Class 12 Maths · Chapter 4
Short answer:
Ye chapter Class 12 Maths ka Chapter 4 hai — Determinants. NCERT ke exercise 4.1 se 4.6 tak total 18 solved questions yahan cover kiye gaye hain: 2×2 aur 3×3 determinant evaluate karna, properties of determinants use karke calculation simplify karna, area of a triangle nikalna, minors aur cofactors, adjoint aur inverse of a matrix (2×2 aur 3×3 dono), aur applications — matrix inverse method + Cramer's rule se linear equations solve karna, aur system ki consistency check karna.
Determinant ek matrix ko ek single number me convert karta hai jo batata hai ki matrix invertible hai ya nahi. Class 11 me matrices padhi thi — ab unka ek 'scalar fingerprint' nikalna seekhte hain jo linear equations solve karne, area nikalne aur inverse find karne me kaam aata hai. Poore chapter ka core idea simple hai: determinant zero ho gaya matlab kuch break ho gaya hai — na inverse banega, na unique solution milega, na independent equations honge.
Chapter 4 Summary — 5 Minute Revision
1. Determinant kya hota hai
Har square matrix A ka ek unique number associate hota hai jise determinant kehte hain, likha jaata hai |A| ya det(A). Sirf square matrices ka determinant define hota hai — rectangular matrix ka nahi.
2. 2×2 determinant
A = [a b; c d] ke liye:
|A| = ad − bc
3. 3×3 determinant — expansion by cofactors
Kisi bhi row ya column se expand kiya ja sakta hai (usually Row 1 se, aur zero waali row/column choose karo agar available ho — kaam kam hota hai).
|A| = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃
Jahan C₁ⱼ cofactor hai (sign ke saath minor).
4. Properties of determinants (calculation shortcut)
- Rows ↔ columns: det(A) = det(Aᵀ)
- Do rows/columns swap → determinant ka sign flip ho jaata hai
- Do rows/columns identical ya proportional → determinant = 0
- Row/column ko constant se multiply → determinant bhi usi constant se multiply hota hai
- Rᵢ → Rᵢ + kRⱼ (ek row me doosri row ka multiple add karna) → determinant unchanged rehta hai — yehi sabse zyada use hone waali property hai
Exam trick: pehle properties se matrix simplify karo (zeros banao), phir expand karo — direct expansion se kaafi kam calculation lagti hai.
5. Area of a triangle using determinants
Vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) ke liye:
Area = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
Absolute value lena zaroori hai kyunki area kabhi negative nahi ho sakta. Agar three points collinear hain to ye determinant zero aata hai — ye collinearity check karne ka bhi formula hai.
6. Minors and cofactors
| Term | Matlab |
|---|---|
| Minor Mᵢⱼ | i-th row aur j-th column delete karke bacha hua determinant |
| Cofactor Cᵢⱼ | Cᵢⱼ = (−1)^(i+j) × Mᵢⱼ — checkerboard sign pattern (+ − + / − + − / + − +) |
↔ Table ko side me swipe karein
7. Adjoint of a matrix
Adjoint = cofactor matrix ka transpose:
adj(A) = [Cᵢⱼ]ᵀ
2×2 ke liye shortcut: A = [a b; c d] → adj(A) = [d −b; −c a] (diagonal swap, off-diagonal sign flip).
8. Inverse of a matrix
A⁻¹ = adj(A) / |A|
Matrix ka inverse tabhi exist karta hai jab determinant zero na ho — isliye har inverse-related question me sabse pehle |A|≠0 check karna zaroori hai. Agar |A|=0 to matrix ko singular kehte hain aur inverse exist hi nahi karega.
9. Applications — solving system of linear equations
Matrix inverse method: equations ko AX = B form me likho, phir X = A⁻¹B se solve karo.
Cramer's rule: x = Dₓ/D, y = D_y/D, z = D_z/D, jahan D coefficient matrix ka determinant hai aur Dₓ me x ki column B se replace hoti hai.
10. Consistency of a system
| Condition | Result |
|---|---|
| D ≠ 0 | Unique solution — consistent |
| D = 0 aur Dₓ = D_y = D_z = 0 | Infinite solutions ya no solution — further check chahiye |
| D = 0 aur koi ek bhi Dₓ/D_y/D_z ≠ 0 | No solution — inconsistent |
↔ Table ko side me swipe karein
Poore Class 12 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q18)
Q1. Evaluate the determinant |2 4; −5 −1|.
2×2 determinant formula |a b; c d| = ad − bc use karte hain.
|A| = (2)(−1) − (4)(−5) = −2 − (−20) = −2 + 20 = 18
Q2. Evaluate the determinant |cosθ −sinθ; sinθ cosθ|.
Direct expansion:
|A| = cosθ·cosθ − (−sinθ)·sinθ = cos²θ + sin²θ = 1
Q3. Evaluate |1 2 3; 4 5 6; 7 8 9| by expanding along Row 1.
Row 1 se expand karte hain:
|A| = 1(5×9 − 6×8) − 2(4×9 − 6×7) + 3(4×8 − 5×7)
= 1(45−48) − 2(36−42) + 3(32−35)
= 1(−3) − 2(−6) + 3(−3) = −3 + 12 − 9 = 0
Note: yahan rows AP me hain (common difference 3 har row me) — is wajah se determinant zero aata hai.
Q4. Using properties of determinants, prove that |1 a a²; 1 b b²; 1 c c²| = (a−b)(b−c)(c−a).
R₂ → R₂ − R₁ aur R₃ → R₃ − R₁ karo (property: iss operation se determinant change nahi hota):
Δ = |1 a a²; 0 (b−a) (b²−a²); 0 (c−a) (c²−a²)|
b²−a² = (b−a)(b+a) aur c²−a² = (c−a)(c+a) factor karke, Row 2 se (b−a) aur Row 3 se (c−a) common nikalo:
Δ = (b−a)(c−a) |1 a a²; 0 1 (b+a); 0 1 (c+a)|
Ab Column 1 se expand karo (do zero hain wahan):
Δ = (b−a)(c−a) × 1 × [(c+a) − (b+a)] = (b−a)(c−a)(c−b)
= (a−b)(b−c)(c−a) [sign adjust karke standard form]
Q5. Without expanding, show that |2 3 4; 5 6 8; 6 9 12| = 0.
Property check karo: R₃ ko dekho — R₃ = (6, 9, 12) aur 3×R₁ = 3×(2,3,4) = (6,9,12).
R₃ = 3R₁
Chunki ek row doosri row ka scalar multiple hai, property ke according determinant zero hoga bina expand kiye.
|A| = 0
Q6. Find the area of the triangle with vertices (3, 8), (−4, 2) and (5, 1).
Formula: Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
Area = ½ |3(2−1) + (−4)(1−8) + 5(8−2)|
= ½ |3(1) + (−4)(−7) + 5(6)| = ½ |3 + 28 + 30|
= ½ × 61 = 30.5 sq units
Q7. Find the value of k if the area of the triangle with vertices (2, −6), (5, 4) and (k, 4) is 35 sq units.
Area formula apply karo:
35 = ½ |2(4−4) + 5(4−(−6)) + k(−6−4)|
35 = ½ |0 + 5(10) + k(−10)| = ½|50 − 10k|
70 = |50 − 10k|
Dono cases lo (mistake: sirf ek case lena):
50 − 10k = 70 ⟹ k = −2
50 − 10k = −70 ⟹ k = 12
k = −2 ya k = 12
Q8. Write the minors and cofactors of the elements of the determinant |1 −2; 4 3|.
2×2 matrix me minor Mᵢⱼ = wahi element jo diagonally opposite bacha rehta hai (row i, column j delete karke).
M₁₁ = 3, M₁₂ = 4, M₂₁ = −2, M₂₂ = 1
Cofactor Cᵢⱼ = (−1)^(i+j) Mᵢⱼ:
C₁₁ = (+1)(3) = 3, C₁₂ = (−1)(4) = −4
C₂₁ = (−1)(−2) = 2, C₂₂ = (+1)(1) = 1
Q9. Write the minors and cofactors of the elements of the determinant |2 −3 5; 6 0 4; 1 5 −7|.
