NCERT Solutions for Class 12 Maths Chapter 4: Determinants (NCERT 2026–27)

These Class 12 Maths Chapter 4 solutions cover Determinants from the latest NCERT textbook (Reprint 2026–27). Every question of Exercises 4.1, 4.2, 4.3, 4.4, 4.5 and the Miscellaneous Exercise is reproduced verbatim and solved step by step, with each numerical answer cross-checked against the book’s answer key — covering second- and third-order determinants, area of a triangle, minors and cofactors, the adjoint and inverse of a matrix, and the matrix method for solving systems of linear equations.

Class: 12 Subject: Mathematics Chapter: 4 – Determinants Exercises: 4.1, 4.2, 4.3, 4.4, 4.5 + Miscellaneous Board: CBSE (NCERT) Session: 2026–27
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Chapter 4 Overview

To every square matrix we can associate a single number called its determinant, which decides whether the matrix is invertible and whether a related system of linear equations has a unique solution. Chapter 4 of Class 12 Maths studies determinants up to order three with real entries. It develops the determinant of orders one, two and three (by expansion along any row or column), the use of a determinant to find the area of a triangle, the ideas of minors and cofactors, the adjoint and the inverse of a matrix, and finally the matrix method for testing consistency and solving systems of linear equations in two or three variables. The Class 12 Maths Chapter 4 solutions below work through every exercise question step by step.

Key Concepts & Definitions

Determinant: a function that associates a unique number (real or complex) with each square matrix A, written |A|, det A or Δ. Only square matrices have determinants.

Minor (Mij): the determinant obtained by deleting the i-th row and j-th column in which aij lies. A minor of an order-n determinant is a determinant of order n−1.

Cofactor (Aij): Aij = (−1)i+j Mij. The value of a determinant equals the sum of products of the elements of any one row (or column) with their corresponding cofactors.

Adjoint (adj A): the transpose of the matrix of cofactors [Aij]. It satisfies A(adj A) = (adj A)A = |A| I.

Singular / non-singular: A is singular if |A| = 0 and non-singular if |A| ≠ 0. A is invertible if and only if it is non-singular, and then A−1 = (1/|A|) adj A.

Consistent / inconsistent system: a system AX = B is consistent if a solution exists, inconsistent otherwise. If |A| ≠ 0 the unique solution is X = A−1B.

Important Formulas (Chapter 4)

Order-2 determinant: |A| = a11a22 − a12a21.

Order-3 (expand along R1): |A| = a11(a22a33 − a23a32) − a12(a21a33 − a23a31) + a13(a21a32 − a22a31).

Scalar multiple: if A = kB (order n), then |A| = kn|B|.

Area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3): Δ = ½ |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|. Points are collinear when this determinant is 0.

Adjoint & inverse: A(adj A) = (adj A)A = |A| I; A−1 = (1/|A|) adj A; |adj A| = |A|n−1; det(A−1) = 1/det(A).

Matrix method: for AX = B with |A| ≠ 0, X = A−1B = (1/|A|)(adj A)B.

Questions below are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

Exercise 4.1 Solutions

Evaluate the determinants in Exercises 1 and 2.

1. Evaluate the determinant 2   4
−5  −1

SOLUTION = 2(−1) − 4(−5) = −2 + 20 = 18.

2. Evaluate the determinants: (i) cosθ  −sinθ
sinθ   cosθ (ii) x2−x+1   x−1
x+1       x+1

SOLUTION (i) = cosθ·cosθ − (−sinθ)·sinθ = cos2θ + sin2θ = 1. (ii) = (x2 − x + 1)(x + 1) − (x − 1)(x + 1). (x2 − x + 1)(x + 1) = x3 + 1 (sum of cubes); (x − 1)(x + 1) = x2 − 1. = (x3 + 1) − (x2 − 1) = x3 − x2 + 2.

3. If A = 1   2
4   2, then show that |2A| = 4|A|.

SOLUTION |A| = 1(2) − 2(4) = 2 − 8 = −6. 2A = 2   4
8   4, so |2A| = 2(4) − 4(8) = 8 − 32 = −24. 4|A| = 4(−6) = −24 = |2A|. Hence |2A| = 4|A| (illustrating |kA| = k2|A| for order 2).

4. If A = 1  0  1
0  1  2
0  0  4, then show that |3A| = 27|A|.

SOLUTION A is upper-triangular, so |A| = product of diagonal = 1·1·4 = 4. 3A = 3  0  3
0  3  6
0  0  12; |3A| = 3·3·12 = 108. 27|A| = 27(4) = 108 = |3A|. Hence |3A| = 27|A| (illustrating |kA| = k3|A| for order 3).

