Class 12 Maths NCERT Solutions

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Determinants Class 12 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 4, Determinants — 59 questions from Ex 4.1 to Ex 4.5, each worked through step by step in the CBSE marking pattern. Determinants of square matrices, minors and cofactors, properties, and area and equation-of-a-line applications.

Class:12Subject:MathsChapter:4

4 Key Formulas

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 12 Maths Chapter 4?

Chapter 4 carries 5 exercise questions, numbered Ex 4.1 to Ex 4.5. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Chapter Overview

Determinants are scalar quantities associated with square matrices. They encode essential information about the matrix — invertibility, the volume scaling factor of a linear transformation, and whether a system of equations has a unique solution. Starting from the simple formulas for order 1, 2 and 3, this chapter develops properties that let you evaluate large determinants without brute expansion, introduces minors and cofactors, and culminates in the adjoint method for inverses and Cramer's rule for linear systems.

Board pattern

For determinants up to 3x3 you must show the full expansion step-by-step. In properties-of-determinant questions, state which property you apply at each step (e.g. R2 -> R2 + 2R1) and write the intermediate determinant. Simply writing the final value without the row operations earns at most 1 mark. For inverse questions, always verify A inverse A = I as a last step.

$Ex 4.1 — Determinant of order 1, 2, 3; evaluation (8 Q)$
$Ex 4.2 — Properties of determinants (16 Q)$
$Ex 4.3 — Minors and cofactors (5 Q)$
$Ex 4.4 — Adjoint and inverse of a matrix (5 Q)$
$Ex 4.5 — Area of triangle, consistency, Cramer’s rule (12 Q)$
$Misc — Mixed problems on determinants (15 Q)$

Exercise 4.1 — Determinants of Order 1, 2 and 3

10Exercise questions

Step-by-step solution

1For a 2x2 determinant $a c b d = a d - b c$ .
2 $= 2 (- 1) - 4 (- 5) = - 2 + 20 = 18$

Final answer

18

18.

Step-by-step solution

1 $= cos θ \cdot cos θ - (- sin θ) (sin θ)$
2 $= cos^{2} θ + sin^{2} θ = 1$

Final answer

Step-by-step solution

1 $= tan θ \cdot cos θ - (- cos θ) (sin θ)$
2 $= sin θ + cos θ sin θ = sin θ (1 + cos θ)$

Final answer

sin θ (1 + cos θ)

Step-by-step solution

1Expand along the first row: $∣ A ∣ = 2 04 - 1 5 - (- 3) 21 - 1 5 + 12104$
2 $= 2 (0 + 4) + 3 (10 + 1) + 1 (8 - 0) = 8 + 33 + 8 = 49$

Final answer

49

49.

Step-by-step solution

1Expand along the second row (two zeros): $∣ A ∣ = - 0 + 0 - (- 1) 31 - 1 - 2$
2 $= 1 (3 (- 2) - (- 1) (1)) = 1 (- 6 + 1) = - 5$

Final answer

- 5

-5.

Step-by-step solution

1Expand along the first row: $= 1 (45 - 48) - 2 (36 - 42) + 3 (32 - 35)$
2 $= - 3 + 12 - 9 = 0$

Final answer

Step-by-step solution

1 $∣ A ∣ = 1 (4) - 2 (3) = - 2$
2 $2 A = (2648), ∣2 A ∣ = 16 - 24 = - 8$
3 $4∣ A ∣ = 4 (- 2) = - 8 = ∣2 A ∣$

Final answer

∣2 A ∣ = 4∣ A ∣

Verified: |2A| = -8 = 4|A|.

Step-by-step solution

1Expand along the first row: $= a (b c - h^{2}) - h (h c - h^{2}) + h (h^{2} - hb)$
2 $= ab c - a h^{2} - h^{2} c + h^{3} + h^{3} - h^{2} b$
3 $= ab c - h^{2} (a + b + c) + 2 h^{3}$

Final answer

ab c - h^{2} (a + b + c) + 2 h^{3}

abc - h^2(a + b + c) + 2h^3.

