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Core Concept Reading Time: 15 min Essential Theory

The Heart of Linear Algebra: Homogeneous Systems & Consistency Theorems

Master the fundamental concepts of homogeneous systems AX=0 and understand when systems have unique, infinite, or no solutions.

AX=0
Homogeneous System
2 Types
of Solutions
5+
Key Theorems
JEE
High Weightage

Why Homogeneous Systems Matter

Homogeneous systems of linear equations form the foundation of linear algebra and appear in every JEE mathematics paper. Understanding these systems is crucial because:

  • They help determine linear independence/dependence of vectors
  • They're essential for understanding eigenvalues and eigenvectors
  • They form the basis for vector spaces and subspaces
  • They appear in differential equations and transformations

🎯 JEE Relevance

Homogeneous systems appear in 2-3 questions per JEE paper, often disguised as:

  • Linear dependence/independence problems
  • Matrix rank questions
  • Eigenvalue problems
  • Vector space problems

🧭 Quick Navigation

Definition Solution Types Key Theorems JEE Examples

1. Homogeneous System: AX = 0

Mathematical Definition

A system of linear equations is called homogeneous if all constant terms are zero.

$$ AX = 0 $$

Where:

  • $A$ is an $m \times n$ coefficient matrix
  • $X$ is an $n \times 1$ column vector of variables
  • $0$ is an $m \times 1$ zero vector

Example: 3×3 Homogeneous System

Consider the system:

$$ \begin{cases} 2x + 3y - z = 0 \\ x - y + 2z = 0 \\ 3x + 2y + z = 0 \end{cases} $$

In matrix form: $AX = 0$ where

$$ A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & -1 & 2 \\ 3 & 2 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad 0 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

2. Trivial vs. Non-Trivial Solutions

Trivial Solution

$$ X = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} $$
  • Always exists for homogeneous systems
  • All variables equal to zero
  • Geometrically: Origin point
  • Exists regardless of matrix A

Non-Trivial Solution

$$ X \neq 0 $$
  • May or may not exist
  • At least one variable ≠ 0
  • Geometrically: Line/plane through origin
  • Exists when system has infinite solutions

Fundamental Theorem

A homogeneous system $AX = 0$ has:

  • Only trivial solution if and only if $\det(A) \neq 0$ (A is invertible)
  • Non-trivial solutions if and only if $\det(A) = 0$ (A is singular)
$$ \det(A) \neq 0 \Rightarrow \text{Unique solution } X = 0 $$ $$ \det(A) = 0 \Rightarrow \text{Infinite solutions} $$

3. Essential Consistency Theorems

Theorem 1: Rank and Solution Space

For an $m \times n$ homogeneous system $AX = 0$:

$$ \text{Number of free variables} = n - \text{rank}(A) $$

Dimension of solution space = Number of linearly independent solutions

Interpretation:

  • If $\text{rank}(A) = n$: Only trivial solution
  • If $\text{rank}(A) < n$: Infinite solutions
  • Solution space dimension = $n - \text{rank}(A)$

Theorem 2: Superposition Principle

If $X_1$ and $X_2$ are solutions of $AX = 0$, then any linear combination is also a solution:

$$ A(c_1X_1 + c_2X_2) = c_1AX_1 + c_2AX_2 = c_1\cdot 0 + c_2\cdot 0 = 0 $$

Key Insight: The solution set forms a vector space (null space of A)

Theorem 3: Relation to Non-Homogeneous Systems

For a non-homogeneous system $AX = B$:

$$ \text{General solution} = \text{Particular solution} + \text{General solution of } AX = 0 $$

Structure Theorem:

If $X_p$ is a particular solution of $AX = B$ and $X_h$ is any solution of $AX = 0$, then:

$$ X = X_p + X_h $$

is also a solution of $AX = B$.

4. JEE Practice Problems

Problem 1: Determining Solution Type

For what value of $k$ does the system have non-trivial solutions?

$$ \begin{cases} x + 2y + 3z = 0 \\ 2x + y + z = 0 \\ x + y + kz = 0 \end{cases} $$

Solution:

Step 1: Non-trivial solutions exist when $\det(A) = 0$

Step 2: Coefficient matrix:

$$ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 1 & k \end{bmatrix} $$

Step 3: Calculate determinant:

$$ \det(A) = 1(1\cdot k - 1\cdot 1) - 2(2\cdot k - 1\cdot 1) + 3(2\cdot 1 - 1\cdot 1) $$ $$ = (k-1) - 2(2k-1) + 3(2-1) $$ $$ = k-1 -4k+2 + 3 = -3k + 4 $$

Step 4: Set $\det(A) = 0$: $-3k + 4 = 0 \Rightarrow k = \frac{4}{3}$

Problem 2: Finding Complete Solution

Find all solutions of the system:

$$ \begin{cases} x + y + z = 0 \\ 2x - y + z = 0 \\ x + 2y + 2z = 0 \end{cases} $$

Solution:

Step 1: Write augmented matrix and reduce:

$$ \begin{bmatrix} 1 & 1 & 1 & | & 0 \\ 2 & -1 & 1 & | & 0 \\ 1 & 2 & 2 & | & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 & | & 0 \\ 0 & -3 & -1 & | & 0 \\ 0 & 1 & 1 & | & 0 \end{bmatrix} $$

Step 2: Continue reduction:

$$ \rightarrow \begin{bmatrix} 1 & 1 & 1 & | & 0 \\ 0 & 1 & 1 & | & 0 \\ 0 & 0 & 2 & | & 0 \end{bmatrix} $$

Step 3: Back substitution gives $z = 0, y = 0, x = 0$

Step 4: Only trivial solution exists: $(x, y, z) = (0, 0, 0)$

Quick Summary & Key Takeaways

Homogeneous Systems AX=0

  • ⚡ Always consistent - trivial solution always exists
  • 🎯 Non-trivial solutions exist when $\det(A) = 0$
  • 📝 Solution space dimension = $n - \text{rank}(A)$

JEE Problem Solving

  • 📥 Check $\det(A)$ for solution type
  • 📤 Use rank to find solution space dimension
  • 💡 Remember superposition principle

📝 Test Your Understanding

Try these problems to reinforce your learning:

1. For what values of $\lambda$ does the system have non-trivial solutions?

$$ \begin{cases} (\lambda-1)x + 2y = 0 \\ 2x + (\lambda-1)y = 0 \end{cases} $$

2. Find the dimension of solution space for:

$$ \begin{cases} x + 2y - z = 0 \\ 2x + 4y - 2z = 0 \\ 3x + 6y - 3z = 0 \end{cases} $$

3. If $X_1$ and $X_2$ are solutions of $AX=0$, prove that $3X_1 - 2X_2$ is also a solution.

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