Question #287337

3x1²+3x2²+3x3²+2x1x2+2x1x3-2x2x3

1
Expert's answer
2022-01-14T11:25:08-0500

Assume the question is as follows:

Reduce the quadratic form 3x12+3x22+3x32+2x1x2+2x1x32x2x33 x_{1}^{2}+3 x_{2}^{2}+3 x_{3}^{2}+2 x_{1} x_{2}+2 x_{1} x_{3}-2 x_{2} x_{3} into 'a sum of squares' by an orthogonal transformation and give the matrix of transformation.

Solution:

On comparing the given quadratic with the general quadratic ax2+by2+cz2+2fyz+2gzx+2hxya x^{2}+b y^{2}+c z^{2}+2 f y z +2 g z x+2 h x y , the matrix is given by

A=[ahghbfgfc]=[311131113]A=\left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right]=\left[\begin{array}{rrr} 3 & 1 & 1 \\ 1 & 3 & -1 \\ 1 & -1 & 3 \end{array}\right]

The desired characteristic equation becomes

AλI=3λ1113λ1113λ=0|A-\lambda I|=\left|\begin{array}{ccc} 3-\lambda & 1 & 1 \\ 1 & 3-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{array}\right|=0

which is a cubic in λ\lambda and has three values viz., 1,4,4.

Hence the desired canonical form i.e., 'a sum of squares' is x2+4y2+4z2x^{2}+4 y^{2}+4 z^{2} . Solving [AλI][X]=0[A-\lambda I][X]=0 for three values of λ\lambda

For λ=1\lambda=1 , we have [211121112]\left[\begin{array}{rrr}2 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right] [x1y1z1]=0 or2x1+y1+z1=0x1+2y1z1=0},i.e.x112=y11+2=z141=k\left[\begin{array}{l}x_{1} \\ y_{1} \\ z_{1}\end{array}\right]=0 \ or \left.\quad \begin{array}{l}2 x_{1}+y_{1}+z_{1}=0 \\ x_{1}+2 y_{1}-z_{1}=0\end{array}\right\}, i.e. \frac{x_{1}}{-1-2}=\frac{y_{1}}{1+2}=\frac{z_{1}}{4-1}=k \\ \therefore

[x1y1z1]=[kkk]=[111]\left[\begin{array}{l} x_{1} \\ y_{1} \\ z_{1} \end{array}\right]=\left[\begin{array}{r} -k \\ k \\ k \end{array}\right]=\left[\begin{array}{r} -1 \\ 1 \\ 1 \end{array}\right]

Similarly for λ=4,[111111111][xyz]=0\lambda=4,\left[\begin{array}{rrr}-1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & -1 & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=0

We have two linearly independent vectors X2=[110],X3=[101]X_{2}=\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right], X_{3}=\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right]

As the transformation has to be an orthogonal one, therefore to obtain ' P ', first divide each elements of a corresponding eigen vector by the square root of sum of the squares of its respective elements and then express as [X Y Z]

Hence the matrix of transformation, P=[1312121312013012]P=\left[\begin{array}{ccc}\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{3}} & 0 & \frac{1}{\sqrt{2}}\end{array}\right]


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