We shall reduce the quadratic form x2+3y2+3z2−2yz to the canonical form through the orthogonal transformation.
NOTE: WE SHALL RE-REPRESENT THE VARAIBALE x, y and z TO VARIABLES x1,x2andx3 respectively.
NOW, using x1,x2andx3
Q(x)=x12+3x22+3x32−2x2x3
The symmetric matrix A with the given quadratic form is
A= ⎣⎡10003−10−13⎦⎤
The characteristic equation of A is
∣(A−λI)∣=λ3−7λ2+14λ−8=0⟹λ=1,2,4
The corresponding eigen values are:
for λ1=1,isX1=k⎣⎡100⎦⎤
for λ2=2,isX2=k⎣⎡011⎦⎤
for λ3=1,isX3=k⎣⎡01−1⎦⎤
Clearly, the eigen vectors X1=⎣⎡100⎦⎤ , X2=⎣⎡011⎦⎤ and X3=⎣⎡01−1⎦⎤ are linearly independent and orthogonal.
The normalized vectors of the above eigen vextors X1,X2,X3 are:
e1=∣∣X1∣∣X1=⎣⎡100⎦⎤ , e2=∣∣X2∣∣X2=⎣⎡02121⎦⎤ , e3=∣∣X3∣∣X3=⎣⎡021−21⎦⎤
Therefore,
The normalized model matrix is
pˉ=⎣⎡10002121021−21⎦⎤ and pˉ−1=pˉT=21⎣⎡20001101−1⎦⎤
pˉ−1Apˉ=21⎣⎡20001101−1⎦⎤⎣⎡ 10003−10−13⎦⎤⎣⎡10002121021−21⎦⎤
=⎣⎡ 10001101−1⎦⎤=D
We know that the quadratic form is Q=XTAX where X=PˉY
then,
Q=(PˉY)A(PˉY)
=(YTPˉT)A(PˉY)=YT(PTAPˉ)Y=YT(Pˉ−1APˉ)YsincePˉT=Pˉ−1=YTDY
Therefore, the canonical form
=[y1y2y3]⎣⎡ 100020004⎦⎤⎣⎡ y1y2y3⎦⎤
=y12+2y22+4y32
which gives the canonical form.
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