Answer to Question #206193 in Analytic Geometry for SATHIYA Str

Question #206193

reduce x1-2y+3z-4x2x3+6x1x3 to canonical forms


1
Expert's answer
2021-06-16T04:53:35-0400

Reduce the quadratic form Q(x,y,z)=x2+5y2+z2+2xy+2yz+6xzQ(x,y,z)=x^2+5y^2+z^2+2xy+2yz+6xz to canonical form.

The given quadratic form is


Q(x,y,z)=x2+5y2+z2+2xy+2yz+6xzQ(x, y, z)=x^2+5y^2+z^2+2xy+2yz+6xz

The matrix of the given quadratic form is

A=(113151311)A=\begin{pmatrix} 1 & 1 & 3 \\ 1 & 5 & 1\\ 3 & 1 & 1 \\ \end{pmatrix}

AλI=(1λ1315λ1311λ)A-\lambda I=\begin{pmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1\\ 3 & 1 & 1-\lambda \\ \end{pmatrix}

det(AλI)=1λ1315λ1311λ\det(A-\lambda I)=\begin{vmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1\\ 3 & 1 & 1-\lambda \\ \end{vmatrix}

=(1λ)5λ111λ1131λ+315λ31=(1-\lambda)\begin{vmatrix} 5-\lambda& 1 \\ 1 & 1-\lambda \end{vmatrix}-\begin{vmatrix} 1 & 1 \\ 3 & 1-\lambda \end{vmatrix}+3\begin{vmatrix} 1 & 5-\lambda \\ 3 & 1 \end{vmatrix}

=(1λ)(55λλ+λ21)(1λ3)=(1-\lambda)(5-5\lambda-\lambda+\lambda^2-1)-(1-\lambda-3)


+3(115+3λ)=46λ+λ24λ+6λ2λ3+3(1-15+3\lambda)=4-6\lambda+\lambda^2-4\lambda+6\lambda^2-\lambda^3

+2+λ42+9λ+2+\lambda-42+9\lambda

=λ3+7λ236=0=-\lambda^3+7\lambda^2-36=0


λ2(λ6)+(λ6)(λ+6)=0-\lambda^2(\lambda-6)+(\lambda-6)(\lambda+6)=0

(λ6)(λ2λ6)=0-(\lambda-6)(\lambda^2-\lambda-6)=0

(λ6)λ3)(λ+2)=0-(\lambda-6)\lambda-3)(\lambda+2)=0

λ1=6,λ=3,λ3=2\lambda_1=6, \lambda=3, \lambda_3=-2

These are the eigenvalues.

λ=6\lambda=6


(1λ1315λ1311λ)=(513111315)\begin{pmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1\\ 3 & 1 & 1-\lambda \\ \end{pmatrix}=\begin{pmatrix} -5 & 1 & 3 \\ 1 & -1 & 1\\ 3 & 1 & -5 \\ \end{pmatrix}

(513111315)(101012000)\begin{pmatrix} -5 & 1 & 3 \\ 1 & -1 & 1\\ 3 & 1 & -5 \\ \end{pmatrix}\to\begin{pmatrix} 1 & 0 & -1 \\ 0 &1 & -2\\ 0 & 0 & 0 \\ \end{pmatrix}

The eigenvector is v=(121)\vec v=\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}


λ=3\lambda=3


(1λ1315λ1311λ)=(213121312)\begin{pmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1\\ 3 & 1 & 1-\lambda \\ \end{pmatrix}=\begin{pmatrix} -2 & 1 & 3 \\ 1 & 2 & 1\\ 3 & 1 & -2 \\ \end{pmatrix}

(213121312)(101011000)\begin{pmatrix} -2 & 1 & 3 \\ 1 & 2 & 1\\ 3 & 1 & -2 \\ \end{pmatrix}\to\begin{pmatrix} 1 & 0 & -1 \\ 0 &1 & 1\\ 0 & 0 & 0 \\ \end{pmatrix}

The eigenvector is u=(111)\vec u=\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}


λ=2\lambda=-2


(1λ1315λ1311λ)=(313171313)\begin{pmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1\\ 3 & 1 & 1-\lambda \\ \end{pmatrix}=\begin{pmatrix} 3 & 1 & 3 \\ 1 & 7 & 1\\ 3 & 1 & 3 \\ \end{pmatrix}

(313171313)(101010000)\begin{pmatrix} 3 & 1 & 3 \\ 1 & 7 & 1\\ 3 & 1 & 3 \\ \end{pmatrix}\to\begin{pmatrix} 1 & 0 & 1 \\ 0 &1 & 0\\ 0 & 0 & 0 \\ \end{pmatrix}

The eigenvector is w=(101)\vec w=\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}

Form the matrix P,P, whose column ii is eigenvector no. ii


P=(111210111)P=\begin{pmatrix} 1 & 1 & -1 \\ 2 & -1 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix}

Form the diagonal matrix D,D, whose element at row i,i, column ii is eigenvalue no. ii


D=(600030002)D=\begin{pmatrix} 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \\ \end{pmatrix}

The matrices PP and DD are such that


P1AP=DP^{-1}AP=D

Hence the quadratic form is reduced to a sum of squeres, i.e. canonical form:


6x~2+3y~22z~26\tilde{x}^2+3\tilde{y}^2-2\tilde{z}^2

P=(111210111)P=\begin{pmatrix} 1 & 1 & -1 \\ 2 & -1 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix} is the matrix of transformation which is an orthogonal matrix.

The canonical form of the quadratic form is


6x~2+3y~22z~26\tilde{x}^2+3\tilde{y}^2-2\tilde{z}^2


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