5. Evaluate the determinants: (i) 3  −1  −2
0   0  −1
3  −5   0 (ii) 3  −4  5
1   1  −2
2   3   1 (iii) 0   1   2
−1  0  −3
−2  3   0 (iv) 2  −1  −2
0   2  −1
3  −5   0

SOLUTION (i) Expand along R2 (two zeros): = −(−1)·(−1)2+3·3  −1
3  −5. Only the entry −1 (position 2,3) contributes: = (−1)(−1)5[3(−5) − (−1)(3)] = (−1)(−1)(−15 + 3) = (1)(−12) = −12. (ii) Expand along R1: = 3(1·1 − (−2)·3) − (−4)(1·1 − (−2)·2) + 5(1·3 − 1·2) = 3(1 + 6) + 4(1 + 4) + 5(3 − 2) = 21 + 20 + 5 = 46. (iii) Expand along R1: = 0 − 1[(−1)(0) − (−3)(−2)] + 2[(−1)(3) − 0(−2)] = −1(0 − 6) + 2(−3) = 6 − 6 = 0. (iv) Expand along R1: = 2(2·0 − (−1)(−5)) − (−1)(0·0 − (−1)(3)) + (−2)(0·(−5) − 2·3) = 2(0 − 5) + 1(0 + 3) − 2(0 − 6) = −10 + 3 + 12 = 5.

6. If A = 1  1  −2
2  1  −3
5  4  −9, find |A|.

SOLUTION Expand along R1: = 1(1·(−9) − (−3)·4) − 1(2·(−9) − (−3)·5) + (−2)(2·4 − 1·5). = 1(−9 + 12) − 1(−18 + 15) − 2(8 − 5) = 3 − (−3) − 6 = 3 + 3 − 6 = 0.

7. Find values of x, if (i) 2  4
5  1 = 2x  4
6   x (ii) 2  3
4  5 = x   3
2x  5

SOLUTION (i) 2(1) − 4(5) = 2x·x − 4·6 ⇒ 2 − 20 = 2x2 − 24 ⇒ −18 = 2x2 − 24 ⇒ 2x2 = 6 ⇒ x2 = 3 ⇒ x = ±√3. (ii) 2(5) − 3(4) = x(5) − 3(2x) ⇒ 10 − 12 = 5x − 6x ⇒ −2 = −x ⇒ x = 2.

8. If x   2
18  x = 6   2
18  6, then x is equal to (A) 6    (B) ±6    (C) −6    (D) 0

SOLUTION x·x − 2·18 = 6·6 − 2·18 ⇒ x2 − 36 = 36 − 36 ⇒ x2 = 36 ⇒ x = ±6. ∴ the correct option is (B) ±6.

Exercise 4.2 Solutions

1. Find area of the triangle with vertices at the point given in each of the following: (i) (1, 0), (6, 0), (4, 3)   (ii) (2, 7), (1, 1), (10, 8)   (iii) (−2, −3), (3, 2), (−1, −8)

SOLUTION Area = ½ |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|. (i) ½|1(0 − 3) + 6(3 − 0) + 4(0 − 0)| = ½|−3 + 18 + 0| = 15/2 sq units. (ii) ½|2(1 − 8) + 1(8 − 7) + 10(7 − 1)| = ½|−14 + 1 + 60| = 47/2 sq units. (iii) ½|−2(2 − (−8)) + 3((−8) − (−3)) + (−1)((−3) − 2)| = ½|−2(10) + 3(−5) − 1(−5)| = ½|−20 − 15 + 5| = ½(30) = 15 sq units.

2. Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.

SOLUTION Points are collinear if the area of the triangle is zero, i.e. the determinant a  b+c  1
b  c+a  1
c  a+b  1 = 0. Apply C2 → C2 + C1: column 2 becomes (a + b + c), (a + b + c), (a + b + c). = a  a+b+c  1
b  a+b+c  1
c  a+b+c  1. Taking the common factor (a + b + c) out of C2 leaves two identical columns (C2 = C3 = all 1s), so the determinant is 0. ∴ the area is 0 and the points are collinear.

3. Find values of k if area of triangle is 4 sq. units and vertices are (i) (k, 0), (4, 0), (0, 2)   (ii) (−2, 0), (0, 4), (0, k)

SOLUTION (i) Area = ½|k(0 − 2) + 4(2 − 0) + 0(0 − 0)| = ½|−2k + 8| = 4 ⇒ |−2k + 8| = 8 ⇒ −2k + 8 = ±8 ⇒ k = 0 or k = 8. So k = 0, 8. (ii) Area = ½|−2(4 − k) + 0(k − 0) + 0(0 − 4)| = ½|−8 + 2k| = 4 ⇒ |2k − 8| = 8 ⇒ 2k − 8 = ±8 ⇒ k = 8 or k = 0. So k = 0, 8.

4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants. (ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

SOLUTION Let P(x, y) lie on the line; then the area of the triangle formed with the two given points is zero. (i) ½x  y  1
1  2  1
3  6  1 = 0 ⇒ x(2 − 6) − y(1 − 3) + 1(6 − 6) = 0 ⇒ −4x + 2y = 0 ⇒ y = 2x. (ii) ½x  y  1
3  1  1
9  3  1 = 0 ⇒ x(1 − 3) − y(3 − 9) + 1(9 − 9) = 0 ⇒ −2x + 6y = 0 ⇒ x − 3y = 0.

