Class 12 Mathematics - NCERT Mathematics Part I
NCERT Solutions for Class 12 Maths Chapter 4: Determinants
Chapter 4 covers determinants, which are scalar values associated with square matrices. Students learn to evaluate 2×2 and 3×3 determinants, properties of determinants, area of a triangle using determinants, minors and cofactors, adjoint and inverse of a matrix, and solving systems of linear equations using the inverse of a matrix and Cramer's rule. This chapter is part of the Algebra unit which carries 10 marks in the CBSE Class 12 board exam 2027 (combined with Chapter 3: Matrices). Determinants typically contribute about 5 marks.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 4 Determinants has 6 exercises plus a Miscellaneous Exercise. Exercise 4.1 (8 Qs) covers evaluation of determinants. Exercise 4.2 (16 Qs) covers properties of determinants. Exercise 4.3 (5 Qs) covers area of a triangle. Exercise 4.4 (5 Qs) covers minors and cofactors. Exercise 4.5 (18 Qs) covers adjoint and inverse. Exercise 4.6 (16 Qs) covers applications of determinants and matrices. The Miscellaneous Exercise has 19 questions. Total 87 questions.
Table of Contents
4.1Exercise 4.1 - Evaluation of Determinants
Q1Evaluate the determinant |2, 4; −5, −1|.
For a 2×2 determinant |a, b; c, d| = ad − bc.
|2, 4; −5, −1| = 2(−1) − 4(−5)
= −2 − (−20)
= −2 + 20
= 18
Q2Evaluate: (i) |cos θ, −sin θ; sin θ, cos θ| (ii) |x² − x + 1, x − 1; x + 1, x + 1|
(i) |cos θ, −sin θ; sin θ, cos θ|
= cos θ × cos θ − (−sin θ) × sin θ
= cos²θ + sin²θ
= 1 (using the Pythagorean identity)
(ii) |x² − x + 1, x − 1; x + 1, x + 1|
= (x² − x + 1)(x + 1) − (x − 1)(x + 1)
Expand first term: (x² − x + 1)(x + 1) = x³ + x² − x² − x + x + 1 = x³ + 1
Expand second term: (x − 1)(x + 1) = x² − 1
= (x³ + 1) − (x² − 1)
= x³ + 1 − x² + 1
= x³ − x² + 2
Q3If A = [[1, 2], [4, 2]], then show that |2A| = 4|A|.
Step 1: Find |A|.
|A| = |1, 2; 4, 2| = 1(2) − 2(4) = 2 − 8 = −6
Step 2: Find 2A and |2A|.
2A = [[2, 4], [8, 4]]
|2A| = |2, 4; 8, 4| = 2(4) − 4(8) = 8 − 32 = −24
Step 3: Verify.
4|A| = 4 × (−6) = −24
Since |2A| = −24 = 4|A|, |2A| = 4|A| is verified. ✓
Note: For a 2×2 matrix, |kA| = k²|A|. Here k = 2, so |2A| = 2²|A| = 4|A|.
Q4If A = [[1, 0, 1], [0, 1, 2], [0, 0, 4]], then show that |3A| = 27|A|.
Step 1: Find |A|.
Expand along R1 (Row 1):
|A| = 1 × |1, 2; 0, 4| − 0 × |0, 2; 0, 4| + 1 × |0, 1; 0, 0|
= 1 × (4 − 0) − 0 + 1 × (0 − 0)
= 4 − 0 + 0 = 4
Step 2: Find 3A and |3A|.
3A = [[3, 0, 3], [0, 3, 6], [0, 0, 12]]
Expand |3A| along R1:
|3A| = 3 × |3, 6; 0, 12| − 0 × |0, 6; 0, 12| + 3 × |0, 3; 0, 0|
= 3 × (36 − 0) − 0 + 3 × (0 − 0)
= 3 × 36 = 108
Step 3: Verify.
27|A| = 27 × 4 = 108
Since |3A| = 108 = 27|A|, |3A| = 27|A| is verified. ✓
Note: For a 3×3 matrix, |kA| = k³|A|. Here k = 3, so |3A| = 3³|A| = 27|A|.
Q5Evaluate 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|
(i) |3, −1, −2; 0, 0, −1; 3, −5, 0|
Expand along R2 (two zeros make it easier):
= −0 × M₂₁ + 0 × M₂₂ + (−1)(−1)^(2+3) × |3, −1; 3, −5|
= 0 − 0 + (−1)(−1) × (3(−5) − (−1)(3))
= 1 × (−15 + 3)
= −12
(ii) |3, −4, 5; 1, 1, −2; 2, 3, 1|
Expand along R1:
= 3(1 + 6) − (−4)(1 + 4) + 5(3 − 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5
= 46
(iii) |0, 1, 2; −1, 0, −3; −2, 3, 0|
Expand along R1:
= 0 × |0, −3; 3, 0| − 1 × |−1, −3; −2, 0| + 2 × |−1, 0; −2, 3|
= 0 − 1(0 − 6) + 2(−3 − 0)
= 0 − 1(−6) + 2(−3)
= 6 − 6
= 0
(iv) |2, −1, −2; 0, 2, −1; 3, −5, 0|
Expand along C1 (Column 1):
= 2(−1)^(1+1) × |2, −1; −5, 0| + 0 + 3(−1)^(3+1) × |−1, −2; 2, −1|
= 2(0 − 5) + 0 + 3(1 + 4)
= 2(−5) + 3(5)
= −10 + 15
= 5
Q6If A = [[1, 1, −2], [2, 1, −3], [5, 4, −9]], find |A|.
Expand along R1:
|A| = 1 × |1, −3; 4, −9| − 1 × |2, −3; 5, −9| + (−2) × |2, 1; 5, 4|
= 1(−9 + 12) − 1(−18 + 15) + (−2)(8 − 5)
= 1(3) − 1(−3) + (−2)(3)
= 3 + 3 − 6
= 0
Note: Since |A| = 0, the matrix A is singular (non-invertible).
