The Hook: The Building Blocks of Determinants
Ever built a complex LEGO structure? You start with small blocks and combine them to create something bigger. Minors and cofactors work exactly the same way!
A $3 \times 3$ determinant is built from smaller $2 \times 2$ determinants (minors). A $4 \times 4$ determinant? Built from $3 \times 3$ minors. A $100 \times 100$ determinant? Built recursively from smaller pieces!
This recursive structure is how:
- Google’s servers compute massive determinants for data analysis
- CAD software checks if 3D transformations are valid
- Physics simulations solve complex systems with thousands of equations
- Machine learning algorithms compute eigenvalues of huge matrices
Why this matters for JEE: Understanding minors and cofactors is essential for:
- Finding inverse matrices (adjoint method)
- Solving determinant problems efficiently
- Understanding Cramer’s rule for linear systems
- JEE Advanced problems involving symbolic determinants
Prerequisites
Before diving into minors and cofactors, you should be comfortable with:
- Matrix Basics — Understanding matrix elements and order
- Determinants Evaluation — Computing $2 \times 2$ and $3 \times 3$ determinants
- Row and column notation
What is a Minor?
Definition
The minor of an element $a_{ij}$ in a matrix $A$, denoted $M_{ij}$, is the determinant of the submatrix obtained by deleting row $i$ and column $j$.
$$\boxed{M_{ij} = \text{det of submatrix after removing row } i \text{ and column } j}$$Visual Understanding
For a $3 \times 3$ matrix:
$$A = \begin{bmatrix} \color{red}{a_{11}} & \color{blue}{a_{12}} & a_{13} \\ \color{blue}{a_{21}} & a_{22} & a_{23} \\ \color{blue}{a_{31}} & a_{32} & a_{33} \end{bmatrix}$$To find $M_{11}$ (minor of $\color{red}{a_{11}}$):
- Delete row 1 (containing $\color{red}{a_{11}}$)
- Delete column 1 (containing $\color{red}{a_{11}}$)
- Find determinant of remaining $2 \times 2$ matrix
Example: Finding All Minors of a 3×3 Matrix
$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$$Finding $M_{11}$: Delete row 1, column 1
$$M_{11} = \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = 45 - 48 = -3$$Finding $M_{12}$: Delete row 1, column 2
$$M_{12} = \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} = 36 - 42 = -6$$Finding $M_{13}$: Delete row 1, column 3
$$M_{13} = \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} = 32 - 35 = -3$$Finding $M_{21}$: Delete row 2, column 1
$$M_{21} = \begin{vmatrix} 2 & 3 \\ 8 & 9 \end{vmatrix} = 18 - 24 = -6$$Finding $M_{22}$: Delete row 2, column 2
$$M_{22} = \begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} = 9 - 21 = -12$$Finding $M_{23}$: Delete row 2, column 3
$$M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} = 8 - 14 = -6$$Finding $M_{31}$: Delete row 3, column 1
$$M_{31} = \begin{vmatrix} 2 & 3 \\ 5 & 6 \end{vmatrix} = 12 - 15 = -3$$Finding $M_{32}$: Delete row 3, column 2
$$M_{32} = \begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix} = 6 - 12 = -6$$Finding $M_{33}$: Delete row 3, column 3
$$M_{33} = \begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} = 5 - 8 = -3$$Matrix of Minors:
$$\begin{bmatrix} M_{11} & M_{12} & M_{13} \\ M_{21} & M_{22} & M_{23} \\ M_{31} & M_{32} & M_{33} \end{bmatrix} = \begin{bmatrix} -3 & -6 & -3 \\ -6 & -12 & -6 \\ -3 & -6 & -3 \end{bmatrix}$$What is a Cofactor?
