數學~行列式單元~特別的題型

(2x-y-z)(2x-y-z)=1
4x^2-2xy-2xz-2xy+y^2+yz-2xz+yz+z^2=1
4x^2+y^2+z^2-4xy-4xz+2yz=1

x(x-y+z)=0=>x^2=xy-xz...(1)
y(x-y+z)=0=>y^2=xy+yz...(2)
z(x-y+z)=0=>z^2=yz-xz...(3)

y^2-x^2=yz+xz...(4) So 2yz=(z^2+y^2-x^2)
4xz=4(yz-z^2)=2(z^2+y^2-x^2)-4z^2=2y^2-2x^2-2z^2
4xy=4(y^2-yz)=4y^2-2(z^2+y^2-x^2)=2y^2+2x^2-2z^2

So 4x^2+y^2+z^2-4xy-4xz+2yz=1 becomes
4x^2+y^2+z^2-(2y^2+2x^2-2z^2)-(2y^2-2x^2-2z^2)+z^2+y^2-x^2=1
4x^2+y^2+z^2-2y^2-2x^2+2z^2-2y^2+2x^2+2z^2+z^2+y^2-x^2=1
3x^2-2y^2+6z^2=1

a+b+c=7

Now

2x - y - z = -1 ... (1)
x - y + z = 0 ... (2)

(1) - (2): x - 2z = -1

x = 2z - 1
y = 3z - 1

a*x^2 + b*y^2 + c*z^2 ≡ 1
a*(2z - 1)^2 + b*(3z - 1)^2 + c*z^2 ≡ 1

(4a + 9b + c)z^2 + (-4a - 6b)z + (a + b) ≡ 1
比較係數得
4a + 9b + c = 0
-4a - 6b = 0
a + b = 1

解得
a = 3, b = -2, c = 6
所以
a + b + c = 7