Q7Find values of x, if: (i) |2, 4; 5, 1| = |2x, 4; 6, x| (ii) |2, 3; 4, 5| = |x, 3; 2x, 5|
(i) |2, 4; 5, 1| = |2x, 4; 6, x|
LHS = 2(1) − 4(5) = 2 − 20 = −18
RHS = 2x(x) − 4(6) = 2x² − 24
Setting LHS = RHS:
2x² − 24 = −18
2x² = 6
x² = 3
x = ±√3
(ii) |2, 3; 4, 5| = |x, 3; 2x, 5|
LHS = 2(5) − 3(4) = 10 − 12 = −2
RHS = x(5) − 3(2x) = 5x − 6x = −x
Setting LHS = RHS:
−x = −2
x = 2
Q8If |x, 2; 18, x| = |6, 2; 18, 6|, then x is equal to: (A) 6 (B) ±6 (C) −6 (D) 0
LHS = |x, 2; 18, x| = x(x) − 2(18) = x² − 36
RHS = |6, 2; 18, 6| = 6(6) − 2(18) = 36 − 36 = 0
Setting LHS = RHS:
x² − 36 = 0
x² = 36
x = ±6
Answer: (B) ±6
4.3Exercise 4.3 - Area of a Triangle
Q1Find the area of the triangle with vertices at the points 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)
Area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is:
Δ = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
(i) Vertices: (1, 0), (6, 0), (4, 3)
Δ = ½|1(0 − 3) + 6(3 − 0) + 4(0 − 0)|
= ½|−3 + 18 + 0|
= ½ × 15
= 15/2 sq. units
(ii) Vertices: (2, 7), (1, 1), (10, 8)
Δ = ½|2(1 − 8) + 1(8 − 7) + 10(7 − 1)|
= ½|2(−7) + 1(1) + 10(6)|
= ½|−14 + 1 + 60|
= ½ × 47
= 47/2 sq. units
(iii) Vertices: (−2, −3), (3, 2), (−1, −8)
Δ = ½|(−2)(2 − (−8)) + 3(−8 − (−3)) + (−1)(−3 − 2)|
= ½|(−2)(10) + 3(−5) + (−1)(−5)|
= ½|−20 − 15 + 5|
= ½ × |−30|
= ½ × 30
= 15 sq. units
Q2Show that the points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.
Three points are collinear if the area of the triangle formed by them is zero.
Area = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
Here (x₁, y₁) = (a, b+c), (x₂, y₂) = (b, c+a), (x₃, y₃) = (c, a+b).
Δ = ½|a((c+a) − (a+b)) + b((a+b) − (b+c)) + c((b+c) − (c+a))|
= ½|a(c+a−a−b) + b(a+b−b−c) + c(b+c−c−a)|
= ½|a(c−b) + b(a−c) + c(b−a)|
= ½|ac − ab + ab − bc + bc − ac|
= ½|0|
= 0
Since the area of the triangle is 0, the points A, B, and C are collinear. ∎
Alternative method: Using the determinant form:
|a, b+c, 1; b, c+a, 1; c, a+b, 1|
Apply C₂ → C₂ + C₁ (add column 1 to column 2):
|a, a+b+c, 1; b, a+b+c, 1; c, a+b+c, 1|
Take (a+b+c) common from C₂:
= (a+b+c)|a, 1, 1; b, 1, 1; c, 1, 1|
Since C₂ = C₃, the determinant = 0.
Hence the points are collinear. ∎
Q3Find the value of k if area of the triangle is 4 sq. units and vertices are: (i) (k, 0), (4, 0), (0, 2) (ii) (−2, 0), (0, 4), (0, k)
(i) Vertices: (k, 0), (4, 0), (0, 2), Area = 4
Δ = ½|k(0 − 2) + 4(2 − 0) + 0(0 − 0)|
= ½|−2k + 8|
= ½|−2k + 8|
Given Δ = 4:
½|−2k + 8| = 4
|−2k + 8| = 8
Case 1: −2k + 8 = 8 → −2k = 0 → k = 0
Case 2: −2k + 8 = −8 → −2k = −16 → k = 8
(ii) Vertices: (−2, 0), (0, 4), (0, k), Area = 4
Δ = ½|(−2)(4 − k) + 0(k − 0) + 0(0 − 4)|
= ½|−8 + 2k|
= ½|2k − 8|
Given Δ = 4:
½|2k − 8| = 4
|2k − 8| = 8
Case 1: 2k − 8 = 8 → 2k = 16 → k = 8
Case 2: 2k − 8 = −8 → 2k = 0 → k = 0
Q4(i) Find the equation of the line joining (1, 2) and (3, 6) using determinants. (ii) Find the equation of the line joining (3, 1) and (9, 3) using determinants.
If (x, y) is a point on the line through (x₁, y₁) and (x₂, y₂), then the three points are collinear. So:
|x₁, y₁, 1; x₂, y₂, 1; x, y, 1| = 0
(i) Points: (1, 2) and (3, 6)
|1, 2, 1; 3, 6, 1; x, y, 1| = 0
Expand along R1:
1(6 − y) − 2(3 − x) + 1(3y − 6x) = 0
6 − y − 6 + 2x + 3y − 6x = 0
2y − 4x = 0
y = 2x
Verification: (1, 2): 2 = 2(1) ✓. (3, 6): 6 = 2(3) ✓.
(ii) Points: (3, 1) and (9, 3)
|3, 1, 1; 9, 3, 1; x, y, 1| = 0
Expand along R1:
3(3 − y) − 1(9 − x) + 1(9y − 3x) = 0
9 − 3y − 9 + x + 9y − 3x = 0
6y − 2x = 0
x = 3y or equivalently x − 3y = 0
Verification: (3, 1): 3 = 3(1) ✓. (9, 3): 9 = 3(3) ✓.
Q5If the area of a triangle with vertices (2, −6), (5, 4), (k, 4) is 35 sq. units, then k is: (A) 12 (B) −2 (C) −12, −2 (D) 12, −2
Δ = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
Vertices: (2, −6), (5, 4), (k, 4)
Δ = ½|2(4 − 4) + 5(4 − (−6)) + k(−6 − 4)|
= ½|2(0) + 5(10) + k(−10)|
= ½|0 + 50 − 10k|
= ½|50 − 10k|
Given Δ = 35:
½|50 − 10k| = 35
|50 − 10k| = 70
Case 1: 50 − 10k = 70
−10k = 20
k = −2
Case 2: 50 − 10k = −70
−10k = −120
k = 12
k = 12 or k = −2
Answer: (D) 12, −2
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Using the properties of determinants, prove that |a−b, b−c, c−a; b−c, c−a, a−b; c−a, a−b, b−c| = 0.