5. If area of triangle is 35 sq units with vertices (2, −6), (5, 4) and (k, 4). Then k is (A) 12    (B) −2    (C) −12, −2    (D) 12, −2

SOLUTION Area = ½|2(4 − 4) + 5(4 − (−6)) + k((−6) − 4)| = ½|0 + 50 − 10k| = 35. |50 − 10k| = 70 ⇒ 50 − 10k = ±70 ⇒ −10k = 20 or −10k = −120 ⇒ k = −2 or k = 12. ∴ the correct option is (D) 12, −2.

Exercise 4.3 Solutions

Write Minors and Cofactors of the elements of following determinants:

1. (i) 2  −4
0   3    (ii) a  c
b  d

SOLUTION (i) Minor of an element = the remaining entry; cofactor Aij = (−1)i+j Mij. M11 = 3, M12 = 0, M21 = −4, M22 = 2; A11 = 3, A12 = 0, A21 = 4, A22 = 2. (ii) M11 = d, M12 = b, M21 = c, M22 = a; A11 = d, A12 = −b, A21 = −c, A22 = a.

2. (i) 1  0  0
0  1  0
0  0  1    (ii) 1  0   4
3  5  −1
0  1   2

SOLUTION (i) For the identity determinant every off-diagonal 2×2 minor is 0 and the diagonal ones are 1: M11 = 1, M12 = 0, M13 = 0, M21 = 0, M22 = 1, M23 = 0, M31 = 0, M32 = 0, M33 = 1. A11 = 1, A12 = 0, A13 = 0, A21 = 0, A22 = 1, A23 = 0, A31 = 0, A32 = 0, A33 = 1. (ii) Deleting rows/columns: M11 = (5·2 − (−1)·1) = 11, M12 = (3·2 − (−1)·0) = 6, M13 = (3·1 − 5·0) = 3. M21 = (0·2 − 4·1) = −4, M22 = (1·2 − 4·0) = 2, M23 = (1·1 − 0·0) = 1. M31 = (0·(−1) − 4·5) = −20, M32 = (1·(−1) − 4·3) = −13, M33 = (1·5 − 0·3) = 5. A11 = 11, A12 = −6, A13 = 3, A21 = 4, A22 = 2, A23 = −1, A31 = −20, A32 = 13, A33 = 5.

3. Using Cofactors of elements of second row, evaluate Δ = 5  3  8
2  0  1
1  2  3.

SOLUTION Cofactors of R2: A21 = (−1)3(3·3 − 8·2) = −(9 − 16) = 7; A22 = (−1)4(5·3 − 8·1) = 15 − 8 = 7; A23 = (−1)5(5·2 − 3·1) = −(10 − 3) = −7. Δ = a21A21 + a22A22 + a23A23 = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7.

4. Using Cofactors of elements of third column, evaluate Δ = 1  x  yz
1  y  zx
1  z  xy.

SOLUTION Cofactors of C3: A13 = (−1)4(1·z − y·1) = z − y; A23 = (−1)5(1·z − x·1) = −(z − x) = x − z; A33 = (−1)6(1·y − x·1) = y − x. Δ = yz(z − y) + zx(x − z) + xy(y − x) = yz2 − y2z + x2z − xz2 + xy2 − x2y. Factorising, this equals (x − y)(y − z)(z − x).

5. If Δ = a11  a12  a13
a21  a22  a23
a31  a32  a33 and Aij is Cofactors of aij, then value of Δ is given by (A) a11A31 + a12A32 + a13A33   (B) a11A11 + a12A21 + a13A31 (C) a21A11 + a22A12 + a23A13   (D) a11A11 + a21A21 + a31A31

SOLUTION Δ equals the sum of products of the elements of any one row or column with their own cofactors. Option (D) is exactly the expansion along the first column (a11A11 + a21A21 + a31A31); the others mix elements with cofactors of a different row/column (which give 0). ∴ the correct option is (D).

Exercise 4.4 Solutions

Find adjoint of each of the matrices in Exercises 1 and 2.

1. 1  2
3  4

SOLUTION For order 2, adj A is obtained by swapping the diagonal and changing the sign of the off-diagonal entries. adj A = 4  −2
−3   1.

2. 1  −1  2
2   3  5
−2  0  1

SOLUTION Compute the nine cofactors: A11 = (3·1 − 5·0) = 3, A12 = −(2·1 − 5·(−2)) = −12, A13 = (2·0 − 3·(−2)) = 6. A21 = −((−1)·1 − 2·0) = 1, A22 = (1·1 − 2·(−2)) = 5, A23 = −(1·0 − (−1)·(−2)) = 2. A31 = ((−1)·5 − 2·3) = −11, A32 = −(1·5 − 2·2) = −1, A33 = (1·3 − (−1)·2) = 5. adj A = transpose of [Aij] = 3   1  −11
−12  5  −1
6   2   5.

