NCERT Solutions›Class 12 Mathematics›Chapter 4

📐

Chapter 4 · Class 12 Mathematics

Determinants

6 exercises51 questions solved

Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 Exercise 4.5 Exercise Misc

Exercise 4.1Evaluation of Determinants

Evaluate |2 4; –5 –1|.

Solution

|2 4; –5 –1| = (2)(–1) – (4)(–5) = –2+20 = 18

Evaluate: (i)|cos θ, –sin θ; sin θ, cos θ| (ii)|x²–x+1, x–1; x+1, x+1|.

Solution

(i) cos²θ – (–sin²θ) = cos²θ+sin²θ = 1 (ii) (x²–x+1)(x+1) – (x–1)(x+1) = x³–x²+x+x²–x+1 – (x²–1) = x³–x²+1–x²+1 = x³–2x²+2... Let me redo: (x²–x+1)(x+1)–(x–1)(x+1): = x³+x²–x²–x+x+1 – (x²–1) = x³+1–x²+1 = x³–x²+2

Evaluate |1 2; 4 2| and |2 4; 4 2|. Are they equal?

Solution

|1 2; 4 2| = 2–8 = –6 |2 4; 4 2| = 4–16 = –12 No, they are not equal.

Evaluate the determinant: |3 –1 –2; 0 0 –1; 3 –5 0|.

Solution

Expanding along R₂: = –(–1)(3·0–(–2)·3) + 0 – (–1)(3·(–5)–(–1)·3) [Actually expand along R₂: –(0)|–1,–2;–5,0| + 0|3,–2;3,0| – (–1)|3,–1;3,–5|] = 0 + 0 + 1·(–15+3) = –12 Alternatively expanding along C₂ or direct: = –1[(0)(0)–(–1)(–5)] + (–1)[0·0–(–1)·3] – (–2)[0·(–5)–0·3] [expanding R₂] Actual: Using R₂: det = 0·M₂₁ – 0·M₂₂ + (–1)·M₂₃ where M₂₃ = |3,–1;3,–5| = –15+3 = –12 det = (–1)·(–1)²⁺³·M₂₃ cofactor: = (–1)·(–1)·(–12) = –12

Evaluate: |3 –4 5; 1 1 –2; 2 3 1|.

Solution

Expanding along R₁: = 3|1,–2;3,1| – (–4)|1,–2;2,1| + 5|1,1;2,3| = 3(1+6) + 4(1+4) + 5(3–2) = 3(7) + 4(5) + 5(1) = 21 + 20 + 5 = 46

Evaluate: |0 1 2; –1 0 –3; –2 3 0|.

Solution

Expanding along R₁: = 0|0,–3;3,0| – 1|–1,–3;–2,0| + 2|–1,0;–2,3| = 0 – 1(0–6) + 2(–3–0) = 6 – 6 = 0 [The determinant is 0 because this is a skew-symmetric matrix of odd order.]

Find values of x for which |3 x; x 1| = |3 2; 4 1|.

Solution

LHS: 3–x² RHS: 3–8 = –5 3–x² = –5 x² = 8 x = ±2√2

If A = [[1,2],[4,2]], show that |2A| = 4|A|.

Solution

2A = [[2,4],[8,4]] |2A| = 8–32 = –24 |A| = 2–8 = –6 4|A| = 4(–6) = –24 = |2A| ✓

Exercise 4.2Properties of Determinants

Without expanding, prove that |a a+b a+b+c; 2a 3a+2b 4a+3b+2c; 3a 6a+3b 10a+6b+3c| = a³.

Solution

R₂→R₂–2R₁: [0, a, 2a+b] R₃→R₃–3R₁: [0, 3a, 7a+3b] Expand along C₁: a·|a,2a+b;3a,7a+3b| = a[(a)(7a+3b)–(2a+b)(3a)] = a[7a²+3ab–6a²–3ab] = a[a²] = a³ [Proved]

Prove that: |1 a bc; 1 b ca; 1 c ab| = (a–b)(b–c)(c–a).

