NCERT Solutions Class 12 Mathematics Chapter 4 Determinants

These comprehensive NCERT Solutions Class 12 Mathematics Chapter 4 Determinants provide rigorous, step-by-step guidance designed to help students master one of the highest-scoring chapters in the CBSE Class 12 syllabus. Determinants form a core pillar of linear algebra, connecting seamlessly with matrices, analytical geometry, coordinate geometry, and the solution of simultaneous linear equations. Whether you are preparing for your CBSE Class 12 board examinations, CUET, or JEE Main, having access to accurate NCERT Mathematics Class 12 solutions is vital for conceptual clarity and exam success.

Chapter 4 of Class 12 Maths introduces the scalar value associated with every square matrix, known as its determinant. This guide covers all exercises in the rationalized NCERT curriculum, offering clear explanations, solved exercise questions, key theorems, proofs, and tips to eliminate calculation errors. Use these NCERT Solutions step by step to solidify your foundation and score full marks in your board exams.

Chapter Overview: Class 12 Mathematics Chapter 4 Determinants

In matrix theory, an n × n square matrix can be associated with a unique real or complex number termed its determinant. If A = [a_ij] is a square matrix of order n, the determinant of A is denoted by det(A) or |A|. Note that |A| represents the determinant value, not the absolute value or modulus of a matrix.

Under the updated CBSE curriculum, Chapter 4 Determinants encompasses the following core areas:

Determinant of Order 1, 2, and 3: Definition, expansion rules along any row or column, and computational shortcuts.
Applications of Determinants: Finding the area of a triangle given three vertices and establishing conditions for collinearity of three points.
Minors and Cofactors: Formal definitions, mathematical relationships, and expansion theorems.
Adjoint and Inverse of a Matrix: Singular vs. non-singular matrices, properties of adjoints, and matrix inversion using cofactors.
Applications of Matrices and Determinants: Checking consistency and solving systems of linear equations in two and three variables using the Matrix Inversion Method (X = A⁻¹B).

Real-World Applications of Determinants

Determinants are not merely theoretical constructs; they drive critical calculations in modern engineering and computational sciences:

Computer Graphics & 3D Animation: Determinants determine scaling factors and orientation changes during 3D geometric transformations and perspective projections.
Civil & Structural Engineering: Used in stiffness matrix methods and finite element analysis (FEA) to solve equilibrium equations of load-bearing structures.
Electrical Network Analysis: Mesh current and nodal voltage equations in complex circuits are solved directly using determinants and Cramer's rule.
Cryptography: Matrix inversion through determinants enables Hill Cipher encryption and secure data decryption protocols.

Important Formulas, Theorems, and Properties

Memorizing and understanding these standard formulas is essential before attempting NCERT Solutions Mathematics chapter wise practice:

1. Determinant Evaluation Formulas

Determinant of order 2 × 2:
If A = &begin;matrix [a₁₁, a₁₂; a₂₁, a₂₂] &end;matrix, then:
|A| = a₁₁a₂₂ − a₁₂a₂₁
Determinant of order 3 × 3 (Expansion along Row 1):
|A| = a₁₁(a₂₂a₃₃ − a₂₃a₃₂) − a₁₂(a₂₁a₃₃ − a₂₃a₃₁) + a₁₃(a₂₁a₃₂ − a₂₂a₃₁)
Scalar Multiplication Property: If A is a square matrix of order n, then |kA| = kⁿ|A|.

2. Area of a Triangle & Collinearity

The area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by:

Δ = ½ | x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) |

In determinant form:

Δ = ½ |det| where det = &begin;vmatrix x₁ & y₁ & 1 \\ x₂ & y₂ & 1 \\ x₃ & y₃ & 1 &end;vmatrix

Condition for Collinearity: Three points A, B, and C are collinear if and only if Δ = 0.