Verify A(adj A) = (adj A)A = |A| I in Exercises 3 and 4.

3. 2   3
−4  −6

SOLUTION |A| = 2(−6) − 3(−4) = −12 + 12 = 0. adj A = −6  −3
4   2. A(adj A) = 2   3
−4  −6−6  −3
4   2 = −12+12  −6+6
24−24  12−12 = 0  0
0  0 = |A| I (since |A| = 0). Similarly (adj A)A = O. Verified.

4. 1  −1   2
3   0  −2
1   0   3

SOLUTION |A| (expand along C2): = −(−1)(3·3 − (−2)·1) + 0 − 0 = 1(9 + 2) = 11. Cofactors: A11 = 0, A12 = −11, A13 = 0; A21 = 3, A22 = 1, A23 = −1; A31 = 2, A32 = 8, A33 = 3. adj A = 0  3  2
−11  1  8
0  −1  3. Multiplying out, A(adj A) = (adj A)A = 11·I = |A| I. Verified.

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

5. 2  −2
4   3

SOLUTION |A| = 2(3) − (−2)(4) = 6 + 8 = 14 ≠ 0. adj A = 3  2
−4  2. A−1 = (1/14)3  2
−4  2.

6. −1  5
−3  2

SOLUTION |A| = (−1)(2) − 5(−3) = −2 + 15 = 13 ≠ 0. adj A = 2  −5
3  −1. A−1 = (1/13)2  −5
3  −1.

7. 1  2  3
0  2  4
0  0  5

SOLUTION |A| = 1·2·5 = 10 ≠ 0 (upper-triangular). Cofactors give adj A = 10  −10  2
0   5  −4
0   0   2. A−1 = (1/10)10  −10  2
0   5  −4
0   0   2.

8. 1  0   0
3  3   0
5  2  −1

SOLUTION |A| = 1·3·(−1) = −3 ≠ 0 (lower-triangular). Cofactors give adj A = −3  0  0
3  −1  0
−9  −2  3. A−1 = (1/(−3))−3  0  0
3  −1  0
−9  −2  3 = −(1/3)−3  0  0
3  −1  0
−9  −2  3.

9. 2   1  3
4  −1  0
−7  2  1

SOLUTION |A| = 2(−1·1 − 0·2) − 1(4·1 − 0·(−7)) + 3(4·2 − (−1)(−7)) = 2(−1) − 1(4) + 3(8 − 7) = −2 − 4 + 3 = −3 ≠ 0. Cofactors give adj A = −1   5   3
−4  23  12
1  −11  −6. A−1 = (1/(−3))−1   5   3
−4  23  12
1  −11  −6 = −(1/3)−1   5   3
−4  23  12
1  −11  −6.

10. 1  −1   2
0   2  −3
3  −2   4

SOLUTION |A| = 1(2·4 − (−3)(−2)) − (−1)(0·4 − (−3)·3) + 2(0·(−2) − 2·3) = 1(8 − 6) + 1(0 + 9) + 2(−6) = 2 + 9 − 12 = −1 ≠ 0. Cofactors give adj A = 2  0  −1
−9  −2  3
−6  −1  2. A−1 = (1/(−1)) adj A = −2  0  1
9  2  −3
6  1  −2.

11. 1   0   0
0  cosα  sinα
0  sinα  −cosα

SOLUTION |A| = 1(cosα(−cosα) − sinα·sinα) = −(cos2α + sin2α) = −1 ≠ 0. Cofactors give adj A = −1  0  0
0  −cosα  −sinα
0  −sinα  cosα. A−1 = (1/(−1)) adj A = 1  0  0
0  cosα  sinα
0  sinα  −cosα (so A is its own inverse).

12. Let A = 3  7
2  5 and B = 6  8
7  9. Verify that (AB)−1 = B−1A−1.

SOLUTION AB = 3·6+7·7  3·8+7·9
2·6+5·7  2·8+5·9 = 67  87
47  61; |AB| = 67·61 − 87·47 = 4087 − 4089 = −2. (AB)−1 = (1/(−2))61  −87
−47  67. |A| = 15 − 14 = 1, A−1 = 5  −7
−2  3; |B| = 54 − 56 = −2, B−1 = (1/(−2))9  −8
−7  6. B−1A−1 = (1/(−2))9  −8
−7  65  −7
−2  3 = (1/(−2))61  −87
−47  67 = (AB)−1. Verified.

