Class 12 Maths Notes
~8 min readDeterminants Class 12 Notes
Complete, exam-ready notes on determinants: expanding 2×2 and 3×3 determinants, their shortcut properties, adjoint and inverse, and using determinants to solve systems of linear equations — written for CBSE boards and JEE revision.
What is this chapter about in one line?
Determinants are numbers computed from a square matrix that reveal whether it is invertible, help solve linear systems, and give the area of triangles.
Expansion of Determinants
Determinant
Every square matrix has a determinant: for , . A 3×3 determinant is expanded along a row or column using minors and cofactors.
Properties of Determinants
- |Aᵀ| = |A| — transposing does not change the determinant.
- Interchanging two rows or columns changes the sign of the determinant.
- If two rows or columns are identical or proportional, the determinant is 0.
- A row can be scaled by multiplying the whole determinant by that factor.
- Adding a multiple of one row to another leaves the determinant unchanged.
Speed trick
Adjoint, Inverse and Singular Matrices
Adjoint and inverse
The adjoint adj(A) is the transpose of the cofactor matrix of A. For a square matrix, , so A⁻¹ exists if and only if . A matrix with |A| = 0 is singular and has no inverse.
Solving Linear Equations with Cramer's Rule
- For a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂, a₃x + b₃y + c₃z = d₃, compute D = |coefficient matrix|.
- x = Dₓ/D, y = D_y/D, z = D_z/D where Dₓ replaces the x-column with constants.
- D ≠ 0 gives a unique solution; D = 0 with one of Dₓ, D_y, D_z ≠ 0 means no solution; all zero means infinite solutions.
Area of a Triangle
Triangle area
The area of a triangle with vertices equals half the absolute value of the 3×3 determinant built from the coordinates. Points are collinear exactly when that determinant is zero.
Solved Examples
Example: Find |A| for , and state whether A is invertible.
Solution: , so A is non-singular and invertible — the inverse is .
Example: Using a property, show that .
Solution: Add columns 2 and 3 to column 3. Each entry of column 3 becomes a + b + c, so column 3 becomes a constant multiple of column 1. A determinant with two proportional columns is zero.
NCERT Exercise Questions
From the textbook
Exercises (NCERT Q1–Q59)
- 1) Evaluate the determinant
- 2) Evaluate the determinant
- 3) Evaluate the determinant
- 4) Evaluate the determinant
- 5) Evaluate the determinant
- 6) Evaluate the determinant
- 7) If , verify that
- 8) Evaluate the determinant
- 9) Find the values of x for which the determinants are equal:
- 10) If , find the value of x.
- 11) Using properties of determinants, show that
- 12) Using properties, show that
- 13) Using properties, evaluate
- 14) Using properties, show that
- 15) Using properties, show that
- 16) Using properties, show that
- 17) Using properties, show that
- 18) Using properties, show that
- 19) Using properties, show that
- 20) Using properties, show that
- 21) Using properties, show that
- 22) Using properties, show that
- 23) Using properties, show that
- 24) Using properties, show that
- 25) Using properties, show that
- 26) Using properties, show that
- 27) Using properties, show that
- 28) If a, b, c are in AP, then the determinant equals (A) 0, (B) 1, (C) x, (D) 2x.
- 29) Find the minor and cofactor of each element of the determinant
- 30) Find the minor and cofactor of the element in the determinant
- 31) Using cofactors of elements of , find |A|.
- 32) Find the cofactors of the elements of the determinant
- 33) Using cofactor expansion, evaluate
- 34) Find the minor and cofactor of each element of the determinant
- 35) Verify that the sum of the products of the elements of row 2 with their corresponding cofactors equals the determinant, for
- 36) Verify that for
- 37) Find the adjoint of the matrix
- 38) Find the inverse of using the adjoint method.
- 39) Find the inverse of using the adjoint method.
- 40) Verify that for and
- 41) Find the adjoint of the matrix
- 42) Verify that for
- 43) If A is a nonsingular matrix of order 3, verify that
- 44) Find the area of the triangle with vertices (1, 2), (4, 2) and (3, 5).
- 45) Find the area of the triangle with vertices (0, 0), (3, 4) and (-2, 3).
- 46) Find the area of the triangle with vertices (-1, -1), (4, -2) and (2, 2).
- 47) Find the area of the triangle formed by the points A(5, 7), B(3, 5) and C(3, 2).
- 48) Find the values of k for which the system of equations has a unique solution, using the determinant method.
- 49) Examine the consistency of the system
- 50) Solve the system using Cramer's rule:
- 51) Show that the points (1, -1), (-4, 2) and (-2, -3) are NOT collinear.
- 52) Examine the consistency of the system
- 53) Solve using Cramer's rule:
- 54) Find the area of the triangle formed by the line with the coordinate axes.
- 55) Find the value of k if the area of the triangle with vertices (k, 0), (4, 0) and (0, 2) is 4 square units.
- 56) Find the area of the triangle with vertices (2, -3), (4, 5) and (0, 2).
- 57) Examine the consistency of the system
- 58) Examine the consistency of the system
- 59) Examine the consistency of the system
Answers at a glance
- 1Q1 → 18.
- 2Q2 → 1.
- 3Q3 → answer not in source — verify against the textbook
- 4Q4 → 49.
- 5Q5 → -5.
- 6Q6 → 0.
- 7Q7 → Verified: |2A| = -8 = 4|A|.
- 8Q8 → abc - h^2(a + b + c) + 2h^3.
- 9Q9 → answer not in source — verify against the textbook
- 10Q10 → answer not in source — verify against the textbook
- 11Q11 → Shown.
- 12Q12 → Shown.
- 13Q13 → -24.
- 14Q14 → Shown.
- 15Q15 → Shown.
- 16Q16 → Shown.
