a11 a12 a13 x1x1 x2 x3 a21 a22 a23 x2 a31 a32 a33 x3

熱心網(wǎng)友

看了好一會(huì),總算看明白了問題.原題應(yīng)該是:令X=(x1,x2,x3), A=(aij), 1≤i,j≤3, 計(jì)算XAX' (其中X'表示X的轉(zhuǎn)置)不用說,這是x1,x2,x3的2次齊式,答案自然是 XAX'=3∑ (aij)(xi)(xj)=i,j=1a11x1^2+a22x2^2+a33x3^2+(a12+a21)x1x2+(a13+a31)x1x3+(a23+a32)x2x3