Further Mathematics

1. 1.

   For A = [[1, 2], [3, 4]], calculate det(A) and A^-1.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. A two-by-two matrix is invertible when ad - bc is nonzero; here it is -2.
   2. Swap the diagonal entries, negate the off-diagonal entries, then divide every entry by -2.
   3. Check by multiplying A by the answer: the diagonal entries become 1 and the other entries 0.

   Marking points

   • det(A) = 4 - 6 = -2.
   • A^-1 = (1/-2)[[4, -2], [-3, 1]].
   • A^-1 = [[-2, 1], [1.5, -0.5]].

   Examiner tip: Divide the entire adjugate by the determinant, not just its diagonal.

2. 2.

   Prove by mathematical induction that 1 + 3 + ... + (2n - 1) = n^2 for every positive integer n.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Start with the smallest allowed integer. The statement is true for 1 because the sum has one term.
   2. Under the induction hypothesis, the first k terms sum to k^2. The next odd number is 2(k + 1) - 1 = 2k + 1.
   3. Adding this term gives precisely the formula for k + 1, completing the logical implication needed for induction.

   Marking points

   • For n = 1, both sides equal 1.
   • Assume the result holds for n = k.
   • For k + 1, add 2k + 1 to obtain k^2 + 2k + 1 = (k + 1)^2.
   • The base case and induction step establish the result for all positive integers.

   Examiner tip: Checking several numerical cases is not a proof for every n. State the induction hypothesis explicitly.

3. 3.

   Let A = [[2, 1], [1, 2]] and P = [[1, 1], [1, -1]]. Show A = P diag(3, 1) P^-1 and deduce a formula for A^n for positive integer n.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Multiplying A by (1, 1) gives 3(1, 1); multiplying by (1, -1) gives (1, -1). Keep this column order in the diagonal matrix.
   2. In repeated products of PDP^-1, every internal P^-1P is the identity, leaving PD^nP^-1.
   3. Multiply out the two-by-two matrices. Substituting n = 1 recovers A, providing a quick check.

   Marking points

   • P^-1 = P/2.
   • The columns of P are eigenvectors with eigenvalues 3 and 1, so AP = P diag(3, 1).
   • A^n = P diag(3^n, 1) P^-1.
   • A^n = (1/2)[[3^n + 1, 3^n - 1], [3^n - 1, 3^n + 1]].

   Examiner tip: A four-point task can require advanced reasoning. Difficulty is not the same as mark count.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.