System of Homogeneous Equations
A homogeneous system is a set of equations in which every equation is homogeneous, meaning that each one equals zero and all its terms share the same total degree.
In simple terms, a homogeneous equation is one where every term is of the same degree and the equation itself equals zero.
How do you find the degree of a monomial? The degree of a monomial is the sum of the exponents of its variables. For example, the monomial xy2 has degree three because x has an exponent of one and y has an exponent of two (1 + 2 = 3).
If all the terms are first-degree, the system is called a homogeneous linear system.
A practical example
Example 1
This system is homogeneous because both equations are homogeneous and equal to zero:
$$ \begin{cases} 3x + 2y = 0 \\ \\ x - y = 0 \end{cases} $$
Let’s look at each equation separately:
- The first equation is homogeneous of the first degree since both terms (3x and 2y) are first-degree monomials.
- The second equation is also homogeneous of the first degree because its terms (x and -y) are first-degree monomials.
Since both are homogeneous, the system itself is homogeneous of the first degree.
Note. In this case, it’s also a homogeneous linear system because both equations are linear.
Example 2
Now consider a system that’s homogeneous of the fourth degree:
$$ \begin{cases} y^2 -2xy -3x^2 = 0 \\ \\ x^2+3xy +2y^2 = 0 \end{cases} $$
Let’s analyze the equations one by one:
- The first equation is homogeneous of the second degree, because all its terms (y2, -2xy, -3x2) are quadratic and it equals zero.
- The second equation is also homogeneous of the second degree, since every term (x2, 3xy, 2y2) is quadratic and the equation equals zero.
Both equations are homogeneous, so the system is homogeneous of the fourth degree.
Note. The degree of a system equals the product of the degrees of its equations. Here, both are quadratic, so 2 × 2 = 4.
How to solve a homogeneous system
The first step is to check for a trivial solution by setting all unknowns to zero.
Then, we look for non-trivial solutions - those that aren’t simply x = 0 and y = 0.
Example
Let’s solve this fourth-degree homogeneous system:
$$ \begin{cases} y^2 -2xy -3x^2 = 0 \\ \\ x^2+3xy +2y^2 = 0 \end{cases} $$
Start by checking the trivial solution, setting x = 0 and y = 0:
$$ \begin{cases} (0)^2 -2(0)(0) -3(0)^2 = 0 \\ \\ (0)^2+3(0)(0) +2(0)^2 = 0 \end{cases} $$
$$ \begin{cases} 0 = 0 \\ \\ 0 = 0 \end{cases} $$
The system clearly admits the trivial solution.
Now, let’s search for other possible solutions.
Let’s set y = tx, introducing a parameter t that expresses y in terms of x.
Substituting into the equations gives:
$$ \begin{cases} (tx)^2 -2x(tx) -3x^2 = 0 \\ \\ x^2+3x(tx) +2(tx)^2 = 0 \end{cases} $$
$$ \begin{cases} t^2x^2 -2tx^2 -3x^2 = 0 \\ \\ x^2+3tx^2 +2t^2x^2 = 0 \end{cases} $$
Dividing both equations by x2 (assuming x ≠ 0) gives:
$$ \begin{cases} t^2 -2t -3 = 0 \\ \\ 2t^2+3t+1 = 0 \end{cases} $$
Now we need to find any values of t that satisfy both equations.
The first equation, t2 - 2t - 3 = 0, has roots t = -1 and t = 3.
The second, 2t2 + 3t + 1 = 0, has roots t = -1 and t = -½.
Both have one root in common: t = -1.
Since y = tx and t = -1, it follows that:
$$ y = -x $$
Therefore, the system has infinitely many solutions of the form (x, y) = (α, -α) for all real values of α.
$$ (x;y) = (α,-α) \ \ \forall \ α \ \in \mathbb{R} $$
Verification. For example, if x = 1 and y = -1, then: $$ \begin{cases} y^2 -2xy -3x^2 = 0 \\ \\ x^2+3xy +2y^2 = 0 \end{cases} $$ $$ \begin{cases} (-1)^2 -2(1)(-1) -3(1)^2 = 0 \\ \\ (1)^2+3(1)(-1) +2(-1)^2 = 0 \end{cases} $$ $$ \begin{cases} 1 +2 -3 = 0 \\ \\ 1-3 +2 = 0 \end{cases} $$ $$ \begin{cases} 0 = 0 \\ \\ 0 = 0 \end{cases} $$ The pair (x, y) = (1, -1) satisfies the system. Other solutions include (2, -2), (-1, 1), (-2, 2), and so on.
