Determinants – Class 12 Maths Chapter 4 Notes & NCERT PDF

Class 12MathematicsChapter 4NCERT book: Mathematics Part I - Class 12

Notes and a simple summary of Chapter 4 of the NCERT Class 12 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 4: Determinants

Introduction

A determinant is a single number attached to every square matrix. If A = [aᵢⱼ] is of order n, its determinant is written |A|, det A or ∆. Determinants were first used to decide whether a system of linear equations has a unique solution, and they remain the quickest test for that today. This chapter (in the rationalised NCERT syllabus) covers determinants up to order 3, area of a triangle, minors and cofactors, adjoint and inverse of a matrix, and solving linear systems by the matrix method.

Only square matrices have determinants. A 2 × 3 matrix has no determinant.

Key Concepts & Theorems

4.2 Determinant of orders 1, 2 and 3

Order 1: if A = [a], then |A| = a.

Order 2: for A = [[a₁₁, a₁₂], [a₂₁, a₂₂]], |A| = a₁₁a₂₂ − a₂₁a₁₂ (product of the main diagonal minus product of the other diagonal).

Order 3: expand along any row or column. Along R₁:

det A = a₁₁(a₂₂a₃₃ − a₃₂a₂₃) − a₁₂(a₂₁a₃₃ − a₃₁a₂₃) + a₁₃(a₂₁a₃₂ − a₃₁a₂₂).

Signs follow the checkerboard pattern (−1)ⁱ⁺ʲ: + − + / − + − / + − +.

Tip from the book: expand along the row or column with the most zeros — it gives the same value with less work.

Useful facts: if A is of order n, then |kA| = kⁿ|A|; |A′| = |A|.

4.3 Area of a triangle

For vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):

∆ = ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|, and we take the absolute value, because area is always positive.

If the area is zero, the three points are collinear.

Replacing one vertex by a general point (x, y) and setting ∆ = 0 gives the equation of a line through two points.

4.4 Minors and cofactors

Minor Mᵢⱼ: the determinant left after deleting the i-th row and j-th column. A minor of a 3 × 3 determinant is a 2 × 2 determinant.

Cofactor Aᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ.

Value of a determinant = sum of (elements of any row/column × their own cofactors).

If elements of one row are multiplied by cofactors of a different row, the sum is 0.

4.5 Adjoint and inverse

adj A = transpose of the matrix of cofactors. For a 2 × 2 matrix [[a, b], [c, d]], adj A = [[d, −b], [−c, a]] (swap the diagonal, change signs of the other two).

Theorem 1: A(adj A) = (adj A)A = |A| I.

Singular matrix: |A| = 0; non-singular: |A| ≠ 0.

Theorem 2: if A, B are non-singular of the same order, AB and BA are non-singular.

Theorem 3: |AB| = |A| |B|. As a consequence, |adj A| = |A|ⁿ⁻¹ for order n.

Theorem 4: A is invertible if and only if A is non-singular, and then A⁻¹ = (1/|A|) adj A.

4.6 Solving linear equations by the matrix method

Write the system as AX = B.

|A| ≠ 0: unique solution X = A⁻¹B — the system is consistent.

|A| = 0: find (adj A)B.

- If (adj A)B ≠ O, there is no solution (inconsistent).

- If (adj A)B = O, the system may have infinitely many solutions or none.

Worked Examples

Area (Example 6). Vertices (3, 8), (−4, 2), (5, 1): ∆ = ½|3(2 − 1) − 8(−4 − 5) + 1(−4 − 10)| = ½|3 + 72 − 14| = ½ × 61 = 61/2 square units.

Line through A(1, 3) and B(0, 0) (Example 7). Set ½|x y 1; 1 3 1; 0 0 1| = 0 ⇒ 3x − y = 0, i.e. y = 3x.

2 × 2 system (Example 16). 2x + 5y = 1, 3x + 2y = 7. |A| = 4 − 15 = −11 ≠ 0, so a unique solution exists. A⁻¹ = −(1/11)[[2, −5], [−3, 2]], giving x = 3, y = −1.

3 × 3 system (Example 17). 3x − 2y + 3z = 8, 2x + y − z = 1, 4x − 3y + 2z = 4. |A| = −17 ≠ 0; X = A⁻¹B gives x = 1, y = 2, z = 3.

Word problem (Example 18). Three numbers add to 6; third × 3 + second = 11; first + third = 2 × second. So x + y + z = 6, y + 3z = 11, x − 2y + z = 0 ⇒ 1, 2, 3.

