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【简答题】
设二次型f(x1,x2,x3)=2(a1x1+a2x2+a3x3)2+(b1x1+b2x2+b3x3)2,记 。 (1)证明二次型f对应的矩阵为2ααT+ββT; (2)若α,β正交且均为单位,证明f在正交变换下的标准形为2y12+y22.
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