Mastering Elementary Operations & Rank of a Matrix
Complete guide to Elementary Row Operations and Matrix Rank with step-by-step methods for JEE Main & Advanced preparation.
Why Elementary Operations & Rank Matter in JEE
Elementary Operations and Matrix Rank are fundamental concepts that appear in 15-20% of JEE Linear Algebra questions. Mastering these will help you solve:
- System of Linear Equations - Consistency and solutions
- Matrix Inversion - Finding inverse using elementary operations
- Determinant Calculations - Simplified computation
- Vector Spaces - Basis and dimension concepts
- Eigenvalue Problems - Characteristic equations
Three Types of Elementary Row Operations
Type 1: Row Switching ($R_i \leftrightarrow R_j$)
Interchange two rows of the matrix. This operation doesn't change the determinant's absolute value (only sign).
Type 2: Row Multiplication ($kR_i \rightarrow R_i$)
Multiply a row by a non-zero scalar $k$. This multiplies the determinant by $k$.
Type 3: Row Addition ($R_i + kR_j \rightarrow R_i$)
Add a multiple of one row to another row. This doesn't change the determinant.
Example:
Apply $R_2 \leftrightarrow R_3$ to matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
Result: $\begin{bmatrix} 1 & 2 & 3 \\ 7 & 8 & 9 \\ 4 & 5 & 6 \end{bmatrix}$
Finding Rank Using Elementary Operations
Reduce the matrix to Row Echelon Form (REF) and count non-zero rows.
Step-by-Step Example:
Given Matrix: $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix}$
Step 1: $R_2 \rightarrow R_2 - 2R_1$
Step 2: $R_3 \rightarrow R_3 - 3R_1$
Step 3: Matrix is in REF with 1 non-zero row
Conclusion: Rank(A) = 1
Rank and System of Equations
For system $AX = B$, where $A$ is coefficient matrix and $[A|B]$ is augmented matrix:
Consistency Conditions:
- Consistent: Rank(A) = Rank([A|B])
- Unique Solution: Rank(A) = Rank([A|B]) = Number of variables
- Infinite Solutions: Rank(A) = Rank([A|B]) < Number of variables
- Inconsistent: Rank(A) ≠ Rank([A|B])
Problem:
Determine consistency of: $$\begin{cases} x + 2y + 3z = 1 \\ 2x + 4y + 6z = 2 \\ 3x + 6y + 9z = 4 \end{cases}$$
Step 1: Coefficient matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix}$
Step 2: Augmented matrix $[A|B] = \begin{bmatrix} 1 & 2 & 3 & | & 1 \\ 2 & 4 & 6 & | & 2 \\ 3 & 6 & 9 & | & 4 \end{bmatrix}$
Step 3: Rank(A) = 1 (from previous example)
Step 4: Apply operations to $[A|B]$:
• $R_2 \rightarrow R_2 - 2R_1$: $\begin{bmatrix} 1 & 2 & 3 & | & 1 \\ 0 & 0 & 0 & | & 0 \\ 3 & 6 & 9 & | & 4 \end{bmatrix}$
• $R_3 \rightarrow R_3 - 3R_1$: $\begin{bmatrix} 1 & 2 & 3 & | & 1 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & | & 1 \end{bmatrix}$
Step 5: Rank([A|B]) = 2 ≠ Rank(A) = 1
Conclusion: System is inconsistent
🚀 Key Properties & Shortcuts
Rank Properties:
- Rank ≤ min(rows, columns)
- Rank(A) = Rank(AT)
- Rank(AB) ≤ min(Rank(A), Rank(B))
- For square matrix: Rank = n ⇔ det(A) ≠ 0
Elementary Operations Tips:
- Always work systematically left to right
- Create zeros below pivots first
- Use fractions carefully to avoid errors
- Check your work by verifying operations
Advanced Concepts Available in Full Version
Includes: Elementary Column Operations, Finding Inverse using E.O., Rank-Nullity Theorem, Special Matrices, and 8+ Practice Problems
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Find rank of $\begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 3 \end{bmatrix}$
2. Using elementary operations, find inverse of $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
3. For what value of k does the system have infinite solutions?
$\begin{cases} x + 2y = 3 \\ 2x + ky = 6 \end{cases}$
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