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Quadratic equation given by

$5{\mathrm{x}}^2+3{\mathrm{y}}^2+2{\mathrm{z}}^2+4\mathrm{xz}=14$

A scalar associated with a given linear transformation of a vector space and having the property that there is some nonzero vector which when multiplied by the scalar is equal to the vector obtained by the transformation operate on the vector especially: a root of the characteristic equation of a matrix.

For the following quadratic equation given by

$5x^2+3y^2+2z^2+4xz=14$

Which represent a quadric surface in the form $A{x}^2+2Hxy+B{y}^2+C{z}^2+2Ryz+2Txz=K$where $A,B,C,H,R,T$and K are constants.

To use the diagonalization process, transform the following quadratic equation given in equation (1) in a matrix form defined by

$\left(\begin{array}{ccc}x& y& z\end{array}\right)\left(\begin{array}{ccc}A& H& T\\ H& B& R\\ T& R& C\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=KOr\left(\begin{array}{ccc}x& y& z\end{array}\right)M\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=K$ Or

HereM represent the$3脳3$ square matrix defined by

$M=\left(\begin{array}{ccc}A& H& T\\ H& B& R\\ T& R& C\end{array}\right)$

Therefore, the following quadratic equation given in equation can be written in a matrix form using equation (2) which reads

$\left(\begin{array}{ccc}x& y& z\end{array}\right)\left(\begin{array}{ccc}5& 0& 2\\ 0& 3& 0\\ 2& 0& 2\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=14\mathrm{Where}M\mathit{=}\left(\begin{array}{ccc}A& H& T\\ H& B& R\\ T& R& C\end{array}\right)\left(\begin{array}{ccc}5& 0& 2\\ 0& 3& 0\\ 2& 0& 2\end{array}\right)$

And K=14

For the following $3脳3$ square matrix M defined by

$M=\left(\begin{array}{ccc}5& 0& 2\\ 0& 3& 0\\ 2& 0& 2\end{array}\right)$

To find its eigenvalues, find the values $位$ of which satisfy the characteristic equation of the matrix M, thus the values of $位$must satisfy $det(M-位I)=|M-位I|=0$where I is the $3脳3$square identity matrix.

Compute the matrix $M-位I$implies that

$$\begin{gathered} M-位I=M=\left(\begin{array}{ccc}5& 0& 2\\ 0& 3& 0\\ 2& 0& 2\end{array}\right)-位\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right) \\ =\left(\begin{array}{ccc}5& 0& 2\\ 0& 3& 0\\ 2& 0& 2\end{array}\right)-\left(\begin{array}{ccc}位& 0& 0\\ 0& 位& 0\\ 0& 0& 位\end{array}\right) \\ =\left(\begin{array}{ccc}5-位& 0& 2\\ 0& 3-位& 0\\ 2& 0& 2-位\end{array}\right) \end{gathered}$$

Here the matrix $M-位I$is just equal to M with $位$subtracted from each element on the main diagonal elements.

Determinants of order three are determined in terms of second-order determinants using the following formula

$\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|=a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|+a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|$

$$\begin{gathered} det\left(M-位I\right)=\left|\begin{array}{ccc}5-位& 0& 2\\ 0& 3-位& 0\\ 2& 0& 2-位\end{array}\right| \\ =\left(5-位\right)\left|\begin{array}{cc}3-位& 0\\ 0& 2-位\end{array}\right|-\left(0\right)\left|\begin{array}{cc}0& 0\\ 2& 2-位\end{array}\right|+\left(2\right)\left|\begin{array}{cc}0& 3-位\\ 2& 0\end{array}\right| \end{gathered}$$

Determinant of order two is defined use the formula

$$\begin{gathered} \left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|=a_{11}{a}_{22}-a_{12}{a}_{21} \\ =\left(5-位\right)\left[\left(3-位\right)\left(2-位\right)-\left(0\right)\left(0\right)\right]-0+2\left[\left(0\right)\left(0\right)-\left(3-位\right)\left(2\right)\right] \\ =\left(5-位\right)\left(3-位\right)\left(2-位\right)-4\left(3-位\right) \\ =\left(3-位\right)\left[\left(5-位\right)\left(2-位\right)-4\right] \\ =\left(3-位\right)\left[10-5位-2位+{位}^2-4\right] \end{gathered}$$

Solve further

$$\begin{gathered} =\left(3-位\right)\left[6-7位+{位}^2\right] \\ =\left(3-位\right)\left[{位}^2-7位+6\right] \end{gathered}$$

Thus,

$det\left(M-位I\right)=\left(3-位\right)\left({位}^2-7位+6\right)$

In the next step find the solution of the quadratic equation ${位}^2-7位+6=0$as follows

To find the roots of the quadratic equation of the form $a{x}^2+bx+c=0$, compute the value of $螖=b^2-4ac,$then find the roots exist and are equal to $x=\frac{-b卤\sqrt{b^2-4ac}}{2a}$, so that

For the quadratic equation

$$\begin{gathered} {位}^2-7位+6=0 \\ 鈭�=b^2-4ac \\ ={\left(-7\right)}^2-4\left(1\right)\left(6\right) \\ =25>0 \\ {位}^2-7位+6=0\mathrm{Implies}\mathrm{that} \\ \mathrm{位}=\frac{7卤\sqrt{{\left(-7\right)}^2-4\left(1\right)\left(6\right)}}{2\left(1\right)} \\ =\frac{7卤\sqrt{25}}{2\left(1\right)} \\ =\frac{7卤5}{2\left(1\right)} \\ \mathrm{位}=\frac{7+5}{2\left(1\right)} \end{gathered}$$

Solve further

=6

And$\begin{array}{c}位=\frac{7-5}{2\left(1\right)}\\ =1\end{array}$

Therefore, computing$det(M-位I)$implies that

$det(M-位I)=(3-位)(位-1)(位-6)$

Thus,

$(3-位)(位-1)(位-6)=0$

Finally, find the solution of$det(M-位I)=0$(of the cubic equation$(3-位)(位-1)(位-6)=0)$as follows

The root of the following cubic equation implies that

$位=1,位=3\mathrm{and}位=6$

Therefore, the eigenvalues of a square matrixM are $位=1,3,6$.

In the next step, choose the principal axes of the quadric surface as our reference axes as follows

Assume that the axes$(x^{\text{'}},y^{\text{'}},z^{\text{'}})$are rotated by an angle$胃$ from the original axes $(x,y,z)$.

Thus, using the relation defined by

$\left(\begin{array}{ccc}{x}^{\text{'}}& y^{\text{'}}& z^{\text{'}}\end{array}\right)C^{-1}MC\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right)=K$

HereC is an orthogonal rotational matrix in three dimensions defined by

$C=\left(\begin{array}{ccc}\cos \left(胃\right)& -\sin \left(胃\right)& 0\\ \sin \left(胃\right)& \cos \left(胃\right)& 0\\ 0& 0& 1\end{array}\right)$

If the matrixC is the matrix which diagonalizes M, then equation (3) represents the equation of the quadric surface relative to its principal axes.

Therefore, choose Cto be the matrix which diagonalizes M, found that

$C^{-1}MC=\left(\begin{array}{ccc}{位}_1& 0& 0\\ 0& {位}_2& 0\\ 0& 0& {位}_3\end{array}\right)$

Here D represent the diagonal matrix whose diagonal elements are the eigen values of $3脳3$the square matrixM andC represent an orthogonal rotational matrix whose columns are the components of the unit eigenvectors.

From the values of the eigenvalues of a square matrix M which are $位=1,3,6$, deduce that

$$\begin{gathered} C^{-1}MC=\left(\begin{array}{ccc}{位}_1& 0& 0\\ 0& {位}_2& 0\\ 0& 0& {位}_3\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& 6\end{array}\right) \end{gathered}$$

Therefore, the quadric surface equation relative to the principal axes$(x^{\text{'}},y^{\text{'}},z^{\text{'}})$becomes

$$\begin{gathered} \left(\begin{array}{ccc}{x}^{\text{'}}& y^{\text{'}}& z^{\text{'}}\end{array}\right)C^{-1}MC\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right)=k \\ \left(\begin{array}{ccc}{x}^{\text{'}}& y^{\text{'}}& z^{\text{'}}\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& 6\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right)=14 \end{gathered}$$

Thus, we finally have shown that the quadric surface equation relative to the principal axes $(x^{\text{'}},y^{\text{'}},z^{\text{'}})$in a matrix form is

$\left(\begin{array}{ccc}{x}^{\text{'}}& y^{\text{'}}& z^{\text{'}}\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& 6\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right)=14$

Now, we must simplify the following equation (4) defined in a matrix form multiplied out gives the quadric surface equation as follows

Suppose that the matrix having 1 row and 3 columns, the matrix $M_{33}$having 3 rows and 3 columns, and the matrix $M_{31}$having 3 rows and 1 column given by

$M_{13}=\left(\begin{array}{ccc}{x}^{\text{'}}& y^{\text{'}}& z^{\text{'}}\end{array}\right),M_{33}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& 6\end{array}\right)$

and

$M_{31}=\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right)$

Firstly, the product $M_{13}{M}_{33}$between matrix $M_{13}$having 1 row and 3 columns and matrix $M_{33}$having 3 rows and 3 columns is the matrix having 1 row and 3 columns given by

$$\begin{gathered} M_{13}{M}_{33}=\left(\left(x^{\text{'}}\right)\left(1\right)+\left(y^{\text{'}}\right)\left(0\right)+\left(z^{\text{'}}\right)\left(0\right)\left(x^{\text{'}}\right)\left(0\right)+\left(y^{\text{'}}\right)\left(3\right)+\left(z^{\text{'}}\right)\left(0\right)\left(x^{\text{'}}\right)\left(0\right)+\left(y^{\text{'}}\right)\left(0\right)+\left(z^{\text{'}}\right)\left(0\right)\right) \\ =\left(\left(x^{\text{'}}\right)\left(1\right)+\left(y^{\text{'}}\right)\left(0\right)+\left(z^{\text{'}}\right)\left(0\right)\left(x^{\text{'}}\right)\left(0\right)+\left(y^{\text{'}}\right)\left(3\right)+\left(z^{\text{'}}\right)\left(0\right)\left(x^{\text{'}}\right)\left(0\right)+\left(y^{\text{'}}\right)\left(0\right)+\left(z^{\text{'}}\right)\left(6\right)\right) \\ =\left(\begin{array}{ccc}{x}^{\text{'}}& 3y^{\text{'}}& 6z^{\text{'}}\end{array}\right) \end{gathered}$$

Thus, shown that the equation (4) defined in a matrix form becomes

$$\begin{gathered} \left(\begin{array}{ccc}{x}^{\text{'}}& y^{\text{'}}& z^{\text{'}}\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& 6\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right)=14 \\ \left(\begin{array}{ccc}{x}^{\text{'}}& 3y^{\text{'}}& 6z^{\text{'}}\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right)=14 \end{gathered}$$

Suppose that the matrix having 1 row and 3 columns given by

$N_{13}=\left(\begin{array}{ccc}{x}^{\text{'}}& 3y^{\text{'}}& 6z^{\text{'}}\end{array}\right)$

Finally, the product $N_{13}{M}_{31}$between matrix $N_{13}$having 1 row and 3 columns and matrix $M_{31}$having 3 rows and 1 column is the matrix having 1 row and 1 column given by

$$\begin{gathered} N_{13}{M}_{31}={\left(\begin{array}{ccc}{x}^{\text{'}}& 3y^{\text{'}}& 6z^{\text{'}}\end{array}\right)}_{1脳3}{\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right)}_{3脳1} \\ =\left(\begin{array}{ccc}{x}^{\text{'}}& 3y^{\text{'}}& 6z^{\text{'}}\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right) \\ =\left(x^{\text{'}}\right)\left(x^{\text{'}}\right)+\left(3y^{\text{'}}\right)\left(y^{\text{'}}\right)+\left(6z^{\text{'}}\right)\left(z^{\text{'}}\right) \end{gathered}$$

Solve further

$$\begin{gathered} =x^{\text{'}2}+3y^{\text{'}2}+6z^{\text{'}2}+0x^{\text{'}}{y}^{\text{'}}+0y^{\text{'}}{z}^{\text{'}}+0x^{\text{'}}{z}^{\text{'}} \\ =鈥�x^{\text{'}2}+3y^{\text{'}2}+6z^{\text{'}2} \end{gathered}$$

Thus, finally have shown that the following equation (4) defined in a matrix form

$$\begin{gathered} \left(\begin{array}{ccc}{x}^{\text{'}}& y^{\text{'}}& z^{\text{'}}\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& 6\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right)=14 \\ 鈬� \\ \mathrm{becomes}\left(\begin{array}{ccc}{\mathrm{x}}^{\text{'}}& 3{\mathrm{y}}^{\text{'}}& 6{\mathrm{z}}^{\text{'}}\end{array}\right)\left(\begin{array}{c}{\mathrm{x}}^{\text{'}}\\ {\mathrm{y}}^{\text{'}}\\ {\mathrm{z}}^{\text{'}}\end{array}\right)=14 \\ 鈬� \\ {\mathrm{x}}^{\text{'}2}+3{\mathrm{y}}^{\text{'}2}+6{\mathrm{z}}^{\text{'}2}=14 \end{gathered}$$

Therefore, simplify the equation (4) defined in a matrix form given by

$\left(\begin{array}{ccc}{x}^{\text{'}}& y^{\text{'}}& z^{\text{'}}\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& 6\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\end{array}\right)=14$

Multiplied out gives the quadric surface equation defined by $x^{\text{'}2}+3y^{\text{'}2}+6^{\text{'}2}=14$

Therefore, shown that the quadric surface equation relative to the principal axes $\left(x^{\text{'}},y^{\text{'}},z^{\text{'}}\right)\mathrm{is}{x}^{\text{'}2}+3y^{\text{'}2}+6z^{\text{'}2}=14$