Har element ka minor nikalte hain (row aur column delete karke bacha 2×2 determinant):
M₁₁ = |0 4; 5 −7| = 0−20 = −20
M₁₂ = |6 4; 1 −7| = −42−4 = −46
M₁₃ = |6 0; 1 5| = 30−0 = 30
M₂₁ = |−3 5; 5 −7| = 21−25 = −4
M₂₂ = |2 5; 1 −7| = −14−5 = −19
M₂₃ = |2 −3; 1 5| = 10+3 = 13
M₃₁ = |−3 5; 0 4| = −12−0 = −12
M₃₂ = |2 5; 6 4| = 8−30 = −22
M₃₃ = |2 −3; 6 0| = 0+18 = 18
Ab checkerboard sign (+ − + / − + − / + − +) laga kar cofactors:
C₁₁ = −20, C₁₂ = 46, C₁₃ = 30
C₂₁ = 4, C₂₂ = −19, C₂₃ = −13
C₃₁ = −12, C₃₂ = 22, C₃₃ = 18
Q10. Find the adjoint of the matrix A = [1 2; 3 4].
2×2 shortcut: adj(A) = [d −b; −c a] — diagonal elements swap, off-diagonal sign flip.
adj(A) = [4 −2; −3 1]
Q11. Find A⁻¹ using adjoint method for A = [2 3; 1 −4].
Pehle |A| check karo — inverse tabhi banega jab |A| ≠ 0:
|A| = (2)(−4) − (3)(1) = −8 − 3 = −11 (≠ 0, inverse exist karega)
Adjoint nikalo:
adj(A) = [−4 −3; −1 2]
A⁻¹ = adj(A)/|A|:
A⁻¹ = (1/−11)[−4 −3; −1 2] = [4/11 3/11; 1/11 −2/11]
Q12. Find adj(A) and A⁻¹ for A = [1 1 1; 1 2 −3; 2 −1 3].
Sabse pehle saare cofactors nikalte hain:
C₁₁ = |2 −3; −1 3| = 6−3 = 3
C₁₂ = −|1 −3; 2 3| = −(3+6) = −9
C₁₃ = |1 2; 2 −1| = −1−4 = −5
C₂₁ = −|1 1; −1 3| = −(3+1) = −4
C₂₂ = |1 1; 2 3| = 3−2 = 1
C₂₃ = −|1 1; 2 −1| = −(−1−2) = 3
C₃₁ = |1 1; 2 −3| = −3−2 = −5
C₃₂ = −|1 1; 1 −3| = −(−3−1) = 4
C₃₃ = |1 1; 1 2| = 2−1 = 1
adj(A) = cofactor matrix ka transpose (rows ↔ columns):
adj(A) = [3 −4 −5; −9 1 4; −5 3 1]
|A| Row 1 se nikalo:
|A| = 1(3) + 1(−9) + 1(−5) = 3 − 9 − 5 = −11
Inverse:
A⁻¹ = (1/−11) [3 −4 −5; −9 1 4; −5 3 1]
Q13. Solve using matrix method: 2x + y = 5, 3x + 2y = 8.
AX = B form: A = [2 1; 3 2], X = [x; y], B = [5; 8]
|A| = (2)(2) − (1)(3) = 4 − 3 = 1 (≠0, unique solution)
adj(A) = [2 −1; −3 2], to A⁻¹ = [2 −1; −3 2] (kyunki |A|=1)
X = A⁻¹B = [2×5 − 1×8; −3×5 + 2×8] = [10−8; −15+16] = [2; 1]
Check: 2(2)+1 = 5 ✓ aur 3(2)+2(1) = 8 ✓ Answer: x = 2, y = 1
Q14. Solve using Cramer's rule: x + y + z = 6, y + 3z = 11, x − 2y + z = 0.
