The Hook: The Detective Number of Matrices
In the movie Transformers, when Optimus Prime transforms, his structure changes dramatically but his “essence” remains the same. Mathematically, this is captured by determinants!
A determinant is like a fingerprint of a matrix:
- If $|A| = 0$: The transformation collapses dimensions (singular/degenerate)
- If $|A| \neq 0$: The transformation is reversible (non-singular)
- $|A|$ tells you how much area/volume gets scaled by the transformation
Real applications:
- Computer graphics: Does this transformation flip the object? Check if $|A| < 0$
- Physics: Is this system solvable? Check if determinant $\neq 0$
- Economics: Is the input-output model consistent? Determinant must be non-zero
- Cryptography: Is the encryption key valid? Requires $|A| \neq 0$ and $\gcd(|A|, 26) = 1$
Why this matters for JEE: Determinants appear in 5-6 questions in JEE Main, and understanding properties is crucial for solving complex 3×3 determinants quickly in JEE Advanced. Speed matters!
Prerequisites
Before diving into determinants, you should be comfortable with:
- Matrix Basics — Understanding square matrices
- Matrix Algebra — Matrix operations
- Basic arithmetic — Sign patterns, multiplication
What is a Determinant?
Definition
A determinant is a scalar value associated with every square matrix.
For a matrix $A$, the determinant is denoted by:
$$|A| \quad \text{or} \quad \det(A) \quad \text{or} \quad \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}$$Key points:
- Only square matrices have determinants
- Result is a single number (not a matrix!)
- Tells if matrix is invertible: $|A| \neq 0$ ↔ $A^{-1}$ exists
Matrix: Square brackets $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$
Determinant: Vertical bars $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$
Don’t confuse the notation!
Determinant of a 2×2 Matrix
Formula
For a $2 \times 2$ matrix:
$$\boxed{\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc}$$Memory trick: “Down-right minus up-right”
- Main diagonal product ($ad$) MINUS
- Off-diagonal product ($bc$)
Visual Representation
$$\begin{vmatrix} \color{blue}{a} & b \\ c & \color{blue}{d} \end{vmatrix} - \begin{vmatrix} a & \color{red}{b} \\ \color{red}{c} & d \end{vmatrix}$$Blue diagonal ($\color{blue}{ad}$) minus red diagonal ($\color{red}{bc}$)
Examples
This matrix is singular (non-invertible) since $|A| = 0$!
Determinant of a 3×3 Matrix
Method 1: Sarrus Rule (Diagonal Method)
Sarrus Rule is a quick method for $3 \times 3$ determinants:
$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$$Step 1: Rewrite first two columns on the right
$$\begin{array}{|ccc|cc} a_1 & b_1 & c_1 & a_1 & b_1 \\ a_2 & b_2 & c_2 & a_2 & b_2 \\ a_3 & b_3 & c_3 & a_3 & b_3 \end{array}$$Step 2: Sum of main diagonal products (↘):
$$\color{blue}{a_1 b_2 c_3 + b_1 c_2 a_3 + c_1 a_2 b_3}$$Step 3: Sum of anti-diagonal products (↙):
$$\color{red}{c_1 b_2 a_3 + a_1 c_2 b_3 + b_1 a_2 c_3}$$Final: Blue sum minus red sum
$$\boxed{|A| = \color{blue}{(a_1 b_2 c_3 + b_1 c_2 a_3 + c_1 a_2 b_3)} - \color{red}{(c_1 b_2 a_3 + a_1 c_2 b_3 + b_1 a_2 c_3)}}$$Positive terms (↘ ↘ ↘):
- Start from column 1, 2, 3 in row 1
- Go diagonally down-right (wrap around)
Negative terms (↙ ↙ ↙):
- Start from column 3, 2, 1 in row 1
- Go diagonally down-left (wrap around)
Mnemonic: “Right is Plus, Left is Minus”
Example using Sarrus Rule
$$\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$$Positive products:
- $1 \cdot 5 \cdot 9 = 45$
- $2 \cdot 6 \cdot 7 = 84$
- $3 \cdot 4 \cdot 8 = 96$
Sum = $45 + 84 + 96 = 225$
Negative products:
- $3 \cdot 5 \cdot 7 = 105$
- $1 \cdot 6 \cdot 8 = 48$
- $2 \cdot 4 \cdot 9 = 72$
Sum = $105 + 48 + 72 = 225$
Determinant: $225 - 225 = 0$
The matrix is singular!