13. If A = 3  1
−1  2, show that A2 − 5A + 7I = O. Hence find A−1.

SOLUTION A2 = 3  1
−1  23  1
−1  2 = 8  5
−5  3. A2 − 5A + 7I = 8  5
−5  3 − 15  5
−5  10 + 7  0
0  7 = 0  0
0  0 = O. Verified. From A2 − 5A + 7I = O, multiply by A−1: A − 5I + 7A−1 = O ⇒ A−1 = (1/7)(5I − A) = (1/7)2  −1
1  3.

14. For the matrix A = 3  2
1  1, find the numbers a and b such that A2 + aA + bI = O.

SOLUTION A2 = 3  2
1  13  2
1  1 = 11  8
4  3. A2 + aA + bI = 11+3a+b  8+2a
4+a  3+a+b = O. From 8 + 2a = 0 ⇒ a = −4; from 4 + a = 0 ⇒ a = −4 (consistent). Then 11 + 3(−4) + b = 0 ⇒ b = 1. So a = −4, b = 1.

15. For the matrix A = 1   1   1
1   2  −3
2  −1   3, show that A3 − 6A2 + 5A + 11I = O. Hence, find A−1.

SOLUTION Computing A2 and A3 and substituting verifies A3 − 6A2 + 5A + 11I = O. (Here |A| = −11.) Multiply the relation by A−1: A2 − 6A + 5I + 11A−1 = O ⇒ A−1 = −(1/11)(A2 − 6A + 5I). Evaluating gives A−1 = (1/11)−3  4  5
9  −1  −4
5  −3  −1.

16. If A = 2  −1   1
−1  2  −1
1  −1   2, verify that A3 − 6A2 + 9A − 4I = O and hence find A−1.

SOLUTION Computing A2 and A3 and substituting verifies A3 − 6A2 + 9A − 4I = O. (Here |A| = 4.) Multiply by A−1: A2 − 6A + 9I − 4A−1 = O ⇒ A−1 = (1/4)(A2 − 6A + 9I). Evaluating gives A−1 = (1/4)3  1  −1
1  3   1
−1  1   3.

17. Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to (A) |A|    (B) |A|2    (C) |A|3    (D) 3|A|

SOLUTION |adj A| = |A|n−1. For n = 3, |adj A| = |A|2. ∴ the correct option is (B) |A|2.

18. If A is an invertible matrix of order 2, then det(A−1) is equal to (A) det(A)    (B) 1/det(A)    (C) 1    (D) 0

SOLUTION Since AA−1 = I, det(A)·det(A−1) = det(I) = 1, so det(A−1) = 1/det(A). ∴ the correct option is (B) 1/det(A).

Exercise 4.5 Solutions

Examine the consistency of the system of equations in Exercises 1 to 6.

1. x + 2y = 2;   2x + 3y = 3

SOLUTION A = 1  2
2  3, |A| = 3 − 4 = −1 ≠ 0. Non-singular ⇒ unique solution. Consistent.

2. 2x − y = 5;   x + y = 4

SOLUTION |A| = 2  −1
1   1 = 2 + 1 = 3 ≠ 0. Consistent.

3. x + 3y = 5;   2x + 6y = 8

SOLUTION |A| = 1  3
2  6 = 6 − 6 = 0. Compute (adj A)B with B = (5, 8)T: adj A = 6  −3
−2  1, (adj A)B = (6·5 − 3·8, −2·5 + 1·8)T = (6, −2)T ≠ O. |A| = 0 and (adj A)B ≠ O ⇒ no solution. Inconsistent.

4. x + y + z = 1;   2x + 3y + 2z = 2;   ax + ay + 2az = 4

SOLUTION |A| = 1  1  1
2  3  2
a  a  2a = 1(6a − 2a) − 1(4a − 2a) + 1(2a − 3a) = 4a − 2a − a = a. For a ≠ 0, |A| ≠ 0 so the system has a unique solution. Consistent.

5. 3x − y − 2z = 2;   2y − z = −1;   3x − 5y = 3

SOLUTION A = 3  −1  −2
0   2  −1
3  −5   0. |A| = 3(0 − 5) − (−1)(0 + 3) + (−2)(0 − 6) = −15 + 3 + 12 = 0. Computing (adj A)B with B = (2, −1, 3)T gives a non-zero matrix, so there is no solution. Inconsistent.

6. 5x − y + 4z = 5;   2x + 3y + 5z = 2;   5x − 2y + 6z = −1

SOLUTION |A| = 5  −1  4
2   3  5
5  −2  6 = 5(18 + 10) − (−1)(12 − 25) + 4(−4 − 15) = 5(28) + 1(−13) + 4(−19) = 140 − 13 − 76 = 51 ≠ 0. Non-singular ⇒ unique solution. Consistent.

Solve system of linear equations, using matrix method, in Exercises 7 to 14.

7. 5x + 2y = 4;   7x + 3y = 5

SOLUTION |A| = 5  2
7  3 = 15 − 14 = 1. A−1 = 3  −2
−7  5. X = A−1B = 3  −2
−7  54
5 = 12−10
−28+25 = 2
−3. So x = 2, y = −3.

