递际可逆个性二换方X=CY,本了 下面介传时求法 设4化二明故f(x1,x2,x3)=r1+2x1x2+2x1x3+2x2+8x2x3+5n3方定满故,并写出递作的 个性二换 若先将含有x1的给配方 f(x1,x2,x3)=x1+2n1(x2+x3)+(x2+x3)2-(x2+x3)2+2x2+8x2x3+5x3 1x2 再对后三给了含有x2的给配方,则 f(x1,x2,x3)=(x1+x2+x3)2+n2+62x3+9n3-5n3=(x1+n2+x3)2+(x2+3x3)2-53 令 Y=BX,本了 013 方可逆矩.故X=B-1Y.即 92 可将阶二明故化方定满关 设5化二明故∫(x1,x2,x3)=x1x2+x1x3+x2x3方定满故,并写出递际的可逆二换矩矩 若即二明故了不含平方给,先做可逆个性二换,使得含有平方给,令 1=y1+y =91-92 则有 f(ar )=2-v2+(+v)+(y-v)= v2=(+y)2-v2 再令 2&Rk
9A+K0 X = CY , s C =   1 − 1 2 −1 1 1 2 −1 0 0 1   . 5}` .  4 J+> f(x1, x2, x3) = x 2 1 + 2x1x2 + 2x1x3 + 2x 2 2 + 8x2x3 + 5x 2 3 0y>￾ <&￾! 9A+K  7XFT x1 !;0 f(x1, x2, x3) = x 2 1 + 2x1(x2 + x3) + (x2 + x3) 2 − (x2 + x3) 2 + 2x 2 2 + 8x2x3 + 5x 2 3 = (x1 + x2 + x3) 2 + x 2 2 + 6x2x3 + 4x 2 3 ^(I;sFT x2 !;0￾a f(x1, x2, x3) = (x1 + x2 + x3) 2 + x 2 2 + 6x2x3 + 9x 2 3 − 5x 2 3 = (x1 + x2 + x3) 2 + (x2 + 3x3) 2 − 5x 2 3 v    y1 = x1 + x2 + x3 y2 = x2 + 3x3 y3 = x3 O Y = BX, s B =   1 1 1 0 1 3 0 0 1   0k
f> X = B−1Y . O    x1 = y1 − y2 + 2y3 x2 = y2 − 3y3 x3 = y3 kX\+>J0y? y 2 1 + y 2 2 − 5y 2 3 .  5 J+> f(x1, x2, x3) = x1x2 + x1x3 + x2x3 0y>￾ <&R!k
+Kff  O+>sF0;￾7k
9A+K￾ FT0;v    x1 = y1 + y2 x2 = y1 − y2 x3 = y3 aT f(x1, x2, x3) = y 2 1 − y 2 2 + (y1 + y2)y3 + (y1 − y2)y3 = y 2 1 + 2y1y3 − y 2 2 = (y1 + y3) 2 − y 2 2 − y 2 3 ^v    z1 = y1 + y3 z2 = y2 z3 = y3 6