用配方法化下列二次型为标准型,并求所作的可逆线性变换 f=2x1x2-6x2x3+2x1x2
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原题中 f = 2x1x2 - 6x2x3 + 2x1x2 应为 f = 2x1x2 - 6x2x3 + 2x1x3 吧。
令 x1 = y1+y2, x2 = y1-y2, x3 = y3
则 f = 2x1x2 - 6x2x3 + 2x1x3
= 2(y1)^2 - 2(y2)^2 - 6(y1-y2)y3 + 2(y1+y2)y3
= 2(y1)^2 - 2(y2)^2 - 4y1y3 + 8y2y3
= 2(y1-y3)^2 - 2(y2)^2 - 2(y3)^2 + 8y2y3
= 2(y1-y3)^2 - 2(y2-2y3)^2 + 10(y3)^2
= 2(z1)^2 - 2(z2)^2 + 10(z3)^2
可激蚂键逆明巧线性变换物亏是
z1 = y1-y3 = (x1+x2)/2 - x3
z2 = y2-2y3 = (x1-x2)/2 - 2y3
z3 = y3 = x3
令 x1 = y1+y2, x2 = y1-y2, x3 = y3
则 f = 2x1x2 - 6x2x3 + 2x1x3
= 2(y1)^2 - 2(y2)^2 - 6(y1-y2)y3 + 2(y1+y2)y3
= 2(y1)^2 - 2(y2)^2 - 4y1y3 + 8y2y3
= 2(y1-y3)^2 - 2(y2)^2 - 2(y3)^2 + 8y2y3
= 2(y1-y3)^2 - 2(y2-2y3)^2 + 10(y3)^2
= 2(z1)^2 - 2(z2)^2 + 10(z3)^2
可激蚂键逆明巧线性变换物亏是
z1 = y1-y3 = (x1+x2)/2 - x3
z2 = y2-2y3 = (x1-x2)/2 - 2y3
z3 = y3 = x3
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