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Chapter 4: Determinants

Class 12 • Mathematics • Chapter 4

Determinants

One number from a square matrix that decides area, invertibility and whether a system can be solved.

CBSE / NCERT Aligned

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Chapter Roadmap

Determinant of a Matrix • Minors and Cofactors • Expansion Along Any Row or Column • Properties of Determinants • Area of a Triangle • Adjoint of a Matrix • Inverse Using the Adjoint • Solving Linear Systems

Why a Single Number Matters So Much

A matrix is a whole grid of numbers, yet for every square matrix we can squeeze out just one number that tells us an enormous amount. That number is the determinant. If it is zero, the matrix has no inverse and the linear system behind it either fails or has infinitely many answers. If it is not zero, the matrix can be inverted and the system has exactly one solution.

The determinant also carries a clear geometric meaning: for a 2 × 2 matrix it measures the area of the parallelogram built from its columns, and a related formula gives the area of a triangle straight from the coordinates of its corners. In this chapter you will learn to compute determinants of any order, simplify them with properties, and use them to find adjoints, inverses and solutions of equations.

Foundation

The determinant of a square matrix A is written det(A) or |A|. For a 2 × 2 matrix [a, b; c, d], the value is ad − bc. A matrix with |A| = 0 is called singular; one with |A| ≠ 0 is non-singular and has an inverse.

Key Terms You Must Know

Term

Meaning

Example

Determinant |A|

A single number attached to every square matrix.

|[3, 1; 2, 4]| = 10

Minor Mᵢⱼ

Determinant left after deleting row i and column j.

Delete a row and a column

Cofactor Aᵢⱼ

The minor with the sign (−1)^(i+j) attached.

A₁₂ = −M₁₂

Singular matrix

A square matrix whose determinant is 0; it has no inverse.

|[1, 2; 2, 4]| = 0

Adjoint adj A

Transpose of the matrix of cofactors.

Used to build the inverse

Inverse A⁻¹

The matrix with A A⁻¹ = I, equal to (1/|A|) adj A.

Exists only if |A| ≠ 0

Core Concepts, Step by Step

1. What a Determinant Is

For a 1 × 1 matrix [a] the determinant is simply a. For a 2 × 2 matrix [a, b; c, d] it is ad − bc, the product of the main diagonal minus the product of the other diagonal. For a 3 × 3 matrix we expand it into three 2 × 2 determinants. The determinant is defined only for square matrices, and it returns one ordinary number that can be positive, negative or zero.

A 3 × 3 matrix whose determinant we will expand

A =

2. Minors and Cofactors

The minor Mᵢⱼ of an entry is the smaller determinant you get by crossing out that entry’s whole row and column. The cofactor Aᵢⱼ is the same minor with a sign attached, following the chessboard pattern of (−1)^(i+j): plus, minus, plus along the top row. Cofactors are the building blocks for expanding a determinant, for the adjoint and for the inverse.

3. Expanding Along Any Row or Column

To evaluate a 3 × 3 determinant, pick any one row or column, multiply each entry by its cofactor, and add the results. Choosing a row or column that already contains zeros saves work, because those terms vanish. A key fact is that you get the same value no matter which row or column you expand along, which is a useful self-check.

4. Properties That Simplify Determinants

Determinants obey neat rules. Swapping two rows (or two columns) flips the sign. If two rows are identical, the determinant is 0. A common factor in any row can be taken outside. Adding a multiple of one row to another leaves the determinant unchanged, which is the property used to create zeros before expanding. The determinant of a triangular matrix is just the product of its diagonal entries, and det(A′) = det(A).

5. Area of a Triangle from Coordinates

If a triangle has corners (x₁, y₁), (x₂, y₂) and (x₃, y₃), its area equals one half of the absolute value of the determinant [x₁, y₁, 1; x₂, y₂, 1; x₃, y₃, 1]. The modulus is taken because area is never negative. A beautiful by-product: if this determinant comes out as 0, the three points lie on one straight line, so they are collinear.

