GELE Mathematics — Advanced Engineering MathematicsExam Answer Templates

Marks

1

Topic

Complex Numbers — Modulus

Difficulty

easy

Template Id

T1

Examiner Tip

For a 1-mark VSA, examiners need only the correct answer. However, showing the formula takes 5 seconds and guarantees you cannot lose the mark to a transcription ambiguity.

Model Answer

Given: z = 5 − 12i, so a = 5, b = −12 |z| = √(a² + b²) = √(5² + (−12)²) = √(25 + 144) = √169 ∴ |z| = 13

Question Type

very_short_answer

Answer Structure

• Line 1: Identify a and b from the rectangular form [setup]
• Line 2: Apply formula |z| = √(a² + b²) and substitute [1 mark — correct answer]

Scoring Breakdown

Common Mark Deductions

• Writing |z| = 5 + 12 = 17 (adding instead of using Pythagorean formula)
• Forgetting to take the square root: leaving answer as √169 without evaluating
• Sign error: using b = +12 but this does not affect |z| since it is squared — however, students sometimes miscopy the sign and compute √(25 − 144), giving an imaginary result

Key Phrases To Include

• |z| = √(a² + b²)
• modulus
• 13

Marks

1

Topic

Matrices and Determinants

Difficulty

easy

Template Id

T2

Examiner Tip

Always write the formula det = ad − bc before substituting. This ensures that even if arithmetic is wrong, the examiner sees you know the method.

Model Answer

det A = (2)(3) − (5)(1) = 6 − 5 ∴ det A = 1

Question Type

very_short_answer

Answer Structure

• Line 1: Apply 2×2 determinant formula det = ad − bc [1 mark — correct evaluation]

Scoring Breakdown

Common Mark Deductions

• Computing (2)(3) + (5)(1) = 11 (adding instead of subtracting)
• Reversing the diagonal: (1)(2) − (3)(5) = −13 (wrong diagonal assignment)

Key Phrases To Include

• ad − bc
• det A = 1

Marks

2

Topic

Complex Numbers — Polar Form Conversion

Difficulty

medium

Template Id

T3

Examiner Tip

The quadrant-check step must be explicitly written — examiners deduct the argument mark even if the reference angle is correct but the quadrant adjustment is missing.

Model Answer

Given: z = −3 + 3i → a = −3, b = +3 (Quadrant II) Step 1 — Modulus: |z| = √((-3)² + 3²) = √(9 + 9) = √18 = 3√2 Step 2 — Argument: Reference angle: α = tan⁻¹(|b|/|a|) = tan⁻¹(3/3) = tan⁻¹(1) = 45° Since a < 0 and b > 0 → Quadrant II: θ = 180° − 45° = 135° ∴ z = 3√2 ∠135°

Question Type

short_answer

Answer Structure

• Line 1: Identify a and b; determine the quadrant [setup, no mark yet]
• Line 2: Compute r = √(a² + b²) = 3√2 [1 mark]
• Line 3: Compute θ using reference angle and quadrant correction = 135° [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using θ = tan⁻¹(3/−3) = −45° without quadrant correction (most common error)
• Giving θ = 45° (ignoring the negative real part entirely)
• Leaving r = √18 without simplifying to 3√2 (minor but noted by careful examiners)

Key Phrases To Include

• Quadrant II
• reference angle
• θ = 180° − α
• 3√2 ∠135°

Marks

2

Topic

Vectors — Dot Product and Perpendicularity

Difficulty

easy

Template Id

T4

Examiner Tip

Always explicitly state 'The dot product yields a scalar result' — this demonstrates conceptual understanding and is rewarded even at short-answer level.

Model Answer

Given: a = (3, −1, 2), b = (4, 2, −1) Step 1 — Dot Product: a · b = (3)(4) + (−1)(2) + (2)(−1) = 12 − 2 − 2 = 8 Step 2 — Perpendicularity Check: For perpendicularity, a · b must equal 0. Since a · b = 8 ≠ 0, the vectors are NOT perpendicular. ∴ a · b = 8 (scalar); the vectors are not perpendicular.

