用配方法将二次型化为标准形并求出所用的可逆变换矩阵f=x_1^2+x_2^2+x_3^2+x_4^2+2x_1 x_2-2x_1 x_4-2x

Question

打漏了。。f=x_1^2+x_2^2+x_3^2+x_4^2+2x_1 x_2-2x_1 x_4-2x_2 x_3+2x_3 x_4

Answer 1

f=x1^2+x2^2+x3^2+x4^2+2x1x2-2x1x4-2x2x3+2x3x4
= (x1+x2-x4)^2+x3^2-2x2x3+2x2x4+2x3x4
= (x1+x2-x4)^2+(x3-x2+x4)^2-x2^2-x4^2+4x2x4
= (x1+x2-x4)^2+(x3-x2+x4)^2-(x2-2x4)^2+3x4^2
= y1^2+y2^2-y3^2+3y4^2

y1=x1+x2-x4
y2=x3-x2+x4
y3=x2-2x4
y4=x4
即
x4=y4
x2=y3+2y4
x3=y2+y3+y4
x1=-y3-y4

所以 C=
1 0 -1 -1
0 0 1 2
0 1 1 1
0 0 0 1