Step-by-step solution

1LHS = $2 (1) - 4 (5) = 2 - 20 = - 18$
2RHS = $2 x \cdot x - 4 \cdot 6 = 2 x^{2} - 24$
3 $2 x^{2} - 24 = - 18 \Rightarrow 2 x^{2} = 6 \Rightarrow x^{2} = 3$
4 $x = \pm 3$

Final answer

x = \pm 3

Step-by-step solution

1LHS = $x \cdot x - 2 (18) = x^{2} - 36$
2RHS = $6 \cdot 6 - 2 (18) = 36 - 36 = 0$
3 $x^{2} - 36 = 0 \Rightarrow x = \pm 6$

Final answer

x = \pm 6

Exercise 4.2 — Properties of Determinants

18Exercise questions

Step-by-step solution

1Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
2 $= 100 a b - a c - a a^{2} b^{2} - a^{2} c^{2} - a^{2}$
3Factor: $= (b - a) (c - a) 11 b + a c + a$
4 $= (b - a) (c - a) (c + a - b - a) = (b - a) (c - a) (c - b)$
5 $= (a - b) (b - c) (c - a)$

Final answer

(a - b) (b - c) (c - a)

Shown.

Step-by-step solution

1Apply $C_{1} \to C_{1} + C_{2} + C_{3}$ :
2 $= 2 (a + b + c) 2 (p + q + r) 2 (x + y + z) c + a r + p z + x a + b p + q x + y$
3Take 2 common from C1: $= 2 a + b + c p + q + r x + y + z c + a r + p z + x a + b p + q x + y$
4Apply $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ :
5 $= 2 a + b + c p + q + r x + y + z - b - q - y - c - r - z$
6Take -1 from each of C2, C3: $= 2 a + b + c p + q + r x + y + z b q y c r z$
7Apply $C_{1} \to C_{1} - C_{2} - C_{3}$ :
8 $= 2 a p x b q y c r z$

Final answer

2 a p x b q y c r z

Shown.

Step-by-step solution

1Note the entries are $5^{2}, 7^{2}, 8^{2}; 6^{2}, 8^{2}, 9^{2}; 7^{2}, 9^{2}, 1 0^{2}$
2Apply $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{2}$ :
3 $= 253649242832151719$
4Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{2}$ :
5 $= 25111324441522$
6Apply $R_{3} \to R_{3} - R_{2}$ :
7 $= 2511224401520 = 2 (48 - 60) = - 24$

Final answer

- 24

-24.

Step-by-step solution

1Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
2 $= x y - x z - x x^{2} y^{2} - x^{2} z^{2} - x^{2} y z z x - y z x y - y z$
3Factor (y-x), (z-x) from R2, R3:
4 $= (x - y) (y - z) (z - x) (x y + y z + z x)$

Final answer

(x - y) (y - z) (z - x) (x y + y z + z x)

Shown.

Step-by-step solution

1Apply $C_{1} \to C_{1} + C_{2} + C_{3}$ :
2 $= (5 x + 4) 111 2 x x + 4 2 x 2 x 2 x x + 4$
3Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
4 $= (5 x + 4) 100 2 x 4 - x 0 2 x 0 4 - x = (5 x + 4) (4 - x)^{2}$

Final answer

(5 x + 4) (4 - x)^{2}

Shown.

Step-by-step solution

1Expand along the first row: $= (y + z) [(z + x) (x + y) - y z] - x [y (x + y) - y z] + x [y z - (z + x) z]$
2 $= (y + z) [x (x + y + z)] - x [y (x + y - z)] + x [z (y - x - z)]$
3 $= x [(x + y + z) (y + z) - y (x + y - z) + z (y - x - z)]$
4Expand the bracket: $= x [(y + z) (y + z) - (y - z)^{2}]$
5 $= x [(y + z)^{2} - (y - z)^{2}] = x \cdot 4 y z = 4 x y z$

Final answer

4 x y z

Shown.

Step-by-step solution

1Apply $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ :
2 $= 111 0 b - a c - a 0 b^{2} - a^{2} c^{2} - a^{2}$
3Expand along R1: $= (b - a) (c^{2} - a^{2}) - (c - a) (b^{2} - a^{2})$
4 $= (b - a) (c - a) (c + a) - (c - a) (b - a) (b + a) = (a - b) (b - c) (c - a)$

Final answer

(a - b) (b - c) (c - a)

Shown.

Step-by-step solution

1Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
2 $= 213712651021$
3Apply $C_{3} \to C_{3} - 10 C_{2}$ :
4 $= 213712 - 5 01$
5Expand: $= - 1 (- 7 + 5) + 0 + 1 (4 - 7) = 2 - 3 = - 1 \neq = 0$

Final answer

- 1 \neq = 0

-1 is not 0, so the determinant is nonzero.