Area of triangle ABC from the determinant of its coordinates

Triangle with vertices A(1,1), B(2,3), C(4,1) plotted on coordinate axes, area found by determinant

6. Adjoint, Inverse and Solving Equations

The adjoint of A is the transpose of its cofactor matrix. It satisfies A (adj A) = (adj A) A = |A| I. From this comes the inverse: when |A| ≠ 0, A⁻¹ = (1/|A|) adj A. Inverses let us solve a system written as AX = B by computing X = A⁻¹ B, giving one unique answer whenever |A| ≠ 0. For a 2 × 2 matrix the determinant even measures the area of the parallelogram formed by its two columns.

For a 2 × 2 matrix, |det| is the area of the parallelogram of its columns

Parallelogram built from column vectors (3,1) and (1,4) with area equal to the determinant value 11

Key Results with Proofs

Result 1: The Determinant of a Product

Statement. For two square matrices of the same order, det(AB) = det(A) det(B).

Proof

The general proof uses row operations; here is a concrete check that shows the pattern.

Take A = [2, 0; 1, 3] and B = [1, 4; 0, 2] as a worked illustration. – choose two matrices

det(A) = 2×3 − 0×1 = 6 and det(B) = 1×2 − 4×0 = 2. – find each determinant

AB

=

[2, 0; 1, 3][1, 4; 0, 2] = [2, 8; 1, 10]

multiply the matrices

det(AB) = 2×10 − 8×1 = 12 = 6 × 2 = det(A) det(B). – the product rule holds

A direct consequence: det(A⁻¹) = 1/det(A), since det(A) det(A⁻¹) = det(I) = 1.

Result 2: The Adjoint Identity

Statement. For any square matrix A, A (adj A) = (adj A) A = |A| I.

Proof

Each diagonal entry of A (adj A) is an expansion of |A|; each off-diagonal entry is an expansion with a repeated row, which is 0.

Take A = [2, 3; 1, 4], so |A| = 8 − 3 = 5. – pick a 2 × 2 matrix

The cofactors give adj A = [4, −3; −1, 2]. – transpose the cofactor matrix

A (adj A)

=

[2, 3; 1, 4][4, −3; −1, 2] = [5, 0; 0, 5]

multiply

[5, 0; 0, 5] = 5 I = |A| I, exactly as claimed. – matches |A| I

This identity is the engine behind the inverse formula in Result 3.

Result 3: The Inverse Formula

Statement. If |A| ≠ 0 then A is invertible and A⁻¹ = (1/|A|) adj A.

Proof

The adjoint identity converts straight into a formula for the inverse.

Start from Result 2: A (adj A) = |A| I. – use the adjoint identity

Since |A| ≠ 0, divide both sides by |A|: A × [(1/|A|) adj A] = I. – divide by the non-zero number |A|

The matrix multiplying A to give I is by definition A⁻¹. – read off the inverse

Therefore A⁻¹ = (1/|A|) adj A. – conclusion

If |A| = 0 no inverse exists, which is why a singular matrix can never be inverted.

Worked Examples

Example 1

Question: Evaluate the determinant of [3, 1; 2, 4].

▶ Show full working

Use ad − bc for a 2 × 2 matrix.

Main diagonal product: 3 × 4 = 12. – ad

Other diagonal product: 1 × 2 = 2. – bc

Subtract: 12 − 2 = 10. – ad − bc

Answer: |[3, 1; 2, 4]| = 10.

Example 2

Question: Evaluate |[1, 2, 3; 0, 4, 5; 1, 0, 6]| by expanding along the first column.

▶ Show full working

Two entries in the first column work; the middle 0 drops out.

Expand along column 1: 1·C₁₁ + 0·C₂₁ + 1·C₃₁. – cofactor expansion

C₁₁ = +|[4, 5; 0, 6]| = 24. C₃₁ = +|[2, 3; 4, 5]| = 10 − 12 = −2. – two 2 × 2 minors with signs

Total: 1×24 + 1×(−2) = 22. – add the surviving terms

Answer: The determinant is 22.

Example 3

Question: Find the minor and cofactor of the entry 6 (position 2,3) in [2, 1, 3; 4, 5, 6; 7, 8, 9].

▶ Show full working

Delete row 2 and column 3, then attach the sign.

