设二次型f(x1,x2,x3)=x12+x22+x32+2ax1x2+2ax1x3+2ax2x3经可逆线性变换 22得g(y1,y2,y3)=y12+y22+4y32+2y1y2. 求可逆矩阵P.

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问题 设二次型f(x1,x2,x3)=x12+x22+x32+2ax1x2+2ax1x3+2ax2x3经可逆线性变换
22得g(y1,y2,y3)=y12+y22+4y32+2y1y2
求可逆矩阵P.

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答案f(x1,x2,x3)=(x1-x2/2一x3/2)2+3(x2-x3)2/4, [*] z=P1x,则 f(x1,x2,x3)=z12+z22由g(y1,y2,y3)=(y1+y2)2+4y32, [*] Z=P2Y,则f(y1,y2,y3)=z12十z22.故P1X=P2Y,即X=P1-1P2Y,所以P=P1-1P2.其中 [*] 所以P=P1-1P2=[*]

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