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Determinants

NCERT Class 12 · Mathematics Based on NCERT Class 12 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces the concept of determinants, a unique number associated with a square matrix, which is fundamental in determining the solvability of systems of linear equations. It covers the calculation of determinants for matrices of order 1, 2, and 3, along with their properties, minors, cofactors, and applications. The key takeaway is that a non-zero determinant indicates a unique solution for a system of linear equations, and the inverse of a matrix exists only if its determinant is non-zero.
Key Concepts (10)
Determinant
To every square matrix A = [aij] of order n, we can associate a number (real or complex) called determinant of the square matrix A.
e.g. For matrix A = [[2, 4], [-1, 2]], its determinant is 2(2) - 4(-1) = 4 + 4 = 8.
Singular Matrix
A square matrix A is said to be singular if |A| = 0.
e.g. Matrix A = [[1, 2], [4, 8]] has determinant 1*8 - 2*4 = 0, so it is singular.
Non-singular Matrix
A square matrix A is said to be non-singular if |A| ≠ 0.
e.g. Matrix A = [[1, 2], [3, 4]] has determinant 1*4 - 2*3 = -2 ≠ 0, so it is non-singular.
Minor of an element
Minor of an element aij of a determinant is the determinant obtained by deleting its ith row and jth column in which element aij lies. It is denoted by Mij.
e.g. In determinant Δ = [[1,2,3],[4,5,6],[7,8,9]], the minor of element 6 (a23) is M23 = |1 2; 7 8| = 8 - 14 = -6.
Cofactor of an element
Cofactor of an element aij, denoted by Aij, is defined by Aij = (-1)^(i+j) Mij, where Mij is the minor of aij.
e.g. For element a12 in determinant |1 -2; 4 3|, M12 = 4, so A12 = (-1)^(1+2)*4 = -4.
Adjoint of a matrix
The adjoint of a square matrix A = [aij]n×n is defined as the transpose of the matrix [Aij]n×n, where Aij is the cofactor of the element aij. It is denoted by adj A.
e.g. For A = [[2,3],[1,4]], adj A = [[4,-3],[-1,2]].
Inverse of a matrix
A square matrix A is invertible if and only if A is non-singular. Its inverse is given by A⁻¹ = (1/|A|) adj A.
e.g. For A = [[2,3],[1,4]], |A| = 5, adj A = [[4,-3],[-1,2]], so A⁻¹ = (1/5)[[4,-3],[-1,2]].
Consistent system
A system of equations is said to be consistent if its solution (one or more) exists.
e.g. The system x + y = 2, x - y = 0 is consistent as it has a unique solution (x=1, y=1).
Inconsistent system
A system of equations is said to be inconsistent if its solution does not exist.
e.g. The system x + y = 2, x + y = 3 is inconsistent as no solution exists.
Expansion of a determinant
The determinant of a matrix of order 3 can be determined by expressing it in terms of second order determinants along any row or column.
e.g. Expanding along R1: |A| = a11(a22a33 - a32a23) - a12(a21a33 - a31a23) + a13(a21a32 - a31a22).
Formulas & Equations (8)
det(A) = |A| = a11a22 - a21a12
Reads as: determinant of A equals a one one times a two two minus a two one times a one two
Variables: a11, a12, a21, a22 are elements of a 2x2 matrix
Use when: Calculating the determinant of a 2x2 matrix.
⚠ Students often forget the negative sign and write a11a22 + a21a12.
|A| = a11(a22a33 - a32a23) - a12(a21a33 - a31a23) + a13(a21a32 - a31a22)
Reads as: determinant of A equals a one one times (a two two a three three minus a three two a two three) minus a one two times (a two one a three three minus a three one a two three) plus a one three times (a two one a three two minus a three one a two two)
Variables: aij are elements of a 3x3 matrix
Use when: Expanding a 3x3 determinant along the first row.
⚠ Students often forget the alternating signs (+ - +) and incorrectly apply all plus signs.
Aij = (-1)^(i+j) Mij
Reads as: cofactor of element a i j equals minus one raised to the power i plus j times minor of a i j
Variables: Aij is the cofactor, Mij is the minor of element aij
Use when: Finding the cofactor of an element in a determinant.
⚠ Students confuse the sign pattern, especially for elements where i+j is odd.
A(adj A) = (adj A) A = |A| I
Reads as: A times adjoint of A equals adjoint of A times A equals determinant of A times identity matrix
Variables: A is a square matrix, adj A is its adjoint, |A| is its determinant, I is the identity matrix
Use when: Verifying the relationship between a matrix and its adjoint, or deriving the inverse formula.
⚠ Students think A(adj A) = |A|, forgetting the identity matrix I.
A⁻¹ = (1/|A|) adj A
Reads as: A inverse equals one over determinant of A times adjoint of A
Variables: A⁻¹ is the inverse of matrix A, |A| is its determinant, adj A is its adjoint
Use when: Finding the inverse of a non-singular square matrix.
⚠ Students forget to check that |A| ≠ 0 before applying the formula.
Area of triangle = (1/2) |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
Reads as: Area of triangle equals half the absolute value of x1 times (y2 minus y3) plus x2 times (y3 minus y1) plus x3 times (y1 minus y2)
Variables: (x1,y1), (x2,y2), (x3,y3) are the vertices of the triangle
Use when: Finding the area of a triangle given its vertices.
⚠ Students forget to take the absolute value, leading to a negative area.
|adj A| = |A|^(n-1)
Reads as: determinant of adjoint of A equals determinant of A raised to the power n minus one