Deleting row 2, column 3 leaves [2, 1; 7, 8]. – strike out the row and column

Minor M₂₃ = 2×8 − 1×7 = 16 − 7 = 9. – evaluate the 2 × 2

Sign is (−1)^(2+3) = −1, so cofactor A₂₃ = −9. – apply the chessboard sign

Answer: Minor = 9, cofactor = −9.

Example 4

Question: Without full expansion, state |[2, 4, 6; 1, 0, 5; 3, 1, 2]| after taking out the common factor of row 1.

▶ Show full working

Row 1 has a common factor of 2.

Take 2 out of row 1: the determinant equals 2 × |[1, 2, 3; 1, 0, 5; 3, 1, 2]|. – factor property

Expanding the bracket gives |[1, 2, 3; 1, 0, 5; 3, 1, 2]| = 24. – evaluate the smaller determinant

So the original determinant is 2 × 24 = 48. – multiply back

Answer: The determinant is 48.

Example 5

Question: Show that swapping two rows changes the sign, using [1, 2; 3, 4].

▶ Show full working

Compute the determinant, swap rows, recompute.

|[1, 2; 3, 4]| = 1×4 − 2×3 = −2. – original value

Swap the rows to get [3, 4; 1, 2]; its determinant is 3×2 − 4×1 = 2. – after the swap

The value changed from −2 to +2, the same size with opposite sign. – sign flipped

Answer: A single row swap multiplies the determinant by −1.

Example 6

Question: Find the area of the triangle with vertices A(1, 1), B(2, 3), C(4, 1).

▶ Show full working

Use one half the absolute value of the coordinate determinant.

Area = ½ |1(3 − 1) + 2(1 − 1) + 4(1 − 3)|. – plug into the formula

= ½ |2 + 0 − 8| = ½ × 6. – simplify inside the bars

= 3 square units. – final value

Answer: Area = 3 square units.

Example 7

Question: Check whether the points (1, 2), (2, 4) and (3, 6) are collinear.

▶ Show full working

They are collinear exactly when the coordinate determinant is 0.

Evaluate 1(4 − 6) + 2(6 − 2) + 3(2 − 4). – set up the determinant

= 1(−2) + 2(4) + 3(−2) = −2 + 8 − 6 = 0. – add the terms

A zero value means zero area, so the points lie on one line. – interpret

Answer: Yes, the three points are collinear.

Example 8

Question: Find the adjoint of A = [2, 3; 1, 4].

▶ Show full working

For a 2 × 2 matrix, swap the main diagonal and negate the other diagonal.

Cofactors: A₁₁=4, A₁₂=−1, A₂₁=−3, A₂₂=2. – compute the four cofactors

adj A is the transpose of the cofactor matrix [4, −1; −3, 2]. – transpose

So adj A = [4, −3; −1, 2]. – write it out

Answer: adj A = [4, −3; −1, 2].

Example 9

Question: Find the inverse of A = [2, 5; 1, 3].

▶ Show full working

Use A⁻¹ = (1/|A|) adj A.

|A| = 2×3 − 5×1 = 1, which is non-zero so the inverse exists. – find the determinant

adj A = [3, −5; −1, 2]. – adjoint of a 2 × 2

A⁻¹ = (1/1)[3, −5; −1, 2] = [3, −5; −1, 2]. – divide by |A|

Answer: A⁻¹ = [3, −5; −1, 2].

Example 10

Question: Solve the system 2x + 3y = 8, x + 4y = 9 by the matrix method.

▶ Show full working

Write AX = B and use X = A⁻¹ B.

A = [2, 3; 1, 4], B = [8; 9], |A| = 8 − 3 = 5. – set up matrices

A⁻¹ = (1/5)[4, −3; −1, 2]. – adjoint over determinant

X = (1/5)[4, −3; −1, 2][8; 9] = (1/5)[5; 10] = [1; 2]. – multiply

Answer: x = 1, y = 2.

Example 11

Question: Find the determinant of the upper-triangular matrix [3, 7, 2; 0, 5, 4; 0, 0, 2].

▶ Show full working

For a triangular matrix the determinant is the product of the diagonal.

The entries below the diagonal are all 0. – recognise triangular form

Multiply the diagonal: 3 × 5 × 2. – product of diagonal

= 30. – result

Answer: The determinant is 30.

