Class 12 Maths Chapter 4: Determinants

Determinants is Chapter 4 of NCERT Class 12 Maths. It shows how to attach a single number, the determinant, to every square matrix and how to use it to find areas of triangles, minors and cofactors, the adjoint and inverse of a matrix, and the solution of a system of linear equations. These notes and solved examples cover every formula you need for the CBSE board exam in a quick, easy-to-revise format.

What you will learn

How to evaluate determinants of order 2 × 2 and 3 × 3
Finding the area of a triangle using determinants
Minors, cofactors and the adjoint of a square matrix
The formula A⁻¹ = (1/|A|) adj A and when a matrix is invertible
Solving a system of linear equations by the matrix method and checking its consistency

Key Concepts and Formulas

Evaluating a Determinant

Every square matrix A has a number linked to it called its determinant, written |A| or det A. Only square matrices have determinants. A 3 × 3 determinant is found by expanding along any one row or column using cofactors.

| a  b |
| c  d | = ad − bc
Expanding a 3 × 3 along row 1:
|A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃
|kA| = kⁿ|A|   (A of order n)
|Aᵀ| = |A|
|AB| = |A| × |B|

Expanding along the row or column with the most zeros saves time.

Area of a Triangle

If the vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), its area can be written as half the absolute value of a determinant.

Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁)
          + x₃(y₁ − y₂)|
Area = ½ | x₁  y₁  1 |
         | x₂  y₂  1 |
         | x₃  y₃  1 |   (take modulus)
Area = 0 ⇒ the points are collinear

Area is always taken as positive, so use the absolute value.

Minors and Cofactors

These are the building blocks of the adjoint and the inverse. The minor Mᵢⱼ is the determinant left after deleting the i-th row and j-th column.

Aᵢⱼ = (−1)^(i + j) × Mᵢⱼ
Sign pattern:  + − +
               − + −
               + − +
a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ = |A|
a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0

Cofactor sign pattern and expansion of a 3 by 3 determinant along row 1, Class 12 Determinants — Cofactor signs and expansion along the first row

Elements of a row times their own cofactors give |A|; times cofactors of a different row give 0.

Adjoint and Inverse of a Matrix

The adjoint of A is the transpose of the matrix of cofactors. It leads directly to the inverse.

adj A = (matrix of cofactors)ᵀ
For A = [ a  b ]   adj A = [  d  −b ]
        [ c  d ]           [ −c   a ]
A (adj A) = (adj A) A = |A| I
A⁻¹ = (1/|A|) adj A,  if |A| ≠ 0
|adj A| = |A|^(n − 1)   (A of order n)

A is singular if |A| = 0 and non-singular if |A| ≠ 0.
A is invertible if and only if it is non-singular.
For a 2 × 2 matrix, adj A is found by swapping the diagonal elements and changing the signs of the other two.

Solving Linear Equations by the Matrix Method

A system of equations can be written as AX = B, where A is the coefficient matrix, X the column of unknowns and B the column of constants.

AX = B
|A| ≠ 0 ⇒ X = A⁻¹B  (unique solution)
|A| = 0, (adj A)B ≠ O ⇒ no solution
|A| = 0, (adj A)B = O ⇒ infinitely many
                        or no solution

A system is consistent if it has at least one solution and inconsistent if it has none.

Solved Examples

Example 1: Evaluate the determinant given in the working.

| 3  −2 |
| 5   4 |
= 3 × 4 − (−2) × 5
= 12 + 10
∴ Value = 22

Example 2: Evaluate the 3 × 3 determinant given in the working.

| 2   1  −1 |
| 0   3   2 |
| 1  −2   4 |
Expand along R₁:
= 2(3×4 − 2×(−2)) − 1(0×4 − 2×1)
  + (−1)(0×(−2) − 3×1)
= 2(16) − 1(−2) − 1(−3)
= 32 + 2 + 3
∴ Value = 37

Example 3: Find the area of the triangle with vertices (1, 2), (4, 6) and (−2, 3).

Area = ½ |1(6 − 3) + 4(3 − 2)
          + (−2)(2 − 6)|
     = ½ |3 + 4 + 8|
     = ½ × 15
∴ Area = 7.5 square units

Triangle with vertices (1, 2), (4, 6) and (−2, 3) with area found using determinants, Class 12 Maths — The triangle in Example 3 has area 7.5 square units

Example 4: Find k if the area of the triangle with vertices (k, 0), (3, 0) and (0, 4) is 6 square units.

½ |k(0 − 4) + 3(4 − 0) + 0(0 − 0)| = 6
|12 − 4k| = 12
12 − 4k = 12   or   12 − 4k = −12
k = 0          or   k = 6
∴ k = 0 or k = 6

Example 5: Find the inverse of the matrix A given in the working.

A = [ 5  −3 ]
    [ 2   7 ]
|A| = 35 − (−6) = 41 ≠ 0
A₁₁ = 7,   A₁₂ = −2
A₂₁ = 3,   A₂₂ = 5
adj A = [  7  3 ]
        [ −2  5 ]
∴ A⁻¹ = (1/41) [  7  3 ]
               [ −2  5 ]

Example 6: Solve 2x + 3y = 8 and x − y = −1 by the matrix method.

A = [ 2   3 ]   X = [ x ]   B = [  8 ]
    [ 1  −1 ]       [ y ]       [ −1 ]
|A| = −2 − 3 = −5 ≠ 0
adj A = [ −1  −3 ]
        [ −1   2 ]
X = (1/|A|)(adj A)B
  = (−1/5) [ −8 + 3 ] = (−1/5) [  −5 ]
           [ −8 − 2 ]          [ −10 ]
  = [ 1 ]
    [ 2 ]
∴ x = 1, y = 2

Example 7: Check the consistency of the system x + 2y = 3 and 2x + 4y = 7.

A = [ 1  2 ]   B = [ 3 ]
    [ 2  4 ]       [ 7 ]
|A| = 4 − 4 = 0
adj A = [  4  −2 ]
        [ −2   1 ]
(adj A)B = [ 12 − 14 ] = [ −2 ]
           [ −6 + 7  ]   [  1 ]
(adj A)B ≠ O
∴ The system is inconsistent (no solution)

Common Mistakes to Avoid

Do not forget the sign pattern of cofactors; a missed minus sign changes the whole adjoint and inverse.
The adjoint is the transpose of the cofactor matrix; placing cofactors without transposing is a frequent error.
Before writing A⁻¹, always check that |A| ≠ 0 and mention it in your answer; CBSE marking schemes award a mark for this step.
In area questions with an unknown, remember the modulus gives two cases, so solve for both values.

Quick Revision

2 × 2 determinant = ad − bc
|kA| = kⁿ|A|   (order n)
Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
Aᵢⱼ = (−1)^(i + j) Mᵢⱼ
A (adj A) = |A| I
A⁻¹ = (1/|A|) adj A,  |A| ≠ 0
AX = B ⇒ X = A⁻¹B