Apply C₁ → C₁ + C₂ + C₃:
C₁ becomes: (a−b)+(b−c)+(c−a) = 0, (b−c)+(c−a)+(a−b) = 0, (c−a)+(a−b)+(b−c) = 0.
Since all elements of C₁ are 0, the determinant = 0. Hence proved.
Solve the system of equations using the matrix method: 2x − y + z = 1, x + 2y − z = 2, 3x − y + 2z = 1.
A = [[2,−1,1],[1,2,−1],[3,−1,2]], X = [[x],[y],[z]], B = [[1],[2],[1]].
|A| = 2(4−1)+1(2+3)+1(−1−6) = 2(3)+1(5)+1(−7) = 6+5−7 = 4 ≠ 0.
Cofactors: A₁₁=3, A₁₂=−5, A₁₃=−7, A₂₁=1, A₂₂=1, A₂₃=−1, A₃₁=−1, A₃₂=3, A₃₃=5.
adj(A) = [[3,1,−1],[−5,1,3],[−7,−1,5]].
A⁻¹ = (1/4)[[3,1,−1],[−5,1,3],[−7,−1,5]].
X = A⁻¹B = (1/4)[[3+2−1],[−5+2+3],[−7−2+5]] = (1/4)[[4],[0],[−4]] = [[1],[0],[−1]].
x = 1, y = 0, z = −1.
Verification: 2(1)−0+(−1) = 1 ✓, 1+0−(−1) = 2 ✓, 3(1)−0+2(−1) = 1 ✓.
Find the value of |a, a², a³−1; b, b², b³−1; c, c², c³−1| if a, b, c are in A.P.
Split the determinant: |a,a²,a³−1; b,b²,b³−1; c,c²,c³−1| = |a,a²,a³; b,b²,b³; c,c²,c³| − |a,a²,1; b,b²,1; c,c²,1|.
First det = abc|1,a,a²; 1,b,b²; 1,c,c²| (taking a,b,c common from R1,R2,R3 and rearranging columns).
Second det = |a,a²,1; b,b²,1; c,c²,1| = (−1)²|1,a,a²; 1,b,b²; 1,c,c²| (swapping columns twice).
So result = (abc−1)|1,a,a²; 1,b,b²; 1,c,c²|.
|1,a,a²; 1,b,b²; 1,c,c²| = (a−b)(b−c)(c−a).
If a,b,c are in A.P., let b = a+d, c = a+2d. Then a−b = −d, b−c = −d, c−a = 2d.
Product = (−d)(−d)(2d) = 2d³.
So the answer = (abc−1) × 2d³ where d is the common difference. If the question requires a specific numerical answer, additional constraints on a,b,c are needed.
Using Cramer's rule, solve: x + y + z = 6, x + 2z = 7, 3x + y + z = 12.
D = |1,1,1; 1,0,2; 3,1,1| = 1(0−2)−1(1−6)+1(1−0) = −2+5+1 = 4.
D₁ = |6,1,1; 7,0,2; 12,1,1| = 6(0−2)−1(7−24)+1(7−0) = −12+17+7 = 12.
D₂ = |1,6,1; 1,7,2; 3,12,1| = 1(7−24)−6(1−6)+1(12−21) = −17+30−9 = 4.
D₃ = |1,1,6; 1,0,7; 3,1,12| = 1(0−7)−1(12−21)+6(1−0) = −7+9+6 = 8.
x = D₁/D = 12/4 = 3, y = D₂/D = 4/4 = 1, z = D₃/D = 8/4 = 2.
Verification: 3+1+2 = 6 ✓, 3+2(2) = 7 ✓, 3(3)+1+2 = 12 ✓.
Evaluate: |1, a, a²; 1, b, b²; 1, c, c²|.
Apply R₂ → R₂ − R₁ and R₃ → R₃ − R₁:
|1, a, a²; 0, b−a, b²−a²; 0, c−a, c²−a²|
= 1 × |b−a, b²−a²; c−a, c²−a²| (expanding along C₁)
= |b−a, (b−a)(b+a); c−a, (c−a)(c+a)|
Take (b−a) common from R₁ and (c−a) from R₂:
= (b−a)(c−a) × |1, b+a; 1, c+a|
= (b−a)(c−a) × [(c+a) − (b+a)]
= (b−a)(c−a)(c−b)
= (a−b)(b−c)(c−a)
This is the Vandermonde determinant.
If A = [[2, 3], [5, −2]], find A⁻¹. Use A⁻¹ to solve: 2x + 3y = 1, 5x − 2y = 11.
|A| = 2(−2) − 3(5) = −4 − 15 = −19 ≠ 0. So A⁻¹ exists.
adj(A) = [[−2, −3], [−5, 2]]. (Swap diagonal elements, negate off-diagonal.)
A⁻¹ = (1/−19)[[−2, −3], [−5, 2]] = [[2/19, 3/19], [5/19, −2/19]].
System: AX = B where A = [[2,3],[5,−2]], X = [[x],[y]], B = [[1],[11]].
X = A⁻¹B = [[2/19, 3/19], [5/19, −2/19]] × [[1],[11]]
Row 1: (2/19)(1) + (3/19)(11) = 2/19 + 33/19 = 35/19.
Row 2: (5/19)(1) + (−2/19)(11) = 5/19 − 22/19 = −17/19.
x = 35/19, y = −17/19.
Verification: 2(35/19) + 3(−17/19) = 70/19 − 51/19 = 19/19 = 1 ✓.
5(35/19) − 2(−17/19) = 175/19 + 34/19 = 209/19 = 11 ✓.
Using properties of determinants, show that |1, 1, 1; a, b, c; a³, b³, c³| = (a−b)(b−c)(c−a)(a+b+c).