Question Type

short_answer

Answer Structure

• Line 1-3: Compute a · b = aₓbₓ + aᵧbᵧ + a_z b_z = 8 [1 mark]
• Line 4: State perpendicularity condition (a · b = 0) and conclude [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing the cross product instead of the dot product
• Giving a vector answer instead of a scalar for the dot product
• Concluding 'not perpendicular' without citing the zero-dot-product criterion

Key Phrases To Include

• a · b = aₓbₓ + aᵧbᵧ + a_z b_z
• scalar
• perpendicular if a · b = 0
• not perpendicular

Marks

3

Topic

Cramer's Rule — 2×2 System

Difficulty

medium

Template Id

T5

Examiner Tip

The verification step costs only 15 seconds and catches arithmetic errors before you submit. If your check fails, re-examine your determinant calculations — the method is almost certainly correct.

Model Answer

Given: 3x + 2y = 7 x − y = 1 Step 1 — Coefficient Matrix A and constant vector b: A = |3 2| b = |7| |1 −1| |1| Step 2 — Compute det A: det A = (3)(−1) − (2)(1) = −3 − 2 = −5 Step 3 — Solve for x (replace column 1 with b): Aₓ = |7 2| |1 −1| det Aₓ = (7)(−1) − (2)(1) = −7 − 2 = −9 x = det Aₓ / det A = (−9)/(−5) = 9/5 = 1.8 Step 4 — Solve for y (replace column 2 with b): Aᵧ = |3 7| |1 1| det Aᵧ = (3)(1) − (7)(1) = 3 − 7 = −4 y = det Aᵧ / det A = (−4)/(−5) = 4/5 = 0.8 ∴ x = 9/5 = 1.8, y = 4/5 = 0.8 Verification: 3(1.8) + 2(0.8) = 5.4 + 1.6 = 7 ✓ | 1.8 − 0.8 = 1 ✓

Question Type

numerical

Answer Structure

• Block 1: Write coefficient matrix A and compute det A = −5 [1 mark]
• Block 2: Form Aₓ, compute det Aₓ, and find x = 9/5 [1 mark]
• Block 3: Form Aᵧ, compute det Aᵧ, and find y = 4/5 [1 mark]
• Optional: Verification step (demonstrates rigor, no extra mark but avoids deduction)

Scoring Breakdown

Common Mark Deductions

• Replacing the wrong column (e.g., replacing column 2 when solving for x)
• Sign error in det A, leading to wrong signs for both unknowns
• Forgetting to check det A ≠ 0 before applying Cramer's rule
• Arithmetic error in any 2×2 determinant — partial credit still given for correct setup

Key Phrases To Include

• Cramer's Rule
• det A
• replace the i-th column with b
• xᵢ = det Aᵢ / det A

Marks

3

Topic

Complex Numbers — De Moivre's Theorem

Difficulty

medium

Template Id

T6

Examiner Tip

Always state De Moivre's Theorem explicitly before applying it. This earns method marks even if exponent arithmetic is wrong. Name the theorem: 'By De Moivre's Theorem, zⁿ = rⁿ ∠nθ'.

Model Answer

Given: z = 1 + i, n = 6 Step 1 — Convert z to polar form: r = √(1² + 1²) = √2 θ = tan⁻¹(1/1) = 45° (Quadrant I, both a, b > 0) z = √2 ∠45° Step 2 — Apply De Moivre's Theorem: zⁿ = rⁿ ∠(nθ) z⁶ = (√2)⁶ ∠(6 × 45°) = (2^(1/2))⁶ ∠270° = 2³ ∠270° = 8∠270° Step 3 — Convert back to rectangular form: z⁶ = 8(cos 270° + i sin 270°) = 8(0 + i(−1)) = 8(−i) ∴ (1 + i)⁶ = −8i (or 0 − 8i in full rectangular form)

Question Type

numerical

Answer Structure

• Block 1: Convert to polar form — r = √2, θ = 45° [1 mark]
• Block 2: Apply De Moivre: r⁶ = 8, nθ = 270° [1 mark]
• Block 3: Convert 8∠270° to rectangular form = −8i [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing (1 + i)⁶ by brute-force expansion instead of De Moivre (wastes time, error-prone)
• Evaluating (√2)⁶ as 2√2 instead of 8
• Giving final answer in polar form (8∠270°) when rectangular form was requested
• Incorrect trig values for 270° (some students write cos 270° = 1)

Key Phrases To Include

• De Moivre's Theorem
• zⁿ = rⁿ ∠nθ
• cos 270° = 0
• sin 270° = −1
• −8i

Marks

3

Topic

Matrices and Determinants — Matrix Inverse

Difficulty

medium

Template Id

T7

Examiner Tip

Always check det A first and state 'since det A ≠ 0, the matrix is invertible.' This single sentence earns examiner confidence and demonstrates conceptual understanding beyond mechanical computation.