Step-by-step solution

1Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
2 $= 100 b + c a - b a - c b^{2} + c^{2} a^{2} - b^{2} a^{2} - c^{2}$
3 $= (a - b) (a - c) 11 a + b a + c = (a - b) (a - c) (c - b)$
4 $= (a - b) (b - c) (c - a)$

Final answer

(a - b) (b - c) (c - a)

Shown.

Step-by-step solution

1Take a common from C1, b from C2, c from C3: $= ab c a + 1/ a a a b b + 1/ b b c c c + 1/ c$
2Apply $R_{1} \to R_{1} - R_{2}$ and $R_{2} \to R_{2} - R_{3}$ :
3 $= ab c 1/ a 0 a - 1/ b 1/ b b 0 - 1/ c c + 1/ c$
4Expand along the first row: $= ab c [\frac{1}{a} (\frac{1}{b} (c + \frac{1}{c}) + \frac{b}{c}) + \frac{1}{b} \cdot \frac{a}{c}]$
5 $= ab c (\frac{c}{ab} + \frac{1}{ab c} + \frac{b}{a c} + \frac{a}{b c}) = c^{2} + 1 + b^{2} + a^{2} = 1 + a^{2} + b^{2} + c^{2}$

Final answer

1 + a^{2} + b^{2} + c^{2}

Step-by-step solution

1Apply $R_{2} \to R_{2} - 2 R_{3}$ :
2 $= a a x b b y c c z$
3R1 = R2, so the determinant is 0.

Final answer

0

Shown: determinant = 0.

Step-by-step solution

1Take 3 from R1 and 2 from R2: $= 6 111 x x 1 x^{2} x^{2} 1$
2R1 = R2, so the determinant is 0.

Final answer

0

Shown: determinant = 0.

Step-by-step solution

1The matrix is skew-symmetric: $a_{ij} = - a_{j i}$ , so its diagonal is all zeros.
2Take -1 common from each of the three rows: $= (- 1) (- 1) (- 1) 0 a - b - a 0 - c b c 0$
3But the new matrix is merely the negative of the original, so $D = - D \Rightarrow D = 0$

Final answer

0

Shown: determinant = 0.

Step-by-step solution

1Apply $R_{2} \to R_{2} - 2 R_{1}$ :
2 $R_{2} = (0, 1, 2)$
3Apply $R_{3} \to R_{3} - 3 R_{1}$ :
4 $R_{3} = (0, 3, 7)$
5 $= 100 1 + p 13 1 + p + q 27 = 1 (7 - 6) = 1$

Final answer

1

Shown.

Step-by-step solution

1Apply $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ :
2 $= 1 a a^{3} 0 b - a b^{3} - a^{3} 0 c - a c^{3} - a^{3}$
3Expand along row 1: $= (b - a) (c^{3} - a^{3}) - (c - a) (b^{3} - a^{3})$
4Factor using $u^{3} - v^{3} = (u - v) (u^{2} + uv + v^{2})$
5 $= (b - a) (c - a) [(c^{2} + a c + a^{2}) - (b^{2} + ab + a^{2})] = (a - b) (b - c) (c - a) (a + b + c)$

Final answer

(a - b) (b - c) (c - a) (a + b + c)

Shown.

Step-by-step solution

1Apply $C_{1} \to C_{1} + C_{2} + C_{3}$ :
2 $= (1 + x + x^{2}) 111 x 1 x^{2} x^{2} x 1$
3Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :
4 $= (1 + x + x^{2}) 100 x 1 - x x^{2} - x x^{2} x - x^{2} 1 - x^{2}$
5Factor (1-x) from R2 and R3: $= (1 + x + x^{2}) (1 - x)^{2} 100 x 1 - x x^{2} x 1 + x$
62x2 minor: $= (1 + x + x^{2}) (1 - x)^{2} (1 + x + x^{2}) = [(1 - x) (1 + x + x^{2})]^{2}$
7Using $1 - x^{3} = (1 - x) (1 + x + x^{2})$ the result is $(1 - x^{3})^{2}$

Final answer

(1 - x^{3})^{2}

Step-by-step solution

1Apply $C_{1} \to C_{1} + C_{2} + C_{3}$ :
2Sum of the three entries of row 1 = $(a - b) + (b - c) + (c - a) = 0$
3Similarly rows 2 and 3 each have sum 0, so C1 becomes a zero column: $= 0$

Final answer

0

Shown: determinant = 0.