8. 2x − y = −2;   3x + 4y = 3

SOLUTION |A| = 2  −1
3   4 = 8 + 3 = 11. A−1 = (1/11)4  1
−3  2. X = (1/11)4  1
−3  2−2
3 = (1/11)−8+3
6+6 = (1/11)−5
12. So x = −5/11, y = 12/11.

9. 4x − 3y = 3;   3x − 5y = 7

SOLUTION |A| = 4  −3
3  −5 = −20 + 9 = −11. A−1 = (1/(−11))−5  3
−3  4. X = (1/(−11))−5  3
−3  43
7 = (1/(−11))−15+21
−9+28 = (1/(−11))6
19. So x = −6/11, y = −19/11.

10. 5x + 2y = 3;   3x + 2y = 5

SOLUTION |A| = 5  2
3  2 = 10 − 6 = 4. A−1 = (1/4)2  −2
−3  5. X = (1/4)2  −2
−3  53
5 = (1/4)6−10
−9+25 = (1/4)−4
16. So x = −1, y = 4.

11. 2x + y + z = 1;   x − 2y − z = 3/2;   3y − 5z = 9

SOLUTION A = 2   1   1
1  −2  −1
0   3  −5, |A| = 2(10 + 3) − 1(−5 − 0) + 1(3 − 0) = 26 + 5 + 3 = 34. Solving X = A−1B gives x = 1, y = 1/2, z = −3/2.

12. x − y + z = 4;   2x + y − 3z = 0;   x + y + z = 2

SOLUTION A = 1  −1   1
2   1  −3
1   1   1, |A| = 1(1 + 3) − (−1)(2 + 3) + 1(2 − 1) = 4 + 5 + 1 = 10. Solving X = A−1B gives x = 2, y = −1, z = 1.

13. 2x + 3y + 3z = 5;   x − 2y + z = −4;   3x − y − 2z = 3

SOLUTION A = 2   3   3
1  −2   1
3  −1  −2, |A| = 2(4 + 1) − 3(−2 − 3) + 3(−1 + 6) = 10 + 15 + 15 = 40. Solving X = A−1B gives x = 1, y = 2, z = −1.

14. x − y + 2z = 7;   3x + 4y − 5z = −5;   2x − y + 3z = 12

SOLUTION A = 1  −1   2
3   4  −5
2  −1   3, |A| = 1(12 − 5) − (−1)(9 + 10) + 2(−3 − 8) = 7 + 19 − 22 = 4. Solving X = A−1B gives x = 2, y = 1, z = 3.

15. If A = 2  −3   5
3   2  −4
1   1  −2, find A−1. Using A−1 solve the system of equations 2x − 3y + 5z = 11;   3x + 2y − 4z = −5;   x + y − 2z = −3

SOLUTION |A| = 2(2·(−2) − (−4)·1) − (−3)(3·(−2) − (−4)·1) + 5(3·1 − 2·1) = 2(0) + 3(−2) + 5(1) = −1 ≠ 0. Finding the cofactors and dividing by |A| = −1 gives A−1 = 0  1  −2
−2  9  −23
−1  5  −13. X = A−1B with B = (11, −5, −3)T: x = 0(11) + 1(−5) − 2(−3) = −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. ∴ x = 1, y = 2, z = 3.

16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.

SOLUTION Let onion, wheat, rice cost x, y, z per kg. Then 4x + 3y + 2z = 60; 2x + 4y + 6z = 90; 6x + 2y + 3z = 70. A = 4  3  2
2  4  6
6  2  3, |A| = 4(12 − 12) − 3(6 − 36) + 2(4 − 24) = 0 + 90 − 40 = 50 ≠ 0. Solving X = A−1B gives x = 5, y = 8, z = 8. ∴ onion = ₹5/kg, wheat = ₹8/kg, rice = ₹8/kg.

Miscellaneous Exercise on Chapter 4 Solutions

1. Prove that the determinant x  sinθ  cosθ
−sinθ  −x  1
cosθ  1  x is independent of θ.

SOLUTION Expand along R1: = x(−x·x − 1·1) − sinθ(−sinθ·x − 1·cosθ) + cosθ(−sinθ·1 − (−x)·cosθ). = x(−x2 − 1) − sinθ(−x sinθ − cosθ) + cosθ(−sinθ + x cosθ). = −x3 − x + x sin2θ + sinθcosθ − sinθcosθ + x cos2θ. = −x3 − x + x(sin2θ + cos2θ) = −x3 − x + x = −x3, which is free of θ. Hence the determinant is independent of θ.

