Given quadratic form is x 1 2 + 2 x 2 x 3 x_1^2+2x_2x_3 x 1 2 + 2 x 2 x 3
compare the given quadratic expression with standard quadratic form,
a x 1 2 + b y 1 2 + c z 1 2 + 2 f y z + 2 g z x + 2 h x y ax_1^2+by_1^2+cz_1^2+2fyz+2gzx+2hxy a x 1 2 + b y 1 2 + c z 1 2 + 2 f yz + 2 g z x + 2 h x y
we get,
a=1,b=0,c=0,f=1,g=0 and h=0
Transforming the quadratic equation into matrix form
So the required matrix is ,
A = [ a h g h b f g f c ] A=\begin{bmatrix}
a & h &g\\
h& b&f\\
g&f&c
\end{bmatrix} A = ⎣ ⎡ a h g h b f g f c ⎦ ⎤ , A = [ 1 0 0 0 0 1 0 1 0 ] A=\begin{bmatrix}
1 & 0 &0\\
0& 0&1\\
0&1&0
\end{bmatrix} A = ⎣ ⎡ 1 0 0 0 0 1 0 1 0 ⎦ ⎤
Using characterstics Equation ( A − λ I ) X = 0 (A-\lambda I)X=0 ( A − λ I ) X = 0
To calculate the values of λ \lambda λ
∣ A − λ I ∣ = 0 |A-\lambda I|=0 ∣ A − λ I ∣ = 0
∣ 1 0 0 0 0 1 0 1 0 ∣ − λ ∣ 1 0 0 0 1 0 0 0 1 ∣ = 0 \begin{vmatrix}
1 & 0 &0\\
0& 0&1\\
0&1&0
\end{vmatrix}-\lambda \begin{vmatrix}
1& 0 &0\\
0& 1&0\\
0&0&1
\end{vmatrix}=0 ∣ ∣ 1 0 0 0 0 1 0 1 0 ∣ ∣ − λ ∣ ∣ 1 0 0 0 1 0 0 0 1 ∣ ∣ = 0
∣ 1 − λ 0 0 0 − λ 1 0 1 − λ ∣ \begin{vmatrix}
1-\lambda & 0 &0\\
0& -\lambda&1\\
0&1&-\lambda
\end{vmatrix} ∣ ∣ 1 − λ 0 0 0 − λ 1 0 1 − λ ∣ ∣ =0
Using Factorization method
( 1 − λ ) ( λ 2 − 1 ) = 0 (1-\lambda)(\lambda^2-1)=0 ( 1 − λ ) ( λ 2 − 1 ) = 0
so λ = 1 , − 1 , 1 \lambda=1,-1,1 λ = 1 , − 1 , 1 (these are the required eigen values)
S0 the required canonical form is
λ 1 x 2 + λ 2 y 2 + λ 3 z 2 = 0 \lambda_1x^2+\lambda_2y^2+\lambda_3z^2=0 λ 1 x 2 + λ 2 y 2 + λ 3 z 2 = 0
x 2 − y 2 + z 2 = 0 x^2-y^2+z^2=0 x 2 − y 2 + z 2 = 0
This is the required canonical form.
Comments
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6x2+3y2+3z2-4xy-2yz+4xz reduce the quadratic from to canonical and find it's nature