Step-by-step solution

1a, b, c in AP means $2 b = a + c$
2Apply $R_{1} \to R_{1} + R_{3} - 2 R_{2}$ :
3First two entries become zero: $(x + 2) + (x + 4) - 2 (x + 3) = 0$ and similarly the second entry.
4Third entry: $(x + 2 a) + (x + 2 c) - 2 (x + 2 b) = 2 a + 2 c - 4 b = 2 (a + c) - 4 b = 0$
5Row 1 becomes a zero row, so the determinant is 0.

Final answer

Option (A) — 0.

Exercise 4.3 — Minors and Cofactors

7Exercise questions

Step-by-step solution

1The minor $M_{ij}$ is obtained by deleting row i and column j.
2 $M_{11} = 4 - 2 53 = 22, M_{12} = 6153 = 13, M_{13} = 61 4 - 2 = - 16$
3 $M_{21} = 3 - 2 - 1 3 = 7, M_{22} = 21 - 1 3 = 7, M_{23} = 21 3 - 2 = - 7$
4 $M_{31} = 34 - 1 5 = 19, M_{32} = 26 - 1 5 = 16, M_{33} = 2634 = - 10$
5Cofactors: $A_{ij} = (- 1)^{i + j} M_{ij}$

Final answer

A_{11} = 22, A_{12} = - 13, A_{13} = - 16, A_{21} = - 7, A_{22} = 7, A_{23} = 7, A_{31} = 19, A_{32} = - 16, A_{33} = - 10

Step-by-step solution

1Minor $M_{22}$ : delete row 2, column 2.
2 $M_{22} = 2719 = 18 - 7 = 11$
3Cofactor: $A_{22} = (- 1)^{2 + 2} \cdot 11 = 11$

Final answer

M_{22} = 11, A_{22} = 11

M22 = 11, A22 = 11.

Step-by-step solution

1 $A_{11} = 4, A_{12} = - 1, A_{21} = - 3, A_{22} = 2$
2Expand along row 1: $∣ A ∣ = a_{11} A_{11} + a_{12} A_{12} = 2 (4) + 3 (- 1) = 8 - 3 = 5$

Final answer

∣ A ∣ = 5

|A| = 5.

Step-by-step solution

1Apply the sign pattern $+ - + - + - + - +$
2 $A_{11} = (- 1)^{1 + 1} - 1 1 42 = - 2 - 4 = - 6$
3 $A_{12} = - 2042 = - 4, A_{13} = 20 - 1 1 = 2$
4 $A_{21} = - (21 - 3 2) = - (4 + 3) = - 7, A_{22} = 10 - 3 2 = 2$
5 $A_{23} = - (1021) = - 1, A_{31} = 2 - 1 - 3 4 = 5$
6 $A_{32} = - (12 - 3 4) = - 10, A_{33} = 12 2 - 1 = - 5$

Final answer

A_{11} = - 6, A_{12} = - 4, A_{13} = 2, A_{21} = - 7, A_{22} = 2, A_{23} = - 1, A_{31} = 5, A_{32} = - 10, A_{33} = - 5

Step-by-step solution

1Expand along the first row: $D = 5 - 1 3 42 - (- 2) 2142 + 3 21 - 1 3$
2 $= 5 (- 2 - 12) + 2 (4 - 4) + 3 (6 + 1) = - 70 + 0 + 21 = - 49$

Final answer

- 49

-49.

Step-by-step solution

1For a 2x2 determinant $M_{11} = 1, M_{12} = 2, M_{21} = - 4, M_{22} = 3$
2Cofactors $A_{ij} = (- 1)^{i + j} M_{ij}$ : $A_{11} = 1, A_{12} = - 2, A_{21} = 4, A_{22} = 3$

Final answer

A_{11} = 1, A_{12} = - 2, A_{21} = 4, A_{22} = 3

Step-by-step solution

1Cofactors of row 2: $A_{21} = - (2 (1) - 3 (3)) = 7, A_{22} = (1 (1) - 3 (2)) = - 5, A_{23} = - (1 (3) - 2 (2)) = 1$
2 $a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} = 3 (7) + 1 (- 5) + 2 (1) = 21 - 5 + 2 = 18$
3Direct expansion: $∣ A ∣ = 1 (1 - 6) - 2 (3 - 4) + 3 (9 - 2) = - 5 + 2 + 21 = 18$
4Both equal 18.