Apply C₁ → C₁ − C₂ and C₂ → C₂ − C₃:
|0, 0, 1; a−b, b−c, c; a³−b³, b³−c³, c³|
Factor: a³−b³ = (a−b)(a²+ab+b²), b³−c³ = (b−c)(b²+bc+c²).
Take (a−b) from C₁ and (b−c) from C₂:
= (a−b)(b−c) × |0, 0, 1; 1, 1, c; a²+ab+b², b²+bc+c², c³|
Expand along R1: = (a−b)(b−c) × 1 × |1, 1; a²+ab+b², b²+bc+c²|
= (a−b)(b−c) × [(b²+bc+c²) − (a²+ab+b²)]
= (a−b)(b−c) × [bc+c²−a²−ab]
= (a−b)(b−c) × [c²−a² + b(c−a)]
= (a−b)(b−c) × [(c−a)(c+a) + b(c−a)]
= (a−b)(b−c)(c−a) × [c+a+b]
= (a−b)(b−c)(c−a)(a+b+c). Hence proved. ∎
If A = [[1, 3, 3], [1, 4, 3], [1, 3, 4]], verify that A(adj A) = |A| × I. Also find A⁻¹.
|A| = 1(16−9) − 3(4−3) + 3(3−4) = 7 − 3 − 3 = 1.
Cofactors: A₁₁=7, A₁₂=−1, A₁₃=−1, A₂₁=−3, A₂₂=1, A₂₃=0, A₃₁=−3, A₃₂=0, A₃₃=1.
adj(A) = [[7,−3,−3],[−1,1,0],[−1,0,1]].
A × adj(A) = [[1,3,3],[1,4,3],[1,3,4]] × [[7,−3,−3],[−1,1,0],[−1,0,1]]
= [[7−3−3, −3+3+0, −3+0+3],[7−4−3, −3+4+0, −3+0+3],[7−3−4, −3+3+0, −3+0+4]]
= [[1,0,0],[0,1,0],[0,0,1]] = I = |A| × I. ✓
Since |A| = 1, A⁻¹ = adj(A) = [[7,−3,−3],[−1,1,0],[−1,0,1]].
Find the value of k for which the system of equations kx + 3y = k − 3 and 12x + ky = k has no solution.
The system has no solution when the determinant of the coefficient matrix is 0 but the system is inconsistent.
D = |k, 3; 12, k| = k² − 36.
For no solution: D = 0 → k² = 36 → k = ±6.
Check k = 6: 6x + 3y = 3 and 12x + 6y = 6 → 2(6x+3y) = 6, consistent (infinite solutions). So k = 6 gives infinite solutions, not no solution.
Check k = −6: −6x + 3y = −9 → 2x − y = 3 and 12x − 6y = −6 → 2x − y = −1. Since 3 ≠ −1, the system is inconsistent.
k = −6.
Evaluate |x+y, y+z, z+x; z, x, y; 1, 1, 1| using properties of determinants.
Apply R₁ → R₁ − R₂:
|x+y−z, y+z−x, z+x−y; z, x, y; 1, 1, 1|
Apply R₁ → R₁ + R₃ (to simplify):
Actually, let's use R₁ → R₁ + R₂:
|x+y+z, x+y+z, x+y+z; z, x, y; 1, 1, 1|
Take (x+y+z) common from R₁:
= (x+y+z)|1, 1, 1; z, x, y; 1, 1, 1|
Since R₁ = R₃, the determinant = 0.
Using determinants, find the area of the triangle with vertices (−1, −8), (−2, −3), (3, 2).
Δ = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
= ½|(−1)(−3−2) + (−2)(2−(−8)) + 3(−8−(−3))|
= ½|(−1)(−5) + (−2)(10) + 3(−5)|
= ½|5 − 20 − 15|
= ½|−30| = ½ × 30 = 15 sq. units.
Solve using Cramer's rule: 3x − 2y = 5, x + 3y = −4.
D = |3, −2; 1, 3| = 9 − (−2) = 9 + 2 = 11.
D₁ = |5, −2; −4, 3| = 15 − 8 = 7.
D₂ = |3, 5; 1, −4| = −12 − 5 = −17.
x = D₁/D = 7/11, y = D₂/D = −17/11.
x = 7/11, y = −17/11.
Verification: 3(7/11) − 2(−17/11) = 21/11 + 34/11 = 55/11 = 5 ✓.
If A = [[2, −1, 1], [−1, 2, −1], [1, −1, 2]], find A⁻¹ and hence solve: 2x − y + z = 3, −x + 2y − z = −4, x − y + 2z = 6.
|A| = 2(4−1)+1(−2+1)+1(1−2) = 2(3)+1(−1)+1(−1) = 6−1−1 = 4 ≠ 0.
Cofactors: A₁₁=3, A₁₂=1, A₁₃=−1, A₂₁=1, A₂₂=3, A₂₃=1, A₃₁=−1, A₃₂=1, A₃₃=3.
adj(A) = [[3,1,−1],[1,3,1],[−1,1,3]].
A⁻¹ = (1/4)[[3,1,−1],[1,3,1],[−1,1,3]].
X = A⁻¹B = (1/4)[[3,1,−1],[1,3,1],[−1,1,3]] × [[3],[−4],[6]]
= (1/4)[[9−4−6],[3−12+6],[−3−4+18]]
= (1/4)[[−1],[−3],[11]].
x = −1/4, y = −3/4, z = 11/4.
Verification: 2(−1/4)−(−3/4)+(11/4) = −2/4+3/4+11/4 = 12/4 = 3 ✓.
Prove using properties: |a², bc, ac+c²; a²+ab, b², ac; ab, b²+bc, c²| = 4a²b²c².
LHS = |a², bc, ac+c²; a²+ab, b², ac; ab, b²+bc, c²|.
Step 1: Take a common from C₁, b from C₂, c from C₃:
= abc × |a, c, a+c; a+b, b, a; b, b+c, c|.
Step 2: Apply R₁ → R₁ − R₂ + R₃:
R₁ becomes: [a−(a+b)+b, c−b+(b+c), (a+c)−a+c] = [0, 2c, 2c].
