Question #288768

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

1
Expert's answer
2022-01-20T08:39:47-0500

2x12+2x1x22x1x3+x224x2x3+x32=0a11=1,a12=2,a13=1,a14=0,a22=1,a23=1,a24=0,a33=2,a34=0,a44=0The variant of the equation when converting coordinates are determinants I1=a11+a22+a33I2=a11a12a12a22+a22a23a23a33+a11a13a13a33I3=a11a12a13a12a22a23a13a23a33I(λ)=a11λa12a13a12a22λa23a13a23a33λk2=a11a14a14a44+a12a24a24a44+a33a34a34a44k3=a11a12a14a12a22a24a14a24a44+a22a23a24a13a33a34a14a34a44+a11a13a14a13a33a34a14a34a44 We haveI1=4I2=1221+1112+1112=1I3=121211112=4I(λ)=1λ2121λ1112λI(λ)=λ3+4λ2+λ4k2=0,k3=0λ34λ2λ+4=0by solving the polynomial, we getλ1=4,λ2=1,λ3=1the canonical form isx12x224x32=02x_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


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