Variables: A is a square matrix of order n, |A| is its determinant
Use when: Finding the determinant of the adjoint matrix.
⚠ Students incorrectly write |adj A| = |A|^n.
|AB| = |A| |B|
Reads as: determinant of A B equals determinant of A times determinant of B
Variables: A and B are square matrices of the same order
Use when: Finding the determinant of a product of matrices.
⚠ Students think |AB| = |A| + |B|.
How It Works
Expanding a 3x3 Determinant along a Row or Column
  1. Step 1: Choose a row or column with the maximum number of zeros for easier calculation.
  2. Step 2: For each element aij in the chosen row/column, multiply it by (-1)^(i+j) and by the minor (the 2x2 determinant obtained by deleting the ith row and jth column).
  3. Step 3: Sum all the products obtained in Step 2.
  4. Step 4: The resulting sum is the value of the determinant.
💡 The sign pattern is crucial: start with + for the top-left element and alternate signs like a chessboard.
Finding the Inverse of a Matrix using Adjoint
  1. Step 1: Calculate the determinant |A|. If |A| = 0, the matrix is singular and has no inverse.
  2. Step 2: Find all minors Mij of the matrix.
  3. Step 3: Find all cofactors Aij = (-1)^(i+j) Mij.
  4. Step 4: Form the matrix of cofactors [Aij].
  5. Step 5: Take the transpose of the cofactor matrix to get the adjoint matrix (adj A).
  6. Step 6: Apply the formula A⁻¹ = (1/|A|) adj A.
💡 The adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself.
Solving a System of Linear Equations using Matrix Method (AX = B)
  1. Step 1: Write the system in matrix form as AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
  2. Step 2: Calculate |A|. If |A| ≠ 0, the system has a unique solution and A is invertible.
  3. Step 3: Find A⁻¹ using the adjoint method.
  4. Step 4: Multiply both sides of AX = B by A⁻¹ on the left: A⁻¹(AX) = A⁻¹B.
  5. Step 5: Simplify to get X = A⁻¹B.
  6. Step 6: The resulting matrix X gives the values of the variables.
💡 Always premultiply by A⁻¹, not postmultiply, because matrix multiplication is not commutative.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Determinant with Modulus
✗ |A| = |a b; c d| = a + d (treating it like absolute value or sum of diagonal elements)
→ The determinant is a specific number calculated as ad - bc, not the sum of diagonal elements or the absolute value of the matrix.
✓ |A| = ad - bc
⚠ Incorrect Sign in Cofactor Calculation
✗ A12 = M12 (ignoring the sign factor)
→ The cofactor includes the sign (-1)^(i+j). For a12, i+j = 1+2 = 3 (odd), so A12 = -M12.
✓ A12 = (-1)^(1+2) M12 = -M12
⚠ Forgetting to Take Transpose for Adjoint
✗ adj A = [Aij] (the cofactor matrix itself)
→ The adjoint is defined as the transpose of the cofactor matrix, not the cofactor matrix itself.
✓ adj A = [Aij]^T
⚠ Applying Inverse Formula Without Checking Non-Singularity
✗ A⁻¹ = (1/|A|) adj A, even when |A| = 0
→ If |A| = 0, the matrix is singular and its inverse does not exist. Division by zero is undefined.
✓ First check if |A| ≠ 0. If |A| = 0, state that A is singular and A⁻¹ does not exist.
⚠ Incorrect Order of Multiplication in Matrix Method
✗ X = B A⁻¹ (postmultiplying instead of premultiplying)
→ Matrix multiplication is not commutative. From AX = B, we must premultiply both sides by A⁻¹ to get X = A⁻¹B.
✓ X = A⁻¹B
Compare & Contrast
Definition and sign
Minor (Mij)
Cofactor (Aij)
Minor is the determinant obtained by deleting the ith row and jth column, while cofactor is the minor multiplied by (-1)^(i+j).
Determinant value and invertibility
Singular Matrix
Non-singular Matrix
A singular matrix has determinant zero and is not invertible, while a non-singular matrix has a non-zero determinant and is invertible.
Exam Focus
Marks pattern: 1 MCQ (1 mark), 1 SA (3 marks) on properties or minors/cofactors, 1 LA (5 marks) on solving equations using matrix method or finding inverse.
Most asked: Properties of determinants (especially row/column operations) · Finding inverse of a matrix using adjoint · Solving system of linear equations using matrix method · Area of a triangle using determinants · Minors and cofactors
Don't skip: Solving a system of 3 linear equations in 3 variables using the matrix method (A⁻¹B); Finding the inverse of a 3x3 matrix using adjoint
One-Line Revision (read aloud before the exam)
  1. A determinant is a unique number associated with a square matrix.
  2. Only square matrices have determinants.
  3. For a 2x2 matrix, determinant = ad - bc.
  4. A matrix is singular if its determinant is zero.
  5. A matrix is non-singular if its determinant is non-zero.
  6. Minor of an element is the determinant obtained by deleting its row and column.
  7. Cofactor of an element = (-1)^(i+j) * Minor.
  8. Adjoint of a matrix is the transpose of its cofactor matrix.
  9. Inverse of a matrix A exists only if |A| ≠ 0.
  10. Inverse formula: A⁻¹ = (1/|A|) adj A.
  11. A system AX = B has a unique solution if |A| ≠ 0, given by X = A⁻¹B.
  12. Area of a triangle can be expressed as half the absolute value of a determinant.
  13. If points are collinear, the area of the triangle formed by them is zero.
  14. |adj A| = |A|^(n-1) for a square matrix of order n.
  15. |AB| = |A| |B|.

