线性代数题急 求一个正交变换X=Py,将二次型f(x1,x2,x3)=5x1^2+5x2^2+2x3^2-8x1x2-4x1x2+4x2x3化为标准型.f(x1,x2,x3)=5x1^2+5x2^2+2x3^2-8x1x2-4x1x3+4x2x3
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![线性代数题急 求一个正交变换X=Py,将二次型f(x1,x2,x3)=5x1^2+5x2^2+2x3^2-8x1x2-4x1x2+4x2x3化为标准型.f(x1,x2,x3)=5x1^2+5x2^2+2x3^2-8x1x2-4x1x3+4x2x3](/uploads/image/z/7795358-62-8.jpg?t=%E7%BA%BF%E6%80%A7%E4%BB%A3%E6%95%B0%E9%A2%98%E6%80%A5+%E6%B1%82%E4%B8%80%E4%B8%AA%E6%AD%A3%E4%BA%A4%E5%8F%98%E6%8D%A2X%3DPy%2C%E5%B0%86%E4%BA%8C%E6%AC%A1%E5%9E%8Bf%28x1%2Cx2%2Cx3%29%3D5x1%5E2%2B5x2%5E2%2B2x3%5E2-8x1x2-4x1x2%2B4x2x3%E5%8C%96%E4%B8%BA%E6%A0%87%E5%87%86%E5%9E%8B.f%28x1%2Cx2%2Cx3%29%3D5x1%5E2%2B5x2%5E2%2B2x3%5E2-8x1x2-4x1x3%2B4x2x3)
线性代数题急 求一个正交变换X=Py,将二次型f(x1,x2,x3)=5x1^2+5x2^2+2x3^2-8x1x2-4x1x2+4x2x3化为标准型.f(x1,x2,x3)=5x1^2+5x2^2+2x3^2-8x1x2-4x1x3+4x2x3
线性代数题急 求一个正交变换X=Py,将二次型f(x1,x2,x3)=5x1^2+5x2^2+2x3^2-8x1x2-4x1x2+4x2x3化为标准型.
f(x1,x2,x3)=5x1^2+5x2^2+2x3^2-8x1x2-4x1x3+4x2x3
线性代数题急 求一个正交变换X=Py,将二次型f(x1,x2,x3)=5x1^2+5x2^2+2x3^2-8x1x2-4x1x2+4x2x3化为标准型.f(x1,x2,x3)=5x1^2+5x2^2+2x3^2-8x1x2-4x1x3+4x2x3
二次型的矩阵 A =
5 -4 -2
-4 5 2
-2 2 2
|A-λE| =
5-λ -4 -2
-4 5-λ 2
-2 2 2-λ
r1+2r3,r2-2r3
1-λ 0 2(1-λ)
0 1-λ -2(1-λ)
-2 2 2-λ
c3+2c2
1-λ 0 2(1-λ)
0 1-λ 0
-2 2 6-λ
= (1-λ)[(1-λ)(6-λ)+4(1-λ)]
= (1-λ)^2(10-λ)
所以 A 的特征值为 λ1=λ2=1,λ3=10.
(A-E)X=0 的基础解系为:a1=(1,1,0)',a2=(1,0,2)'
正交化得:b1=(1,1,0)',b2=(1/2)(1,-1,4)'
单位化得:c1=(1/√2,1/√2,0)',c2=(1/√18,-1/√18,4/√18)'
(A-10E)X=0 的基础解系为:a3=(-2,2,1)'
单位化得:c3=(-2/3,2/3,1/3)'
令P=(c1,c2,c3)=
1/√2 1/√18 -2/3
1/√2 -1/√18 2/3
0 4/√18 1/3
则 P为正交矩阵
X=PY是正交变换,使
f = y1^2+y2^2+10y3^2