Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Overview
This chapter applies matrices and determinants to solve systems of linear equations. Key tools: adjoint, inverse, rank, Cramer's rule, and Gaussian elimination.
1. Adjoint and Inverse
adj(A) = transpose of the cofactor matrix
A⁻¹ = (1/|A|) · adj(A), provided |A| ≠ 0
A is non-singular if |A| ≠ 0 (only then A⁻¹ exists)
2. Solving AX = B
- Matrix inversion method: X = A⁻¹B.
- Cramer's rule: xᵢ = Δᵢ/Δ.
- Gaussian elimination: reduce the augmented matrix [A|B] to echelon form.
3. Rank and Consistency
ρ(A) = ρ(A|B) = n unknowns → unique solution
ρ(A) = ρ(A|B) < n → infinitely many solutions
ρ(A) ≠ ρ(A|B) → no solution (inconsistent)
Worked Examples
Example 1 — Inverse of a 2×2 matrix
A = [[2, 1], [7, 4]]. Find A⁻¹.
|A| = (2)(4) − (1)(7) = 8 − 7 = 1
A⁻¹ = (1/1)[[4, −1], [−7, 2]] = [[4, −1], [−7, 2]]
Example 2 — Cramer's rule
Solve 2x + y = 5, 3x − y = 0.
Δ = |2 1; 3 −1| = −2 − 3 = −5
Δ_x = |5 1; 0 −1| = −5 − 0 = −5 → x = −5/−5 = 1
Δ_y = |2 5; 3 0| = 0 − 15 = −15 → y = −15/−5 = 3
Solution: x = 1, y = 3
Example 3 — Inverse method
Solve using X = A⁻¹B: 2x + y = 5, 7x + 4y = 18.
A = [[2,1],[7,4]], |A| = 1, A⁻¹ = [[4,−1],[−7,2]]
X = A⁻¹B = [[4,−1],[−7,2]] [5;18] = [20−18 ; −35+36] = [2 ; 1]
x = 2, y = 1
Example 4 — Consistency by rank
For x + y = 2, 2x + 2y = 5 — state consistency.
ρ(A) = 1 (rows proportional), ρ(A|B) = 2 → ρ(A) ≠ ρ(A|B)
→ inconsistent, no solution.
Common Mistakes
- Trying to invert a singular matrix (|A| = 0).
- Forgetting the 1/|A| factor in A⁻¹.
- Replacing the wrong column when forming Δᵢ in Cramer's rule.
- Mixing up adjoint (transpose of cofactors) with the cofactor matrix.
Exam Tips & Revision Checklist
- Always check |A| before using inverse or Cramer's rule.
- Use rank for "discuss the consistency" questions.
- Practised: 2×2 inverse, Cramer's rule, inverse method, rank/consistency.
Frequently Asked Questions
A⁻¹ = (1/|A|) · adj(A), provided |A| ≠ 0. adj(A) is the transpose of the cofactor matrix.