线性代数 05.07 用合同变换法化二次型为标准形

§第五章第七节用合同变换法化二次型为标准形 \color{blue}{\S 第五章第七节用合同变换法化二次型为标准形}

定义10.若对方阵A作一次初等行变换,接着对所得矩阵作一次同种的初等列变换,就称对A进行一次合同变换. 定义10.若对方阵A作一次初等行变换,接着对所得矩阵\\ 作一次同种的初等列变换,就称对A进行一次合同变换.
用合同变换法化二次型为标准形的实质是:利用可逆线性变换x=Cy,把f=x T Ax化为标准形,即f=x T Ax=(Cy) T ACy=y T C T ACy=y T Λy只需C T AC=Λ.又因C=P 1 P 2 ⋯P s ,其中P 1 ,P 2 ,⋯,P s 均为初等方阵.所以(P 1 P 2 ⋯P s ) T AP 1 P 2 ⋯P s =Λ即P T s ⋯P T 2 P T 1 AP 1 P 2 ⋯P s =Λ(1)而P T s ⋯P T 2 P T 1 =P T s ⋯P T 2 P T 1 E=C T (2)结合(1)和(2),得出将Λ化成对角矩阵,同时求出可逆矩阵C: 用合同变换法化二次型为标准形的实质是: \\ 利用可逆线性变换x = Cy,把 f = x^TAx化为标准形,即\\ f = x^TAx = (Cy)^TACy = y^TC^T ACy = y^T \Lambda y \\ 只需C^TAC = \Lambda. \\ 又因 C = P_1P_2 \cdots P_s,其中P_1, P_2, \cdots, P_s均为初等方阵. \\ 所以\\ (P_1P_2 \cdots P_s)^T A P_1 P_2 \cdots P_s = \Lambda \\ 即 \\ P_s^T \cdots P_2^T P_1^T A P_1 P_2 \cdots P_s = \Lambda \quad (1) \\ 而 P_s^T \cdots P_2^T P_1^T = P_s^T \cdots P_2^T P_1^T E = C^T \quad (2) \\ 结合(1)和(2),得出将\Lambda化成对角矩阵,同时求出可逆矩阵C:
(A|E)A合同变换⟶E作行变换 (Λ|C T ) (A | E) \left. \begin{array}{l} A合同变换 \\ \quad \longrightarrow \\ E作行变换 \end{array} \right. (\Lambda | C^T)
求出C T ,作可逆线性变换x=Cy,则该变换将f化为标准形.f=k 1 y 2 1 +k 2 y 2 2 +⋯+k r y 2 r . 求出C^T,作可逆线性变换x = Cy,则该变换将f化为标准形. \\ f = k_1y_1^2 + k_2y_2^2 + \cdots + k_ry_r^2.

例3.利用合同变换将f=2x 1 x 2 +2x 1 x 3 −2x 1 x 4 −2x 2 x 3 +2x 2 x 4 +2x 3 x 4 化为标准形,并写出所用的可逆线性变换. 例3.利用合同变换将 \\ f = 2x_1x_2 + 2x_1x_3 -2x_1x_4 - 2x_2x_3 + 2x_2x_4 + 2x_3x_4 \\ 化为标准形,并写出所用的可逆线性变换.