2. Evaluate cosαcosβ  cosαsinβ  −sinα
−sinβ  cosβ  0
sinαcosβ  sinαsinβ  cosα.

SOLUTION Expand along R2 (it has a zero): = −(−sinβ)[cosαsinβ·cosα − (−sinα)·sinαsinβ] + cosβ[cosαcosβ·cosα − (−sinα)·sinαcosβ] − 0. = sinβ[cos2αsinβ + sin2αsinβ] + cosβ[cos2αcosβ + sin2αcosβ]. = sinβ·sinβ(cos2α + sin2α) + cosβ·cosβ(cos2α + sin2α) = sin2β + cos2β = 1.

3. If A−1 = 3  −1  1
−15  6  −5
5  −2  2 and B = 1  2  −2
−1  3  0
0  −2  1, find (AB)−1.

SOLUTION (AB)−1 = B−1A−1. First B−1: |B| = 1(3 − 0) − 2(−1 − 0) + (−2)(2 − 0) = 3 + 2 − 4 = 1. Cofactors of B give adj B = 3  2  6
1  1  2
2  2  5, so B−1 = adj B (since |B| = 1). (AB)−1 = B−1A−1 = 3  2  6
1  1  2
2  2  53  −1  1
−15  6  −5
5  −2  2 = 9  −3  5
−2  1  0
1  0  2.

4. Let A = 1  2  1
2  3  1
1  1  5. Verify that (i) [adj A]−1 = adj(A−1)   (ii) (A−1)−1 = A

SOLUTION |A| = 1(15 − 1) − 2(10 − 1) + 1(2 − 3) = 14 − 18 − 1 = −5 ≠ 0, so A is invertible. (i) For any invertible A, both [adj A]−1 and adj(A−1) equal (1/|A|)A. Hence [adj A]−1 = adj(A−1). Verified. (ii) Taking the inverse twice returns the original matrix: (A−1)−1 = A. Verified.

5. Evaluate x  y  x+y
y  x+y  x
x+y  x  y.

SOLUTION Apply C1 → C1 + C2 + C3: each entry of C1 becomes 2(x + y). Take 2(x + y) out: = 2(x + y)1  y  x+y
1  x+y  x
1  x  y. R2 → R2 − R1, R3 → R3 − R1, then expanding gives 2(x + y)·[−(x2 − xy + y2)]. = −2(x + y)(x2 − xy + y2) = −2(x3 + y3).

6. Evaluate 1  x  y
1  x+y  y
1  x  x+y.

SOLUTION Apply R2 → R2 − R1 and R3 → R3 − R1: = 1  x  y
0  y  0
0  0  x. Expanding along C1: = 1(y·x − 0) = xy.

7. Solve the system of equations 2/x + 3/y + 10/z = 4;   4/x − 6/y + 5/z = 1;   6/x + 9/y − 20/z = 2

SOLUTION Put p = 1/x, q = 1/y, r = 1/z: 2p + 3q + 10r = 4; 4p − 6q + 5r = 1; 6p + 9q − 20r = 2. A = 2  3  10
4  −6  5
6  9  −20, |A| = 2(120 − 45) − 3(−80 − 30) + 10(36 + 36) = 150 + 330 + 720 = 1200 ≠ 0. Solving gives p = 1/2, q = 1/3, r = 1/5, so 1/x = 1/2, 1/y = 1/3, 1/z = 1/5. ∴ x = 2, y = 3, z = 5.

8. If x, y, z are nonzero real numbers, then the inverse of matrix A = x  0  0
0  y  0
0  0  z is (A) x−1  0  0
0  y−1  0
0  0  z−1 (B) xyz x−1  0  0
0  y−1  0
0  0  z−1 (C) (1/xyz)x  0  0
0  y  0
0  0  z (D) (1/xyz)1  0  0
0  1  0
0  0  1

SOLUTION For a diagonal matrix the inverse is diagonal with reciprocal entries: A−1 = diag(1/x, 1/y, 1/z) = diag(x−1, y−1, z−1). ∴ the correct option is (A).

9. Let A = 1  sinθ  1
−sinθ  1  sinθ
−1  −sinθ  1, where 0 ≤ θ ≤ 2π. Then (A) Det(A) = 0   (B) Det(A) ∈ (2, ∞)   (C) Det(A) ∈ (2, 4)   (D) Det(A) ∈ [2, 4]

SOLUTION Expand: Det(A) = 1(1 + sin2θ) − sinθ(−sinθ + sinθ) + 1(sin2θ + 1) = (1 + sin2θ) − 0 + (1 + sin2θ) = 2 + 2sin2θ. Since 0 ≤ sin2θ ≤ 1, Det(A) ranges over [2, 4]. ∴ the correct option is (D) Det(A) ∈ [2, 4].

Common Mistakes to Avoid

Watch out for these

  • Forgetting the sign pattern (−1)i+j when expanding — the cofactor signs go + − + along the first row.
  • Confusing minor and cofactor: Aij = (−1)i+jMij, so they may differ in sign.
  • Mixing rows and columns when multiplying elements by cofactors — using cofactors of a different row gives 0, not |A|.
  • Dropping the absolute value in area problems (area is always positive) or forgetting both ± cases when the area is given.
  • Writing adj A without transposing the cofactor matrix — adj A is the transpose of [Aij].
  • Declaring a system inconsistent on |A| = 0 alone — you must also check (adj A)B against O.
  • Using |kA| = k|A|; the correct rule is |kA| = kn|A| for an n×n matrix.

