Question #197041

Reduce the quadratic form to a canonical form and find its nature

1) 2xy+2yz+2zx


1
Expert's answer
2021-06-01T05:56:22-0400

Given question is 2xy+2yz+2zx

The general equation a1x2+a2y2+a3z2+a4xy+a5yz+a6zxa_1x^2+a_2y^2+a_3z^2+a_4 xy+a_5yz+a_6 zx

For the given equation, a1=a2=a3=0a_1=a_2=a_3=0

a4=2,a5=2,a6=2a_4=2, a_5=2, a_6=2

So, matrix A=[011101110]A=\begin{bmatrix} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix}

After the matrix operation,

Let λ\lambda be the eigen value of the equation, so AλI=0A-\lambda I=0

Given question is 2xy+2yz+2zx

The general equation a1x2+a2y2+a3z2+a4xy+a5yz+a6zxa_1x^2+a_2y^2+a_3z^2+a_4 xy+a_5yz+a_6 zx

For the given equation, a1=a2=a3=0a_1=a_2=a_3=0

a4=2,a5=2,a6=2a_4=2, a_5=2, a_6=2

So, matrix A=[011101110]A=\begin{bmatrix} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix}

Let the eigen value for the equation be λ,\lambda, From the characteristic equation,

AλI=0|A-\lambda I|=0

[λ111λ111λ]=0\begin{bmatrix} -\lambda & 1 & 1\\ 1 & -\lambda & 1 \\ 1 & 1 & -\lambda \\ \end{bmatrix}=0

After the expansion of the matrix,

λ3+3λ2=0\Rightarrow -\lambda^3+3\lambda -2=0

λ33λ+2=0\Rightarrow \lambda^3-3\lambda +2=0

(λ+1)(λ+1)(λ2)=0(\lambda +1)(\lambda +1)(\lambda -2)=0

λ=1,λ=1,λ=2\lambda =-1, \lambda =-1, \lambda =2

Now, substituting the eigen values,

λ=2,\lambda = 2, x1=[111]x_1=\begin{bmatrix} 1\\ 1 \\ 1 \\ \end{bmatrix}

λ=1,x2=[110]\lambda=-1, x_2=\begin{bmatrix} -1\\ 1 \\ 0 \\ \end{bmatrix}

λ=1,x3=[011]\lambda=-1, x_3=\begin{bmatrix} 0\\ 1 \\ -1 \\ \end{bmatrix}


Now, P=(y1,y2,y2)P=(y_1,y_2,y_2)

Applying the transpose of matrix P,

Pt=[y1y2y3]P^t=\begin{bmatrix} y_1\\ y_2 \\ y_3 \end{bmatrix}

Now, multiplying the P , A and transpose matrix of P

PAPt=[y1y2y3][211111111][y1y2y3]PAP^t=\begin{bmatrix} y_1 & y_2 & y_3 \\ \end{bmatrix}\begin{bmatrix} -2 & 1 & 1\\ 1 & -1 & 1 \\ 1 & 1 & -1 \\ \end{bmatrix}\begin{bmatrix} y_1\\ y_2 \\ y_3 \end{bmatrix}

=[2y1+y2+y3y1y2+y3y1+y2y3][y1y2y3]=\begin{bmatrix} -2y_1 +y_2 + y_3 &y_1-y_2+y_3 & y_1+y_2-y_3 \end{bmatrix}\begin{bmatrix} y_1\\ y_2 \\ y_3 \end{bmatrix}

=2y12+y1y2+y1y3+y1y2y22+y2y3+y1y3+y2y3y32=-2y_1^2+y_1y_2+y_1y_3+y_1y_2-y_2^2+y_2y_3+y_1y_3+y_2y_3-y_3^2

=(y12+y22+y322y1y22y2y32y3y1)=-(y_1^2+y_2^2+y_3^2-2y_1y_2-2y_2y_3-2y_3y_1)

=(y1y2y3)2=-(y_1-y_2-y_3)^2

The above equation is in the canonical form.

Rank =3

Index = 1



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS