1.34. Reduce the equation of the surface to canonical form

$$
3 x^{2}+4 y^{2}+5 z^{2}+4 x y-4 y z+2 x-8 y+14 z=16
$$
Solution. The quadratic form defining the given equation has the form

$$
F(x, y)=3 x^{2}+4 y^{2}+5 z^{2}+4 x y-4 y z
$$

The corresponding matrix $\tilde{A}$ of the quadratic form is:

$$
\tilde{A}=\left(\begin{array}{ccc}
3 & 2 & 0 \\
2 & 4 & -2 \\
0 & -2 & 5
\end{array}\right)
$$

To find the eigenvalues, we form the characteristic equation:

$$
\left|\begin{array}{ccc}
(3-\lambda) & 3 & 0 \\
2 & (4-\lambda) & -2 \\
0 & -2 & (5-\lambda)
\end{array}\right|=0 \Rightarrow \lambda^{3}-12 \lambda^{2}+39 \lambda-28=0 \Rightarrow
$$

We will look for the roots of this equation among the divisors of the free term 28. We have, for $\lambda_{1}=1, f(\lambda)=\lambda^{3}-12 \lambda^{2}+39 \lambda-28=0$. Dividing the polynomial:

$$
\left(\lambda^{3}-12 \lambda^{2}+39 \lambda-28\right):(\lambda-1)=\lambda^{2}-11 \lambda+28
$$

from which $\lambda^{2}-11 \lambda+28=0$ and $\lambda_{2}=4, \lambda_{3}=7$. The quadratic form in the new coordinate system $O x^{\prime} y^{\prime} z^{\prime}$ will be written as follows: $F\left(x^{\prime}, y^{\prime}, z^{\prime}\right)=\left(x^{\prime}\right)^{2}+4\left(y^{\prime}\right)^{2}+7\left(z^{\prime}\right)^{2}$. Let's find the eigenvector for $\lambda_{1}=1$. We form the system:

$$
\left\{\begin{array} { l } 
{ ( 5 - 1 ) x ^ { \prime } + 2 y ^ { \prime } + 0 \cdot z ^ { \prime } = 0 , } \\
{ 2 x ^ { \prime } + ( 4 - 1 ) y ^ { \prime } - 2 z ^ { \prime } = 0 , } \\
{ 0 \cdot x ^ { \prime } - 2 y ^ { \prime } + ( 5 - 1 ) z ^ { \prime } = 0 , }
\end{array} \Rightarrow \left\{\begin{array}{l}
x^{\prime}+y^{\prime}=0 \\
2 x^{\prime}+3 y^{\prime}-2 z^{\prime}=0 \\
y^{\prime}-2 z^{\prime}=0
\end{array}\right.\right.
$$

Since the determinant of the system is zero, and the minor formed by the elements of the first and last rows

$$
\left|\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right|=1 \neq 0
$$

then the second equation is a consequence of the first and third equations. The free unknown here is $y^{\prime}$, and the basic ones are $x^{\prime}, z^{\prime}$. Discarding the second equation and solving the resulting system of the remaining equations with respect to $x^{\prime}, z^{\prime}$, after moving $y^{\prime}$ to the right-hand sides, we get $x^{\prime}=-y^{\prime}$, $z^{\prime}=0.5 y^{\prime}$. Setting $y^{\prime}=m$, where $m$ is any non-zero real number, we find the coordinates of the first eigenvector: $x^{(1)}=-m \vec{i}+m \vec{j}+0.5 m \vec{k}$. Similarly, for $\lambda_{2}=1$ and $\lambda_{3}=7$, we get the eigenvectors $x^{(2)}=m(\vec{i}+0.5 \vec{j}+\vec{k}), x^{(3)}=m(\vec{i}+2 \vec{j}-2 \vec{k})$. We normalize them by dividing by their magnitudes:

$$
\begin{aligned}
& \left|x^{(1)}\right|=m \sqrt{1+1+0.25}=1.5 m ; \quad \tilde{x}^{(1)}=\frac{x^{(1)}}{\left|x^{(1)}\right|}=-\frac{2}{3} \vec{i}+\frac{2}{3} \vec{j}+\frac{1}{3} \vec{k} \\
& \left|x^{(2)}\right|=m \sqrt{1+0.25+1}=1.5 m ; \quad \tilde{x}^{(2)}=\frac{x^{(2)}}{\left|x^{(2)}\right|}=\frac{2}{3} \vec{i}+\frac{1}{3} \vec{j}+\frac{2}{3} \vec{k} \\
& \left|x^{(3)}\right|=m \sqrt{1+4+4}=3 m ; \quad \tilde{x}^{(3)}=\frac{x^{(3)}}{\left|x^{(3)}\right|}=\frac{1}{3} \vec{i}+\frac{2}{3} \vec{j}-\frac{2}{3} \vec{k} .
\end{aligned}
$$

We write down the matrix of the linear transformation and the transformation itself:

$$
\tilde{B}=\left(\begin{array}{ccc}
-\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\
\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\
\frac{1}{3} & \frac{2}{3} & -\frac{2}{3}
\end{array}\right), \begin{aligned}
& x=-\frac{2}{3} x^{\prime}+\frac{2}{3} y^{\prime}+\frac{1}{3} z^{\prime} \\
& y=\frac{2}{3} x^{\prime}+\frac{1}{3} y^{\prime}+\frac{2}{3} z^{\prime} \\
& z=\frac{1}{3} x^{\prime}+\frac{2}{3} y^{\prime}-\frac{2}{3} z^{\prime}
\end{aligned}
$$

Substitute these dependencies into the given equation of the surface:

$$
\begin{aligned}
& \left(x^{\prime}\right)^{2}+4\left(y^{\prime}\right)^{2}+7\left(z^{\prime}\right)^{2}+2\left(-\frac{2}{3} x^{\prime}+\frac{2}{3} y^{\prime}+\frac{1}{3} z^{\prime}\right)- \\
& -8\left(\frac{2}{3} x^{\prime}+\frac{1}{3} y^{\prime}+\frac{2}{3} z^{\prime}\right)+14\left(\frac{1}{3} x^{\prime}+\frac{2}{3} y^{\prime}-\frac{2}{3} z^{\prime}\right)=16
\end{aligned}
$$

We complete the squares in the obtained equation:

$$
\begin{gathered}
\left(\left(x^{\prime}\right)^{2}-2 x^{\prime}+1-1\right)+4\left(\left(y^{\prime}\right)^{2}+2 y^{\prime}+1-1\right)+ \\
+7\left(\left(z^{\prime}\right)^{2}-2 z^{\prime}+1-1\right)=16 .
\end{gathered}
$$

The formulas $x^{\prime \prime}=x^{\prime}-1, y^{\prime \prime}=y^{\prime}+1, z^{\prime \prime}=z^{\prime}-1$ allow us to perform a parallel translation of the coordinate axes to the new origin of coordinates $(1 ;-1 ; 1)$. Finally, we obtain the following canonical equation of the surface:

$$
\frac{\left(x^{\prime \prime}\right)^{2}}{28}+\frac{\left(y^{\prime \prime}\right)^{2}}{7}+\frac{\left(z^{\prime \prime}\right)^{2}}{4}=1,
$$

which is the equation of an ellipsoid.

## 1.9 .  PROBLEMS