Re: [中學] 克拉瑪
a_1 b_1 c_1 1 0 0 d_1
令 A = [ a_2 b_2 c_2] B = [3 1 0] d = [d_2]
a_3 b_3 c_3 0 -2 1 d_3
1 -1 1
原方程式： A [2] = d => A d = [2]
3 3
x
新方程式： AB [y] = 5d
z
x 1
故 B [y] = 5[2]
z 3
x -1 1 5
[y] = 5 B [2] = [-5].
z 3 5
要記得提醒學生，解一般聯立方程組時，
矩陣列運算 優於 克拉瑪公式，階數越高越有感覺...
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