HKDSE Maths M2 study materials
HKDSE Maths M2 study resources

HKDSE Maths M2: Algebra (Extended Part Module 2): Proof, Binomial Coefficients, Matrices and Vectors

HKDSE Maths M2 · Work through your exam, one topic at a time.

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Algebra (Extended Part Module 2): Proof, Binomial Coefficients, Matrices and Vectors

Algebra (Extended Part Module 2)

The Algebra domain of HKDSE Mathematics Extended Part Module 2 tests symbolic manipulation, proof, and structure. Questions often combine topics-mathematical induction with binomial coefficients, logarithms with exponential equations, or matrices with systems of linear equations. Precise notation matters: n!, C(n,r), M^(-1), det M, and the scalar product a·b.

Core rules and concepts

  • Binomial theorem: (a+b)^n = Σ_{r=0}^{n} C(n,r) a^(n-r) b^r. The general term is T_(r+1) = C(n,r) a^(n-r) b^r; use it to find one coefficient without expanding everything.
  • Mathematical induction: prove P(1); assume P(k); show P(k+1). The induction step must use the assumption explicitly.
  • Exponential and logarithmic functions: a^x a^y = a^(x+y); log_a(xy) = log_a x + log_a y; log_a b = log_c b / log_c a. State domains: log arguments must be positive.
  • Matrices: for a 2×2 matrix M = [[a,b],[c,d]], det M = ad − bc and M^(-1) = (1/det M)[[d,−b],[−c,a]] when det M ≠ 0. The system Mx = b has a unique solution iff det M ≠ 0.
  • Vectors: a·b = |a||b|cos θ; perpendicular vectors have zero scalar product. In 3D, the vector product gives a vector perpendicular to both.

Worked example

Prove by induction that 7^n − 1 is divisible by 6 for every positive integer n.

For n = 1, 7^1 − 1 = 6, divisible by 6. Assume 7^k − 1 = 6m for some integer m. Then 7^(k+1) − 1 = 7·7^k − 1 = 7(7^k − 1) + 6 = 7(6m) + 6 = 6(7m + 1). Since 7m + 1 is an integer, 7^(k+1) − 1 is divisible by 6. By induction, the statement holds for all positive integers n. The key move is rewriting 7^(k+1) − 1 in terms of 7^k − 1; do not merely substitute n = k+1 into the assumed formula.

Traps that cost marks

  • Forgetting the base case or checking only n = 1 and n = 2 without a general induction step.
  • Misplacing the exponent in the binomial general term: check whether r starts at 0 or 1.
  • Using M^(-1) when det M = 0; the inverse does not exist.
  • Assuming AB = BA for matrices; order matters.
  • Omitting the angle condition in a·b or forgetting the absolute value in a magnitude.