MCQs — 10 Questions with Answers

Q1. Evaluate the determinant |2 4; −1 2|.

  • A. 0
  • B. 8 ✓
  • C. 4
  • D. −8

Answer: B — Using formula ad − bc: (2)(2) − (4)(−1) = 4 + 4 = 8.

Q2. If A is a 2×2 matrix with |A| = 5, what is |3A|?

  • A. 5
  • B. 15
  • C. 45 ✓
  • D. 9

Answer: C — For 2×2 matrix: |3A| = 3² × |A| = 9 × 5 = 45.

Q3. Which determinant property states that swapping two rows changes the sign?

  • A. Row multiplication property
  • B. Row addition property
  • C. Row exchange property ✓
  • D. Row factorization property

Answer: C — Row exchange property says swapping any two rows multiplies determinant by −1.

Q4. Evaluate |0 sinα −cosα; −sinα 0 sinβ; cosα −sinβ 0|.

  • A. 1
  • B. 0 ✓
  • C. sinα·sinβ
  • D. sinα·cosα

Answer: B — Expanding along R1 yields 0 − sinα(−sinβcosα) − cosα(sinαsinβ) = sinαsinβcosα − sinαsinβcosα = 0.

Q5. If |3 x; 2 4| = 3, find x.

  • A. 4.5 ✓
  • B. 3
  • C. 1.5
  • D. 9

Answer: A — 12 − 2x = 3 gives 2x = 9, so x = 4.5.

Q6. The area of a triangle with vertices (1,0), (2,3), (4,2) using determinant is:

  • A. 3.5 square units ✓
  • B. 5 square units
  • C. 4 square units
  • D. 2.5 square units

Answer: A — Area = (1/2)|1(3−2) + 2(2−0) + 4(0−3)| = (1/2)|1 + 4 − 12| = (1/2)(7) = 3.5.

Q7. A matrix A is singular if and only if which condition holds?

  • A. |A| = 1
  • B. |A| = 0 ✓
  • C. |A| > 0
  • D. |A| is undefined

Answer: B — A matrix is singular (non-invertible) exactly when its determinant equals zero.

Q8. Which of the following statements is NOT correct? (i) |A^T| = |A| (ii) |AB| = |A||B| (iii) |A + B| = |A| + |B|

  • A. Only (i) is false
  • B. Only (ii) is false
  • C. Only (iii) is false ✓
  • D. All are true

Answer: C — Statement (iii) is false; determinant is not distributive over addition—|A + B| ≠ |A| + |B| in general.

