Question

线性代数配方

又请您出山了！发烦您能给我讲讲这类题都怎么配方么

第二题





Answer 1

这题从形式上看是标准形, 但是变换是不可逆的
1 1 0
0 1 -1
1 0 1
行列式等于0
所以要打开括号重新配方
f =
2x1^2+2x2^2+2x3^2+2x1x2+2x1x3-2x2x3
= 2(x1+(1/2)x2+(1/2)x3)^2+(3/2)x2^2+(3/2)x3^2-3x2x3
= 2(x1+(1/2)x2+(1/2)x3)^2+(3/2)(x2-x3)^2
= 2y1^2+(3/2)y2^2
所以 p=2, q=0

Answer 2

1.     f = 2(x1)^2+2(x2)^2+2(x3)^2+2x1x2-2x2x3+2x1x3
     = 2[x1+(1/2)x2+(1/2)x3]^2+(3/2)(x2)^2+(3/2)(x3)^2-3x2x3
     = 2[x1+(1/2)x2+(1/2)x3]^2+(3/2)(x2-x3)^2
     =2(y1)^2+(3/2)(y2)^2,     则 正惯性指数 p=2， 负惯性指数 q=0.

2. 没有参数 a  啊 ？ 题目不完全