Method 2: Expansion by First Row
$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$$Expand along Row 1 with alternating signs $(+, -, +)$:
$$\boxed{= a_1 \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} - b_1 \begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1 \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}}$$In expanded form:
$$= a_1(b_2 c_3 - b_3 c_2) - b_1(a_2 c_3 - a_3 c_2) + c_1(a_2 b_3 - a_3 b_2)$$Sign pattern for expansion:
$$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$$This is the checkerboard pattern starting with $+$ at position $(1,1)$.
Example using Row Expansion
$$\begin{vmatrix} 2 & 3 & 1 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$$Expand along Row 1:
$$= 2\begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 3\begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 1\begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}$$ $$= 2(45 - 48) - 3(36 - 42) + 1(32 - 35)$$ $$= 2(-3) - 3(-6) + 1(-3)$$ $$= -6 + 18 - 3 = 9$$Answer: $|A| = 9$
Sarrus Rule:
- ✓ Fast for numerical $3 \times 3$ matrices
- ✓ Good for quick calculations
- ✗ Only works for $3 \times 3$ (not generalizable)
- ✗ Easy to make sign errors
Row/Column Expansion:
- ✓ Works for any size square matrix
- ✓ Systematic with clear sign pattern
- ✓ Useful when row/column has zeros
- ✗ Slower for purely numerical matrices
JEE Strategy: Use Sarrus for quick $3 \times 3$. Use expansion when:
- Matrix has many zeros
- Matrix has symbolic entries
- Need to prove properties
Expansion Along Any Row or Column
You can expand along any row or column (not just first row!)
General Formula
For element $a_{ij}$, the cofactor sign is:
$$\boxed{(-1)^{i+j}}$$Sign pattern:
$$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$$Expansion Along Row $i$
$$|A| = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} M_{ij}$$where $M_{ij}$ is the minor (determinant after removing row $i$ and column $j$).
Expansion Along Column $j$
$$|A| = \sum_{i=1}^{n} (-1)^{i+j} a_{ij} M_{ij}$$Always expand along the row/column with the most zeros!
This minimizes calculations.
Example:
$$\begin{vmatrix} 1 & 2 & 3 \\ 0 & 0 & 5 \\ 4 & 0 & 6 \end{vmatrix}$$Expand along Row 2 (has two zeros):
$$= -0 \cdot M_{21} + 0 \cdot M_{22} - 5 \cdot M_{23}$$ $$= -5 \begin{vmatrix} 1 & 2 \\ 4 & 0 \end{vmatrix} = -5(0 - 8) = -5(-8) = 40$$Only one $2 \times 2$ determinant needed!
Special Determinants
1. Diagonal Matrix
$$\begin{vmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{vmatrix} = abc$$Rule: Product of diagonal elements.
2. Triangular Matrix (Upper or Lower)
$$\begin{vmatrix} a & * & * \\ 0 & b & * \\ 0 & 0 & c \end{vmatrix} = abc$$Rule: Product of diagonal elements (regardless of other entries).
3. Identity Matrix
$$|I_n| = 1$$4. Zero Matrix
$$|O| = 0$$5. Matrix with Zero Row or Column
If any row or column is entirely zeros:
$$|A| = 0$$6. Matrix with Two Identical Rows/Columns
If two rows (or columns) are identical:
$$|A| = 0$$Example:
$$\begin{vmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = 0$$(Row 1 = Row 2)
7. Matrix with Proportional Rows/Columns
If one row is a scalar multiple of another:
$$|A| = 0$$Example:
$$\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 7 & 8 & 9 \end{vmatrix} = 0$$(Row 2 = 2 × Row 1)
Before calculating a complex determinant, check:
- Is any row/column all zeros? → $|A| = 0$
- Are two rows/columns identical? → $|A| = 0$
- Is one row a multiple of another? → $|A| = 0$
- Is it triangular/diagonal? → Product of diagonal
These checks can save 2-3 minutes on JEE Advanced!