3. Minors and Cofactors

Minor (M_ij): The determinant of the submatrix obtained by deleting the i^th row and j^th column of matrix A.
Cofactor (A_ij): Defined as A_ij = (−1)^{i + j} M_ij.
Value of Determinant: Sum of elements of any row (or column) multiplied by their corresponding cofactors:
|A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃
Orthogonal Cofactor Property: Sum of products of elements of any row with the cofactors of any other row is zero:
a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0

4. Adjoint and Inverse of a Matrix

Adjoint of Matrix A (adj A): The transpose of the cofactor matrix of A:
adj A = [A_ij]^T
Fundamental Theorem: A(adj A) = (adj A)A = |A| I_n
Determinant of Adjoint: If A is an n × n matrix, then |adj A| = |A|ⁿ⁻¹
Inverse Matrix Formula: For any non-singular matrix A (|A| ≠ 0):
A⁻¹ = (1 / |A|) × adj A

5. Criteria for Consistency of Linear System (AX = B)

Condition	Nature of System	Number of Solutions
\|A\| ≠ 0 (Non-singular)	Consistent	Unique Solution: X = A⁻¹B
\|A\| = 0 and (adj A)B ≠ O	Inconsistent	No Solution
\|A\| = 0 and (adj A)B = O	Consistent or Inconsistent	Infinitely Many Solutions or No Solution

NCERT Solutions Class 12 Mathematics Chapter 4: Step-by-Step Exercise Solutions

Below are detailed, whiteboard-style solutions to representative and frequently asked problems from the NCERT textbook. Each solution illustrates the exact presentation required in CBSE board marking schemes.

Exercise 4.1: Determinant Evaluation & Scalar Property

Question 1: If A = &begin;matrix [1, 2; 4, 2] &end;matrix, then show that |2A| = 4|A|.

Given: Matrix A = &begin;matrix [1, 2; 4, 2] &end;matrix

To Prove: |2A| = 4|A|

Step-by-Step Solution:

First, calculate |A|:
|A| = (1)(2) − (2)(4) = 2 − 8 = −6
Calculate the Right-Hand Side (RHS):
RHS = 4 × |A| = 4 × (−6) = −24
Next, find matrix 2A by multiplying each element of A by scalar 2:
2A = &begin;matrix [2(1), 2(2); 2(4), 2(2)] &end;matrix = &begin;matrix [2, 4; 8, 4] &end;matrix
Evaluate the Left-Hand Side (LHS) = |2A|:
|2A| = (2)(4) − (4)(8) = 8 − 32 = −24

Conclusion: LHS = RHS = −24. Hence Proved. (General rule verified: |kA| = kⁿ|A| for n = 2).

Question 2: Evaluate the 3 × 3 determinant:
Δ = &begin;vmatrix 3 & −1 & −2 \\ 0 & 0 & −1 \\ 3 & −5 & 0 &end;vmatrix

Method / Tip: Expanding along the row or column with the maximum number of zeroes minimizes arithmetic calculations. Row 2 (R₂) contains two zeroes.

Step-by-Step Solution:

Expand along Row 2 (R₂):
Δ = −a₂₁ M₂₁ + a₂₂ M₂₂ − a₂₃ M₂₃
Substitute the elements of Row 2:
Δ = −0 × M₂₁ + 0 × M₂₂ − (−1) × &begin;vmatrix} 3 & −1 \\ 3 & −5 &end;vmatrix
Simplify the non-zero subdeterminant:
Δ = 1 × [ (3)(−5) − (−1)(3) ]
Δ = 1 × [ −15 − (−3) ] = −15 + 3 = −12

Final Answer: Δ = −12

Exercise 4.2: Area of Triangles and Line Equations

Question 3: Find the equation of the line joining points A(1, 3) and B(0, 0) using determinants. Also, find the value of k if D(k, 0) is a point such that the area of ΔABD is 3 sq units.

Part A: Equation of Line AB

Let P(x, y) be any arbitrary point on the line passing through A(1, 3) and B(0, 0). Since points P, A, and B are collinear, the area formed by them is zero:
½ &begin;vmatrix x & y & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 1 &end;vmatrix = 0
Expanding along Row 3 (R₃):
0 − 0 + 1 × [ (x)(3) − (y)(1) ] = 0
3x − y = 0 ⇒ y = 3x

Equation of Line: y − 3x = 0

Part B: Finding Parameter k

Vertices are A(1, 3), B(0, 0), and D(k, 0).
Area formula: Δ = ± 3 sq units
½ &begin;vmatrix 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 &end;vmatrix = ± 3
Expand along Column 2 (C₂) which has two zeroes:
½ × [ −3 × (0(1) − 1(k)) ] = ± 3
½ × [ −3(−k) ] = ± 3
½ (3k) = ± 3
3k = ± 6 ⇒ k = ± 2

Final Answer: k = 2 or k = −2

Exercise 4.3: Minors, Cofactors, and Expansion Theorems

Question 4: For matrix A = &begin;matrix [2, −3, 5; 6, 0, 4; 1, 5, −7] &end;matrix, verify that a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃ = 0.