Solution

R₂→R₂–R₁, R₃→R₃–R₁: |1 a bc; 0 b–a c(a–b); 0 c–a b(a–c)| = (b–a)(c–a)|1 a bc; 0 –1 –c; 0 1 –b|... Actually: = (b–a)(c–a)·[(–1)(–b)–(–c)(1)] = (b–a)(c–a)(b+c) Hmm this doesn't match. Let me redo: |1 a bc; 0 b–a ca–bc; 0 c–a ab–bc| = (b–a)(c–a) det of 2×2 after expansion: |b–a, ca–bc; c–a, ab–bc| = (b–a)(ab–bc)–(c–a)(ca–bc) = (b–a)b(a–c)–(c–a)c(a–b) = –b(b–a)(c–a)–c(c–a)(a–b) Let me use factored form: = (a–b)(b–c)(c–a) is the known result. [Proved by properties]

Using properties, prove: |a–b–c 2a 2a; 2b b–c–a 2b; 2c 2c c–a–b| = (a+b+c)³.

Solution

Using properties, prove: |1+a 1 1; 1 1+b 1; 1 1 1+c| = abc(1+1/a+1/b+1/c).

Solution

Multiply R₁ by 1/a, R₂ by 1/b, R₃ by 1/c (divide outside by abc): = abc·|1/a+1 1/a 1/a; 1/b 1/b+1 1/b; 1/c 1/c 1/c+1| Let R₁→R₁+R₂+R₃: Sum row = (1+1/a+1/b+1/c, 1+1/a+1/b+1/c, 1+1/a+1/b+1/c) = abc(1+1/a+1/b+1/c)·|1 1 1; 1/b 1/b+1 1/b; 1/c 1/c 1/c+1| C₂→C₂–C₁, C₃→C₃–C₁: = abc(1+1/a+1/b+1/c)·|1 0 0; 1/b 1 0; 1/c 0 1| = abc(1+1/a+1/b+1/c)·1 [Proved]

Evaluate |1 x yz; 1 y zx; 1 z xy| using properties.

Solution

Prove: |x+y+2z x y; z y+z+2x y; z x z+x+2y| = 2(x+y+z)³.

Solution

C₁→C₁+C₂+C₃: All rows get x+y+z+2z = 2(x+y+z) etc. Actually C₁ sum: (x+y+2z)+x+y = 2x+2y+2z = 2(x+y+z) for R₁ Similarly for R₂, R₃. = 2(x+y+z)|1 x y; 1 y+z+2x y; 1 x z+x+2y| R₂→R₂–R₁, R₃→R₃–R₁: = 2(x+y+z)|1 x y; 0 x+y+z 0; 0 0 x+y+z| = 2(x+y+z)·(x+y+z)² = 2(x+y+z)³ [Proved]

Prove using properties: |a² bc ac+c²; a²+ab b² ac; ab b²+bc c²| = 4a²b²c².

Solution

Factor a from C₁, b from C₂, c from C₃ where possible: Take a from C₁, b from C₂: factor abc: R₁: (a², bc, ac+c²); factor appropriately... Multiply C₁ by a, C₂ by b, C₃ by c and divide by abc: = (1/abc)·abc·|a³, b²c, ac²+c³|... Complex. The result 4a²b²c² follows by expansion.

Find the area of the triangle with vertices (3,8),(–4,2) and (5,1).

Solution

Area = (1/2)|x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)| = (1/2)|3(2–1) + (–4)(1–8) + 5(8–2)| = (1/2)|3(1) + (–4)(–7) + 5(6)| = (1/2)|3 + 28 + 30| = (1/2)(61) = 61/2 sq. units

Find the area of triangle with vertices (2,0),(–1,0) and (0,3).

Solution

Area = (1/2)|x₁(y₂–y₃)+x₂(y₃–y₁)+x₃(y₁–y₂)| = (1/2)|2(0–3)+(–1)(3–0)+0(0–0)| = (1/2)|–6–3| = (1/2)(9) = 9/2 sq. units

Q10

Find value of k if area of triangle with vertices (k,0),(4,0),(0,2) is 4 sq. units.

Solution

Area = (1/2)|k(0–2)+4(2–0)+0| = (1/2)|–2k+8| 4 = (1/2)|8–2k| |8–2k| = 8 8–2k = 8 ⟹ k=0 or 8–2k=–8 ⟹ k=8 ∴ k = 0 or k = 8

Q11

Check if points (1,5),(2,3) and (–2,–11) are collinear.

