Answer to Question #288768 in Linear Algebra for pree

Question

find the canonical form of the quadratic form 2x₁² +x₂² + x₃² +2x₁x₂ – 2x₁x₃ – 4x₂x₃

Answer 1

$2x_1^2+2x_1x_2-2x_1x_3+x_2^2-4x_2x_3+x_3^2=0\\ a_{11}=1,a_{12}=-2,a_{13}=-1,a_{14}=0,a_{22}=1,\\a_{23}=1,a_{24}=0,a_{33}=2,a_{34}=0,a_{44}=0\\ \text{The variant of the equation when converting coordinates are determinants }\\ I_1=a_{11}+a_{22}+a_{33}\\ I_2=\begin{vmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{vmatrix}+\begin{vmatrix} a_{22} & a_{23} \\ a_{23} & a_{33} \end{vmatrix}+\begin{vmatrix} a_{11} & a_{13} \\ a_{13} & a_{33} \end{vmatrix}\\ I_3=\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{vmatrix}\\ I(\lambda)=\begin{vmatrix} a_{11}-\lambda & a_{12} & a_{13} \\ a_{12} & a_{22}-\lambda & a_{23} \\ a_{13} & a_{23} & a_{33}-\lambda \end{vmatrix}\\ k_2=\begin{vmatrix} a_{11} & a_{14} \\ a_{14} & a_{44} \end{vmatrix}+\begin{vmatrix} a_{12} & a_{24} \\ a_{24} & a_{44} \end{vmatrix}+\begin{vmatrix} a_{33} & a_{34} \\ a_{34} & a_{44} \end{vmatrix}\\ k_3=\begin{vmatrix} a_{11} & a_{12} & a_{14} \\ a_{12} & a_{22} & a_{24} \\ a_{14} & a_{24} & a_{44} \end{vmatrix}+\begin{vmatrix} a_{22} & a_{23} & a_{24} \\ a_{13} & a_{33} & a_{34} \\ a_{14} & a_{34} & a_{44} \end{vmatrix}+\begin{vmatrix} a_{11} & a_{13} & a_{14} \\ a_{13} & a_{33} & a_{34} \\ a_{14} & a_{34} & a_{44} \end{vmatrix}\\~We~ have\\ I_1=4\\ I_2=\begin{vmatrix} 1 & -2 \\ -2 & 1 \end{vmatrix}+\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix}+\begin{vmatrix} 1 & -1 \\ -1 & 2 \end{vmatrix}=-1\\ I_3=\begin{vmatrix} 1 & -2 & -1 \\ -2 &1 & 1 \\ -1 & 1 & 2 \end{vmatrix}=-4\\ I(\lambda)=\begin{vmatrix} 1-\lambda & -2 & -1 \\ -2 &1-\lambda & 1 \\ -1 & 1 & 2-\lambda \end{vmatrix}\\ I(\lambda)=-\lambda^3+4\lambda^2+\lambda-4\\ k_2=0,k_3=0\\ \lambda^3-4\lambda^2-\lambda+4=0\\ by~solving~the~polynomial,~we~get\\ \lambda_1=4 ,\lambda_2=1, \lambda_3=-1\\ the~canonical~form~is\\ x_1^2-x_2^2-4x_3^2=0$