Homogeneous Equations

Chapter 1 • Section 1-3

"What did the zero say to the equation? 'I'm always the solution!' 😉"

0️⃣ What is a Homogeneous System?

A homogeneous linear equation is one where the constant term is ZERO.

$$ a_1 x_1 + a_2 x_2 + \dots + a_n x_n = 0 $$

A homogeneous system is a system where every equation is homogeneous.

The Trivial Solution 👶

Setting all variables to zero ($x_1=0, \dots, x_n=0$) ALWAYS works. This is called the trivial solution.

Nontrivial Solutions 🦄

We are interested in finding solutions where at least one variable is NOT zero. These are the interesting ones!

Theorem: Existence of Nontrivial Solutions

If a homogeneous system has more variables than equations ($n > m$), then it has a nontrivial solution (in fact, infinitely many).

Why? Because there must be at least one free variable!

🧠 Quick Check Win $10

Does a homogeneous system ALWAYS have a solution?

Solving a Homogeneous System

Let's solve this system:

$$ \begin{cases} x_1 + x_2 - x_3 + 3x_4 = 0 \\ -x_1 + 4x_2 + 5x_3 - 2x_4 = 0 \\ x_1 + 6x_2 + 3x_3 + 4x_4 = 0 \end{cases} $$

Augmented Matrix $\to$ RREF:

$$ \left[\begin{array}{rrrr|r} 1 & 1 & -1 & 3 & 0 \\ -1 & 4 & 5 & -2 & 0 \\ 1 & 6 & 3 & 4 & 0 \end{array}\right] \to \dots \to \left[\begin{array}{rrrr|r} 1 & 0 & -9/5 & 14/5 & 0 \\ 0 & 1 & 4/5 & 1/5 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$

We have 2 leading variables ($x_1, x_2$) and 2 free variables ($x_3, x_4$). Let $x_3 = s$ and $x_4 = t$.

General Solution:

$$ \begin{cases} x_1 = \frac{9}{5}s - \frac{14}{5}t \\ x_2 = -\frac{4}{5}s - \frac{1}{5}t \\ x_3 = s \\ x_4 = t \end{cases} $$

Linear Combinations & Basic Solutions

We can rewrite the solution vector as a linear combination of vectors.

$$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = s \begin{bmatrix} 9/5 \\ -4/5 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} -14/5 \\ -1/5 \\ 0 \\ 1 \end{bmatrix} $$

Basic Solutions

The vectors multiplied by the parameters ($s, t$) are called basic solutions.

$$ X_1 = \begin{bmatrix} 9/5 \\ -4/5 \\ 1 \\ 0 \end{bmatrix}, \quad X_2 = \begin{bmatrix} -14/5 \\ -1/5 \\ 0 \\ 1 \end{bmatrix} $$

Pro Tip 💡

Any nonzero multiple of a basic solution is ALSO a basic solution! We can multiply by 5 to clear fractions:

$$ 5X_1 = \begin{bmatrix} 9 \\ -4 \\ 5 \\ 0 \end{bmatrix} $$

Theorem: Structure of Solutions

Let $A$ be an $m \times n$ matrix of rank $r$. Then:

  1. The system has exactly $n-r$ basic solutions.
  2. Every solution is a linear combination of these basic solutions.

🤔 Concept Check Win $15

If a homogeneous system has more variables ($n$) than equations ($m$), i.e., $n > m$, what can we say?

Hint: Think about free variables!

Rank and Solutions

For a homogeneous system with $n$ variables and rank $r$:

♾️

If $r < n$:

There are $n-r$ free variables (parameters). $\implies$ Infinitely many solutions.

1️⃣

If $r = n$:

There are 0 free variables. $\implies$ Unique solution (only the trivial one $\mathbf{0}$).

Advanced Problem: Parameter Hunt 🕵️‍♂️

Find all values of $a$ for which the system has nontrivial solutions, and determine the solutions.

$$ \begin{cases} x + y = 0 \\ ay + z = 0 \\ x + y + az = 0 \end{cases} $$

Solution Steps

  1. Augmented Matrix:
    $$ \left[\begin{array}{rrr|r} 1 & 1 & 0 & 0 \\ 0 & a & 1 & 0 \\ 1 & 1 & a & 0 \end{array}\right] $$
  2. Row Reduce: Subtract Row 1 from Row 3:
    $$ \to \left[\begin{array}{rrr|r} 1 & 1 & 0 & 0 \\ 0 & a & 1 & 0 \\ 0 & 0 & a & 0 \end{array}\right] $$
  3. Analyze for Nontrivial Solutions:

    For nontrivial solutions, we need a free variable. This happens if the rank is less than the number of variables ($3$).

    Looking at the matrix, we need a row of zeros (or a missing pivot).

    • If $a \neq 0$, we have 3 pivots ($1, a, a$). Rank = 3. Unique solution (Trivial).
    • If $a = 0$, the matrix becomes:
      $$ \left[\begin{array}{rrr|r} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$
      Rank = 2. Free variable exists!
  4. Find Solutions when $a=0$:

    From RREF: $x+y=0 \implies x=-y$. And $z=0$. Let $y=s$.

    $$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = s \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} $$

🏆 Final Challenge Win $20

Consider the system below. For what value of $a$ does it have nontrivial solutions?

$$ \begin{cases} x + y = 0 \\ ay + z = 0 \\ x + y + az = 0 \end{cases} $$