Given question is 2xy+2yz+2zx
The general equation a1x2+a2y2+a3z2+a4xy+a5yz+a6zx
For the given equation, a1=a2=a3=0
a4=2,a5=2,a6=2
So, matrix A=⎣⎡011101110⎦⎤
After the matrix operation,
Let λ be the eigen value of the equation, so A−λI=0
Given question is 2xy+2yz+2zx
The general equation a1x2+a2y2+a3z2+a4xy+a5yz+a6zx
For the given equation, a1=a2=a3=0
a4=2,a5=2,a6=2
So, matrix A=⎣⎡011101110⎦⎤
Let the eigen value for the equation be λ, From the characteristic equation,
∣A−λI∣=0
⎣⎡−λ111−λ111−λ⎦⎤=0
After the expansion of the matrix,
⇒−λ3+3λ−2=0
⇒λ3−3λ+2=0
(λ+1)(λ+1)(λ−2)=0
λ=−1,λ=−1,λ=2
Now, substituting the eigen values,
λ=2, x1=⎣⎡111⎦⎤
λ=−1,x2=⎣⎡−110⎦⎤
λ=−1,x3=⎣⎡01−1⎦⎤
Now, P=(y1,y2,y2)
Applying the transpose of matrix P,
Pt=⎣⎡y1y2y3⎦⎤
Now, multiplying the P , A and transpose matrix of P
PAPt=[y1y2y3]⎣⎡−2111−1111−1⎦⎤⎣⎡y1y2y3⎦⎤
=[−2y1+y2+y3y1−y2+y3y1+y2−y3]⎣⎡y1y2y3⎦⎤
=−2y12+y1y2+y1y3+y1y2−y22+y2y3+y1y3+y2y3−y32
=−(y12+y22+y32−2y1y2−2y2y3−2y3y1)
=−(y1−y2−y3)2
The above equation is in the canonical form.
Rank =3
Index = 1
Comments