Pehle D (coefficient determinant) nikalo:
D = |1 1 1; 0 1 3; 1 −2 1| = 1(1+6) − 1(0−3) + 1(0−1) = 7+3−1 = 9
Dₓ — x ki column ko B se replace karo:
Dₓ = |6 1 1; 11 1 3; 0 −2 1| = 6(1+6) − 1(11−0) + 1(−22−0) = 42−11−22 = 9
D_y — y ki column replace karo:
D_y = |1 6 1; 0 11 3; 1 0 1| = 1(11−0) − 6(0−3) + 1(0−11) = 11+18−11 = 18
D_z — z ki column replace karo:
D_z = |1 1 6; 0 1 11; 1 −2 0| = 1(0+22) − 1(0−11) + 6(0−1) = 22+11−6 = 27
Cramer's rule apply karo:
x = Dₓ/D = 9/9 = 1, y = D_y/D = 18/9 = 2, z = D_z/D = 27/9 = 3
Q15. Check the consistency of the system: x + 2y = 3, 2x + 4y = 7.
D nikalo pehle:
D = |1 2; 2 4| = (1)(4) − (2)(2) = 4 − 4 = 0
D = 0 hai, isliye Dₓ check karo (agar ye non-zero nikla to system inconsistent hai):
Dₓ = |3 2; 7 4| = 12 − 14 = −2 (≠ 0)
Chunki D = 0 lekin Dₓ ≠ 0 — system inconsistent hai, koi solution nahi hai.
Q16. Check the consistency of the system: 2x + y = 5, 4x + 2y = 10.
D nikalo:
D = |2 1; 4 2| = 4 − 4 = 0
Dₓ aur D_y bhi check karo:
Dₓ = |5 1; 10 2| = 10 − 10 = 0
D_y = |2 5; 4 10| = 20 − 20 = 0
D = Dₓ = D_y = 0 — system consistent hai lekin infinitely many solutions honge (dono equations actually same line hain: 4x+2y=10 is just 2×(2x+y=5)).
Q17. Using properties of determinants, evaluate |1 2; 3 4| by first applying R₂ → R₂ − 3R₁, and verify the value stays same as direct expansion.
Direct expansion se:
|A| = (1)(4) − (2)(3) = 4 − 6 = −2
Ab R₂ → R₂ − 3R₁ apply karo: new R₂ = (3−3×1, 4−3×2) = (0, −2)
Naya determinant = |1 2; 0 −2| = (1)(−2) − (2)(0) = −2
Dono answers same (−2) hain — is se property confirm hoti hai ki Rᵢ → Rᵢ + kRⱼ operation se determinant change nahi hota.
Q18. If A is a 3×3 matrix with |A| = 4, find |adj A| and |A⁻¹|.
Property: for n×n matrix, |adj A| = |A|^(n−1). Yahan n = 3:
|adj A| = |A|^(3−1) = 4² = 16
Property: |A⁻¹| = 1/|A|:
|A⁻¹| = 1/4
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| 2×2 determinant | |a b; c d| = ad − bc |
| 3×3 determinant (cofactor expansion, Row 1) | |A| = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃ |
| Minor Mᵢⱼ | Determinant left after deleting i-th row and j-th column |
| Cofactor Cᵢⱼ | Cᵢⱼ = (−1)^(i+j) × Mᵢⱼ |
| Area of triangle | Area = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| |
| Collinearity of 3 points | Area determinant = 0 |
| Adjoint | adj(A) = [Cᵢⱼ]ᵀ (transpose of cofactor matrix) |
| 2×2 adjoint shortcut | A = [a b; c d] → adj(A) = [d −b; −c a] |
| Inverse | A⁻¹ = adj(A) / |A|, valid only when |A| ≠ 0 |
| Matrix method solution | AX = B ⟹ X = A⁻¹B |
| Cramer's rule | x = Dₓ/D, y = D_y/D, z = D_z/D |
| Consistency (D ≠ 0) | Unique solution — consistent |
| Consistency (D = 0, all Dₓ,D_y,D_z = 0) | Infinite solutions or no solution |
| Consistency (D = 0, any one of Dₓ,D_y,D_z ≠ 0) | No solution — inconsistent |
| |adj A| for n×n matrix | |adj A| = |A|^(n−1) |
| |A⁻¹| | |A⁻¹| = 1/|A| |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Cofactor sign galat lagana. Checkerboard pattern (+ − + / − + − / + − +) yaad na hone se students minor ko hi cofactor samajh kar use kar lete hain — Cᵢⱼ = (−1)^(i+j)Mᵢⱼ hamesha likho, phir substitute karo.
