Determinants

1. Introduction

A determinant is a scalar value associated with a square matrix. It provides essential information about the matrix — whether it is invertible, the volume scaling factor of its linear transformation, and more.

2. Determinant of a Square Matrix

For a 2×2 matrix A = [[a, b], [c, d]], |A| = ad - bc.

For a 3×3 matrix, expansion along any row or column using signs (+ - +, - + -, + - +).

3. Properties of Determinants

1. |Aᵀ| = |A|
2. If two rows (or columns) are identical or proportional, |A| = 0.
3. If two rows are interchanged, the determinant changes sign.
4. If a row is multiplied by k, the determinant is multiplied by k.
5. The value of a determinant is unchanged if a multiple of one row is added to another row.
6. |AB| = |A|·|B|
7. If A is triangular, |A| = product of diagonal entries.

'Property 5 is the most powerful for simplifying determinants. Use it to create zeros before expanding.'

4. Minors and Cofactors

Minor Mᵢⱼ: Determinant obtained by deleting row i and column j. Cofactor Cᵢⱼ = (-1)^{i+j} Mᵢⱼ.

The determinant can be expanded along row i: |A| = Σ aᵢⱼ Cᵢⱼ for j = 1 to n.

5. Adjoint and Inverse

The adjoint of A, adj(A), is the transpose of the cofactor matrix.

Inverse using adjoint: A^{-1} = adj(A) / |A|, provided |A| ≠ 0.

If |A| = 0, the matrix is singular and has no inverse.

6. Solution of Linear Equations

6.1 Cramer's Rule

For a system of n equations in n unknowns: xᵢ = Dᵢ/D, where D = |A| and Dᵢ is |A| with column i replaced by the constants.

'Cramer's rule works only when D ≠ 0 (unique solution). If D = 0, the system may have no solution or infinitely many solutions.'

6.2 Matrix Method (AX = B)

X = A^{-1}B, provided A is non-singular.

7. Area of Triangle

Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| = (1/2)|determinant formed by coordinates|

'Area is always taken as positive, so use absolute value.'

8. Worked Problems

Problem 1: Using properties, prove that |[[1, a, a²], [1, b, b²], [1, c, c²]]| = (a-b)(b-c)(c-a). Solution: Apply R₂ → R₂ - R₁, R₃ → R₃ - R₁. Take (b-a) and (c-a) common. Then expand.

Problem 2: Solve using Cramer's rule: x + y + z = 6, 2x + 3y - z = 5, x - y + 2z = 3. Solution: D = |[[1, 1, 1], [2, 3, -1], [1, -1, 2]]| = 1(6-1) - 1(4+1) + 1(-2-3) = 5-5-5 = -5. D₁ = |[[6, 1, 1], [5, 3, -1], [3, -1, 2]]| = 6(6-1) - 1(10+3) + 1(-5-9) = 30-13-14 = 3. Similarly D₂ = 5, D₃ = 22. So x = -3/5, y = -1, z = -22/5.

Problem 3: Find the inverse of A = [[2, -1, 1], [-1, 2, -1], [1, -1, 2]] using adjoint method. Solution: |A| = 4 ≠ 0. Find cofactors, form adj(A), then A^{-1} = adj(A)/4.

9. Area of Triangle — Detailed Explanation

The area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) can be computed as:

Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

This is exactly half the absolute value of the determinant: |[[x₁, y₁, 1], [x₂, y₂, 1], [x₃, y₃, 1]]|

If three points are collinear, the area is zero (determinant = 0). This gives a useful test for collinearity.

'When using the determinant formula for area, the points must be taken in order (clockwise or anticlockwise). If taken in the opposite order, the determinant gives the same magnitude but opposite sign — the absolute value gives the area.'

10. Consistency of Linear Systems

For AX = B (n equations, n unknowns):

• If |A| ≠ 0: Unique solution exists (consistent system).
• If |A| = 0 and (adj A)B = 0: Infinitely many solutions (consistent).
• If |A| = 0 and (adj A)B ≠ 0: No solution (inconsistent).

This is known as the consistency condition and is frequently tested in ISC examinations.

11. Common Mistakes

'Students often forget the alternating sign pattern (+ - +) when expanding determinants. Mark the signs clearly before computing.'

'When using Cramer's rule, replace the correct column — column 1 for x, column 2 for y, column 3 for z.'

12. ISC Exam Focus

'Students often forget the alternating sign pattern (+ - +) when expanding determinants. Mark the signs clearly before computing.'

'When using Cramer's rule, replace the correct column — column 1 for x, column 2 for y, column 3 for z.'

10. ISC Exam Focus

11. Self-Test Questions

1. Prove that |[[x, y, z], [y, z, x], [z, x, y]]| = (x + y + z)(xy + yz + zx - x² - y² - z²).
2. Find the inverse of [[1, 2, 3], [2, 4, 5], [3, 5, 6]] using adjoint.
3. Solve by matrix method: x + 2y - z = 7, 2x - y + 3z = 9, 3x + y + 2z = 14.
4. Find area of triangle with vertices (1, 2), (4, 5), (3, 8).
5. Using properties, evaluate |[[a+b, b+c, c+a], [b+c, c+a, a+b], [c+a, a+b, b+c]]|.