= abc × |0, 2c, 2c; a+b, b, a; b, b+c, c|.
Step 3: Take 2c common from R₁:
= 2abc² × |0, 1, 1; a+b, b, a; b, b+c, c|.
Step 4: Apply C₁ → C₁ − C₂ + C₃:
C₁ becomes: [0−1+1, (a+b)−b+a, b−(b+c)+c] = [0, 2a, 0].
= 2abc² × |0, 1, 1; 2a, b, a; 0, b+c, c|.
Step 5: Expand along C₁:
= 2abc² × [2a × (−1)^(2+1) × |1, 1; b+c, c|]
= 2abc² × [2a × (−1) × (c − b − c)]
= 2abc² × [2a × (−1) × (−b)]
= 2abc² × 2ab = 4a²b²c² = RHS. Hence proved. ∎
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): If A is a 3×3 matrix, then |3A| = 27|A|.
Reason (R): For an n×n matrix A, |kA| = kⁿ|A|.
Answer: (A)
Both are TRUE and the reason correctly explains the assertion. For a 3×3 matrix (n=3), |3A| = 3³|A| = 27|A|. Each of the 3 rows gets multiplied by 3, contributing a factor of 3 each time.
AR2
Assertion (A): If any two rows (or columns) of a determinant are identical, then the value of the determinant is zero.
Reason (R): Interchanging two rows (or columns) of a determinant changes its sign.
Answer: (A)
Both are TRUE and the reason explains the assertion. If two rows are identical and we swap them, the determinant should change sign: |A| = −|A|. This means 2|A| = 0, so |A| = 0.
AR3
Assertion (A): A square matrix A is invertible if and only if |A| ≠ 0.
Reason (R): A⁻¹ = (1/|A|) × adj(A).
Answer: (A)
Both are TRUE and the reason explains the assertion. The formula A⁻¹ = (1/|A|) × adj(A) requires |A| ≠ 0 (division by zero is undefined). If |A| = 0, the matrix is singular and has no inverse.
AR4
Assertion (A): |AB| = |A| + |B| for square matrices A and B of the same order.
Reason (R): The determinant of a product equals the product of determinants: |AB| = |A| × |B|.
Answer: (D)
Assertion is FALSE: The correct relation is |AB| = |A| × |B| (product, not sum). Reason is TRUE: It correctly states the multiplicative property of determinants.
AR5
Assertion (A): The area of a triangle with collinear vertices is zero.
Reason (R): If three points are collinear, the determinant |x₁, y₁, 1; x₂, y₂, 1; x₃, y₃, 1| = 0.
Answer: (A)
Both are TRUE and the reason explains the assertion. Collinear points lie on the same straight line and cannot form a triangle with positive area. The determinant formula gives area = ½ × |det| = ½ × 0 = 0.
AR6
Assertion (A): If |A| = 0, then the system AX = B has infinitely many solutions.
Reason (R): When |A| = 0, the matrix A is singular.
Answer: (D)
Assertion is FALSE: When |A| = 0, the system may have infinitely many solutions OR no solution (inconsistent). We cannot conclude infinitely many solutions without checking consistency. Reason is TRUE: |A| = 0 does mean A is singular (non-invertible).
Case Study Questions
A surveyor maps three plots of land on a coordinate plane. Plot A has vertices at P(0, 0), Q(4, 0), R(2, 3). Plot B has vertices at S(1, 1), T(5, 1), U(3, 5). The surveyor also needs to check if three boundary markers at (1, 5), (2, 3), and (−2, 11) are collinear.
(i) Find the area of Plot A using the determinant formula. (1 mark)
Area = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
= ½|0(0−3) + 4(3−0) + 2(0−0)|
= ½|0 + 12 + 0| = ½ × 12 = 6 sq. units.
(ii) Find the area of Plot B. (2 marks)
Area = ½|1(1−5) + 5(5−1) + 3(1−1)|
= ½|1(−4) + 5(4) + 3(0)|
= ½|−4 + 20 + 0|
= ½ × 16 = 8 sq. units.
(iii)(a) Check if the three boundary markers at (1, 5), (2, 3), (−2, 11) are collinear. (2 marks)
Area = ½|1(3−11) + 2(11−5) + (−2)(5−3)|
= ½|1(−8) + 2(6) + (−2)(2)|
= ½|−8 + 12 − 4|
= ½|0| = 0.
Since area = 0, the three markers are collinear.
(iii)(b) Find the equation of the line passing through (1, 5) and (2, 3) using determinants. (2 marks)
Using |x, y, 1; 1, 5, 1; 2, 3, 1| = 0:
x(5−3) − y(1−2) + 1(3−10) = 0
2x + y − 7 = 0
2x + y = 7 (or y = −2x + 7).
Verification: (1,5): 2+5=7 ✓. (2,3): 4+3=7 ✓. (−2,11): −4+11=7 ✓.
A factory produces three products X, Y, and Z. Each product requires processing on three machines M₁, M₂, and M₃. The processing time (in hours) per unit is given by the matrix: A = [[1, 2, 1], [2, 1, 3], [1, 1, 2]] where rows represent machines and columns represent products. The total available hours on M₁, M₂, and M₃ are 10, 19, and 13 respectively. The factory wants to find how many units of each product to produce.
(i) Write the system of equations and find the determinant |A|. (1 mark)
System: x + 2y + z = 10, 2x + y + 3z = 19, x + y + 2z = 13.
|A| = 1(2−3) − 2(4−3) + 1(2−1) = 1(−1) − 2(1) + 1(1) = −1 − 2 + 1 = −2.
Since |A| ≠ 0, the system has a unique solution.
(ii) Find A⁻¹ using the adjoint method. (2 marks)
Cofactors: A₁₁ = (2−3) = −1, A₁₂ = −(4−3) = −1, A₁₃ = (2−1) = 1.
A₂₁ = −(4−1) = −3, A₂₂ = (2−1) = 1, A₂₃ = −(1−2) = 1.
A₃₁ = (6−1) = 5, A₃₂ = −(3−2) = −1, A₃₃ = (1−4) = −3.
adj(A) = [[−1,−3,5],[−1,1,−1],[1,1,−3]].
