A + B + C + D + E = 1 A

2022-12-25 01:18| 来源: 网络整理| 查看: 265

detailed information：The input equation set is: A + B + C + D + E = 1 (1) A -1B + C -1D + E = -1 (2) A + 2B + 3C + 4D + 5E = 6 (3) 6A + 5B + 4C + 3D + 2E = 1 (4) A + B + 2C + 3D + 5E = 8 (5)Question solving process:

Subtract both sides of equation (1) from both sides of equation (2) ,the equations are reduced to:

A + B + C + D + E = 1 (1)-2B -2D = -2 (2) A + 2B + 3C + 4D + 5E = 6 (3) 6A + 5B + 4C + 3D + 2E = 1 (4) A + B + 2C + 3D + 5E = 8 (5)Subtract both sides of equation (1) from both sides of equation (3) ,the equations are reduced to: A + B + C + D + E = 1 (1)-2B -2D = -2 (2) B + 2C + 3D + 4E = 5 (3) 6A + 5B + 4C + 3D + 2E = 1 (4) A + B + 2C + 3D + 5E = 8 (5)Multiply both sides of equation (1) by 6, the equation can be obtained: 6A + 6B + 6C + 6D + 6E = 6 (6), then subtract both sides of equation (6) from both sides of equation (4), the equations are reduced to: A + B + C + D + E = 1 (1)-2B -2D = -2 (2) B + 2C + 3D + 4E = 5 (3)-1B -2C -3D -4E = -5 (4) A + B + 2C + 3D + 5E = 8 (5)Subtract both sides of equation (1) from both sides of equation (5) ,the equations are reduced to: A + B + C + D + E = 1 (1)-2B -2D = -2 (2) B + 2C + 3D + 4E = 5 (3)-1B -2C -3D -4E = -5 (4) C + 2D + 4E = 7 (5)Divide the two sides of equation (2) by 2, the equation can be obtained: -1B -1D = -1 (7), then add the two sides of equation (7) to both sides of equation (3), the equations are reduced to: A + B + C + D + E = 1 (1)-2B -2D = -2 (2) 2C + 2D + 4E = 4 (3)-1B -2C -3D -4E = -5 (4) C + 2D + 4E = 7 (5)Divide the two sides of equation (2) by 2, the equation can be obtained: -1B -1D = -1 (8), then subtract both sides of equation (8) from both sides of equation (4), the equations are reduced to: A + B + C + D + E = 1 (1)-2B -2D = -2 (2) 2C + 2D + 4E = 4 (3)-2C -2D -4E = -4 (4) C + 2D + 4E = 7 (5)Add both sides of equation (3) to both sides of equation (4) ,the equations are reduced to: A + B + C + D + E = 1 (1)-2B -2D = -2 (2) 2C + 2D + 4E = 4 (3)0 = 0 (4) C + 2D + 4E = 7 (5)Divide the two sides of equation (3) by 2, the equation can be obtained: C + D + 2E = 2 (9), then subtract both sides of equation (9) from both sides of equation (5), the equations are reduced to: A + B + C + D + E = 1 (1)-2B -2D = -2 (2) 2C + 2D + 4E = 4 (3)0 = 0 (4) D + 2E = 5 (5)交After the exchange of equation (4) and equation (5)， the equation system becomes: A + B + C + D + E = 1 (1)-2B -2D = -2 (2) 2C + 2D + 4E = 4 (3) D + 2E = 5 (4)0 = 0 (5)Multiply both sides of equation (4) by 2, get the equation： 2D + 4E = 10 (10), then subtract both sides of equation (10) from both sides of equation (3), get the equation： A + B + C + D + E = 1 (1)-2B -2D = -2 (2) 2C = -6 (3) D + 2E = 5 (4)0 = 0 (5)Multiply both sides of equation (4) by 2, get the equation： 2D + 4E = 10 (11), then add the two sides of equation (11) to both sides of equation (2), get the equation： A + B + C + D + E = 1 (1)-2B + 4E = 8 (2) 2C = -6 (3) D + 2E = 5 (4)0 = 0 (5)Subtract both sides of equation (4) from both sides of equation (1), get the equation： A + B + C -1E = -4 (1)-2B + 4E = 8 (2) 2C = -6 (3) D + 2E = 5 (4)0 = 0 (5)Divide both sides of equation (3) by 2, get the equation： C = -3 (12), then subtract both sides of equation (12) from both sides of equation (1), get the equation： A + B -1E = -1 (1)-2B + 4E = 8 (2) 2C = -6 (3) D + 2E = 5 (4)0 = 0 (5)Divide both sides of equation (2) by 2, get the equation： -1B + 2E = 4 (13), then add the two sides of equation (13) to both sides of equation (1), get the equation： A + E = 3 (1)-2B + 4E = 8 (2) C = -3 (3) D + 2E = 5 (4)0 = 0 (5)The coefficient of the unknown number is reduced to 1, and the equations are reduced to： A + E = 3 (1) B -2E = -4 (2) C = -3 (3) D + 2E = 5 (4)0 = 0 (5)

Therefore, the solution of the equation set is:

A = 3 - 1EB = -4 + 2EC = -3D = 5 - 2EWhere: 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》

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