Inverse (Example 13 style). For A = [[1, 3, 3], [1, 4, 3], [1, 3, 4]], |A| = 1 and adj A = [[7, −3, −3], [−1, 1, 0], [−1, 0, 1]], so A⁻¹ = adj A. Check: A(adj A) = I = |A| I.

Formula Sheet

2 × 2: |A| = ad − bc

Aᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ

A(adj A) = (adj A)A = |A| I

A⁻¹ = (1/|A|) adj A, only when |A| ≠ 0

|AB| = |A||B|; |kA| = kⁿ|A|; |adj A| = |A|ⁿ⁻¹; |A⁻¹| = 1/|A|

Area = ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|, taken positive

AX = B ⇒ X = A⁻¹B when |A| ≠ 0

Board Exam Focus

1-mark: value of a 2 × 2 determinant; find x when two determinants are equal; |kA| or |adj A| for a given |A|.

2–3 marks: area of a triangle or collinearity; minors and cofactors of all elements; adj A and verifying A(adj A) = |A| I.

5 marks: solve a 3-variable system by the matrix method, often as a word problem (like Example 18). Outline: (i) write AX = B, (ii) show |A| ≠ 0, (iii) find all nine cofactors, (iv) form adj A and A⁻¹, (v) compute X = A⁻¹B, (vi) state the answer and, if time allows, substitute back to check.

Key Terms

Determinant (सारणिक)

Square matrix (वर्ग आव्यूह)

Minor (उपसारणिक)

Cofactor (सहखंड)

Adjoint (सहखंडज)

Inverse matrix (व्युत्क्रम आव्यूह)

Singular matrix (अव्युत्क्रमणीय / अव्युत्क्रमी आव्यूह)

Non-singular matrix (व्युत्क्रमणीय आव्यूह)

Consistent system (संगत निकाय)

Inconsistent system (असंगत निकाय)

Collinear points (संरेख बिंदु)

Area of triangle (त्रिभुज का क्षेत्रफल)

Expansion (प्रसार)

Common Mistakes

Forgetting the (−1)ⁱ⁺ʲ sign when turning a minor into a cofactor.

Writing adj A as the cofactor matrix without transposing it.

Using |kA| = k|A| instead of kⁿ|A| (for a 3 × 3 matrix, |2A| = 8|A|).

Leaving the area negative — always take the absolute value.

Trying to use X = A⁻¹B when |A| = 0; check the determinant first.

Arithmetic slips in the 9 cofactors — verify with A(adj A) = |A| I before moving on.

💡 Key Learning Points

✓Evaluate determinants of order 1, 2 and 3 by expanding along any row or column, choosing the one with most zeros.
✓Use the determinant formula for the area of a triangle, test collinearity and find the equation of a line through two points.
✓Find minors Mᵢⱼ and cofactors Aᵢⱼ = (−1)ⁱ⁺ʲMᵢⱼ, and form adj A as the transpose of the cofactor matrix.
✓Apply A(adj A) = |A|I and A⁻¹ = (1/|A|) adj A, knowing A is invertible exactly when |A| ≠ 0.
✓Solve systems of linear equations by the matrix method X = A⁻¹B and decide consistency using |A| and (adj A)B.

👨‍🏫 Teaching Tips

→Draw the (−1)ⁱ⁺ʲ checkerboard of signs on the board and keep it visible throughout the chapter.
→Have students expand the same 3 × 3 determinant along two different rows to see the value does not change.
→Use a grid-paper triangle: compute its area by counting squares and by the determinant formula, then compare.
→Teach the 2 × 2 adjoint shortcut (swap diagonal, negate off-diagonal) before the full 3 × 3 cofactor method.
→Insist that students verify A(adj A) = |A|I after computing cofactors; it catches most sign errors.
→Turn Example 18 into a class activity where groups write their own three-number puzzle and solve a partner's.
→Give a quick 'singular or not?' round with several matrices before introducing the matrix method.

📋 Assessment Questions

Can the student evaluate a 3 × 3 determinant correctly and choose an efficient row or column?
Can the student find all cofactors, form adj A and verify A(adj A) = |A|I?
Can the student apply |kA| = kⁿ|A| and |adj A| = |A|ⁿ⁻¹ in one-mark questions?
Can the student solve a word problem by writing AX = B and computing X = A⁻¹B?
Can the student explain what happens when |A| = 0?

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