设f(x1x2x3)=x1²+X2²+X3²+4X1X2+4X1X3+4X2X3(1)求一正交变换化f为标准型(2)判定f的正定性
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![设f(x1x2x3)=x1²+X2²+X3²+4X1X2+4X1X3+4X2X3(1)求一正交变换化f为标准型(2)判定f的正定性](/uploads/image/z/9449658-18-8.jpg?t=%E8%AE%BEf%EF%BC%88x1x2x3%EF%BC%89%3Dx1%26%23178%3B%2BX2%26%23178%3B%2BX3%26%23178%3B%2B4X1X2%2B4X1X3%2B4X2X3%EF%BC%881%EF%BC%89%E6%B1%82%E4%B8%80%E6%AD%A3%E4%BA%A4%E5%8F%98%E6%8D%A2%E5%8C%96f%E4%B8%BA%E6%A0%87%E5%87%86%E5%9E%8B%EF%BC%882%EF%BC%89%E5%88%A4%E5%AE%9Af%E7%9A%84%E6%AD%A3%E5%AE%9A%E6%80%A7)
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设f(x1x2x3)=x1²+X2²+X3²+4X1X2+4X1X3+4X2X3(1)求一正交变换化f为标准型(2)判定f的正定性
设f(x1x2x3)=x1²+X2²+X3²+4X1X2+4X1X3+4X2X3
(1)求一正交变换化f为标准型
(2)判定f的正定性
设f(x1x2x3)=x1²+X2²+X3²+4X1X2+4X1X3+4X2X3(1)求一正交变换化f为标准型(2)判定f的正定性
f的矩阵A=
1 2 2
2 1 2
2 2 1
|A-λE| = (5-λ)(1+λ)^2.
所以A的特征值为 5,-1,-1
(A-5E)X = 0 的基础解系为:a1 = (1,1,1)'
(A+E)X = 0 的基础解系为:a2 = (1,-1,0)',a3 = (1,0,-1)'
将 a2,a3 正交化得 b2 = (1,-1,0)',b3 = (1/2,1/2,-1)'
单位化得
c1 = (1/√3,1/√3,1/√3)',
c2 = (1/√2,-1/√2,0)',
c3 = (1/√6,1/√6,-2/√6)'
令矩阵P = (c1,c2,c3),则P为正交矩阵,
且 P^(-1)AP = diag(5,-1,-1).
正交变换 X=PY,f = 5y1^2-y2^2-y3^2.
f不是正定的,也不负定的.
=x1²等于X6