Example 12

Question: Determine whether A = [1, 2; 2, 4] has an inverse.

▶ Show full working

A matrix is invertible only when its determinant is non-zero.

|A| = 1×4 − 2×2 = 4 − 4 = 0. – compute the determinant

Because the determinant is 0, the matrix is singular. – interpret

A singular matrix has no inverse. – conclusion

Answer: No, A is singular and has no inverse.

Where You Meet This in Real Life

Computer graphics

Determinants tell graphics engines whether a transformation flips or squashes a shape, and scale factors for areas and volumes come straight from them.

Engineering and physics

Solving networks of equations for currents, forces or stresses relies on determinants to confirm a unique solution exists.

Surveying and GPS

Areas of irregular land plots are found from corner coordinates using the same triangle determinant formula.

Economics

Input-output models that balance many industries are solved with matrix inverses, which depend on a non-zero determinant.

Cryptography

Some classic ciphers encode messages with matrices and can only be decoded when the key matrix has a determinant that is invertible.

Practice Sets A–D

Practice Set A – Basics

A1. Evaluate |[5, 2; 3, 4]|.

▶ Reveal full working

Use ad − bc.

5×4 − 2×3 = 20 − 6 = 14.

Answer: 14.

A2. Evaluate |[2, 0; 7, 3]|.

▶ Reveal full working

Use ad − bc.

2×3 − 0×7 = 6.

Answer: 6.

A3. Write the sign attached to the cofactor at position (2, 1).

▶ Reveal full working

Use (−1)^(i+j).

(−1)^(2+1) = (−1)^3 = −1.

Answer: Negative.

A4. State the determinant of the identity matrix [1, 0; 0, 1].

▶ Reveal full working

Apply ad − bc.

1×1 − 0×0 = 1.

Answer: 1.

Practice Set B – Conceptual

B1. Why does a matrix with two equal rows have determinant 0?

▶ Reveal full working

Link it to the row-swap property.

Swapping the two equal rows changes the sign but leaves the matrix unchanged. – swap identical rows

So the determinant equals its own negative, which forces it to be 0. – value = −value

Answer: Because the determinant must equal its own negative, so it is 0.

B2. What does |A| = 0 tell you about solving AX = B?

▶ Reveal full working

Connect to invertibility.

A zero determinant means A has no inverse. – singular

So AX = B cannot be solved by X = A⁻¹ B; the system has no unique solution. – no unique answer

Answer: The system has no unique solution.

B3. If det(A) = 4, what is det(A⁻¹)?

▶ Reveal full working

Use det(A) det(A⁻¹) = 1.

Since the product of the determinants is det(I) = 1, det(A⁻¹) = 1/4. – reciprocal

Answer: det(A⁻¹) = 1/4.

B4. A 3 × 3 matrix is multiplied throughout by 2. By what factor does its determinant change?

▶ Reveal full working

Each of the three rows contributes one factor.

Taking 2 out of each of the 3 rows gives 2 × 2 × 2. – factor per row

So the determinant is multiplied by 2³ = 8. – total factor

Answer: It is multiplied by 8.

Practice Set C – Application / Numerical

C1. Evaluate |[1, 2, 1; 0, 3, 1; 2, 1, 0]| by expanding along the first column.

▶ Reveal full working

Use the entries 1 and 2 in column 1.

1·+|[3, 1; 1, 0]| = 1×(0 − 1) = −1. – first cofactor

2·+|[2, 1; 3, 1]| = 2×(2 − 3) = −2. – third cofactor, sign +

Sum: −1 + (−2) = −3. – add

Answer: −3.

C2. Find the area of the triangle with vertices (0, 0), (4, 0) and (0, 3).

▶ Reveal full working

Use the coordinate determinant formula.

Area = ½|0(0 − 3) + 4(3 − 0) + 0(0 − 0)|. – substitute

= ½|12| = 6. – simplify

Answer: 6 square units.

C3. Find the inverse of [4, 7; 2, 6].