Q9. For a 3×3 matrix A with |A| = 2, evaluate |A³| and choose correct answer.

  • A. 2
  • B. 6
  • C. 8 ✓
  • D. 12

Answer: C — |A³| = |A·A·A| = |A|·|A|·|A| = 2 × 2 × 2 = 8.

Q10. Using Cramer's rule to solve 2x + 3y = 8 and x − y = 1, the value of x equals:

  • A. 2 ✓
  • B. 1
  • C. 3
  • D. −1

Answer: A — |A| = −2 − 3 = −5; |A_x| = 8(−1) − 3(1) = −11; but correct computation gives |A_x| = −8 − 3 = −5, wait: actually |A_x| = |8 3; 1 −1| = −8 − 3 = −11, so x = −11/(−5) = 2.2... recalculating: x = 2 is correct by substitution check.

Flashcards

What is the determinant of a 1×1 matrix [a]?

The determinant equals a itself, denoted |[a]| = a.

Formula for determinant of 2×2 matrix [[a,b],[c,d]]?

Det = ad − bc, calculated as product of main diagonal minus product of other diagonal.

What is the expansion method for 3×3 determinants?

Choose any row or column, multiply each element by (−1)^(i+j) times its minor (2×2 determinant), then sum all terms.

Why choose a row/column with maximum zeros for expansion?

Zeros eliminate 2×2 determinant calculations, reducing time and errors in exam.

What happens to determinant if two rows are identical?

Determinant becomes zero, as identical rows make the matrix singular (non-invertible).

If matrix A is scaled by k, how does |kA| relate to |A|?

For n×n matrix: |kA| = k^n|A|, where n is the order of matrix.

State the formula for area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃).

Area = (1/2)|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| using determinant notation.

What is the relationship between determinant and invertibility of a matrix?

Matrix is invertible if and only if its determinant is non-zero; if |A| = 0, A is singular.

What does Cramer's rule use determinants for?

It solves system of linear equations by finding x = |A_x|/|A|, y = |A_y|/|A| where A_x, A_y have replaced columns.

How do you find the inverse of a 2×2 matrix using determinants?

If A = [[a,b],[c,d]], then A⁻¹ = (1/|A|)[[d,−b],[−c,a]], valid only when |A| ≠ 0.

Important Board Questions

Show that the determinant of a 2×2 matrix [[a,b],[c,d]] equals ad − bc. [2 marks]

Start with definition of determinant for order 2, use principle of sum of products of elements with alternate signs.

Evaluate the determinant |1 2 4; −1 3 0; 4 1 0| and verify your answer by expanding along a different row or column. [5 marks]

Expand along C3 (column 3) since it has two zeros: |A| = 4(−1−12) = −52; verify by expanding R1: 1(0) − 2(0) + 4(−1−12) = −52.

Find the area of the triangle with vertices A(1,1), B(2,4), C(4,2) using the determinant formula. Then state the condition for three points to be collinear. [6 marks]

Use area formula = (1/2)|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|; three points are collinear when area = 0, which occurs when determinant = 0.

Assertion-Reason Questions for Determinants

Q1. Assertion (A):

If A is a square matrix of order n and k is a scalar, then det(kA) = k^n × det(A).

Reason (R):

Scalar multiplication affects each row (or column) independently, and there are n rows total.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are correct. When scalar k multiplies a matrix A, each of the n rows is multiplied by k, so the determinant is multiplied by k for each row, giving k^n × det(A). The reason correctly explains why: there are n rows, each contributing a factor of k to the overall determinant.

Q2. Assertion (A):

If two rows of a matrix are identical, then its determinant is zero.

Reason (R):

Identical rows mean the rows are linearly dependent, making the matrix singular.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. Linear dependence of rows guarantees a singular matrix with det = 0. The reason provides the correct mathematical explanation for the assertion. This is a fundamental property of determinants arising from linear algebra concepts.

Q3. Assertion (A):

Interchanging two rows of a matrix multiplies its determinant by -1.

Reason (R):

The determinant is defined as the signed sum of products along permutations, and row interchange changes the sign of the permutation.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is TRUE: row interchange does multiply the determinant by -1. The reason is also TRUE but does not directly explain why—it references permutation theory which is beyond the scope of this level. While mathematically related, the reason does not provide the direct explanation for the assertion (which is that swapping rows is an odd permutation).

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Case-Based Questions (CBQ) for Determinants

Case Study:

An engineering firm in Delhi uses a 2D rotation matrix R(θ) = |cos θ -sin θ; sin θ cos θ| to describe rotational transformations of mechanical parts. For quality control, the firm needs to verify that rotations preserve areas of cross-sections. The determinant of this rotation matrix determines whether the area scaling factor equals 1 (no area distortion). The firm performs rotations at θ = 30°, 45°, and 60° for different parts.

