求非退化线性替换，把二次型 f(χ1,χ2,χ3,χ4)=3χ1²+χ2²-2χ3²-6χ1χ2+12χ2χ3-2χ2 20
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最后一项是 -2x2x4 吧
f(x1,x2,x3,x4)=3x1^2+x2^2-2x3^2-6x1x2+12x2x3-2x2x4
= 3(x1-x2)^2 -2x2^2-2x3^2+12x2x3-2x2x4
= 3(x1-x2)^2 -2(x2-3x3+1/2x4)^2+16x3^2+1/2x4^2 -6x3x4
= 3(x1-x2)^2 -2(x2-3x3+1/2x4)^2+(4x3-3/4x4)^2-1/16x4^2
= 3y1^2 -2y2^2+y3^2-1/16y4^2
y1=x1-x2
y2=x2-3x3+1/2x4
y3=4x3-3/4x4
y4=x4
f(x1,x2,x3,x4)=3x1^2+x2^2-2x3^2-6x1x2+12x2x3-2x2x4
= 3(x1-x2)^2 -2x2^2-2x3^2+12x2x3-2x2x4
= 3(x1-x2)^2 -2(x2-3x3+1/2x4)^2+16x3^2+1/2x4^2 -6x3x4
= 3(x1-x2)^2 -2(x2-3x3+1/2x4)^2+(4x3-3/4x4)^2-1/16x4^2
= 3y1^2 -2y2^2+y3^2-1/16y4^2
y1=x1-x2
y2=x2-3x3+1/2x4
y3=4x3-3/4x4
y4=x4
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