二次型f(x1,x2,x3)=5x12+5x22+cx32-2x1x2-6x2x3+6x1x3的秩为2,求c及此二次型的规范形,并写出相应的变换.

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问题 二次型f(x1,x2,x3)=5x12+5x22+cx32-2x1x2-6x2x3+6x1x3的秩为2,求c及此二次型的规范形,并写出相应的变换.

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答案二次型矩阵A=[*],由二次型的秩为2,即矩阵A的秩r(A)=2,则有 |A|=24(c-3)=0 => c=3. 用配方法求规范形和所作变换. f(x1,x2,x3)=5x12+5x22+3x32-2x1x2+6x1x3-6x2x3 =3(x3+x1-x2)2-3(x1-x2)2+5x12+5x22-2x1x2 =3(x1-x2+x3)2+2x12+2x22+4x1x2 =3(x1-x2+x3)2+2(x1+x2)2 [*] 则f(x1,x2,x3)=y12+y22,为规范二次型. [*]

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