A⁻¹ = (1/−2)[[−1,−3,5],[−1,1,−1],[1,1,−3]] = [[1/2, 3/2, −5/2],[1/2, −1/2, 1/2],[−1/2, −1/2, 3/2]].
(iii)(a) Find the number of units of each product to be produced. (2 marks)
X = A⁻¹B = [[1/2, 3/2, −5/2],[1/2, −1/2, 1/2],[−1/2, −1/2, 3/2]] × [[10],[19],[13]].
Row 1: (1/2)(10) + (3/2)(19) + (−5/2)(13) = 5 + 57/2 − 65/2 = 5 + (57−65)/2 = 5 − 4 = 1.
Row 2: (1/2)(10) + (−1/2)(19) + (1/2)(13) = 5 − 19/2 + 13/2 = 5 + (−19+13)/2 = 5 − 3 = 2.
Row 3: (−1/2)(10) + (−1/2)(19) + (3/2)(13) = −5 − 19/2 + 39/2 = −5 + 20/2 = −5 + 10 = 5.
X = [[1],[2],[5]].
x = 1 unit of X, y = 2 units of Y, z = 5 units of Z.
Verification: 1+2(2)+5 = 1+4+5 = 10 ✓, 2(1)+2+3(5) = 2+2+15 = 19 ✓, 1+2+2(5) = 1+2+10 = 13 ✓.
(iii)(b) If the available hours on M₃ increase to 15, find the new production quantities. (2 marks)
New B' = [[10],[19],[15]]. X = A⁻¹B' = [[1/2, 3/2, −5/2],[1/2, −1/2, 1/2],[−1/2, −1/2, 3/2]] × [[10],[19],[15]].
Row 1: (1/2)(10) + (3/2)(19) + (−5/2)(15) = 5 + 57/2 − 75/2 = 5 + (57−75)/2 = 5 − 9 = −4.
Row 2: (1/2)(10) + (−1/2)(19) + (1/2)(15) = 5 − 19/2 + 15/2 = 5 + (−19+15)/2 = 5 − 2 = 3.
Row 3: (−1/2)(10) + (−1/2)(19) + (3/2)(15) = −5 − 19/2 + 45/2 = −5 + 26/2 = −5 + 13 = 8.
X = [[−4],[3],[8]].
Since x = −4 is negative, this means the system gives a mathematically valid but physically infeasible solution. The factory cannot meet these constraints with non-negative production quantities. The production manager should reconsider the available hours or explore adding capacity to machine M₃.
MCQ Practice
MCQ1. If A is a 3×3 matrix and |A| = 5, then |adj(A)| is:
(A) 5
(B) 25
(C) 125
(D) 1/5
Answer: (B) - For an n×n matrix, |adj(A)| = |A|^(n−1). Here n=3, so |adj(A)| = 5² = **25**.
MCQ2. If |2x, 5; 8, x| = |6, 5; 8, 3|, then the value of x is:
(A) 3
(B) ±3
(C) ±6
(D) 6
Answer: (B) - LHS = 2x² − 40. RHS = 18 − 40 = −22. So 2x² − 40 = −22 → 2x² = 18 → x² = 9 → **x = ±3**.
MCQ3. If A is a non-singular matrix of order 3, then |A⁻¹| is:
(A) |A|
(B) 1/|A|
(C) |A|²
(D) |A|³
Answer: (B) - Since AA⁻¹ = I, |AA⁻¹| = |I| = 1. So |A| × |A⁻¹| = 1, giving **|A⁻¹| = 1/|A|**.
MCQ4. The value of the determinant |a−b, b+c, a; b−c, c+a, b; c−a, a+b, c| is:
(A) a³+b³+c³
(B) 3abc
(C) a³+b³+c³−3abc
(D) 0
Answer: (C) - Apply C₁ → C₁ + C₃: |a−b+a, b+c, a; b−c+b, c+a, b; c−a+c, a+b, c| = |2a−b, b+c, a; 2b−c, c+a, b; 2c−a, a+b, c|. After further operations (C₂→C₂−C₃), systematic simplification yields **a³+b³+c³−3abc**.
MCQ5. If A = [[1, 2], [4, 2]], then |2A| is:
(A) −12
(B) −24
(C) 24
(D) −6
Answer: (B) - |A| = 2 − 8 = −6. For a 2×2 matrix, |2A| = 2²|A| = 4(−6) = **−24**.
MCQ6. Three points (1, −1), (2, 1), and (4, 5) are:
(A) vertices of a right triangle
(B) vertices of an equilateral triangle
(C) collinear
(D) vertices of an isosceles triangle
Answer: (C) - Area = ½|1(1−5)+2(5+1)+4(−1−1)| = ½|−4+12−8| = ½|0| = 0. Since area = 0, the points are **collinear**.
MCQ7. If A is a square matrix of order 3 such that |adj(A)| = 64, then |A| is:
(A) ±8
(B) ±4
(C) 64
(D) ±64
Answer: (A) - |adj(A)| = |A|^(n−1) = |A|². So |A|² = 64, giving **|A| = ±8**.
MCQ8. The minor M₂₁ of the determinant |1, 2, 3; 4, 5, 6; 7, 8, 9| is:
(A) −6
(B) 6
(C) −3
(D) −9
Answer: (A) - M₂₁ is obtained by deleting row 2 and column 1: |2, 3; 8, 9| = 18 − 24 = **−6**.
MCQ9. If A is a singular matrix, then |A| is:
(A) 1
(B) −1
(C) 0
(D) non-zero
Answer: (C) - By definition, a singular matrix is one whose determinant is **0**. It has no inverse.
MCQ10. The system of equations x + y = 2, 2x + 2y = 5 has:
(A) a unique solution
(B) no solution
(C) infinitely many solutions
(D) exactly two solutions
Answer: (B) - D = |1,1; 2,2| = 2−2 = 0. The equations are x+y=2 and x+y=5/2. Since 2 ≠ 5/2, the system is inconsistent. **No solution** exists.