解:(A|E)=⎛ ⎝ ⎜ ⎜ ⎜ 011−1 10−11 1−101 −1110 1000 0100 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 111−1 10−11 0−101 0110 1000 1100 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2100 10−11 0−101 0110 1000 1100 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2200 10−11 0−201 0210 1000 1200 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2200 20−22 0−201 0210 1000 1200 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2000 2−2−22 0−201 0210 1−100 1100 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2000 0−2−22 0−201 0210 1−100 1100 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2000 0−202 0−221 02−10 1−110 11−10 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2000 0−202 002−1 02−10 1−110 11−10 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2000 0−200 002−1 02−12 1−11−1 11−11 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2000 0−200 002−1 00−12 1−11−1 11−11 0010 0001 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2000 0−200 002−2 00−28 1−11−2 11−12 0010 0002 ⎞ ⎠ ⎟ ⎟ ⎟ ∼⎛ ⎝ ⎜ ⎜ ⎜ 2000 0−200 0020 0006 1−11−1 11−11 0011 0002 ⎞ ⎠ ⎟ ⎟ ⎟ 于是,求得:C T =⎛ ⎝ ⎜ ⎜ ⎜ 1−11−1 11−11 0011 0002 ⎞ ⎠ ⎟ ⎟ ⎟ ,C=⎛ ⎝ ⎜ ⎜ ⎜ 1100 −1100 1−110 −1112 ⎞ ⎠ ⎟ ⎟ ⎟ 作可逆变换x=Cy,即⎛ ⎝ ⎜ ⎜ ⎜ x 1 x 2 x 3 x 4 ⎞ ⎠ ⎟ ⎟ ⎟ =⎛ ⎝ ⎜ ⎜ ⎜ 1100 −1100 1−110 −1112 ⎞ ⎠ ⎟ ⎟ ⎟ ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ y 1 y 2 y 3 y 4 ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ,把f化成标准形:f=2y 2 1 −2y 2 2 +2y 2 3 +6y 2 4 . 解:\\ (A | E) = \begin{pmatrix} 0 & 1 & 1 & -1 & 1 & 0 & 0 & 0 \\ 1 & 0 & -1 & 1 & 0 & 1 & 0 & 0 \\ 1 & -1 & 0 & 1 & 0 & 0 & 1 & 0 \\ -1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & -1 & 1 & 0 & 1 & 0 & 0 \\ 1 & -1 & 0 & 1 & 0 & 0 & 1 & 0 \\ -1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & -1 & 1 & 0 & 1 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 2 & 0 & -2 & 2 & 0 & 2 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 2 & 0 & 0 & 1 & 1 & 0 & 0 \\ 2 & 0 & -2 & 2 & 0 & 2 & 0 & 0 \\ 0 & -2 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 2 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & -2 & 2 & -1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & -2 & 2 & -1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & -2 & 2 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -1 & 1 & -1 & 1 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 2 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -1 & 1 & -1 & 1 & 0 \\ 0 & 2 & -1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 2 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -1 & 1 & -1 & 1 & 0 \\ 0 & 0 & -1 & 2 & -1 & 1 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -1 & 1 & -1 & 1 & 0 \\ 0 & 0 & -1 & 2 & -1 & 1 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -2 & 1 & -1 & 1 & 0 \\ 0 & 0 & -2 & 8 & -2 & 2 & 0 & 2 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 & 1 & -1 & 1 & 0 \\ 0 & 0 & 0 & 6 & -1 & 1 & 1 & 2 \\ \end{pmatrix} \\ 于是,求得: \\ C^T = \begin{pmatrix} 1 & 1 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 1 & -1 & 1 & 0 \\ -1 & 1 & 1 & 2 \\ \end{pmatrix} , C = \begin{pmatrix} 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} \\ 作可逆变换x = Cy,即 \\ \begin{pmatrix}x_1 \\ x_2 \\ x_3 \\x_4 \end{pmatrix} = \begin{pmatrix} 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} \begin{pmatrix}y_1 \\ y_2 \\ y_3 \\y_4 \end{pmatrix}, \\ 把f化成标准形: \\ f = 2y_1^2 - 2y_2^2 + 2y_3^2 + 6y_4^2 .

例4.设二次型f=4x 2 1 +3x 2 2 +3x 2 3 +2x 2 x 3 1)用正交变换化f为标准形;2)用配方法化f为标准形;3)用合同法化f为标准形. 例4.设二次型 \\ f = 4x_1^2 + 3x_2^2 + 3x_3^2 + 2x_2x_3 \\ 1) 用正交变换化f为标准形;\\ 2) 用配方法化f为标准形;\\ 3) 用合同法化f为标准形.