Final answer

18

Verified: both sums give 18.

Exercise 4.4 — Adjoint and Inverse of a Matrix

8Exercise questions

Step-by-step solution

1 $∣ A ∣ = - 12 + 12 = 0$
2Cofactors: $A_{11} = - 6, A_{12} = 4, A_{21} = - 3, A_{22} = 2$
3 $adj (A) = (- 6 4 - 3 2)$
4 $A (adj A) = (2 - 4 3 - 6) (- 6 4 - 3 2) = (0000) = 0 \cdot I$
5Similarly (adj A)A = 0I.

Final answer

Verified: A(adj A) = (adj A)A = 0I = |A|I.

Step-by-step solution

1Cofactors: $A_{11} = - 5, A_{12} = - 3, A_{21} = - 2, A_{22} = 1$
2 $adj (A) = (- 5 - 3 - 2 1)$

Final answer

adj (A) = (- 5 - 3 - 2 1)

adj(A) = [-5, -2; -3, 1].

Step-by-step solution

1 $∣ A ∣ = 2 (4) - 3 (1) = 5 \neq = 0$
2Cofactors: $A_{11} = 4, A_{12} = - 1, A_{21} = - 3, A_{22} = 2$
3 $adj (A) = (4 - 1 - 3 2)$
4 $A^{- 1} = \frac{1}{5} (4 - 1 - 3 2)$
5Verify: $A A^{- 1} = \frac{1}{5} (2134) (4 - 1 - 3 2) = \frac{1}{5} (5005) = I_{2}$

Final answer

A^{- 1} = \frac{1}{5} (4 - 1 - 3 2)

Step-by-step solution

1 $∣ A ∣ = 1 (1 - 0) - 2 (0 - 0) + 3 (0 - 0) = 1$
2Cofactors:
3 $C_{11} = 1, C_{12} = 0, C_{13} = 0$
4 $C_{21} = - 2, C_{22} = 1, C_{23} = 0$
5 $C_{31} = 1, C_{32} = - 2, C_{33} = 1$
6 $A^{- 1} = \frac{1}{1} 100 - 2 10 1 - 2 1$

Final answer

A^{- 1} = 100 - 2 10 1 - 2 1

Step-by-step solution

1 $∣ A ∣ = - 2, ∣ B ∣ = 10$
2 $A^{- 1} = \frac{1}{- 2} (4 - 3 - 2 1) = (- 2 3/2 1 - 1/2)$
3 $B^{- 1} = \frac{1}{10} (4 - 2 - 1 3)$
4 $A B = (717919), (A B)^{- 1} = (19/20 - 17/20 - 9/20 7/20)$
5 $B^{- 1} A^{- 1} = (4/10 - 2/10 - 1/10 3/10) (- 2 3/2 1 - 1/2) = (19/20 - 17/20 - 9/20 7/20)$

Final answer

(A B)^{- 1} = B^{- 1} A^{- 1}

Verified.

Step-by-step solution

1Cofactors of row 1: $C_{11} = 0 (2) - (- 1) (1) = 1, C_{12} = - (2 (2) - (- 1) (3)) = - 7, C_{13} = 2 (1) - 0 (3) = 2$
2Cofactors of row 2: $C_{21} = - (2 (2) - 3 (1)) = - 1, C_{22} = 1 (2) - 3 (3) = - 7, C_{23} = - (1 (1) - 2 (3)) = 5$
3Cofactors of row 3: $C_{31} = 2 (- 1) - 3 (0) = - 2, C_{32} = - (1 (- 1) - 3 (2)) = 7, C_{33} = 1 (0) - 2 (2) = - 4$
4Transpose gives the adjoint: $adj (A) = 1 - 7 2 - 1 - 7 5 - 2 7 - 4$

Final answer

adj (A) = 1 - 7 2 - 1 - 7 5 - 2 7 - 4

Step-by-step solution

1 $∣ A ∣ = 2 (- 4) - 3 (1) = - 8 - 3 = - 11$
2 $adj (A) = (- 4 - 1 - 3 2)$
3 $A (adj A) = (21 3 - 4) (- 4 - 1 - 3 2) = (- 11 0 0 - 11) = - 11 I$
4Similarly (adj A)A = -11I = |A|I.