Definition
The cofactor of element $a_{ij}$, denoted $C_{ij}$ or $A_{ij}$, is the minor with an appropriate sign:
$$\boxed{C_{ij} = (-1)^{i+j} \cdot M_{ij}}$$In words: Cofactor = (sign factor) × (minor)
Sign Pattern
The sign $(-1)^{i+j}$ follows a checkerboard pattern:
$$\begin{bmatrix}
- & - & + & - & \cdots \
- & + & - & + & \cdots \
- & - & + & - & \cdots \
- & + & - & + & \cdots \ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}$$
For $3 \times 3$ matrix:
$$\begin{bmatrix}
- & - & + \
- & + & - \
- & - & + \end{bmatrix} = \begin{bmatrix} (-1)^{1+1} & (-1)^{1+2} & (-1)^{1+3} \ (-1)^{2+1} & (-1)^{2+2} & (-1)^{2+3} \ (-1)^{3+1} & (-1)^{3+2} & (-1)^{3+3} \end{bmatrix}$$
Position $(1,1)$ is always POSITIVE (+)
Then alternate like a checkerboard:
- Move right → sign changes
- Move down → sign changes
- Move diagonally → sign stays same
Quick check: $i + j$ even → positive, $i + j$ odd → negative
Examples:
- $(1,1)$: $1+1=2$ (even) → $+$
- $(1,2)$: $1+2=3$ (odd) → $-$
- $(2,2)$: $2+2=4$ (even) → $+$
Example: Finding All Cofactors
Using the same matrix:
$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$$We already found minors. Now apply signs:
$$C_{11} = (-1)^{1+1} M_{11} = (+1)(-3) = -3$$ $$C_{12} = (-1)^{1+2} M_{12} = (-1)(-6) = 6$$ $$C_{13} = (-1)^{1+3} M_{13} = (+1)(-3) = -3$$ $$C_{21} = (-1)^{2+1} M_{21} = (-1)(-6) = 6$$ $$C_{22} = (-1)^{2+2} M_{22} = (+1)(-12) = -12$$ $$C_{23} = (-1)^{2+3} M_{23} = (-1)(-6) = 6$$ $$C_{31} = (-1)^{3+1} M_{31} = (+1)(-3) = -3$$ $$C_{32} = (-1)^{3+2} M_{32} = (-1)(-6) = 6$$ $$C_{33} = (-1)^{3+3} M_{33} = (+1)(-3) = -3$$Cofactor Matrix:
$$\begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} = \begin{bmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \end{bmatrix}$$Cofactor Expansion of Determinants
Expansion Along a Row
The determinant of $A$ can be found by expanding along any row $i$:
$$\boxed{|A| = \sum_{j=1}^{n} a_{ij} \cdot C_{ij}}$$In words: Sum of (element × its cofactor) for all elements in row $i$
Expansion Along a Column
The determinant of $A$ can be found by expanding along any column $j$:
$$\boxed{|A| = \sum_{i=1}^{n} a_{ij} \cdot C_{ij}}$$In words: Sum of (element × its cofactor) for all elements in column $j$
Example: Expansion Along Row 1
$$A = \begin{bmatrix} 2 & 3 & 1 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$$Expand along Row 1:
$$|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$$ $$C_{11} = (-1)^{1+1} \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = 45 - 48 = -3$$ $$C_{12} = (-1)^{1+2} \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} = -(36 - 42) = 6$$ $$C_{13} = (-1)^{1+3} \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} = 32 - 35 = -3$$ $$|A| = 2(-3) + 3(6) + 1(-3) = -6 + 18 - 3 = 9$$Example: Expansion Along Column with Zeros
$$A = \begin{bmatrix} 1 & 0 & 3 \\ 2 & 0 & 5 \\ 4 & 7 & 6 \end{bmatrix}$$Smart choice: Expand along Column 2 (has two zeros!)
$$|A| = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32}$$ $$= 0 \cdot C_{12} + 0 \cdot C_{22} + 7 \cdot C_{32}$$ $$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 3 \\ 2 & 5 \end{vmatrix} = -(5 - 6) = 1$$ $$|A| = 7(1) = 7$$Only one $2 \times 2$ determinant needed!