解:1)用正交变换化f为标准形.①把f写成矩阵形式:f(x 1 ,x 2 ,x 3 )=⎛ ⎝ ⎜ 400 031 013 ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ y 1 y 2 y 3  ⎞ ⎠ ⎟ ②通过|A−λE|=0,求A的全部特征值.|A−λE|=∣ ∣ ∣ ∣ 4−λ00 03−λ1 013−λ ∣ ∣ ∣ ∣ =(4−λ)[(3−λ) 2 −1]=−(λ−2)(λ−4) 2 =0解得A的特征值为:λ 1 =2,λ 2 =λ 3 =4.③通过(A−λE)x=0,求A的特征向量.当λ 1 =2时，(A−2E)x=0,由A−2E=⎛ ⎝ ⎜ 200 011 011 ⎞ ⎠ ⎟ ∼⎛ ⎝ ⎜ 100 010 010 ⎞ ⎠ ⎟ 解得基础解系ξ 1 =⎛ ⎝ ⎜ 0−11 ⎞ ⎠ ⎟ ,单位化p 1 =12  √  ⎛ ⎝ ⎜ 0−11 ⎞ ⎠ ⎟ 当λ 2 =λ 3 =4时,(A−4E)x=0,由A−4E=⎛ ⎝ ⎜ 000 0−11 01−1 ⎞ ⎠ ⎟ ∼⎛ ⎝ ⎜ 000 010 0−10 ⎞ ⎠ ⎟ 解得基础解系:ξ 2 =⎛ ⎝ ⎜ 100 ⎞ ⎠ ⎟ ,ξ 3 =⎛ ⎝ ⎜ 011 ⎞ ⎠ ⎟ ,因为ξ 2 与ξ 3 正交,直接单位化得:p 1 =⎛ ⎝ ⎜ 100 ⎞ ⎠ ⎟ ,p 2 =12  √  ⎛ ⎝ ⎜ 011 ⎞ ⎠ ⎟ ,④把p 1 ,p 2 ,p 3 拼成正交变换矩阵P,作正交变换x=Py,即P=(p 1 ,p 2 ,p 3  )=⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ 0−12  √  12  √   100 012  √  12  √   ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎛ ⎝ ⎜ x 1 x 2 x 3  ⎞ ⎠ ⎟ =⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ 0−12  √  12  √   100 012  √  12  √   ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎛ ⎝ ⎜ y 1 y 2 y 3  ⎞ ⎠ ⎟ ⑤把f化为标准形:f=2y 2 1 +4y 2 2 +4y 2 3 2)用配方法化f为标准形f=4x 2 1 +3x 2 2 +3x 2 3 +2x 2 x 3 =4x 2 1 +2(x 2 +x 3 ) 2 +(x 2 −x 3 ) 2   解: 1) 用正交变换化f为标准形. ①把f写成矩阵形式: \\ f(x_1, x_2, x_3) = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 1 & 3 \\ \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \\ ②通过 |A - \lambda E | = 0,求A的全部特征值. \\ |A - \lambda E | = \begin{vmatrix} 4 - \lambda & 0 & 0 \\ 0 & 3 - \lambda & 1 \\ 0 & 1 & 3 - \lambda \\ \end{vmatrix} \\ = (4 - \lambda)[(3 - \lambda)^2 - 1] \\ = -(\lambda - 2)(\lambda - 4)^2 = 0 解得A的特征值为: \lambda_1 = 2, \lambda_2 = \lambda_3 = 4. \\ ③通过(A - \lambda E) x = 0, 求A的特征向量. \\ 当\lambda_1 = 2时，(A - 2E) x = 0,由 \\ A - 2E = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix} \\ 解得基础解系 \xi_1 = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}, 单位化 p_1 = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} \\ 当\lambda_2 = \lambda_3 = 4时, (A - 4E) x = 0,由 \\ A - 4E = \begin{pmatrix} 0 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \\ \end{pmatrix} \sim \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{pmatrix} \\ 解得基础解系: \\ \xi_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \xi_3 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, 因为\xi_2与\xi_3正交,直接单位化得: \\ p_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, p_2 = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \\ ④ 把p_1,p_2,p_3拼成正交变换矩阵P, 作正交变换x = Py,即\\ P = \begin{pmatrix} p_1 , p_2, p_3 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ -\dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \end{pmatrix} \\ \begin{pmatrix} x_1 \\ x_2 \\x _3 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ -\dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \\y _3 \end{pmatrix} \\ ⑤ 把f化为标准形: \\ f = 2y_1^2 + 4y_2^2 + 4y_3^2 \\ 2) 用配方法化f为标准形 \\ f = 4x_1^2 + 3x_2^2 + 3x_3^2 + 2x_2x_3 \\ = 4x_1^2 + 2(x_2 + x_3)^2 + (x_2 - x_3)^2
令⎧ ⎩ ⎨ ⎪ ⎪ y 1 =x 1 y 2 =x 2 +x 3 y 3 =x 2 −x 3    令 \left \{ \begin{array}{l} y_1 = x_1 \\ y_2 = x_2 + x_3 \\ y_3 = x_2 - x_3 \end{array} \right.
即⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x 1 =y 1 x 2 =12 y 2 +12 y 3 x 3 =12 y 2 −12 y 3    即\left \{ \begin{array}{l} x_1 = y_1 \\ x_2 = \dfrac{1}{2}y_2 + \dfrac{1}{2} y_3 \\ x_3 = \dfrac{1}{2}y_2 - \dfrac{1}{2}y_3 \end{array} \right.
即(x 1  x 2  x 3  )=⎛ ⎝ ⎜ ⎜ ⎜ ⎜ 100 012 12  012 −12  ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎛ ⎝ ⎜ y 1 y 2 y 3  ⎞ ⎠ ⎟ f为标准形为:f=4y 2 1 +2y 2 2 +y 2 3 3)用合同法化f为标准形.(A|E)=⎛ ⎝ ⎜ 400 031 013 100 010 001 ⎞ ⎠ ⎟ ∼⎛ ⎝ ⎜ 400 033 019 100 010 003 ⎞ ⎠ ⎟ ∼⎛ ⎝ ⎜ 400 033 0327 100 010 003 ⎞ ⎠ ⎟ ∼⎛ ⎝ ⎜ 400 030 0324 100 01−1 003 ⎞ ⎠ ⎟ ∼⎛ ⎝ ⎜ 400 030 0024 100 01−1 003 ⎞ ⎠ ⎟ 于是,得A=⎛ ⎝ ⎜ 400 030 0024 ⎞ ⎠ ⎟ C T =⎛ ⎝ ⎜ 100 01−1 003 ⎞ ⎠ ⎟ ,C=⎛ ⎝ ⎜ 100 010 0−13 ⎞ ⎠ ⎟ ,从而c=Cy,即⎛ ⎝ ⎜ x 1 x 2 x 3  ⎞ ⎠ ⎟ =⎛ ⎝ ⎜ 100 010 0−13 ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ y 1 y 2 y 3  ⎞ ⎠ ⎟ ,则该变换将f化成标准形为:f=4y 2 1 +3y 2 2 +24y 2 3   即 \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \dfrac{1}{2} & \dfrac{1}{2} \\ 0 & \dfrac{1}{2} & -\dfrac{1}{2} \\ \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \\ f为标准形为: f = 4y_1^2 + 2y_2^2 + y_3^2 \\ 3) 用合同法化f为标准形. \\ (A | E) = \begin{pmatrix} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 & 1 & 0 \\ 0 & 1 & 3 & 0 & 0 & 1 \\ \end{pmatrix} \sim \begin{pmatrix} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 & 1 & 0 \\ 0 & 3 & 9 & 0 & 0 & 3 \\ \end{pmatrix} \sim \begin{pmatrix} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 3 & 0 & 1 & 0 \\ 0 & 3 & 27 & 0 & 0 & 3 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 3 & 0 & 1 & 0 \\ 0 & 0 & 24 & 0 & -1 & 3 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 & 1 & 0 \\ 0 & 0 & 24 & 0 & -1 & 3 \\ \end{pmatrix} \\ 于是,得 \\ A = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 24 \\ \end{pmatrix} \\ C^T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 3 \\ \end{pmatrix}, C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 3 \\ \end{pmatrix}, \\ 从而c = Cy, 即 \\ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 3 \\ \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}, \\ 则该变换将f化成标准形为: \\ f = 4y_1^2 + 3y_2^2 + 24y_3^2