3阶实对称矩阵A=,B=,其中k1,k2,k3为大于0的任意常数,证明A与B合同,并写出可逆矩阵C,使得CTAC=B.

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问题 3阶实对称矩阵A=,B=,其中k1,k2,k3为大于0的任意常数,证明A与B合同,并写出可逆矩阵C,使得CTAC=B.

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答案A所对应的二次型为 f(x1,x2,x3)=xTAX =(a1+a2+a3)x12+(a2+a3)x22+a3x32+2(a2+a3)x1x2+2a3x1x3+2a3x2x3 =a3(x12+2x1x2+x22+2x1x3+2x2x3+x32)+a2(x12+2x1x2+x22)+a1x12 =a1x12+a2(x1+x2)2+a3(x1+x2+x3)2 [*] 则f(x1,x2,x3)=k3a1y12+k2a2y22+k1a3y32 =(y1,y2,y3)[*] 并将该式记为x=Cy,因∣C∣=[*]≠0,所以C为可逆矩阵,且CTAC=B,故A与B合同,所以所求的可逆矩阵为 C=[*]

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