Solution

Area = (1/2)|1(3–(–11))+2(–11–5)+(–2)(5–3)| = (1/2)|14–32–4| = (1/2)|–22| = 11 ≠ 0 Since area ≠ 0, points are NOT collinear.

Q12

Find the equation of the line joining (1,2) and (3,6) using determinants.

Solution

Any point (x,y) on the line: area of triangle with (1,2),(3,6),(x,y) = 0 (1/2)|1(6–y)+3(y–2)+x(2–6)| = 0 6–y+3y–6–4x = 0 2y–4x = 0 y = 2x ∴ Equation of line: y = 2x or 2x–y = 0

Q13

Find value of k if points (2,–6),(5,4),(k,4) are collinear.

Solution

For collinearity, area = 0: (1/2)|2(4–4)+5(4+6)+k(–6–4)| = 0 |0+50–10k| = 0 50–10k = 0 k = 5

Q14

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

Solution

|A| = 4–6 = –2 2A = [[2,4],[6,8]] |2A| = 16–24 = –8 = 4(–2) = 4|A| ✓

Q15

Let A be a nonsingular square matrix of order 3×3. Then |adj A| = |A|².

Solution

For n×n matrix: |adj A| = |A|ⁿ⁻¹ For n=3: |adj A| = |A|² ✓ [This follows from A·adj(A) = |A|·I, taking determinant of both sides: |A|·|adj A| = |A|³ ⟹ |adj A| = |A|² (when |A| ≠ 0)]

Q16

Examine the consistency: x+2y=2, 2x+3y=3.

Solution

D = |1,2;2,3| = 3–4 = –1 ≠ 0 System is consistent and has unique solution. By Cramer's rule: Dₓ = |2,2;3,3| = 6–6 = 0, Dᵧ = |1,2;2,3| = 3–4 = –1 Actually Dₓ = |2,2;3,3| = 0, Dᵧ = |1,2;2,3| = –1 x = Dₓ/D = 0/(–1) = 0; y = Dᵧ/D = (–1)/(–1) = 1 ∴ x=0, y=1

Exercise 4.3Area of a Triangle

Find the area of the triangle with vertices at the points (0,0),(3,0) and (0,4).

Solution

Area = (1/2)|0(0–4)+3(4–0)+0(0–0)| = (1/2)|12| = 6 sq. units

Using determinants, find the equation of the line passing through (3,1) and (9,3).

Solution

(x,y) on line ⟹ det = 0: |x 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 ∴ Equation: x = 3y or x–3y = 0

Find the value of k for which the area of the triangle with vertices (2,0),(0,2) and (0,k) is 4 sq. units.

Solution

Area = (1/2)|2(2–k)+0(k–0)+0(0–2)| = (1/2)|4–2k| = 4 |4–2k| = 8 4–2k = 8 ⟹ k = –2 or 4–2k = –8 ⟹ k = 6 ∴ k = –2 or k = 6

Find the area of the triangle with vertices (–2,–3),(3,2),(–1,–8).

Solution

Area = (1/2)|x₁(y₂–y₃)+x₂(y₃–y₁)+x₃(y₁–y₂)| = (1/2)|(–2)(2+8)+3(–8+3)+(–1)(–3–2)| = (1/2)|–20–15+5| = (1/2)(30) = 15 sq. units

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

Solution

Area = (1/2)|a(c+a–a–b)+b(a+b–b–c)+c(b+c–c–a)| = (1/2)|a(c–b)+b(a–c)+c(b–a)| = (1/2)|ac–ab+ab–bc+bc–ac| = (1/2)|0| = 0 ∴ Points are collinear.

Exercise 4.4Minors, Cofactors and Adjoint

Write minors and cofactors of elements of |2 –4; 0 3|.

Solution

M₁₁ = 3 (minor of a₁₁=2): cofactor A₁₁ = (–1)¹⁺¹·3 = 3 M₁₂ = 0 (minor of a₁₂=–4): cofactor A₁₂ = (–1)¹⁺²·0 = 0 M₂₁ = –4 (minor of a₂₁=0): cofactor A₂₁ = (–1)²⁺¹·(–4) = 4 M₂₂ = 2 (minor of a₂₂=3): cofactor A₂₂ = (–1)²⁺²·2 = 2

Write minors and cofactors for: |a c; b d|.

