MHB Solve System of Equations: 4x^2+25y^2+9z^2-10xy-15yz-6xz=0

[kaliprasad]

Solve the system of equations
$4x^2+25y^2 +9z^2 - 10xy -15yz - 6xz = 0\cdots(1)$
$x+y+z=5\cdots(2)$

 

[Albert1]

Solve the system of equations
$4x^2+25y^2 +9z^2 - 10xy -15yz - 6xz = 0\cdots(1)$
$x+y+z=5\cdots(2)$

my solution:

Spoiler

rearrange (1) we have :
$2x(2x-5y)+5y(5y-3z)+3z(3z-2x)=0--(3)$
for any $2x=5y=3z, (3)$ will be zero
so we have :$x:y:z=15:6:10---(4)$
from $(2)$ we have :$x=\dfrac{75}{31},y=\dfrac{30}{31},z=\dfrac{50}{31}$

 

[kaliprasad]

my solution:

Spoiler

rearrange (1) we have :
$2x(2x-5y)+5y(5y-3z)+3z(3z-2x)=0--(3)$
for any $2x=5y=3z, (3)$ will be zero
so we have :$x:y:z=15:6:10---(4)$
from $(2)$ we have :$x=\dfrac{75}{31},y=\dfrac{30}{31},z=\dfrac{50}{31}$

Spoiler

Above is a right solution. kindly find other solutions or prove that no other solution exists

 

[Albert1]

Solve the system of equations
$4x^2+25y^2 +9z^2 - 10xy -15yz - 6xz = 0\cdots(1)$
$x+y+z=5\cdots(2)$

no other solution exists

Spoiler

$(1)\times 2$ we have:
$8x^2+50y^2+18z^2-20xy-30yz-12zx=0$
or $(2x-5y)^2+(5y-3z)^2+(3z-2x)^2=0$
$\therefore 2x=5y=3z$ must hold

 

[I like Serena]

Solve the system of equations
$4x^2+25y^2 +9z^2 - 10xy -15yz - 6xz = 0\cdots(1)$
$x+y+z=5\cdots(2)$

Spoiler

The first equation is a quadric.
The first step is to analyze it and categorize it.
Let:
$$A = \begin{bmatrix}4& -5 & -3 \\ -5 & 25 & -15/2 \\ -3 & -15/2 & 9 \end{bmatrix}$$
Then (1) can be written as:
$$\mathbf x^T A \mathbf x = 0$$

$A$ is a real symmetric matrix, meaning it is diagonalizable with real eigenvalues and an orthonormal set of eigenvectors.
Its characteristic equation is:
$$\lambda^3 - 38\lambda^2 + \frac{1083}{4}\lambda = 0$$
This implies we have eigenvalues $\lambda_1 = 0$, and $\lambda_2, \lambda_3 > 0$.
So:
$$\mathbf x^T A \mathbf x = \mathbf x^T B^T \begin{bmatrix}0& 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix} B \mathbf x = 0$$
where $B$ is an orthogonal matrix formed by the corresponding eigenvectors.

Consequently, we're looking at a degenerate quadric that is similar to:
$$\lambda_2 (y')^2 + \lambda_3 (z')^2 = 0$$
This is a line through the origin, since $\lambda_2, \lambda_3 > 0$.

The direction of the line is given by the eigenvector for $0$, and the line is also the kernel of $A$.
Solving $A\mathbf x=0$, we find:
$$\mathbf x = t\begin{pmatrix}15 \\ 6\\ 10\end{pmatrix}$$

Combining with (2) tells us that:
$$x=\frac{75}{31},y=\frac{30}{31},z=\frac{50}{31}$$

Just like Albert found. ;)