1. Chapter at a Glance
Every square matrix has a single number attached to it, its determinant. This number tells us whether the matrix has an inverse, whether a system of linear equations has a unique solution, and even the area of a triangle from its vertices. Determinants first appeared in the work of the Japanese mathematician Seki Takakazu and of Leibniz in the late 17th century, in connection with solving equations. This chapter covers determinants of order 2 and 3, area of a triangle, minors and cofactors, the adjoint, the inverse of a matrix, and solving linear systems by the matrix method.
- Determinant of 1 × 1, 2 × 2 and 3 × 3 matrices; expansion along any row/column.
- Area of a triangle using determinants; collinearity.
- Minors and cofactors.
- Adjoint of a matrix; A(adj A) = |A| I.
- Singular and non-singular matrices; inverse A⁻¹ = adj A / |A|.
- Consistency of a system and solving AX = B by X = A⁻¹B.
2. Determinant of a Matrix
Signs follow the checkerboard pattern + − + / − + − / + − +. Expanding along any row or column gives the same value — choose the one with most zeros.
Evaluate Δ = |3 −1 −2; 0 0 −1; 3 −5 0|.
Solution: Expand along row 2 (two zeros): only the −1 at position (2, 3), sign (−1)²⁺³ = −1. Δ = −(−1) × |3 −1; 3 −5| = 1 × (−15 + 3) = −12.
- |A′| = |A|.
- For an n × n matrix: |kA| = kⁿ|A|.
- |AB| = |A||B|.
- If two rows (or columns) are identical or proportional, the determinant is 0.
- Interchanging two rows changes the sign.
3. Area of a Triangle
For vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):
Take the absolute value. Area = 0 ⇔ the points are collinear.
Find k if the area of the triangle with vertices (k, 0), (4, 0), (0, 2) is 4.
Solution: ½|k(0 − 2) − 0 + 1(8 − 0)| = 4 → |8 − 2k| = 8 → k = 0 or k = 8.
Find the line through (1, 2) and (3, 6) using determinants.
Solution: For (x, y) on the line, |x y 1; 1 2 1; 3 6 1| = 0 → x(2 − 6) − y(1 − 3) + (6 − 6) = 0 → −4x + 2y = 0 → y = 2x.
4. Minors and Cofactors
- Minor Mᵢⱼ: determinant left after deleting row i and column j.
- Cofactor Aᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ.
- |A| = sum of (elements of any row) × (their cofactors).
- Sum of elements of a row × cofactors of a different row = 0.
5. Adjoint and Inverse
adj A = transpose of the cofactor matrix. For 2 × 2: adj [a b; c d] = [d −b; −c a] (swap the diagonal, negate the other two).
A(adj A) = (adj A)A = |A| I.
- A is singular if |A| = 0, non-singular if |A| ≠ 0.
- A is invertible ⇔ A is non-singular, and then A⁻¹ = (1/|A|) adj A.
- |adj A| = |A|ⁿ⁻¹ for an n × n matrix; |A⁻¹| = 1/|A|.
Find the inverse of A = [2 3; 1 4].
Solution: |A| = 8 − 3 = 5. adj A = [4 −3; −1 2]. A⁻¹ = (1/5)[4 −3; −1 2].
Find the inverse of A = [1 3 3; 1 4 3; 1 3 4].
Solution: |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⁻¹ = [7 −3 −3; −1 1 0; −1 0 1].
6. Solving Linear Systems
The system a₁x + b₁y + c₁z = d₁, … can be written as AX = B.
| Case | Conclusion |
|---|---|
| |A| ≠ 0 | Unique solution X = A⁻¹B (consistent) |
| |A| = 0, (adj A)B ≠ O | No solution (inconsistent) |
| |A| = 0, (adj A)B = O | Infinitely many or no solutions — check further |
Solve x + y + z = 6, y + 3z = 11, x − 2y + z = 0.
Solution: A = [1 1 1; 0 1 3; 1 −2 1], |A| = 1(1 + 6) − 1(0 − 3) + 1(0 − 1) = 9. adj A = [7 −3 2; 3 0 −3; −1 3 1]. X = (1/9)[7 −3 2; 3 0 −3; −1 3 1][6; 11; 0] = (1/9)[9; 18; 27] → x = 1, y = 2, z = 3.
7. Solved Problems (Board Pattern)
Verify for A = [2 3; −4 −6].
Solution: |A| = −12 + 12 = 0; adj A = [−6 −3; 4 2]. A(adj A) = [−12 + 12, −6 + 6; 24 − 24, 12 − 12] = O = 0 × I ✓. (A is singular, so it has no inverse.)
Is the system x + 3y = 5, 2x + 6y = 8 consistent?
Solution: |A| = 6 − 6 = 0. adj A = [6 −3; −2 1]; (adj A)B = [30 − 24; −10 + 8] = [6; −2] ≠ O → inconsistent (parallel lines).
If A is 3 × 3 with |A| = 4, find |2A|, |adj A| and |A⁻¹|.
Solution: |2A| = 2³ × 4 = 32; |adj A| = 4² = 16; |A⁻¹| = 1/4.