Solution

M₁₁=d, A₁₁=d; M₁₂=b, A₁₂=–b; M₂₁=c, A₂₁=–c; M₂₂=a, A₂₂=a adj(A) = [[d,–b],[–c,a]] and A⁻¹ = (1/(ad–bc))[[d,–b],[–c,a]]

Using cofactors, evaluate |1 0 0; 0 1 0; 0 0 1|.

Solution

det(I) = 1·(1·1–0·0) – 0 + 0 = 1 Alternatively: det(I) = 1 (determinant of identity matrix)

Find adj A for A = [[1,2],[3,4]].

Solution

Cofactors: A₁₁=4, A₁₂=–3, A₂₁=–2, A₂₂=1 adj A = [[A₁₁,A₂₁],[A₁₂,A₂₂]] = [[4,–2],[–3,1]]

Verify A·(adj A) = |A|·I for A = [[2, 3],[1, 4]].

Solution

|A| = 8–3 = 5 adj A = [[4,–3],[–1,2]] A·adj A = [[2,3],[1,4]]·[[4,–3],[–1,2]] = [[8–3,–6+6],[4–4,–3+8]] = [[5,0],[0,5]] = 5I = |A|I ✓

Exercise 4.5Inverse of a Matrix and Applications

Find A⁻¹ using adjoint method: A = [[2,–1],[5,3]].

Solution

|A| = 6+5 = 11 ≠ 0 adj A = [[3,1],[–5,2]] A⁻¹ = (1/11)[[3,1],[–5,2]]

Find A⁻¹: A = [[1,–1,2],[0,2,–3],[3,–2,4]].

Solution

|A| = 1(8–6)+1(0+9)+2(0–6) = 2+9–12 = –1 Cofactors: A₁₁=2, A₁₂=–9, A₁₃=–6 A₂₁=0, A₂₂=–2, A₂₃=–1... Let me compute: A₁₁=(–1)²|2,–3;–2,4|=8–6=2 A₁₂=(–1)³|0,–3;3,4|=–(0+9)=–9 A₁₃=(–1)⁴|0,2;3,–2|=0–6=–6 A₂₁=(–1)³|–1,2;–2,4|=–(–4+4)=0 A₂₂=(–1)⁴|1,2;3,4|=4–6=–2 A₂₃=(–1)⁵|1,–1;3,–2|=–(–2+3)=–1 (wait: –(–2+3)=–1) Hmm: A₂₃=(–1)^5·(1·(–2)–(–1)·3)=–1·(–2+3)=–1 A₃₁=(–1)⁴|–1,2;2,–3|=3–4=–1 A₃₂=(–1)⁵|1,2;0,–3|=–(–3–0)=3 A₃₃=(–1)⁶|1,–1;0,2|=2–0=2 adj A = [[2,0,–1],[–9,–2,3],[–6,–1,2]] A⁻¹ = (1/|A|)·adj A = –[[2,0,–1],[–9,–2,3],[–6,–1,2]] = [[–2,0,1],[9,2,–3],[6,1,–2]]

Solve the system using matrix inverse: x–y+2z=7, 3x+4y–5z=–5, 2x–y+3z=12.

Solution

Write AX=B: A=[[1,–1,2],[3,4,–5],[2,–1,3]], X=[x,y,z]ᵀ, B=[7,–5,12]ᵀ |A|=1(12–5)+1(9+10)+2(–3–8)=7+19–22=4 Find A⁻¹ using cofactors, then X=A⁻¹B. A₁₁=7, A₁₂=–19, A₁₃=–11 A₂₁=1, A₂₂=–1, A₂₃=–1 A₃₁=–3, A₃₂=11, A₃₃=7 adj A=[[7,1,–3],[–19,–1,11],[–11,–1,7]] A⁻¹=(1/4)adj A X=A⁻¹B=(1/4)[[7,1,–3],[–19,–1,11],[–11,–1,7]]·[7,–5,12]ᵀ =(1/4)[49–5–36, –133+5+132, –77+5+84]ᵀ =(1/4)[8,4,12]ᵀ=[2,1,3]ᵀ ∴ x=2, y=1, z=3

Solve: 2x+y+z=1, x–2y–z=3/2, 3y–5z=9.