Common Mistakes to Avoid
Wrong: Using Sarrus rule for $4 \times 4$ or larger matrices
Right: Sarrus rule ONLY works for $3 \times 3$ matrices!
For $n \geq 4$, you must use cofactor expansion or properties.
Wrong: Forgetting the alternating signs in expansion
Right: Always use the checkerboard pattern:
$$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$$Position $(i,j)$ has sign $(-1)^{i+j}$
Wrong: Finding determinant of a $2 \times 3$ matrix
Right: Determinants are only defined for square matrices!
$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$ has no determinant.
Wrong:
$$\begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = \begin{bmatrix} -2 & 2 \\ 3 & -1 \end{bmatrix}$$Right:
$$\begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = -2 \text{ (a scalar, not a matrix!)}$$Determinant is always a single number, not a matrix!
Wrong: Not recognizing proportional rows before calculating
Example:
$$\begin{vmatrix} 2 & 4 & 6 \\ 1 & 2 & 3 \\ 5 & 7 & 9 \end{vmatrix}$$Row 1 = 2 × Row 2, so $|A| = 0$ immediately!
Don’t waste time calculating!
Quick Calculation Techniques for JEE
Technique 1: Factor Out Common Factors
If a row/column has a common factor, take it out:
$$\begin{vmatrix} 2 & 4 & 6 \\ 1 & 3 & 5 \\ 7 & 8 & 9 \end{vmatrix} = 2 \begin{vmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 7 & 8 & 9 \end{vmatrix}$$Technique 2: Create Zeros Using Row Operations
Use the property: $R_i \to R_i + kR_j$ doesn’t change the determinant!
$$\begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 3 & 7 & 11 \end{vmatrix}$$$R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$:
$$= \begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 1 & 2 \end{vmatrix}$$Expand along Column 1:
$$= 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = 2 - 1 = 1$$Technique 3: Pattern Recognition
$$\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = (b-a)(c-a)(c-b)$$This is a Vandermonde determinant — memorize it for JEE!
More patterns in Properties of Determinants.
Practice Problems
Level 1: Foundation (NCERT)
Evaluate: $\begin{vmatrix} 3 & -2 \\ 4 & 5 \end{vmatrix}$
Solution:
$$= (3)(5) - (-2)(4) = 15 + 8 = 23$$Answer: 23
Find: $\begin{vmatrix} 6 & 3 \\ 10 & 5 \end{vmatrix}$
Solution:
$$= (6)(5) - (3)(10) = 30 - 30 = 0$$The matrix is singular!
Answer: 0
Evaluate using Sarrus rule: $\begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix}$
Solution:
This is a diagonal matrix!
$$= 1 \cdot 2 \cdot 3 = 6$$Answer: 6
Level 2: JEE Main
Evaluate: $\begin{vmatrix} 2 & 3 & 5 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$
Solution:
Using Sarrus rule:
Positive: $2(5)(9) + 3(6)(7) + 5(4)(8) = 90 + 126 + 160 = 376$
Negative: $5(5)(7) + 2(6)(8) + 3(4)(9) = 175 + 96 + 108 = 379$
$$|A| = 376 - 379 = -3$$Answer: $-3$
Without expanding, show that: $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 5 & 7 & 9 \end{vmatrix} = 0$
Solution:
Notice Row 2 = 2 × Row 1
When one row is a scalar multiple of another, determinant = 0.
Answer: $|A| = 0$ (no calculation needed!)
Evaluate by choosing the best row/column: $\begin{vmatrix} 1 & 0 & 3 \\ 2 & 0 & 5 \\ 4 & 0 & 6 \end{vmatrix}$
Solution:
Column 2 is all zeros, so $|A| = 0$ immediately!
Alternatively, expand along Column 2:
$$= -0 + 0 - 0 = 0$$Answer: 0
Level 3: JEE Advanced
If $\begin{vmatrix} x & 2 & 3 \\ 4 & x & 5 \\ 6 & 7 & x \end{vmatrix} = 0$, find the sum of all possible values of $x$.