MCQ11. If A is an invertible matrix and A² = I, then A⁻¹ is:
(A) 2A
(B) A
(C) A²
(D) A/2
Answer: (B) - Given A² = I. Multiply both sides by A⁻¹: A²A⁻¹ = IA⁻¹ → A(AA⁻¹) = A⁻¹ → A(I) = A⁻¹ → **A⁻¹ = A**.
MCQ12. If |a, b, c; d, e, f; g, h, i| = Δ, then |d, e, f; a, b, c; g, h, i| is:
(A) Δ
(B) −Δ
(C) 0
(D) 2Δ
Answer: (B) - The second determinant is obtained by interchanging R₁ and R₂ of the first. Swapping two rows changes the sign of the determinant. So the value is **−Δ**.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Evaluate |cos 15°, sin 15°; sin 75°, cos 75°|.
= cos 15° × cos 75° − sin 15° × sin 75° = cos(15° + 75°) = cos 90° = **0**.
If A is a 3×3 matrix with |A| = 9, find |2A|.
|2A| = 2³ × |A| = 8 × 9 = **72**.
If A is a non-singular 3×3 matrix, find |A × adj(A)|.
A × adj(A) = |A| × I. So |A × adj(A)| = ||A| × I| = |A|³ × |I| = |A|³. If |A| = k, then answer is **k³**.
What is the value of the determinant |0, 1, 2; 0, 3, 4; 0, 5, 6|?
Since all elements of C₁ are zero, the determinant = **0**.
If |x, 2; 3, x| = |x, 2; x, 3|, find x.
LHS = x² − 6. RHS = 3x − 2x = x. So x² − 6 = x → x² − x − 6 = 0 → (x−3)(x+2) = 0. **x = 3 or x = −2.**
2-3 Mark Questions
Using properties, evaluate |1, a, a²; 1, b, b²; 1, c, c²|.
Apply R₂→R₂−R₁, R₃→R₃−R₁: |1,a,a²; 0,b−a,b²−a²; 0,c−a,c²−a²|. Factor: = (b−a)(c−a)|1, a, a²; 0, 1, b+a; 0, 1, c+a|. = (b−a)(c−a)(c+a−b−a) = (a−b)(b−c)(c−a).
Find the equation of the line through (3, 2) and (−1, 4) using determinants.
|x,y,1; 3,2,1; −1,4,1| = 0. x(2−4) − y(3+1) + 1(12+2) = 0 → −2x − 4y + 14 = 0 → x + 2y = 7.
Show that |1, 1, 1; a, b, c; a², b², c²| = (a−b)(b−c)(c−a) using properties.
Apply C₁→C₁−C₂, C₂→C₂−C₃: |0,0,1; a−b, b−c, c; a²−b², b²−c², c²|. Factor to get (a−b)(b−c)|0,0,1; 1,1,c; a+b, b+c, c²|. Expand along R1: (a−b)(b−c)[(b+c)−(a+b)] = (a−b)(b−c)(c−a).
Find the area of a triangle with vertices (1, −1), (−4, 6), and (−3, −5).
Δ = ½|1(6+5) + (−4)(−5+1) + (−3)(−1−6)| = ½|11 + 16 + 21| = ½ × 48 = 24 sq. units.
3-4 Mark Questions
Find the inverse of A = [[1, 2, 3], [0, 2, 4], [0, 0, 5]] and verify AA⁻¹ = I.
|A| = 1(10−0) − 2(0) + 3(0) = 10 ≠ 0. Cofactors: A₁₁=10, A₁₂=0, A₁₃=0, A₂₁=−10, A₂₂=5, A₂₃=0, A₃₁=2, A₃₂=−4, A₃₃=2. adj(A) = [[10,−10,2],[0,5,−4],[0,0,2]]. A⁻¹ = (1/10)[[10,−10,2],[0,5,−4],[0,0,2]]. Verify: AA⁻¹ = I ✓.
Solve using matrix method: x − y + 2z = 7, 3x + 4y − 5z = −5, 2x − y + 3z = 12.
A = [[1,−1,2],[3,4,−5],[2,−1,3]]. |A| = 1(12−5)+1(9+10)+2(−3−8) = 7+19−22 = 4. Find adj(A), then X = A⁻¹B = (1/4) × adj(A) × [[7],[−5],[12]]. Solution: x = 2, y = 1, z = 3.
Using Cramer's rule, solve: 5x − y + 4z = 5, 2x + 3y + 5z = 2, 5x − 2y + 6z = −1.
D = |5,−1,4; 2,3,5; 5,−2,6| = 5(18+10)+1(12−25)+4(−4−15) = 140−13−76 = 51. D₁ = |5,−1,4; 2,3,5; −1,−2,6| = 5(28)+1(17)+4(−1) = 140+17−4 = 153. x = 153/51 = 3. Similarly find y and z.
Key Concepts
Important Formulas
Study Plan - Determinants
Evaluation of Determinants (2×2 and 3×3)
Read Sections 4.1-4.2. Understand the definition and evaluation of 2×2 and 3×3 determinants by cofactor expansion. Complete Exercise 4.1 (all 8 questions). Practice expanding along different rows and columns. Focus on choosing the optimal row/column with the most zeros.
Properties of Determinants
Read Section 4.3. Study all properties thoroughly: interchanging rows/columns, identical rows, scalar multiplication, sum property, and row/column operations. Complete Exercise 4.2 (16 questions). Focus on using properties to simplify determinants before evaluation.
Area of Triangle, Minors and Cofactors
Read Sections 4.4-4.5. Learn the area formula using determinants and the condition for collinearity. Understand minors and cofactors with the sign convention. Complete Exercise 4.3 (5 Qs) and Exercise 4.4 (5 Qs). Practice finding all 9 cofactors for a 3×3 matrix.
Adjoint, Inverse, and Systems of Equations
Read Sections 4.6-4.7. Master the adjoint and inverse formulas. Learn the matrix method (AX = B → X = A⁻¹B) and Cramer's rule. Complete Exercise 4.5 (18 Qs) and Exercise 4.6 (16 Qs). These are the highest-weightage topics in board exams.