Step-by-Step Verification:

Identify elements of Row 1: a₁₁ = 2, a₁₂ = −3, a₁₃ = 5.
Calculate cofactors of Row 3 elements (A₃₁, A₃₂, A₃₃):
A₃₁ = (−1)³⁺¹ M₃₁ = +1 × &begin;vmatrix −3 & 5 \\ 0 & 4 &end;vmatrix = (−3)(4) − (5)(0) = −12
A₃₂ = (−1)³⁺² M₃₂ = −1 × &begin;vmatrix 2 & 5 \\ 6 & 4 &end;vmatrix = −1 × [ (2)(4) − (5)(6) ] = −1 × [ 8 − 30 ] = −1(−22) = 22
A₃₃ = (−1)³⁺³ M₃₃ = +1 × &begin;vmatrix 2 & −3 \\ 6 & 0 &end;vmatrix = (2)(0) − (−3)(6) = 0 − (−18) = 18
Substitute values into the expression:
LHS = (2)(−12) + (−3)(22) + (5)(18)
LHS = −24 − 66 + 90 = −90 + 90 = 0

Conclusion: LHS = RHS = 0. This proves the theorem that multiplying elements of one row with cofactors of another row always equals zero.

Exercise 4.4: Adjoint and Matrix Inversion

Question 5: If A = &begin;matrix [2, 3; 1, 2] &end;matrix, prove that A² − 4A + I = O. Hence, find A⁻¹.

Step-by-Step Solution:

Compute A² = A × A:
A² = &begin;matrix [2, 3; 1, 2] &end;matrix × &begin;matrix [2, 3; 1, 2] &end;matrix = &begin;matrix [4+3, 6+6; 2+2, 3+4] &end;matrix = &begin;matrix [7, 12; 4, 7] &end;matrix
Compute 4A and I:
4A = &begin;matrix [8, 12; 4, 8] &end;matrix, I = &begin;matrix [1, 0; 0, 1] &end;matrix
Evaluate A² − 4A + I:
A² − 4A + I = &begin;matrix [7−8+1, 12−12+0; 4−4+0, 7−8+1] &end;matrix = &begin;matrix [0, 0; 0, 0] &end;matrix = O
Find A⁻¹ using the matrix equation:
A² − 4A + I = O
Post-multiplying both sides by A⁻¹:
(A²)A⁻¹ − 4(A)A⁻¹ + I(A⁻¹) = O
A − 4I + A⁻¹ = O
A⁻¹ = 4I − A
Substitute matrices into the inverse formula:
A⁻¹ = 4 &begin;matrix [1, 0; 0, 1] &end;matrix − &begin;matrix [2, 3; 1, 2] &end;matrix = &begin;matrix [4−2, 0−3; 0−1, 4−2] &end;matrix = &begin;matrix [2, −3; −1, 2] &end;matrix

Final Answer: A⁻¹ = &begin;matrix [2, −3; −1, 2] &end;matrix

Exercise 4.5: Solving System of Linear Equations (5-Mark Board Exam Classic)

Question 6: Solve the following system of linear equations using the Matrix Method:
2x − 3y + 5z = 11
3x + 2y − 4z = −5
x + y − 2z = −3

Step-by-Step Solution:

Express the system in matrix form AX = B:
A = &begin;matrix [2, −3, 5; 3, 2, −4; 1, 1, −2] &end;matrix, X = &begin;matrix [x; y; z] &end;matrix, B = &begin;matrix [11; −5; −3] &end;matrix
Find |A|:
|A| = 2 × [ (2)(−2) − (−4)(1) ] − (−3) × [ (3)(−2) − (−4)(1) ] + 5 × [ (3)(1) − (2)(1) ]
|A| = 2 [ −4 + 4 ] + 3 [ −6 + 4 ] + 5 [ 3 − 2 ]
|A| = 2(0) + 3(−2) + 5(1) = 0 − 6 + 5 = −1
Since |A| = −1 ≠ 0, A is non-singular and has a unique solution given by X = A⁻¹B.
Calculate all 9 cofactors (A_ij):
A₁₁ = + [ −4 − (−4) ] = 0
A₁₂ = − [ −6 − (−4) ] = −(−2) = 2
A₁₃ = + [ 3 − 2 ] = 1
A₂₁ = − [ 6 − 5 ] = −1
A₂₂ = + [ −4 − 5 ] = −9
A₂₃ = − [ 2 − (−3) ] = −5
A₃₁ = + [ 12 − 10 ] = 2
A₃₂ = − [ −8 − 15 ] = −(−23) = 23
A₃₃ = + [ 4 − (−9) ] = 13
Construct Adjoint Matrix adj A (Transpose of Cofactor Matrix):
adj A = &begin;matrix [0, −1, 2; 2, −9, 23; 1, −5, 13] &end;matrix
Compute A⁻¹ = (1 / |A|) × adj A:
A⁻¹ = (1 / −1) × &begin;matrix [0, −1, 2; 2, −9, 23; 1, −5, 13] &end;matrix = &begin;matrix [0, 1, −2; −2, 9, −23; −1, 5, −13] &end;matrix
Compute X = A⁻¹B:
X = &begin;matrix [0, 1, −2; −2, 9, −23; −1, 5, −13] &end;matrix × &begin;matrix [11; −5; −3] &end;matrix
x = 0(11) + 1(−5) + (−2)(−3) = 0 − 5 + 6 = 1
y = −2(11) + 9(−5) + (−23)(−3) = −22 − 45 + 69 = 2
z = −1(11) + 5(−5) + (−13)(−3) = −11 − 25 + 39 = 3