8. Master Formula Table
| Result | Formula |
|---|---|
| 2 × 2 determinant | ad − bc |
| Cofactor | Aᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ |
| Triangle area | ½|det[x y 1]| |
| Adjoint | (cofactor matrix)′ |
| Key identity | A(adj A) = |A| I |
| Inverse | A⁻¹ = adj A / |A| |
| Scaling | |kA| = kⁿ|A| |
| Adjoint determinant | |adj A| = |A|ⁿ⁻¹ |
| System | X = A⁻¹B if |A| ≠ 0 |
9. Master Summary Table
| Task | Method | Check |
|---|---|---|
| Evaluate |A| | Expand along row/column with zeros | Sign pattern |
| Area / collinearity | ½|det| | Absolute value |
| Inverse | Cofactors → transpose → divide by |A| | A A⁻¹ = I |
| Solve system | X = A⁻¹B | Substitute back |
| Consistency | |A| and (adj A)B | Interpret geometrically |
10. Mind Map
- ad − bc
- Expansion
- ½|det|
- Collinear if 0
- Cofactors, adj
- adj A/|A|
- AX = B
- Consistency
11. Tips and Tricks
- Expand along the row or column with the most zeros.
- For 2 × 2 inverses: swap the diagonal, change signs of the off-diagonal, divide by the determinant.
- After finding A⁻¹, multiply one row by A to check you get a row of I.
- Forgetting to transpose the cofactor matrix to get adj A.
- Sign errors in cofactors (check the checkerboard).
- |2A| = 2|A| for a 3 × 3 matrix — it is 8|A|.
12. Important One-liners
- Only square matrices have determinants.
- A matrix is invertible iff its determinant is non-zero.
- A(adj A) = (adj A)A = |A| I.
- Three points are collinear iff the area determinant is 0.
- |AB| = |A||B| and |A′| = |A|.
13. Exam-pattern Questions with Detailed Answers
|cos θ −sin θ; sin θ cos θ| equals:
- 0
- 1
- cos 2θ
- −1
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cos²θ + sin²θ = 1.
Answer: (B)
Find x if |2 4; 5 1| = |2x 4; 6 x|.
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2 − 20 = 2x² − 24 → 2x² = 6 → x = ±√3.
Answer: x = ±√3
If A is a 3 × 3 matrix with |A| = 5, find |3A|.
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3³ × 5 = 135.
Answer: 135
Find the area of the triangle with vertices (−2, −3), (3, 2), (−1, −8).
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Using ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| = ½|−2(2 + 8) + 3(−8 + 3) + (−1)(−3 − 2)| = ½|−20 − 15 + 5| = 15.
Answer: 15 sq units
Show that (a, b + c), (b, c + a), (c, a + b) are collinear.
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Add column 1 to column 2: column 2 becomes a + b + c in every row, proportional to column 3 (all 1s) → determinant 0 → collinear.
Answer: Proved
Find adj A and A⁻¹ for A = [−1 5; −3 2].
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|A| = −2 + 15 = 13. adj A = [2 −5; 3 −1]. A⁻¹ = (1/13)[2 −5; 3 −1].
Answer: As above
Write the minors and cofactors of row 1 of [1 0 0; 0 1 0; 0 0 1].
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M₁₁ = 1, M₁₂ = 0, M₁₃ = 0; A₁₁ = 1, A₁₂ = 0, A₁₃ = 0.
Answer: As above
Solve 2x + 5y = 1, 3x + 2y = 7 by the matrix method.
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|A| = 4 − 15 = −11. A⁻¹ = (−1/11)[2 −5; −3 2]. X = (−1/11)[2 − 35; −3 + 14] = (−1/11)[−33; 11] = [3; −1].
Answer: x = 3, y = −1
If A = [2 −1 1; −1 2 −1; 1 −1 2], verify A³ − 6A² + 9A − 4I = O and hence find A⁻¹.
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A² = [6 −5 5; −5 6 −5; 5 −5 6]; A³ = [22 −21 21; −21 22 −21; 21 −21 22]. A³ − 6A² + 9A − 4I = O (check entrywise, e.g. 22 − 36 + 18 − 4 = 0). Multiply by A⁻¹: A² − 6A + 9I = 4A⁻¹ → A⁻¹ = ¼(A² − 6A + 9I) = ¼[3 1 −1; 1 3 1; −1 1 3].
Answer: A⁻¹ = ¼[3 1 −1; 1 3 1; −1 1 3]
Solve 2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y − 2z = −3.
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|A| = 2(−4 + 4) + 3(−6 + 4) + 5(3 − 2) = −1. adj A = [0 −1 2; 2 −9 23; 1 −5 13]. X = A⁻¹B = −adj A · B = −[0 + 5 − 6; 22 + 45 − 69; 11 + 25 − 39] = −[−1; −2; −3] = [1; 2; 3].
Answer: x = 1, y = 2, z = 3
Practice data: the cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60; of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90; of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70. Find the price per kg of each.
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4x + 3y + 2z = 60, 2x + 4y + 6z = 90, 6x + 2y + 3z = 70. |A| = 4(12 − 12) − 3(6 − 36) + 2(4 − 24) = 0 + 90 − 40 = 50. Solving: x = 5, y = 8, z = 8. Check: 20 + 24 + 16 = 60 ✓; 10 + 32 + 48 = 90 ✓; 30 + 16 + 24 = 70 ✓.
Answer: Onion ₹5, wheat ₹8, rice ₹8 per kg
If A and B are invertible matrices of the same order, prove that (AB)⁻¹ = B⁻¹A⁻¹.
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(AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AIA⁻¹ = I, and similarly (B⁻¹A⁻¹)(AB) = I. By uniqueness of inverses, (AB)⁻¹ = B⁻¹A⁻¹. ∎
Answer: Proved
14. Quick Revision Summary
- |a b; c d| = ad − bc; expand 3 × 3 along any row/column with signs + − +.
- Area = ½|det[x y 1]|; zero ⇔ collinear.
- Cofactor Aᵢⱼ = (−1)ⁱ⁺ʲMᵢⱼ; adj A = (cofactors)′.
- A⁻¹ = adj A/|A| when |A| ≠ 0.
- AX = B: unique solution X = A⁻¹B if |A| ≠ 0; else test (adj A)B.