Miscellaneous Exercise and Revision
Complete the Miscellaneous Exercise (19 Qs). Solve PYQs from 2019-2026. Revise all formulas and properties. Practice MCQs, assertion-reason, and case study questions. Time yourself on 5-mark inverse/system-of-equations problems (target: under 10 minutes each).
Common Mistakes to Avoid
- ✗Forgetting the sign pattern in cofactor expansion: the signs alternate as (+, −, +, −, ...) starting from (−1)^(i+j). A wrong sign in one cofactor changes the entire determinant value
- ✗Confusing minors with cofactors: Minor Mᵢⱼ is the determinant of the submatrix; Cofactor Aᵢⱼ = (−1)^(i+j) × Mᵢⱼ includes the sign factor
- ✗Applying |kA| = k|A| instead of |kA| = kⁿ|A| for an n×n matrix. Each row gets multiplied by k, so for a 3×3 matrix, |3A| = 27|A|, not 3|A|
- ✗Forgetting to take the absolute value when computing area of a triangle. The determinant can be negative, but area is always non-negative: Area = ½|det|
- ✗Computing adj(A) incorrectly by not transposing the cofactor matrix. adj(A) = (cofactor matrix)ᵀ, not just the cofactor matrix itself
- ✗Assuming |A| = 0 means the system AX = B has infinitely many solutions. When |A| = 0, the system could have either no solution or infinitely many solutions - further investigation is needed
Tips for Board Exam
- ✓Choose the row or column with the most zeros for cofactor expansion - this minimizes computation. For example, if R2 has two zeros, expand along R2 to calculate only one cofactor
- ✓Use properties of determinants to simplify before expanding: add/subtract rows or columns to create zeros. This is especially useful for 'prove that' type questions where direct expansion is tedious
- ✓For solving systems of equations, check |A| first. If |A| ≠ 0, proceed with matrix method or Cramer's rule. If |A| = 0, state that the system is either inconsistent or has infinitely many solutions
- ✓Memorize the formula A⁻¹ = (1/|A|) × adj(A) and the steps: (1) Find |A|, (2) Find all 9 cofactors, (3) Write cofactor matrix, (4) Transpose to get adj(A), (5) Multiply by 1/|A|
- ✓Always verify your answer by substituting back into the original equations. This takes 30 seconds but catches arithmetic mistakes that cost marks
Practice Determinants with AI Tutor
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Start Determinants - Choose Your ModeAbout Determinants - Class 12 Mathematics
Chapter 4 Determinants introduces the concept of a determinant as a unique scalar value associated with every square matrix. The chapter begins with evaluation of 2×2 determinants using the formula ad−bc and extends to 3×3 determinants using cofactor expansion along any row or column. Students learn six key properties of determinants: interchanging rows/columns changes the sign, identical rows give zero determinant, scalar multiplication of a row multiplies the determinant by that scalar, a determinant with a row expressed as a sum can be split into two determinants, and adding a scalar multiple of one row to another leaves the determinant unchanged. These properties are essential tools for simplifying complex determinants without direct expansion.
The chapter then covers applications including finding the area of a triangle using vertex coordinates and testing collinearity of three points. Minors and cofactors provide the foundation for computing the adjoint matrix, which leads to the formula for the inverse: A⁻¹ = (1/|A|) × adj(A). The most important application is solving systems of linear equations using two methods: the matrix method (X = A⁻¹B) and Cramer's rule. For the CBSE 2027 board exam, students should prioritize the 5-mark question on solving 3×3 systems of equations (appears in every paper), properties-based proofs (frequently asked as 3-mark questions), and evaluation of determinants using properties. This chapter, combined with Chapter 3 Matrices, forms the Algebra unit worth 10 marks.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 4 Determinants?
There are 6 exercises (4.1 to 4.6) plus a Miscellaneous Exercise, totalling 87 questions. Exercise 4.1 has 8 questions, Exercise 4.2 has 16, Exercise 4.3 has 5, Exercise 4.4 has 5, Exercise 4.5 has 18, Exercise 4.6 has 16, and the Miscellaneous Exercise has 19 questions.
How many marks does Determinants carry in the CBSE board exam?
Determinants is part of the Algebra unit (10 marks total, combined with Matrices). Typically, Determinants contributes about 5 marks - one MCQ (1 mark), one short answer on evaluation or properties (2-3 marks), and one long answer on solving systems of equations (5 marks).
What is the difference between a minor and a cofactor?
The minor Mᵢⱼ is the determinant of the submatrix obtained by deleting row i and column j. The cofactor Aᵢⱼ = (−1)^(i+j) × Mᵢⱼ includes a sign factor. The signs alternate in a checkerboard pattern: +, −, +, − across rows and down columns.
What is Cramer's rule and when do we use it?
Cramer's rule is a method to solve a system of n linear equations in n unknowns using determinants. For a system AX = B, the solution is x = D₁/D, y = D₂/D, z = D₃/D, where D = |A| (must be non-zero) and D₁, D₂, D₃ are obtained by replacing the 1st, 2nd, 3rd columns of A with the constants B.
When does a system of linear equations have no solution?
A system AX = B has no solution when |A| = 0 AND the system is inconsistent (the equations contradict each other). When |A| = 0 and the system is consistent, it has infinitely many solutions. When |A| ≠ 0, it always has a unique solution.
How do I find the inverse of a 3×3 matrix?
Step 1: Compute |A|. If |A| = 0, A has no inverse. Step 2: Find all 9 cofactors Aᵢⱼ. Step 3: Write the cofactor matrix. Step 4: Transpose it to get adj(A). Step 5: A⁻¹ = (1/|A|) × adj(A). Always verify by checking AA⁻¹ = I.
What is the relationship between determinants and area of a triangle?
The area of a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) equals ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|. This can also be written as ½|det| where det = |x₁,y₁,1; x₂,y₂,1; x₃,y₃,1|. If area = 0, the points are collinear.
Is NCERT enough for Determinants in the CBSE board exam?
Yes, NCERT is sufficient for the board exam. All board questions are based on NCERT concepts and patterns. Focus on all exercises (especially 4.5 and 4.6 for inverse and systems of equations), the Miscellaneous Exercise, and solved examples. Practice PYQs for the 5-mark system-of-equations questions, as they appear in every board paper.