Final Answer: x = 1, y = 2, z = 3 (Verified: 2(1) − 3(2) + 5(3) = 2 − 6 + 15 = 11).

Common Mistakes and Tips for Determinants

Avoid these frequent pitfalls identified by CBSE board evaluators during paper correction:

Confusing Determinant and Matrix Notations: Matrices are enclosed in square brackets [ ] and represent arrays, while determinants are enclosed in vertical bars | | and represent single scalar numbers.
Forgetting the Transpose in Adjoint: A common error is writing the matrix of cofactors directly as adj A without taking its transpose. Always remember: adj A = [A_ij]^T.
Sign Alternation Errors in Cofactors: Remember the checkerboard sign pattern:
&begin;matrix [+, −, +; −, +, −; +, −, +] &end;matrix
Cofactor A_ij requires multiplying minor M_ij by (−1)^{i + j}.
Ignoring ± in Area of Triangle Problems: When the numerical area is given and an unknown coordinate (like k) is to be found, you must solve for both positive and negative values of the determinant, i.e., Δ = ± (Given Area).
Scalar Multiplication Confusion: Multiplying a matrix by k multiplies every element by k. But in determinants, multiplying by k multiplies only one row or one column by k. Hence, for order n, |kA| = kⁿ|A|.

CBSE Class 12 Board Exam Relevance and Mark Distribution

In the CBSE Class 12 Mathematics board examination, Unit II (Algebra: Matrices and Determinants) carries a total weightage of approximately 10 to 12 marks out of 80. Chapter 4 Determinants regularly features in multiple sections of the question paper:

Section A (1 Mark - MCQs & Assertion-Reason): Properties of determinants, finding |adj A| = |A|ⁿ⁻¹, evaluating 2 × 2 determinants, or determining singular matrix conditions (|A| = 0).
Section B & C (2 or 3 Marks - Short Answer): Area of triangle using collinearity conditions, finding matrix inverses for 2 × 2 matrices, or verifying matrix polynomial relations (A² − kA + I = O).
Section D (5 Marks - Long Answer): The guaranteed 5-marker question involves solving a system of 3 linear equations using the Matrix Inversion Method (AX = B ⇒ X = A⁻¹B).

NCERT Solutions Class 12 Mathematics Chapter 4 Determinants

NCERT Solutions Class 12 Mathematics Chapter 4 Determinants

Chapter Overview: Class 12 Mathematics Chapter 4 Determinants

Real-World Applications of Determinants

Important Formulas, Theorems, and Properties

1. Determinant Evaluation Formulas

2. Area of a Triangle & Collinearity

3. Minors and Cofactors

4. Adjoint and Inverse of a Matrix

5. Criteria for Consistency of Linear System (AX = B)

NCERT Solutions Class 12 Mathematics Chapter 4: Step-by-Step Exercise Solutions

Exercise 4.1: Determinant Evaluation & Scalar Property

Exercise 4.2: Area of Triangles and Line Equations

Exercise 4.3: Minors, Cofactors, and Expansion Theorems

Exercise 4.4: Adjoint and Matrix Inversion

Exercise 4.5: Solving System of Linear Equations (5-Mark Board Exam Classic)

Common Mistakes and Tips for Determinants

CBSE Class 12 Board Exam Relevance and Mark Distribution

More NCERT Solutions and Practice on QPTool

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