sides of equation (1) by 2, the equation can be obtained: | | 2 | A | | - | 5 2 | B | + | | C | | - | 1 2 | D | = | | 0 | (6) | , then add the two sides of equation (6) to both sides of equation (3), the equations are reduced to:
| | | | | | 15 4 | B | + | | | 13 2 | C | | - | 5 4 | D | = | | 0 | | (2) |
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Divide the two sides of equation (1) by 4, the equation can be obtained: | | A | | - | 5 4 | B | + | | | 1 2 | C | | - | 1 4 | D | = | | 0 | (7) | , then subtract both sides of equation (7) from both sides of equation (4), the equations are reduced to:
| | | | | | 15 4 | B | + | | | 13 2 | C | | - | 5 4 | D | = | | 0 | | (2) |
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Multiply both sides of equation (2) by 2 Divide the two sides of equation (2) by 5, the equation can be obtained: , then subtract both sides of equation (8) from both sides of equation (3), the equations are reduced to:
| | | | | | 15 4 | B | + | | | 13 2 | C | | - | 5 4 | D | = | | 0 | | (2) |
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Multiply both sides of equation (2) by 3 Divide the two sides of equation (2) by 5, the equation can be obtained: | | | 9 4 | B | + | | | 39 10 | C | | - | 3 4 | D | = | | 0 | (9) | , then subtract both sides of equation (9) from both sides of equation (4), the equations are reduced to:
| | | | | | 15 4 | B | + | | | 13 2 | C | | - | 5 4 | D | = | | 0 | | (2) |
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Multiply both sides of equation (3) by 17 Divide the two sides of equation (3) by 7, the equation can be obtained: , then add the two sides of equation (10) to both sides of equation (4), the equations are reduced to:
| | | | | | 15 4 | B | + | | | 13 2 | C | | - | 5 4 | D | = | | 0 | | (2) |
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Multiply both sides of equation (4) by 7 Divide both sides of equation (4) by 10, get the equation:, then subtract both sides of equation (11) from both sides of equation (3), get the equation:
| | | | | | 15 4 | B | + | | | 13 2 | C | | - | 5 4 | D | = | | 0 | | (2) |
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Multiply both sides of equation (4) by 7 Divide both sides of equation (4) by 8, get the equation:, then subtract both sides of equation (12) from both sides of equation (2), get the equation:
Multiply both sides of equation (4) by 7 Divide both sides of equation (4) by 10, get the equation:, then subtract both sides of equation (13) from both sides of equation (1), get the equation:
Multiply both sides of equation (3) by 65 Divide both sides of equation (3) by 14, get the equation:, then subtract both sides of equation (14) from both sides of equation (2), get the equation:
Multiply both sides of equation (3) by 10 Divide both sides of equation (3) by 7, get the equation:, then subtract both sides of equation (15) from both sides of equation (1), get the equation:
Multiply both sides of equation (2) by 4 Divide both sides of equation (2) by 3, get the equation:, then add the two sides of equation (16) 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:
Convert the solution of the equation set to decimals:
解方程组的详细方法请参阅:《多元一次方程组的解法》 |