Solution:
Expand using first row:
$$x(x^2 - 35) - 2(4x - 30) + 3(28 - 6x) = 0$$ $$x^3 - 35x - 8x + 60 + 84 - 18x = 0$$ $$x^3 - 61x + 144 = 0$$By sum of roots of cubic $ax^3 + bx^2 + cx + d = 0$:
Sum of roots $= -\frac{b}{a} = 0$ (since coefficient of $x^2$ is 0)
Answer: Sum = 0
Prove that: $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (b-a)(c-a)(c-b)$
Solution:
Apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$$= \begin{vmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ a^2 & b^2-a^2 & c^2-a^2 \end{vmatrix}$$ $$= \begin{vmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ a^2 & (b-a)(b+a) & (c-a)(c+a) \end{vmatrix}$$Factor out from columns 2 and 3:
$$= (b-a)(c-a) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 1 \\ a^2 & b+a & c+a \end{vmatrix}$$Expand along Row 1:
$$= (b-a)(c-a) \begin{vmatrix} 1 & 1 \\ b+a & c+a \end{vmatrix}$$ $$= (b-a)(c-a)[(c+a) - (b+a)]$$ $$= (b-a)(c-a)(c-b)$$Proved!
Evaluate: $\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix}$ where $a, b, c \neq 0$
Solution:
Apply $R_1 \to R_1 - R_3$ and $R_2 \to R_2 - R_3$:
$$= \begin{vmatrix} a & 0 & -c \\ 0 & b & -c \\ 1 & 1 & 1+c \end{vmatrix}$$Expand along Row 1:
$$= a \begin{vmatrix} b & -c \\ 1 & 1+c \end{vmatrix} - 0 + (-c) \begin{vmatrix} 0 & b \\ 1 & 1 \end{vmatrix}$$ $$= a[b(1+c) + c] - c[0 - b]$$ $$= a[b + bc + c] + bc$$ $$= ab + abc + ac + bc$$ $$= abc + ab + bc + ca$$Answer: $abc + ab + bc + ca$
Quick Revision Box
| Concept | Formula/Rule |
|---|---|
| $2 \times 2$ determinant | $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$ |
| Sarrus rule | 3 positive diagonals - 3 negative diagonals |
| Expansion signs | Checkerboard: $(-1)^{i+j}$ |
| Diagonal matrix | Product of diagonal elements |
| Triangular matrix | Product of diagonal elements |
| Zero row/column | $\|A\| = 0$ |
| Identical rows | $\|A\| = 0$ |
| Proportional rows | $\|A\| = 0$ |
| Identity matrix | $\|I\| = 1$ |
| Best expansion | Choose row/column with most zeros |
Related Topics
Within Matrices & Determinants Chapter
- Matrix Basics — Foundation of square matrices
- Matrix Algebra — Operations before determinants
- Minors and Cofactors — Building blocks of expansion
- Properties of Determinants — Shortcuts and patterns
- Adjoint and Inverse — Using $|A|$ to find inverse
Math Connections
- Linear Equations — Cramer’s rule uses determinants
- Coordinate Geometry — Area of triangle formula
- Vector Algebra — Scalar triple product
Real-World Applications
- Linear transformations — Determinant as scaling factor
- Computer graphics — Check if transformation preserves orientation
- Systems of equations — Solvability condition
- Cryptography — Key validity check
Teacher’s Summary
- Determinant is a scalar associated with square matrices only
- 2×2 formula: $ad - bc$ (main diagonal minus off-diagonal)
- 3×3 methods: Sarrus rule (quick) or expansion (systematic)
- Sign pattern: Checkerboard starting with $+$ at $(1,1)$: $(-1)^{i+j}$
- Smart expansion: Choose row/column with most zeros
- Quick checks: Zero row, identical rows, proportional rows → $|A| = 0$
- Triangular/diagonal: Just multiply diagonal elements
- Sarrus ONLY for 3×3 — doesn’t work for larger matrices!
“Before calculating, always check for zeros, identical rows, or proportional rows — save precious JEE time!”
Exam Strategy:
- For $2 \times 2$: Direct formula (5 seconds)
- For $3 \times 3$ numerical: Sarrus rule (30 seconds)
- For $3 \times 3$ with zeros: Expand along best row/column (45 seconds)
- For symbolic: Use properties and row operations (next chapter!)