Further Mathematics

1. 1.

   Column vectors are first reflected in y = x by S = [[0,1],[1,0]], then stretched parallel to x by D = [[2,0],[0,1]]. Find the combined matrix and image of (3,-1).

   [3 marks] · no calculator

   Answer explanation

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

   1. Reflection swaps the coordinates to (-1,3); the stretch doubles only the first coordinate.
   2. Multiplying DS by the original column confirms the same image.

   Marking points

   • The product is DS, not SD.
   • DS = [[0,2],[1,0]].
   • The image is (-2,3).

   Examiner tip: For column vectors the rightmost transformation acts first.

2. 2.

   A triangle of area 7 is transformed by [[2,1],[1,-1]]. Find its image area and state whether orientation is reversed.

   [3 marks] · no calculator

   Answer explanation

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

   1. Areas scale by the absolute determinant, so a negative signed scale cannot produce negative area.
   2. The sign records orientation separately from the area scale: |-3| times 7.

   Marking points

   • The determinant is -3.
   • Image area is 21 square units.
   • Orientation is reversed because the determinant is negative.

   Examiner tip: Use |det A| for area and det A's sign for orientation.

3. 3.

   Find the eigenvalues and corresponding eigenvector directions of A = [[3,1],[1,3]].

   [3 marks] · no calculator

   Answer explanation

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

   1. A vector along y = x has both components multiplied by four.
   2. Along y = -x the components are multiplied by two; direct multiplication verifies both directions.

   Marking points

   • Characteristic equation: (3 - lambda)^2 - 1 = 0.
   • Eigenvalues are 4 and 2.
   • Directions are (1,1) for 4 and (1,-1) for 2.

   Examiner tip: Associate each direction with its own eigenvalue; do not list unordered vectors.

4. 4.

   For A = [[k,2],[3,6]], find k for singularity. At that k, describe the image of the whole plane and the vectors sent to zero.

   [3 marks] · no calculator

   Answer explanation

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

   1. At k = 1 the second row is three times the first, so two-dimensional inputs collapse onto one line.
   2. To find the kernel set the first output to zero; the second then vanishes automatically.

   Marking points

   • 6k - 6 = 0 gives k = 1.
   • Outputs are (x + 2y, 3x + 6y), so the image is the line Y = 3X.
   • The kernel is x + 2y = 0, or multiples of (-2,1).

   Examiner tip: The image line and the kernel line are different objects.

5. 5.

   For A = [[3,1],[1,3]], derive A^n for positive integers n using its eigenvectors.

   [3 marks] · no calculator

   Answer explanation

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

   1. The two eigenvector columns form a basis. Repeated action multiplies their coefficients by 4^n and 2^n respectively.
   2. Adjacent P^-1 P factors cancel in a repeated product; multiply the remaining three matrices to obtain the expression.

   Marking points

   • Use P = [[1,1],[1,-1]] and P^-1 = P/2.
   • A = P diag(4,2) P^-1, hence A^n = P diag(4^n,2^n) P^-1.
   • A^n = (1/2)[[4^n + 2^n,4^n - 2^n],[4^n - 2^n,4^n + 2^n]].

   Examiner tip: A matrix power is not an entrywise power; diagonalise before raising entries.

6. 6.

   Let B = [[2,1],[0,2]]. Show that B is not diagonalisable and derive B^n for n >= 1.

   [3 marks] · no calculator

   Answer explanation

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

   1. A repeated eigenvalue alone does not prevent diagonalisation; here the failure is the absence of a second independent eigenvector.
   2. All terms containing N^2 or a higher power vanish, leaving 2^n I + n*2^(n-1)N.

   Marking points

   • The only eigenvalue is 2 and its eigenspace is y = 0, of dimension one.
   • Write B = 2I + N with N^2 = 0.
   • The commuting binomial expansion gives B^n = [[2^n,n*2^(n-1)],[0,2^n]].

   Examiner tip: Justify the binomial method by noting that I and N commute.

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