配方法化二次型为标准型技巧

羊显枯3555用配方法化二次型为标准型 f(x1,x2,x3)=2x1x2+4x1x3 -
叶慧禄15151089579 ______[答案] f = 2(y1+y2)(y1-y2) + 4(y1+y2)y3 = 2y1^2 - 2y2^2 + 4y1y3 + 4y2y3 = 2(y1+y3)^2 -2y2^2 - 2y3^2 + 4y2y3 = 2(y1+y3)^2 -2(y2-y3)^2 = 2z1^2 - 2z2^2.

羊显枯3555用配方法将下列二次型化为标准形f(x1,x2,…,x2n)=x1x2n+x2x2n - 1+…+xnxn+1. -
叶慧禄15151089579 ______[答案] 令 x1=y1-y2n,x2=y2-y2n-1,…,xn=yn-yn-1x2n=y1+y2n,x2n-1=y2+y2n-1,…,xn+1=yn+yn+1 则 f(x1,x2,…,x2n)=x1x2n+x2x2n-1+…+xnxn+1 =(y1-y2n)(y1+y2n)+(y2-y2n-1)(y2+y2n-1)+…+(yn-yn+1)(yn+yn+1) =y12+…+yn2-yn+12-…-y2n2 所以得解.

羊显枯3555用配方法化二次型为标准型:f=2x1x2 - 2x1x3 - 6x2x3 -
叶慧禄15151089579 ______[答案] f=2x1x2-2x1x3-6x2x3 = 2(y1+y2)(y1-y2)-2(y1+y2)y3-6(y1-y2)y3 = 2y1^2 - 8y1y3 - 2y2^2 + 4y2y3 = 2(y1-2y3)^2 - 2y2^2 + 4y2y3 - 8y3^2 = 2(y1-2y3)^2 - 2(y2-y3)^2 - 6y3^2 = 2z1^2-2z2^2-6z3^2

羊显枯3555线代里怎么把二次型用配方法变换成标准形?应该怎么配,总不会每次都要算一下过度矩阵是不是可逆吧 -
叶慧禄15151089579 ______[答案] 一般来讲配成第k项含有x_k但不含x_1,...,x_{k-1}的形式,换句话说就是每配一项就要消去一个变量(本质上就是Gauss消去法),这样自动满足可逆的要求

羊显枯3555一道二次型题用配方法化下列二次型为标准形,并写出所用的替换矩阵:f(x1,x2,x3)=x1^2+2x3^2+2x1x3 - 2x2x3 -
叶慧禄15151089579 ______[答案] f(x1,x2,x3)=x1^2+2x3^2+2x1x3-2x2x3= (x1+x3)^2 + x3^2-2x2x3= (x1+x3)^2 + (x2-x3)^2 - x2^2= y1^2 + y2^2 - y3^2其中y1=x1+x3y2=x2-x3y3=x2即x1=y1-y3+y2x2=y3x3=y3-y2X=PY, P= 1 1 -1 0 0 1 0 -1 1

羊显枯3555线性代数的题,如图所示.请问如何用配方法把这个二次型化为标准形? -
叶慧禄15151089579 ______ f = 5[(x1)^2+(14/5)x1x2+(4/5)x1x3] + 5(x2)^2 - 4(x3)^2 - 4x2x3= 5[x1+(7/5)x2+(2/5)x3]^2 - (24/5)(x2)^2 - (36/5)(x3)^2 - 4x2x3= 5[x1+(7/5)x2+(2/5)x3]^2 - (24/5)[x2+(5/12)x3]^2 - (191/30)(x3)^2= 5(y1)^2 - (24/5)(y2)^2 - (191/30)(y3)^2

羊显枯3555用配方法化二次型f(x1,x2,x3)=x1平方 - x2平方+x1x2为标准型 -
叶慧禄15151089579 ______[答案] f = (x1+(1/2)x2)^2 - (1/4)x2^2 -x2^2 = (x1+(1/2)x2)^2 - (5/4)x2^2 = y1^2 - 5/4y2^2

羊显枯3555用配方法化二次型为标准型 -
叶慧禄15151089579 ______ =(x1+x2+x3+x4)²

羊显枯3555二次型中含有四个未知数(2X1X2 - 2X3X4)怎么把它化为标准型啊 用配方法 -
叶慧禄15151089579 ______[答案] 2X1X2-2X3X4 = 2(y1+y2)(y1-y2) -2 (y3+y4)(y3-y4) = 2y1^2-2y2^2-2y3^2+2y4^2

羊显枯3555用配方法化二次型为标准形刘老师您好:f(x1,x2,x3)=x1x2+x1x3+5x2x3答案是这样的: x1=y1+y2解 先作如下变换使其出现平方项,令 x2=y1 - y2 x3=y3我不清楚... -
叶慧禄15151089579 ______[答案] 是的.如果a11=0,就可以这样变换出现平方项.这样变换以后就相当于a11=0了,然后配方.再类似的变换使a22=0,最后就变换成标准型.