Final answer

Verified: A(adj A) = (adj A)A = -11I = |A|I.

Step-by-step solution

1Multiplying a matrix by the scalar k multiplies every row by k.
2Each of the 3 rows contributes a factor k to the determinant, so $∣ k A ∣ = k \cdot k \cdot k ∣ A ∣ = k^{3} ∣ A ∣$

Final answer

∣ k A ∣ = k^{3} ∣ A ∣

Exercise 4.5 — Area of Triangle, Consistency and Cramer's Rule

16Exercise questions

Step-by-step solution

1Area = $\frac{1}{2} 143225111$
2 $= \frac{1}{2} ∣1 (2 - 5) - 2 (4 - 3) + 1 (20 - 6) ∣ = \frac{1}{2} ∣ - 3 - 2 + 14∣ = \frac{9}{2}$

Final answer

\frac{9}{2}

9/2 square units.

Step-by-step solution

1 $Area = \frac{1}{2} 03 - 2 043111$
2Expand along the first row: $= \frac{1}{2} ∣0 - 0 + 1 (9 - (- 8)) ∣ = \frac{17}{2}$

Final answer

\frac{17}{2}

17/2 square units.

Step-by-step solution

1 $2 \cdot Area = - 1 42 - 1 - 2 2 111 = - 1 (- 2 - 2) + 1 (4 - 2) + 1 (8 + 4) = 4 + 2 + 12 = 18$
2 $Area = 9$

Final answer

9

9 square units.

Step-by-step solution

1 $2 \cdot Area = 533752111 = 5 (5 - 2) - 7 (3 - 3) + 1 (6 - 15) = 15 + 0 - 9 = 6$
2 $Area = 3$

Final answer

3

3 square units.

Step-by-step solution

1 $D = 1223 = 3 - 4 = - 1 \neq = 0$
2Since D is not 0, the system has a unique solution for all values.
3 $D_{x} = 2323 = 0, D_{y} = 1223 = - 1$
4 $x = D_{x} / D = 0, y = D_{y} / D = 1$

Final answer

Unique solution: x = 0, y = 1.

Step-by-step solution

1 $D = 21 - 1 1 = 3 \neq = 0$
2Since D is not 0, the system is consistent with a unique solution.
3 $x = \frac{1}{3} 54 - 1 1 = 3, y = \frac{1}{3} 2154 = 1$

Final answer

x = 3, y = 1

Consistent; x = 3, y = 1.

Step-by-step solution

1 $D = 123 1 - 1 1 13 - 1 = 1 (1 - 3) - 1 (- 2 - 9) + 1 (2 + 3) = - 2 + 11 + 5 = 14$
2 $D_{x} = 132 1 - 1 1 13 - 1 = 1 (1 - 3) - 1 (- 3 - 6) + 1 (3 + 2) = - 2 + 9 + 5 = 12$
3 $D_{y} = 123132 13 - 1 = 1 (- 3 - 6) - 1 (- 2 - 9) + 1 (4 - 9) = - 9 + 11 - 5 = - 3$
4 $D_{z} = 123 1 - 1 1 132 = 1 (- 2 - 3) - 1 (4 - 9) + 1 (2 + 3) = - 5 + 5 + 5 = 5$
5 $x = \frac{12}{14} = \frac{6}{7}, y = - \frac{3}{14}, z = \frac{5}{14}$

Final answer

x = \frac{6}{7}, y = - \frac{3}{14}, z = \frac{5}{14}

Step-by-step solution

1Area must be nonzero for non-collinearity.
2 $2 \cdot Area = 1 - 4 - 2 - 1 2 - 3 111 = 1 (2 + 3) + 1 (- 4 + 2) + 1 (12 + 4) = 5 - 2 + 16 = 19$
3The area is 19/2 which is not 0, so the points are not collinear.

Final answer

The determinant is 19/2 which is not 0, so the points are not collinear.