- |A|≠0 check kiye bina inverse dhoondhna. Agar determinant zero nikal aaye aur phir bhi adj(A)/|A| formula laga do, division by zero ki galti hoti hai. Har inverse question me sabse pehle |A| calculate karo.
- Area of triangle me absolute value bhool jaana. Determinant se area nikalte waqt formula ke aage mod (| |) laga hua hai — agar andar ka value negative aaye aur bina mod ke likh do to negative area answer aa jaata hai jo galat hai.
- Adjoint nikalte waqt transpose lena bhool jaana. Cofactor matrix banane ke baad usko transpose karna zaroori hai — bahut students cofactor matrix ko hi seedha adjoint bol dete hain, jo sirf symmetric matrices me coincidentally sahi lagta hai.
- 3×3 determinant expand karte waqt arithmetic slip. Beech ke terms me minus sign miss ho jaana (khaaskar Row 1 ke second term me −a₁₂M₁₂ hota hai, students +a₁₂M₁₂ likh dete hain) — expansion likhte waqt sign explicitly likho, mental math mat karo.
- Cramer's rule me D=0 hone par bhi x=Dₓ/D likh dena. D zero hote hi division undefined ho jaata hai — pehle consistency check karo (Dₓ, D_y, D_z bhi zero hain ya nahi), tabhi decide karo system consistent hai ya inconsistent.
Board-Style Important Questions
- 1 mark: Evaluate the determinant |5 3; −7 −4|.
- 2 marks: If A is a square matrix of order 3 and |A| = 5, find |adj A|.
- 2 marks: Find the area of the triangle whose vertices are (2, 7), (1, 1) and (10, 8).
- 4 marks: Using properties of determinants, prove that |a a+b a+b+c; 2a 3a+2b 4a+3b+2c; 3a 6a+3b 10a+6b+3c| = a³.
- 4 marks: Find the inverse of the matrix A = [2 −3 5; 3 2 −4; 1 1 −2] using the adjoint method.
- 5 marks: Solve the following system of equations by matrix method: x − y + 2z = 7, 3x + 4y − 5z = −5, 2x − y + 3z = 12.
Aksar Poochhe Jaane Wale Sawaal
Determinant aur matrix me kya farak hai?
Matrix numbers ka rectangular arrangement hai, lekin determinant ek single number hai jo sirf square matrix se nikalta hai. Determinant matrix ke kai important properties batata hai — jaise inverse exist karega ya nahi.
Minor aur cofactor me kya difference hai?
Minor Mij wo determinant hai jo i-th row aur j-th column delete karne ke baad bachta hai. Cofactor Cij wahi minor hai lekin sign ke saath — Cij = (−1)^(i+j) Mij. Sirf sign ka fark hai.
Adjoint aur inverse same cheez hain kya?
Nahi. Adjoint sirf cofactor matrix ka transpose hai — koi bhi square matrix ka adjoint bana sakte ho, chahe determinant zero ho ya na ho. Inverse ke liye adjoint ko |A| se divide karna padta hai, aur ye tabhi possible hai jab |A| ≠ 0.
Singular matrix kya hoti hai?
Jis matrix ka determinant zero hota hai use singular matrix kehte hain. Singular matrix ka inverse exist nahi karta, isliye usse related linear equations ka unique solution bhi nahi hota.
Matrix method aur Cramer's rule me konsa better hai?
Dono same result dete hain — bas approach alag hai. Matrix method (X = A⁻¹B) tab convenient hai jab same coefficient matrix A ke saath multiple B values solve karne hon. Cramer's rule direct ratio of determinants deta hai aur chhote systems (2 ya 3 variables) ke liye jaldi solve ho jaata hai.
Area of triangle determinant se negative kyun aa sakta hai?
Determinant expansion me vertices ke order (clockwise ya anticlockwise) ki wajah se sign negative bhi aa sakta hai — isliye formula ke aage absolute value (mod) lagaya jaata hai, kyunki actual area kabhi negative nahi hota.
Class 12 Maths — Saare Chapters
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