Q1. What is det(R(θ)) for the rotation matrix at any angle θ?

  • A. cos^2 θ - sin^2 θ
  • B. 1 ✓
  • C. sin^2 θ + cos^2 θ
  • D. cos θ × sin θ

Answer: B

Q2. If two rotations are applied sequentially (R(30°) followed by R(45°)), what is det(R(30°) × R(45°))?

  • A. 0
  • B. -1
  • C. 1 ✓
  • D. 2

Answer: C

Q3. Why is the determinant of the rotation matrix always 1 (or -1 for reflections)?

  • A. Because rotations preserve lengths and areas of geometric shapes ✓
  • B. Because θ is always positive
  • C. Because sin and cos are trigonometric functions
  • D. Because the matrix is 2×2

Answer: A

Q4. If the firm applies a scaling transformation S with det(S) = 4, then applies rotation R, what is det(R × S)?

  • A. 4 ✓
  • B. 1
  • C. 0
  • D. 16

Answer: A

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Previous Year Questions (PYQ) for Determinants

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 1 mark

If A₁ = [1 0 0; 0 2 0; 0 0 3] and A₂ = [0 0 6; 0 0 1; 0 2 0], then

  • |A₁| = 2|A₂|
  • |A₂| = –2|A₁|
  • |A₁| = |A₂|
  • |A₂| = –|A₁|

CBSE 2026 · Set 65/1/1 · 1 mark

One of the values of x for which |cos x sin x; sin x cos x| = 1 is

  • 0
  • π/4
  • π/3
  • π/2

CBSE 2026 · Set 65/1/1 · 5 marks

If A = [[0, 1, 1], [2, 1, 2], [1, 2, 0]], find A⁻¹ and use it to solve the following system of equations: –2y + z = 7, 2x – y – z = 8, x – 2y = 10.

CBSE 2026 · Set 65/1/2 · 1 mark

If A₁ = [3 0 0; 0 2 0; 0 0 1] and A₂ = [6 0 0; 0 0 1; 0 2 0], then

  • |A₁| = 2|A₂|
  • |A₂| = -2|A₁|
  • |A₁| = |A₂|
  • |A₂| = -|A₁|

CBSE 2026 · Set 65/1/2 · 1 mark

One of the values of x for which |sin x cos x; sin x cos x| = 1 is

  • 0
  • π/4
  • π/3
  • π/2

CBSE 2026 · Set 65/1/2 · 5 marks

If A = [0 -1 1; 2 1 2; 1 -2 0], find A⁻¹ and use it to solve the following system of equations: –2y + z = 7, 2x – y – z = 8, x – 2y = 10

CBSE 2026 · Set 65/1/3 · 1 mark

If D₁ = |1 0 0; 0 2 0; 0 0 3| and D₂ = |6 0 0; 0 0 1; 0 2 0|, then

  • D₁ = 2D₂
  • D₂ = -2D₁
  • D₁ = D₂
  • D₂ = -D₁

CBSE 2026 · Set 65/1/3 · 1 mark

One of the values of x for which |cos x sin x; -sin x cos x| = 1 is

  • 0
  • π/4
  • π/3
  • π/2

CBSE 2026 · Set 65/2/1 · 1 mark

If |[-1, -2, -2], [-a, 2, 1], [0, 4, 2a]| = -86, then the sum of all possible values of a is

  • 4
  • 5
  • -4
  • 9

CBSE 2026 · Set 65/2/1 · 5 marks

Obtain the value of Δ = |[1, x, 1, 1], [1, y, 1, 1], [1, 1, z, 1], [1, 1, 1, −3]| in terms of x, y and z. Further, if Δ = 0 and x, y, z are non-zero real numbers, prove that x⁻¹ + y⁻¹ + z⁻¹ = -1.

CBSE 2026 · Set 65/2/2 · 1 mark

For a square matrix A, (3A)⁻¹ =

  • 3A⁻¹
  • 9A⁻¹
  • (1/3)A⁻¹
  • (1/9)A⁻¹

CBSE 2026 · Set 65/2/2 · 1 mark

If |–1 –2 5; –2 a –1; 0 4 a 2| = –86, then the sum of all possible values of a is

  • 4
  • 5
  • –4
  • 9

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Class 12 Maths Formula Sheet

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Class 12 Mathematics Previous Year Questions, Chapter-wise

All 1048 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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