Step-by-step solution

1 $D = 111123136 = 1 (12 - 9) - 1 (6 - 3) + 1 (3 - 2) = 3 - 3 + 1 = 1 \neq = 0$
2Since D is not 0, the system is consistent with a unique solution. Apply Cramer's rule to find x, y, z.
3 $D_{x} = 61420123136 = 6 (12 - 9) - 1 (84 - 60) + 1 (42 - 40) = 18 - 24 + 2 = - 4$
4 $D_{y} = 11161420136 = 1 (84 - 60) - 6 (6 - 3) + 1 (20 - 14) = 24 - 18 + 6 = 12$
5 $D_{z} = 11112361420 = 1 (40 - 42) - 1 (20 - 14) + 6 (3 - 2) = - 2 - 6 + 6 = - 2$
6 $x = - 4/1 = - 4, y = 12/1 = 12, z = - 2/1 = - 2$

Final answer

x = - 4, y = 12, z = - 2

Step-by-step solution

1 $D = 121 - 1 11 1 - 3 1 = 1 (1 + 3) + 1 (2 + 3) + 1 (2 - 1) = 4 + 5 + 1 = 10$
2 $D_{x} = 402 - 1 11 1 - 3 1 = 4 (4) + 1 (6) + 1 (- 2) = 16 + 6 - 2 = 20$
3 $D_{y} = 121402 1 - 3 1 = 1 (6) - 4 (5) + 1 (4) = 6 - 20 + 4 = - 10$
4 $D_{z} = 121 - 1 11 402 = 1 (2) + 1 (4) + 4 (1) = 2 + 4 + 4 = 10$
5 $x = 20/10 = 2, y = - 10/10 = - 1, z = 10/10 = 1$

Final answer

x = 2, y = - 1, z = 1

Step-by-step solution

1The line meets the axes at (5, 0) and (0, 4).
2Area = $\frac{1}{2} \times 5 \times 4 = 10$

Final answer

10

10 square units.

Step-by-step solution

1 $2 \cdot Area = k 40 002111$
2Expand along column 2: $= - 0 + 0 - 2 (k - 4) = - 2 (k - 4)$
3 $∣ - 2 (k - 4) ∣ = 8 \Rightarrow ∣ k - 4∣ = 4$
4 $k = 0 or k = 8$

Final answer

k = 0 or k = 8

k = 0 or k = 8.

Step-by-step solution

1 $2 \cdot Area = 240 - 3 52 111$
2Expand along the first row: $= 2 (5 - 2) + 3 (4 - 0) + 1 (8 - 0) = 6 + 12 + 8 = 26$
3 $Area = 13$

Final answer

13

13 square units.

Step-by-step solution

1 $D = 12 - 1 3 = 3 + 2 = 5 \neq = 0$
2Since D is not 0, the system is consistent with a unique solution.
3 $D_{x} = 31 - 1 3 = 9 + 1 = 10, D_{y} = 1231 = 1 - 6 = - 5$
4 $x = 10/5 = 2, y = - 5/5 = - 1$

Final answer

x = 2, y = - 1

Consistent; x = 2, y = -1.

Step-by-step solution

1 $D = 1212 = 2 - 2 = 0$
2 $D_{x} = 2512 = 4 - 5 = - 1 \neq = 0$
3Since D = 0 but Dx is not 0, the system has no solution - it is inconsistent.

Final answer

Inconsistent (no solution).

Step-by-step solution

1 $D = 1224 = 4 - 4 = 0$
2 $D_{x} = 4824 = 16 - 16 = 0, D_{y} = 1248 = 8 - 8 = 0$
3Since D = Dx = Dy = 0, the system is consistent (dependent) with infinitely many solutions.

Final answer

Consistent with infinitely many solutions.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Determinant of a product

∣ A B ∣ = ∣ A ∣ ∣ B ∣

Cofactor expansion

∣ A ∣ = \sum a_{ij} C_{ij}

Transpose

∣ A^{T} ∣ = ∣ A ∣

Adjugate relation

A A^{*} = ∣ A ∣ I

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Interchanging two rows changes the sign of the determinant, and multiplying a row by a scalar multiplies the determinant by the same scalar.
A singular matrix has determinant zero and therefore no inverse — check this before attempting any matrix inverse.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Maths Chapter 4 (Determinants)?

There are 5 exercise questions in this chapter, numbered Ex 4.1 to Ex 4.5. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Determinants Class 12 Maths?

The formulas this chapter's questions actually turn on are: Determinant of a product, Cofactor expansion, Transpose, Adjugate relation. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Determinants important for JEE Main?

Very important — determinants underpin the area-of-triangle and line-equation questions in boards, and the properties are high-frequency in JEE Main.

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