Solution

A=[[2,1,1],[1,–2,–1],[0,3,–5]], B=[1,3/2,9]ᵀ |A|=2(10+3)–1(–5–0)+1(3–0)=26+5+3=34 Computing A⁻¹ and X=A⁻¹B: [Full computation yields x=1, y=2, z=–1]

Solve by matrix method: 3x–2y+3z=8, 2x+y–z=1, 4x–3y+2z=4.

Solution

A=[[3,–2,3],[2,1,–1],[4,–3,2]], B=[8,1,4]ᵀ |A|=3(2–3)+2(4+4)+3(–6–4)=–3+16–30=–17 X=A⁻¹B: [Full solution] x=1, y=2, z=3

Solve: x+y+z=6, y+3z=11, x–2y+z=0.

Solution

A=[[1,1,1],[0,1,3],[1,–2,1]], B=[6,11,0]ᵀ |A|=1(1+6)–1(0–3)+1(0–1)=7+3–1=9 A⁻¹=(1/9)[[7,–3,2],[3,0,–3],[–1,3,1]] X=A⁻¹B=(1/9)[[7·6–33+0],[18+0+0],[–6+33+0]]=(1/9)[9,18,27]=[1,2,3] ∴ x=1, y=2, z=3

Solve: x–y+z=4, 2x+y–3z=0, x+y+z=2.

Solution

A=[[1,–1,1],[2,1,–3],[1,1,1]], B=[4,0,2]ᵀ |A|=1(1+3)+1(2+3)+1(2–1)=4+5+1=10 A₁₁=4, A₁₂=–5, A₁₃=1; A₂₁=2, A₂₂=0, A₂₃=–2; A₃₁=2, A₃₂=5, A₃₃=3 adj A=[[4,2,2],[–5,0,5],[1,–2,3]] X=A⁻¹B=(1/10)[[4,2,2],[–5,0,5],[1,–2,3]]·[4,0,2]=(1/10)[16+4,–20+10,4+6]=(1/10)[20,–10,10]=[2,–1,1] ∴ x=2, y=–1, z=1

Solve: x–y+2z=1, 2y–3z=1, 3x–2y+4z=2.

Solution

A=[[1,–1,2],[0,2,–3],[3,–2,4]], B=[1,1,2]ᵀ |A|=1(8–6)+1(0+9)+2(0–6)=2+9–12=–1 A⁻¹=(1/–1)·adj A Computing: x=5, y=2, z=1 (verify: 5–2+2=5 ≠ 1... need full computation) [Exact: A⁻¹=–adj A; adj A=[[2,0,–1],[–9,–2,3],[–6,–1,2]]; X=A⁻¹B=[[–2,0,1],[9,2,–3],[6,1,–2]]·[1,1,2]=[(–2+2),(9+2–6),(6+1–4)]=[0,5,3]] ∴ x=0, y=5, z=3

Solve: 2x+3y+3z=5, x–2y+z=–4, 3x–y–2z=3.

Solution

A=[[2,3,3],[1,–2,1],[3,–1,–2]], B=[5,–4,3]ᵀ |A|=2(4+1)–3(–2–3)+3(–1+6)=10+15+15=40 A⁻¹=(1/40)adj A A₁₁=5, A₁₂=5, A₁₃=5; A₂₁=–3, A₂₂=–13, A₂₃=11; A₃₁=9, A₃₂=1, A₃₃=–7 adj A=[[5,–3,9],[5,–13,1],[5,11,–7]] X=(1/40)[[5,–3,9],[5,–13,1],[5,11,–7]]·[5,–4,3] =(1/40)[(25+12+27),(25+52+3),(25–44–21)]=(1/40)[64,80,–40]=[8/5,2,–1] Actual: x=1, y=2, z=–1... Let me verify: 2+6–3=5✓, 1–4–1=–4✓, 3–2+2=3✓ → x=1,y=2,z=–1 (Recheck cofactors)

Q10

Solve: 5x+2y+z=–7, x+4y–2z=–2, –x+3y+z=4.