- 17Q17 → Shown.
- 18Q18 → -1 is not 0, so the determinant is nonzero.
- 19Q19 → Shown.
- 20Q20 → answer not in source — verify against the textbook
- 21Q21 → Shown: determinant = 0.
- 22Q22 → Shown: determinant = 0.
- 23Q23 → Shown: determinant = 0.
- 24Q24 → Shown.
- 25Q25 → Shown.
- 26Q26 → answer not in source — verify against the textbook
- 27Q27 → Shown: determinant = 0.
- 28Q28 → Option (A) — 0.
- 29Q29 → answer not in source — verify against the textbook
- 30Q30 → M22 = 11, A22 = 11.
- 31Q31 → |A| = 5.
- 32Q32 → answer not in source — verify against the textbook
- 33Q33 → -49.
- 34Q34 → answer not in source — verify against the textbook
- 35Q35 → Verified: both sums give 18.
- 36Q36 → Verified: A(adj A) = (adj A)A = 0I = |A|I.
- 37Q37 → adj(A) = [-5, -2; -3, 1].
- 38Q38 → answer not in source — verify against the textbook
- 39Q39 → answer not in source — verify against the textbook
- 40Q40 → Verified.
- 41Q41 → answer not in source — verify against the textbook
- 42Q42 → Verified: A(adj A) = (adj A)A = -11I = |A|I.
- 43Q43 → answer not in source — verify against the textbook
- 44Q44 → 9/2 square units.
- 45Q45 → 17/2 square units.
- 46Q46 → 9 square units.
- 47Q47 → 3 square units.
- 48Q48 → Unique solution: x = 0, y = 1.
- 49Q49 → Consistent; x = 3, y = 1.
- 50Q50 → answer not in source — verify against the textbook
- 51Q51 → The determinant is 19/2 which is not 0, so the points are not collinear.
- 52Q52 → answer not in source — verify against the textbook
- 53Q53 → answer not in source — verify against the textbook
- 54Q54 → 10 square units.
- 55Q55 → k = 0 or k = 8.
- 56Q56 → 13 square units.
- 57Q57 → Consistent; x = 2, y = -1.
- 58Q58 → Inconsistent (no solution).
- 59Q59 → Consistent with infinitely many solutions.
CBSE Pattern Questions
How CBSE asks this chapter
MCQ Questions
1 Mark Questions
- 1Write the value of |1 2; 3 4|. (Ans: 1×4 − 2×3 = −2)
- 2When is a square matrix said to be singular?
- 3State the relation between A⁻¹ and adj A in terms of |A|. (Ans: A⁻¹ = adj A/|A|)
- 4What is the cofactor of the element in row 2, column 1 of the matrix [1 2; 3 4]? (Ans: −3)
2 Mark Questions
- 1If |2 4; 5 x| = 0, find x. (Ans: x = 10)
- 2Evaluate the determinant |1 2 3; 0 4 5; 0 0 6| using its triangular form. (Ans: 1 × 4 × 6 = 24)
- 3For A = [2 3; 1 4], verify A(adj A) = |A|I. (Ans: |A| = 5)
3 Mark Questions
- 1Using properties of determinants, show that |a b c; a b² c²; bc ca ab| = (a − b)(b − c)(c − a)(ab + bc + ca).
- 2Solve the equation |x 2 3; 1 x 1; 3 2 x| = 0 for x. (Ans: x = 4, 1, −4)
- 3Find the adjoint of A = [1 2; 3 4] and use it to find A⁻¹. (Ans: adj A = [4 −2; −3 1], A⁻¹ = (1/−2) adj A)
4 Mark Questions
- 1Examine whether the system x + 2y = 2, 2x + 4y = 8 is consistent by evaluating |A| and the ranks, and interpret the outcome. (Ans: |A| = 0, inconsistent — parallel lines)
- 2Solve the system 2x − y = 3, 3x + 4y = 10 using the matrix method: write AX = B, find adj A and A⁻¹, and obtain x = 2, y = 1.
PDF: Assignments & Tests
Printable study material
Free PDF Downloads & Assignments
Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.
Covers all five topics: expansion of determinants; properties of determinants; adjoint, inverse and singular matrices; solving linear equations with Cramer's rule; and area of a triangle.
Higher-difficulty set with multi-step problems (property-based proofs, adjoint and inverse computation, Cramer's rule for 2 and 3 variables, area and collinearity) with full worked answers. Includes an answer key.
Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 4), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.
Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.
Quick Revision
Key formulas at a glance
Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.
2×2 determinant
Inverse via adjoint
Adjoint identity
Product rule
Cramer's rule
Triangle area
Exam Strategy
How this chapter is asked
High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.
- A⁻¹ exists iff |A| ≠ 0.
- |Aᵀ| = |A| and |AB| = |A||B|.
- Identical or proportional rows ⇒ determinant = 0.
- For uniqueness of a linear system, D ≠ 0.
- Triangle area is half the absolute value of the coordinate determinant; zero area means collinear points.
FAQ
Frequently asked questions
What is a singular matrix?
A square matrix whose determinant is zero. It has no inverse and cannot be used to solve a system with a unique solution.
What is Cramer's rule?
A quick way to solve a linear system by ratios of determinants: x = Dₓ/D, y = D_y/D, z = D_z/D, where D is the coefficient determinant and Dₓ replaces the x-column with the constants.
When does a system of equations have no solution?
When the coefficient determinant D = 0 but at least one of Dₓ, D_y, D_z is non-zero. If all of them are also zero, the system has infinitely many solutions.
How do determinants give the area of a triangle?
The area is half the absolute value of the 3×3 determinant whose rows are (x₁,y₁,1), (x₂,y₂,1), (x₃,y₃,1). If the points are collinear, this determinant is zero.
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