a + 2b + 4c + 5d = 3 2a + 3b + 5c + 6d = 4 4a + 5b + 7c + 8d = 6 5a + 6b + 8c + 9d = 7,求解a,b,c,d
时间: 2023-11-21 22:33:46 浏览: 165
一个有趣的东西,但是他是纯代码写成的...三维空间?
根据题目中给的四个等式,可以列出一个由未知数a、b、c、d组成的4元一次方程组,如下所示:
2a + 4b + 5c + d = 3
2a + 3b + 5c + 6d = 4
4a + 5b + 7c + 8d = 6
5a + 6b + 8c + 9d = 7
可以通过高斯消元法求解这个方程组。首先对方程组进行初等行变换,将其转化为阶梯形式。逐行令该行的第一个非零元为1,然后将该列的其余元素变为0。具体步骤如下:
2a + 4b + 5c + d = 3 (1)
2a + 3b + 5c + 6d = 4 (2)
4a + 5b + 7c + 8d = 6 (3)
5a + 6b + 8c + 9d = 7 (4)
(1) - (2):b - d = -1
(1) - (3):-3a - b - 2c - 3d = -3
(1) - (4):-3a - 2b - 3c - 4d = -4
-3a - b - 2c - 3d = -3 (5)
b - d = -1 (6)
-3a - 2b - 3c - 4d = -4 (7)
(5) - 2*(6):-3a - 4c - 5d = 1
(5) + 3*(7):-12a - 11d = 5
(6):b - d = -1
-3a - 4c - 5d = 1 (8)
-12a - 11d = 5 (9)
b - d = -1
(8) + 4*(9):-59a = -59
a = 1
将a = 1代入(9),解出d = -1
将a = 1, d = -1代入(6),解出b = -2
将a = 1, b = -2, d = -1代入(5),解出c = 1
因此,方程组的解为a = 1, b = -2, c = 1, d = -1。
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