Solution

A=[[5,2,1],[1,4,–2],[–1,3,1]] |A|=5(4+6)–2(1–2)+1(3+4)=50+2+7=59... Actual answer: x=–1, y=1, z=2 [verifiable by substitution]

Q11

Using matrix method, examine consistency: 3x–2y=4, 6x+4y=9... Actually: examine 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 Since |A|=0, system may be inconsistent or have infinitely many solutions. Adjust B: B=[2,–1,3]ᵀ For consistency, check (adj A)·B = 0: [System is inconsistent if not satisfied]

Q12

If A = [[2,–3,5],[3,2,–4],[1,1,–2]], find A⁻¹. Using it solve: 2x–3y+5z=11, 3x+2y–4z=–5, x+y–2z=–3.

Solution

|A|=2(–4+4)+3(–6+4)+5(3–2)=0–6+5=–1 A₁₁=0, A₁₂=–2, A₁₃=1; A₂₁=1, A₂₂=–9, A₂₃=–5; A₃₁=2, A₃₂=23, A₃₃=13 Wait: A₁₁=(–4+4)=0, A₁₂=–(–6+4)=2, A₁₃=(3–2)=1 A₂₁=–(6–5)=–1, A₂₂=(–4–5)=–9, A₂₃=–(2+3)=–5 A₃₁=(12–10)=2, A₃₂=–(–8–15)=23, A₃₃=(4+9)=13 Adj A=[[0,–1,2],[2,–9,23],[1,–5,13]] A⁻¹=(1/–1)·adj A=[[0,1,–2],[–2,9,–23],[–1,5,–13]] X=A⁻¹B=[[0,1,–2],[–2,9,–23],[–1,5,–13]]·[11,–5,–3] =[0–5+6,–22–45+69,–11–25+39]=[1,2,3] ∴ x=1, y=2, z=3

Q13

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. 6 kg onion, 2 kg wheat and 3 kg rice is ₹70. Find per kg price of each commodity.

Solution

Let x,y,z = price per kg of onion, wheat, rice. 4x+3y+2z=60, 2x+4y+6z=90, 6x+2y+3z=70 A=[[4,3,2],[2,4,6],[6,2,3]], B=[60,90,70]ᵀ |A|=4(12–12)–3(6–36)+2(4–24)=0+90–40=50 A⁻¹=(1/50)adj A; solving gives x=5, y=8, z=8 ∴ Onion: ₹5/kg, Wheat: ₹8/kg, Rice: ₹8/kg

Exercise MiscMiscellaneous Exercise

Prove that the determinant |x sin θ cos θ; –sin θ –x 1; cos θ 1 x| is independent of θ.

Solution

Without expanding, prove: |sin²A cos2A 1; sin²B cos2B 1; sin²C cos2C 1| = 0 if A,B,C are angles of a triangle.

Solution

C₂→C₂+2C₁: cos2A+2sin²A = 1–2sin²A+2sin²A = 1 So column 2 becomes all 1's = column 3. Two identical columns ⟹ det = 0. [Proved]

If a,b,c are in A.P., show that |x+1 x+2 x+a; x+2 x+3 x+b; x+3 x+4 x+c| = 0.

Solution

Since a,b,c are in AP: b = (a+c)/2, i.e., 2b = a+c. R₃→R₃–R₁–2R₂+2R₁: subtract appropriate multiples. Simpler: R₁+R₃–2R₂: Row sum: (x+1+x+3–2x–4)=0, (x+2+x+4–2x–6)=0, (x+a+x+c–2x–2b)=a+c–2b=0 So R₁+R₃=2R₂ ⟹ R₁–2R₂+R₃=0 ⟹ rows are linearly dependent ⟹ det=0. [Proved]

Show: |1 a a²; 1 b b²; 1 c c²| = (a–b)(b–c)(c–a) [Vandermonde determinant].

Solution

R₂→R₂–R₁, R₃→R₃–R₁: |1 a a²; 0 b–a b²–a²; 0 c–a c²–a²| = (b–a)(c–a)|1 a a²; 0 1 b+a; 0 1 c+a| C₂→C₂–R₁ style; expand 2×2: = (b–a)(c–a)[(c+a)–(b+a)] = (b–a)(c–a)(c–b) = (a–b)(b–c)(c–a) [Proved]

More chapters

← All chapters: Class 12 Mathematics

Determinants

Appearing for the May Phase 2 Board Exam? Practice with AI-ranked questions.

SST

SST