▶ Reveal full working

Use (1/|A|) adj A.

|A| = 4×6 − 7×2 = 24 − 14 = 10. – determinant

adj A = [6, −7; −2, 4]. – adjoint

A⁻¹ = (1/10)[6, −7; −2, 4]. – divide by 10

Answer: A⁻¹ = (1/10)[6, −7; −2, 4].

C4. Solve x + 2y = 4, 3x + y = 7 using determinants.

▶ Reveal full working

Use ratios of determinants (Cramer’s style).

D = |[1, 2; 3, 1]| = 1 − 6 = −5. – main determinant

Dₓ = |[4, 2; 7, 1]| = 4 − 14 = −10, so x = −10/−5 = 2. – x value

Dₑ = |[1, 4; 3, 7]| = 7 − 12 = −5, so y = −5/−5 = 1. – y value

Answer: x = 2, y = 1.

Practice Set D – HOTS / Multi-step

D1. If A = [2, 0; 1, 3] and B = [1, 4; 0, 2], verify det(AB) = det(A) det(B).

▶ Reveal full working

Compute both sides.

det(A) = 6, det(B) = 2, so the product is 12. – right side

AB = [2, 8; 1, 10], det(AB) = 20 − 8 = 12. – left side

Both sides equal 12. – verified

Answer: Both equal 12, so the rule holds.

D2. Find the value of k for which the points (1, 2), (3, k) and (5, 8) are collinear.

▶ Reveal full working

Set the coordinate determinant to 0.

1(k − 8) + 3(8 − 2) + 5(2 − k) = 0. – collinearity condition

k − 8 + 18 + 10 − 5k = 0, so −4k + 20 = 0. – simplify

k = 5. – solve

Answer: k = 5.

D3. For A = [2, 3; 1, 4], verify A (adj A) = |A| I.

▶ Reveal full working

Multiply A by its adjoint.

|A| = 5 and adj A = [4, −3; −1, 2]. – preliminaries

A (adj A) = [2, 3; 1, 4][4, −3; −1, 2] = [5, 0; 0, 5]. – multiply

[5, 0; 0, 5] = 5 I = |A| I. – matches

Answer: A (adj A) = 5 I = |A| I, verified.

D4. Solve 2x + y = 5 and 4x + 2y = 11 using determinants, and explain the result.

▶ Reveal full working

Check the main determinant first.

D = |[2, 1; 4, 2]| = 4 − 4 = 0. – main determinant is 0

A zero determinant means there is no unique solution. – interpret

Here the lines are parallel but distinct, so the system is inconsistent (no solution). – describe the geometry

Answer: D = 0, so the system has no unique solution; in fact it is inconsistent.

Chapter Summary

Everything in One Glance

Determinant

One number from a square matrix; 2 × 2 is ad − bc, larger orders expand by cofactors.

Minor & Cofactor

Minor is the smaller determinant after deleting a row and column; cofactor adds the (−1)^(i+j) sign.

Properties

Row swap flips sign; equal rows give 0; common factors come out; row operations keep the value.

Area & Collinearity

Triangle area = ½|coordinate determinant|; a value of 0 means the points are collinear.

Adjoint & Inverse

adj A is the transpose of cofactors; A⁻¹ = (1/|A|) adj A when |A| ≠ 0.

Solving Systems

AX = B has the unique solution X = A⁻¹ B exactly when |A| ≠ 0.

Are You Exam-Ready?

8-Point Exam Quick-Check

Evaluate |[6, 2; 4, 3]|.

State the sign of the cofactor at position (3, 2).

What does a determinant of 0 tell you about the inverse?

Find the area of the triangle with vertices (0, 0), (6, 0), (0, 4).

Write the adjoint of [1, 2; 3, 4].

If det(A) = 5, find det(A⁻¹).

Find the determinant of the diagonal matrix [2, 0; 0, 7].

Are (1, 1), (2, 2), (3, 3) collinear? Justify in one line.

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Class 12 Mathematics Chapter 4: Determinants, Complete Notes and Practice

These free Class 12 Maths Determinants notes follow the NCERT 2026 to 27 syllabus and cover determinant evaluation, minors and cofactors, properties, area of a triangle, adjoint, inverse and solving linear systems with twelve worked examples and sixteen graded practice questions. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.