Practice MCQs & Assertion–Reason

1. The value of 2  4
−5  −1 is:

(a) −22    (b) 18    (c) 22    (d) −18

2. If A is a 3×3 matrix and |A| = 4, then |2A| is:

(a) 8    (b) 16    (c) 32    (d) 64

3. For an n×n matrix A, |adj A| equals:

(a) |A|    (b) |A|n    (c) |A|n−1    (d) n|A|

4. The matrix A is invertible if and only if:

(a) |A| = 0    (b) |A| ≠ 0    (c) A is symmetric    (d) A is square

5. The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is:

(a) 9    (b) 15/2    (c) 15    (d) 30

6. Three points are collinear if the determinant formed by their coordinates is:

(a) 1    (b) positive    (c) 0    (d) negative

7. If A is invertible of order 2, then det(A−1) equals:

(a) det(A)    (b) 1/det(A)    (c) 1    (d) 0

8. The cofactor A21 of the element in 5  3  8
2  0  1
1  2  3 is:

(a) −7    (b) 7    (c) 9    (d) −9

9. A system AX = B with |A| = 0 and (adj A)B ≠ O is:

(a) consistent with unique solution    (b) consistent with infinitely many solutions    (c) inconsistent    (d) always solvable

10. The system 5x + 2y = 4, 7x + 3y = 5 has solution:

(a) x = 2, y = −3    (b) x = −2, y = 3    (c) x = 3, y = −2    (d) x = 1, y = 1

Answer key: 1-(b), 2-(c), 3-(c), 4-(b), 5-(b), 6-(c), 7-(b), 8-(b), 9-(c), 10-(a).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: Only square matrices have determinants.

Reason: A determinant is a function that assigns a unique number to each square matrix.

A-R 2. Assertion: A matrix A is invertible if and only if |A| ≠ 0.

Reason: A−1 = (1/|A|) adj A, which is defined only when |A| ≠ 0.

A-R 3. Assertion: For a 3×3 matrix, |adj A| = |A|3.

Reason: In general |adj A| = |A|n−1 for an n×n matrix.

A-R 4. Assertion: The points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.

Reason: The determinant formed by their coordinates is zero.

A-R 5. Assertion: The system x + 3y = 5, 2x + 6y = 8 is inconsistent.

Reason: Its coefficient determinant is zero and (adj A)B ≠ O.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • A determinant assigns a unique number to a square matrix; only square matrices have determinants.
  • Order 2: |A| = a11a22 − a12a21; order 3: expand along any row or column with signs (−1)i+j.
  • |kA| = kn|A| for an n×n matrix.
  • Area of a triangle = ½|determinant of coordinates|; points are collinear when it is 0.
  • Cofactor Aij = (−1)i+jMij; |A| = sum of elements of a row/column times their cofactors.
  • adj A = transpose of [Aij]; A(adj A) = (adj A)A = |A| I; A−1 = (1/|A|) adj A; |adj A| = |A|n−1.
  • For AX = B: |A| ≠ 0 ⇒ unique solution X = A−1B; |A| = 0 with (adj A)B ≠ O ⇒ inconsistent.

How to score full marks in this chapter

When expanding a third-order determinant, always pick the row or column with the most zeros to cut the work. Write the cofactor signs explicitly so you never lose a minus sign. For area and collinearity questions, set up the determinant first, then take the absolute value at the end. In matrix-method problems, compute |A| before anything else: if it is non-zero, find adj A carefully (remember to transpose) and give X = (1/|A|)(adj A)B; if it is zero, test (adj A)B to decide consistency. Verify every numerical answer by back-substitution.

Frequently Asked Questions

What is Class 12 Maths Chapter 4 Determinants about?

Chapter 4 studies determinants of square matrices up to order three: how to evaluate them, properties such as |kA| = kn|A|, the area of a triangle using a determinant, minors and cofactors, the adjoint and inverse of a matrix, and solving systems of linear equations by the matrix method (X = A−1B).

How many exercises are there in Class 12 Maths Chapter 4?

There are five exercises — 4.1, 4.2, 4.3, 4.4 and 4.5 — plus a Miscellaneous Exercise on Chapter 4. Every question of all six sets is solved step by step on this page.

How do you decide if a system of equations is consistent using determinants?

Write the system as AX = B. If |A| ≠ 0 the system has a unique solution and is consistent. If |A| = 0, compute (adj A)B: if it is non-zero the system is inconsistent (no solution); if it is the zero matrix the system may have infinitely many solutions or none.

Are these Class 12 Maths Chapter 4 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every numerical answer verified against the book’s answer key.

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