標題:
Cramer's rule
發問:
Use Cramer s rule to solve each system of equations. If D=0, use another method to determine the solution set. -2x-2y+3z=4 -- (1) 5x+7y-z=2 -- (2) 2x+2y-3z=-4 -- (3) 我知道(1) 同埋 (3) 係一樣, 不過應該點計落去呢? 用Gauss Jordon method?定係...? Correct answer: x= (-32+19z)/4 y= (24-13z)/4 究竟點計ga?
最佳解答:
Gauss Jordon method: -2 -2 3 | 4 5 7 -1 | 2 2 2 -3 | -4 1 1 -3/2 | -2 ~ 5 7 -1 | 2 0 0 0 | 0 1 1 -3/2 | -2 ~ 0 2 13/2 | 12 0 0 0 | 0 1 1 -3/2 | -2 ~ 0 1 13/4 | 6 0 0 0 | 0 1 0 -19/4 | -8 ~ 0 1 13/4 | 6 0 0 0 | 0 let z = t (where t is arbitrary) then x = -8 + 19z/4 y = 6 - 13z/4 x= (-32+19z)/4 y= (24-13z)/4 in this case |-2 -2 3 | | 5 7 -1 | | 2 2 -3 | =0 >>>cannot use Cramer's rule
Cramer's rule can be used if there is a unique solution in this case determinant=0
Cramer's rule
發問:
Use Cramer s rule to solve each system of equations. If D=0, use another method to determine the solution set. -2x-2y+3z=4 -- (1) 5x+7y-z=2 -- (2) 2x+2y-3z=-4 -- (3) 我知道(1) 同埋 (3) 係一樣, 不過應該點計落去呢? 用Gauss Jordon method?定係...? Correct answer: x= (-32+19z)/4 y= (24-13z)/4 究竟點計ga?
最佳解答:
Gauss Jordon method: -2 -2 3 | 4 5 7 -1 | 2 2 2 -3 | -4 1 1 -3/2 | -2 ~ 5 7 -1 | 2 0 0 0 | 0 1 1 -3/2 | -2 ~ 0 2 13/2 | 12 0 0 0 | 0 1 1 -3/2 | -2 ~ 0 1 13/4 | 6 0 0 0 | 0 1 0 -19/4 | -8 ~ 0 1 13/4 | 6 0 0 0 | 0 let z = t (where t is arbitrary) then x = -8 + 19z/4 y = 6 - 13z/4 x= (-32+19z)/4 y= (24-13z)/4 in this case |-2 -2 3 | | 5 7 -1 | | 2 2 -3 | =0 >>>cannot use Cramer's rule
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其他解答:Cramer's rule can be used if there is a unique solution in this case determinant=0
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