Answer to Question #206193 in Analytic Geometry for SATHIYA Str

Question

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

Answer 1

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

The given quadratic form is

$Q(x, y, z)=x^2+5y^2+z^2+2xy+2yz+6xz$

The matrix of the given quadratic form is

$A=\begin{pmatrix} 1 & 1 & 3 \\ 1 & 5 & 1\\ 3 & 1 & 1 \\ \end{pmatrix}$

$A-\lambda I=\begin{pmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1\\ 3 & 1 & 1-\lambda \\ \end{pmatrix}$

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

$=(1-\lambda)\begin{vmatrix} 5-\lambda& 1 \\ 1 & 1-\lambda \end{vmatrix}-\begin{vmatrix} 1 & 1 \\ 3 & 1-\lambda \end{vmatrix}+3\begin{vmatrix} 1 & 5-\lambda \\ 3 & 1 \end{vmatrix}$

$=(1-\lambda)(5-5\lambda-\lambda+\lambda^2-1)-(1-\lambda-3)$

$+3(1-15+3\lambda)=4-6\lambda+\lambda^2-4\lambda+6\lambda^2-\lambda^3$

$+2+\lambda-42+9\lambda$

$=-\lambda^3+7\lambda^2-36=0$

$-\lambda^2(\lambda-6)+(\lambda-6)(\lambda+6)=0$

$-(\lambda-6)(\lambda^2-\lambda-6)=0$

$-(\lambda-6)\lambda-3)(\lambda+2)=0$

$\lambda_1=6, \lambda=3, \lambda_3=-2$

These are the eigenvalues.

$\lambda=6$

$\begin{pmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1\\ 3 & 1 & 1-\lambda \\ \end{pmatrix}=\begin{pmatrix} -5 & 1 & 3 \\ 1 & -1 & 1\\ 3 & 1 & -5 \\ \end{pmatrix}$

$\begin{pmatrix} -5 & 1 & 3 \\ 1 & -1 & 1\\ 3 & 1 & -5 \\ \end{pmatrix}\to\begin{pmatrix} 1 & 0 & -1 \\ 0 &1 & -2\\ 0 & 0 & 0 \\ \end{pmatrix}$

The eigenvector is $\vec v=\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$

$\lambda=3$

$\begin{pmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1\\ 3 & 1 & 1-\lambda \\ \end{pmatrix}=\begin{pmatrix} -2 & 1 & 3 \\ 1 & 2 & 1\\ 3 & 1 & -2 \\ \end{pmatrix}$

$\begin{pmatrix} -2 & 1 & 3 \\ 1 & 2 & 1\\ 3 & 1 & -2 \\ \end{pmatrix}\to\begin{pmatrix} 1 & 0 & -1 \\ 0 &1 & 1\\ 0 & 0 & 0 \\ \end{pmatrix}$

The eigenvector is $\vec u=\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$

$\lambda=-2$

$\begin{pmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1\\ 3 & 1 & 1-\lambda \\ \end{pmatrix}=\begin{pmatrix} 3 & 1 & 3 \\ 1 & 7 & 1\\ 3 & 1 & 3 \\ \end{pmatrix}$

$\begin{pmatrix} 3 & 1 & 3 \\ 1 & 7 & 1\\ 3 & 1 & 3 \\ \end{pmatrix}\to\begin{pmatrix} 1 & 0 & 1 \\ 0 &1 & 0\\ 0 & 0 & 0 \\ \end{pmatrix}$

The eigenvector is $\vec w=\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$

Form the matrix $P,$ whose column $i$ is eigenvector no. $i$

$P=\begin{pmatrix} 1 & 1 & -1 \\ 2 & -1 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix}$

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

$D=\begin{pmatrix} 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \\ \end{pmatrix}$

The matrices $P$ and $D$ are such that

$P^{-1}AP=D$

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

$6\tilde{x}^2+3\tilde{y}^2-2\tilde{z}^2$

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

The canonical form of the quadratic form is

$6\tilde{x}^2+3\tilde{y}^2-2\tilde{z}^2$