解方程组a-b+c=0,4a+2b+c=12,2a+5b+c=9
问题描述:
解方程组a-b+c=0,4a+2b+c=12,2a+5b+c=9
答
a - b +c = 0(1)4a+2b + c =12(2)2a +5b + c = 9(3)(2)-(1)得到3a + 3b = 12(4)(2)-(3)得到2a - 3b = 3(5)(4)+(5)得到5a = 15,因此a= 3将a = 3代入(5)得到b = 1将a = 3,b= 1代入到(1)得到c = b-a = -2即a = 3,b=1,c ...