Always expand along the row/column with the MOST ZEROS!
Why? Each zero eliminates one $2 \times 2$ determinant calculation.
Example ranking:
- Row with 2 zeros → Best (only 1 calculation)
- Row with 1 zero → Good (2 calculations)
- Row with 0 zeros → Slowest (3 calculations)
Time saved: Can reduce 3 minutes to 30 seconds on complex problems!
Properties of Minors and Cofactors
Property 1: Sum of Products (Row)
For any row $i$:
$$\sum_{j=1}^{n} a_{ij} C_{ij} = |A|$$This is the cofactor expansion formula!
Property 2: Sum of Products (Different Row)
For row $i$ and cofactors of different row $k$ (where $i \neq k$):
$$\boxed{\sum_{j=1}^{n} a_{ij} C_{kj} = 0}$$In words: Sum of (elements of one row) × (cofactors of a different row) = 0
Example:
$$a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23} = 0$$This is a crucial property for finding adjoints!
Property 3: Similar Property for Columns
For column $j$ and cofactors of different column $l$ (where $j \neq l$):
$$\sum_{i=1}^{n} a_{ij} C_{il} = 0$$Applications of Cofactors
1. Finding Determinants
As we’ve seen, cofactor expansion is a systematic way to find determinants of any size.
2. Finding Adjoint (Next Topic)
The adjoint of a matrix is the transpose of the cofactor matrix:
$$\text{adj}(A) = [C_{ij}]^T$$3. Finding Inverse
The inverse formula uses cofactors:
$$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{|A|} [C_{ij}]^T$$4. Cramer’s Rule
Solving linear systems uses cofactors and determinants.
Working with Larger Matrices
4×4 Determinant Using Cofactors
For a $4 \times 4$ matrix, each cofactor is a $3 \times 3$ determinant:
$$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}$$Expand along Row 1:
$$|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} + a_{14}C_{14}$$Where each $C_{1j}$ is $(-1)^{1+j}$ times a $3 \times 3$ determinant.
Determinant calculation grows EXPONENTIALLY with size:
- $2 \times 2$: 2 multiplications
- $3 \times 3$: 6 multiplications (Sarrus) or 9 (expansion)
- $4 \times 4$: About 40 multiplications
- $5 \times 5$: About 200 multiplications
- $n \times n$: About $n!$ operations (factorial growth!)
For JEE: You’ll rarely see beyond $4 \times 4$, and usually with special structure (zeros, patterns) to make it manageable.
Real-world: Computers use specialized algorithms (LU decomposition, QR factorization) that are much faster than cofactor expansion!
Common Mistakes to Avoid
Wrong: $C_{ij} = M_{ij}$ (forgetting sign factor)
Right: $C_{ij} = (-1)^{i+j} M_{ij}$
Example: For position $(2,3)$:
- Minor: $M_{23} = 5$
- Cofactor: $C_{23} = (-1)^{2+3} \cdot 5 = -5$ (NOT 5!)
Wrong: To find $M_{23}$, delete row 2 and column 2
Right: To find $M_{23}$, delete row 2 and column 3
The subscripts $(i,j)$ tell you exactly which row and column to delete!
Wrong: Assuming position $(2,2)$ has negative sign
Right: Position $(2,2)$: $(-1)^{2+2} = (-1)^4 = +1$ (POSITIVE!)
Middle element of $3 \times 3$ is always positive, not negative.
Check: $2+2=4$ (even) → positive sign
Wrong: $|A| = a_{11}C_{11} + a_{22}C_{22} + a_{33}C_{33}$
Right: When expanding, use all elements from ONE row or ONE column!
You can’t mix elements from different rows (unless using special properties).
Correct: $|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$ (all from row 1)
Wrong: Adjoint = Cofactor matrix
Right: Adjoint = Transpose of cofactor matrix
$$\text{adj}(A) = [C_{ij}]^T$$The cofactor matrix must be transposed!
