Question

The Canonical form of the quadratic
from 3x^2+3y^2+3z^2+2xy+2xz-2yz​

Answer 1

Answer:

Reduce 3x2+5y2+3z2-2yz+2zx-2xy to its canonical form through anorthogonal transformationandfind the rank, signature, index and the nature

Step-by-step explanation:

Answer 2

Answer:

The canonical form of the given quadratic equation is $y_1^{2} +4y_2^{2} +4y_3^{2}$.

Step-by-step explanation:

Given the quadratic equation $3x^2+3y^2+3z^2+2xy+2xz-2yz$

A quadratic equation in matrix form can be written as $X^TAT=0$

$\mbox{where}~ A=\left[\begin{array}{ccc}3&1&1\\1&3&-1\\1&-1&3\end{array}\right], ~X=\left[\begin{array}{c}x_1&x_2&x_3\end{array}\right] ~\mbox{and} ~X^T=\left[\begin{array}{ccc}x_1&x_2&x_3\end{array}\right]$  

The characteristic equation of A can be written as    

$\lambda^{3} -D_1\lambda^{2}+D_2\lambda-D_3=0$ where

the sum of diagonals, $D_1=9$

$D_{2}=\left|\begin{array}{cc}3&1\\1&3\end{array}\right| +\left|\begin{array}{cc}3&-1\\-1&3\end{array}\right| +\left|\begin{array}{cc}3&1\\1&3\end{array}\right|$

     $=8+8+8=24$

And the determinant of  $|A|=D_3=3(9-1)-1(3+1)+1(-1-3)$

$24-4-4=16$

Substituting the values, the characteristic equation is

$\lambda^{3} -9\lambda^{2}+24\lambda-16=0$

Solving the equation, the roots are $\lambda=1,4,4$.

These are called the eigenvalues of the equation.

Now to find the corresponding eigenvectors of each eigenvalue,

let $X=\left[\begin{array}{c}x&y&z\end{array}\right]$ is the eigenvector corresponding to an eigenvalue λ,

satisfies the condition given by $(A-\lambda I)X=0$

where $I$ is the identity matrix.

For the matrix A we have, the condition can be written as

$\left[\begin{array}{ccc}3-\lambda &1&1\\1&3-\lambda&-1\\1&-1&3-\lambda\end{array}\right] \left[\begin{array}{c}x&y&z\end{array}\right] =\left[\begin{array}{c}0&0&0\end{array}\right]$

When $\lambda =1$

$\left[\begin{array}{ccc}2 &1&1\\1&2&-1\\1&-1&2\end{array}\right] \left[\begin{array}{c}x&y&z\end{array}\right] =\left[\begin{array}{c}0&0&0\end{array}\right]$

$\implies 2x+y+z=0~~ \mbox{or}~~x+2y-z=0~~ \mbox{or}~~x-y+2z=0$

Solving for $x, y ~\mbox{and}~z$, we get  $\dfrac{x}{-3} = \dfrac{y}{-(-3)} = \dfrac{z}{3}$

So $X_1=\left(\begin{array}{c}1&-1&-1\end{array}\right)$ is an eigenvector corresponding to $\lambda =1$

Similarly when $\lambda =4$

$\left[\begin{array}{ccc}-1 &1&1\\1&-1&-1\\1&-1&-1\end{array}\right] \left[\begin{array}{c}x&y&z\end{array}\right] =\left[\begin{array}{c}0&0&0\end{array}\right]$

$\implies -x+y+z=0~~ \mbox{or}~~x-y-z=0$

Solving for $x, y ~\mbox{and}~z$, we get $x=0$, $y=1$ and $z=-1$

So, $X_2=\left(\begin{array}{c}0&1&-1\end{array}\right)$ is an eigenvector corresponding to $\lambda =4$

To find the another vector with same $\lambda =4$,

Let us take $X_3$ an orthogonal to $X_2$, so that $X_2X_1^T=0$

that gives $\dfrac{x}{-2} = \dfrac{y}{-1} = \dfrac{z}{-1}$

Hence $X_3=\left(\begin{array}{c}2&1&1\end{array}\right)$ is an another eigenvector corresponding to $\lambda =4$

Modal matrix $N=\left[\begin{array}{ccc}\dfrac{1}{\sqrt{3} } &0&\dfrac{2}{\sqrt{6} }\\\dfrac{-1}{\sqrt{3} }&\dfrac{1}{\sqrt{2} }&\dfrac{1}{\sqrt{6} }\\\dfrac{-1}{\sqrt{3} }&\dfrac{-1}{\sqrt{2} }&\dfrac{1}{\sqrt{6} }\end{array}\right]$

$N^TAN=\left[\begin{array}{ccc}\dfrac{1}{\sqrt{3} } &\dfrac{-1}{\sqrt{3} }&\dfrac{-1}{\sqrt{3} }\\0&\dfrac{1}{\sqrt{2} }&\dfrac{-1}{\sqrt{2} }\\\dfrac{2}{\sqrt{6} }&\dfrac{1}{\sqrt{6} }&\dfrac{1}{\sqrt{6} }\end{array}\right]\left[\begin{array}{ccc}3&1&1\\1&3&-1\\1&-1&3\end{array}\right]\left[\begin{array}{ccc}\dfrac{1}{\sqrt{3} } &0&\dfrac{2}{\sqrt{6} }\\\dfrac{-1}{\sqrt{3} }&\dfrac{1}{\sqrt{2} }&\dfrac{1}{\sqrt{6} }\\\dfrac{-1}{\sqrt{3} }&\dfrac{-1}{\sqrt{2} }&\dfrac{1}{\sqrt{6} }\end{array}\right]$

$=\left[\begin{array}{ccc}1&0&0\\0&4&0\\0&0&4\end{array}\right] =D$

Set $X=NY$ where $Y=\left[\begin{array}{c}y_1&y_2&y_3\end{array}\right]$

Then the quadratic form $Q=X^TAX$

$(NY)^TA(NY)$

$=Y^T(N^TAN)Y$

$=Y^TDY$

$=\left[\begin{array}{ccc}y_1&y_2&y_3\end{array}\right]\left[\begin{array}{ccc}1&0&0\\0&4&0\\0&0&4\end{array}\right] \left[\begin{array}{c}y_1&y_2&y_3\end{array}\right]$

$y_1^{2} +4y_2^{2} +4y_3^{2}$

is the canonical form of the given quadratic equation.