Practice Problems
Level 1: Foundation (NCERT)
For $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, find $M_{11}$ and $C_{11}$.
Solution:
$M_{11}$ = determinant after deleting row 1, column 1 = just the element $4$
$$M_{11} = 4$$ $$C_{11} = (-1)^{1+1} M_{11} = (+1)(4) = 4$$Answer: $M_{11} = 4$, $C_{11} = 4$
For $A = \begin{bmatrix} 2 & 3 & 1 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$, find $M_{23}$ and $C_{23}$.
Solution:
Delete row 2, column 3:
$$M_{23} = \begin{vmatrix} 2 & 3 \\ 7 & 8 \end{vmatrix} = 16 - 21 = -5$$ $$C_{23} = (-1)^{2+3} M_{23} = (-1)(-5) = 5$$Answer: $M_{23} = -5$, $C_{23} = 5$
Write the sign pattern for a $4 \times 4$ matrix.
Solution:
$$\begin{bmatrix}
- & - & + & - \
- & + & - & + \
- & - & + & - \
- & + & - & + \end{bmatrix}$$
Answer: Checkerboard starting with $+$ at $(1,1)$
Level 2: JEE Main
Find the determinant by expanding along the best row/column:
$$\begin{vmatrix} 2 & 0 & 0 \\ 3 & 5 & 0 \\ 1 & 7 & 9 \end{vmatrix}$$Solution:
This is a lower triangular matrix!
For triangular matrices: $|A|$ = product of diagonal elements
$$|A| = 2 \cdot 5 \cdot 9 = 90$$Alternatively, expand along Row 1 (has two zeros):
$$|A| = 2 \cdot C_{11} + 0 \cdot C_{12} + 0 \cdot C_{13}$$ $$C_{11} = (+1) \begin{vmatrix} 5 & 0 \\ 7 & 9 \end{vmatrix} = 45$$ $$|A| = 2(45) = 90$$Answer: 90
For $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}$, find the cofactor matrix.
Solution:
$$C_{11} = (+1) \begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} = 24$$ $$C_{12} = (-1) \begin{vmatrix} 0 & 5 \\ 0 & 6 \end{vmatrix} = 0$$ $$C_{13} = (+1) \begin{vmatrix} 0 & 4 \\ 0 & 0 \end{vmatrix} = 0$$ $$C_{21} = (-1) \begin{vmatrix} 2 & 3 \\ 0 & 6 \end{vmatrix} = -12$$ $$C_{22} = (+1) \begin{vmatrix} 1 & 3 \\ 0 & 6 \end{vmatrix} = 6$$ $$C_{23} = (-1) \begin{vmatrix} 1 & 2 \\ 0 & 0 \end{vmatrix} = 0$$ $$C_{31} = (+1) \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = -2$$ $$C_{32} = (-1) \begin{vmatrix} 1 & 3 \\ 0 & 5 \end{vmatrix} = -5$$ $$C_{33} = (+1) \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} = 4$$Answer: $\begin{bmatrix} 24 & 0 & 0 \\ -12 & 6 & 0 \\ -2 & -5 & 4 \end{bmatrix}$
Verify that for $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$:
$$a_{11}C_{21} + a_{12}C_{22} = 0$$Solution:
$$C_{21} = (-1)^{2+1} M_{21} = (-1)(2) = -2$$ $$C_{22} = (-1)^{2+2} M_{22} = (+1)(1) = 1$$ $$a_{11}C_{21} + a_{12}C_{22} = 1(-2) + 2(1) = -2 + 2 = 0$$✓
Verified! (Property: different row cofactors sum to zero)
Level 3: JEE Advanced
If $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix}$, show that the cofactor of each element in the second row equals zero.
Solution:
Notice that Row 3 = Row 1 + Row 2
This means the rows are linearly dependent, so $|A| = 0$.
$$C_{21} = (-1)^{2+1} \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = -(10-12) = 2$$Wait, this is NOT zero! Let me recalculate…
Actually, Row 2 = average of Row 1 and Row 3, so let’s find all cofactors:
$$C_{21} = (-1) \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = -(10-12) = 2$$ $$C_{22} = (+1) \begin{vmatrix} 1 & 3 \\ 3 & 5 \end{vmatrix} = 5-9 = -4$$ $$C_{23} = (-1) \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = -(4-6) = 2$$Now verify expansion: $|A| = a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23}$
$$= 2(2) + 3(-4) + 4(2) = 4 - 12 + 8 = 0$$✓
Individual cofactors are NOT zero, but their expansion gives zero (because $|A| = 0$).
Answer: The cofactor expansion equals zero, not individual cofactors.
Find the determinant using cofactor expansion:
$$\begin{vmatrix} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 0 & 0 & 0 & 10 \end{vmatrix}$$Solution:
This is an upper triangular matrix!
$$|A| = 1 \cdot 5 \cdot 8 \cdot 10 = 400$$Verification by expansion along Column 1:
$$|A| = 1 \cdot C_{11} = 1 \cdot \begin{vmatrix} 5 & 6 & 7 \\ 0 & 8 & 9 \\ 0 & 0 & 10 \end{vmatrix}$$The $3 \times 3$ minor is also triangular:
$$= 1 \cdot (5 \cdot 8 \cdot 10) = 400$$✓
Answer: 400
Quick Revision Box
| Concept | Formula/Definition |
|---|---|
| Minor $M_{ij}$ | Det after deleting row $i$, column $j$ |
| Cofactor $C_{ij}$ | $(-1)^{i+j} M_{ij}$ |
| Sign pattern | Checkerboard: $(i+j)$ even → $+$, odd → $-$ |
| Row expansion | $\|A\| = \sum_j a_{ij} C_{ij}$ |
| Column expansion | $\|A\| = \sum_i a_{ij} C_{ij}$ |
| Different row | $\sum_j a_{ij} C_{kj} = 0$ if $i \neq k$ |
| Best expansion | Choose row/column with most zeros |
| Triangular matrix | $\|A\|$ = product of diagonal (fast!) |
Related Topics
Within Matrices & Determinants Chapter
- Matrix Basics — Foundation of matrix elements
- Determinants Evaluation — Using cofactors to find determinants
- Adjoint and Inverse — Using cofactors to build adjoint
- Properties of Determinants — Shortcuts using properties
- Linear Equations — Cramer’s rule uses cofactors
Math Connections
- Recursion — Determinants defined recursively
- Permutations — Sign of permutation in determinant
- Linear Algebra — Geometric interpretation of cofactors
Real-World Applications
- Inverse calculation — Essential for solving systems
- Cramer’s rule — Direct solution method
- Eigenvalue problems — Characteristic equation uses cofactors
Teacher’s Summary
- Minor $M_{ij}$: Determinant after deleting row $i$ and column $j$
- Cofactor $C_{ij}$: Minor with sign $(-1)^{i+j}$ (checkerboard pattern)
- Sign pattern: Start with $+$ at $(1,1)$, alternate like checkerboard
- Determinant expansion: Sum of (element × cofactor) along any row/column
- Smart strategy: Expand along row/column with most zeros
- Different row property: $\sum a_{ij} C_{kj} = 0$ if $i \neq k$ (crucial for adjoint!)
- Triangular matrices: Just multiply diagonal (skip all cofactor work!)
- Cofactor matrix: Foundation for adjoint and inverse
“Choose wisely where you expand — one zero can save you 2 minutes of calculation!”
Exam Strategy:
- Step 1: Check if triangular/diagonal → direct answer
- Step 2: Find row/column with most zeros
- Step 3: Expand there using cofactors
- Step 4: Be careful with signs (checkerboard!)