Model Answer

Given: A = |1 3| |2 7| Step 1 — Compute det A: det A = (1)(7) − (3)(2) = 7 − 6 = 1 Since det A = 1 ≠ 0, the inverse exists. Step 2 — Apply the 2×2 inverse formula: A⁻¹ = (1/det A) × | d −b| |−c a| A⁻¹ = (1/1) × | 7 −3| |−2 1| Step 3 — Final answer: A⁻¹ = | 7 −3| |−2 1| Verification: A × A⁻¹ = |1·7 + 3·(−2) 1·(−3) + 3·1| = |7−6 −3+3| = |1 0| = I ✓ |2·7 + 7·(−2) 2·(−3) + 7·1| |14−14 −6+7| |0 1|

Question Type

numerical

Answer Structure

• Block 1: Compute and state det A = 1; confirm inverse exists [1 mark]
• Block 2: Write the 2×2 inverse formula with a,b,c,d identified [1 mark]
• Block 3: Correctly assemble A⁻¹ with proper signs [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting the (1/det A) scalar factor
• Not negating off-diagonal elements b and c (writing | d b; c a| instead)
• Not stating that det A ≠ 0 ensures invertibility

Key Phrases To Include

• det A = ad − bc
• A⁻¹ = (1/det A) |d −b; −c a|
• inverse exists since det A ≠ 0

Marks

3

Topic

Vectors — Cross Product

Difficulty

medium

Template Id

T8

Examiner Tip

Explicitly write out the 3×3 determinant with i, j, k headings. Examiners follow your cofactor expansion row by row — if your setup is visible, partial credit is awarded even if individual components have arithmetic errors.

Model Answer

Given: a = (2, 1, −1), b = (−3, 4, 2) Step 1 — Cross product via cofactor expansion: a × b = |i j k | |2 1 −1 | |−3 4 2 | i-component: (1)(2) − (−1)(4) = 2 + 4 = 6 j-component: −[(2)(2) − (−1)(−3)] = −[4 − 3] = −1 k-component: (2)(4) − (1)(−3) = 8 + 3 = 11 ∴ a × b = (6, −1, 11) [vector result] Step 2 — Magnitude of cross product: |a × b| = √(6² + (−1)² + 11²) = √(36 + 1 + 121) = √158 ≈ 12.57 ∴ a × b = 6i − j + 11k; |a × b| = √158 ≈ 12.57

Question Type

numerical

Answer Structure

• Block 1: Set up 3×3 determinant with i, j, k in top row [1 mark]
• Block 2: Correctly compute each component (note the sign flip on j) [1 mark]
• Block 3: Compute magnitude |a × b| = √158 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting the negative sign on the j-component (most frequent error in cross product)
• Giving a scalar result instead of a vector for a × b
• Switching the row order of a and b, reversing the sign of the result

Key Phrases To Include

• cross product
• 3×3 determinant
• j-component sign reversal
• |a × b| = |a||b|sinθ
• vector perpendicular to both

Marks

2

Topic

Complex Numbers — Multiplication in Polar Form

Difficulty

easy

Template Id

T9

Examiner Tip

State the rule explicitly: 'In polar form, multiplication means: magnitudes multiply and arguments add.' One line saves the mark.

Model Answer

Given: z₁ = 4∠30°, z₂ = 3∠60° Step 1 — Multiply in polar form (multiply magnitudes, add angles): z₁ × z₂ = (4 × 3)∠(30° + 60°) = 12∠90° Step 2 — Convert to rectangular form: z = 12(cos 90° + i sin 90°) = 12(0 + i·1) = 12i ∴ z₁z₂ = 12∠90° = 0 + 12i

Question Type

numerical

Answer Structure

• Line 1: Multiply magnitudes (4 × 3 = 12) and add angles (30° + 60° = 90°) → 12∠90° [1 mark]
• Line 2: Convert to rectangular using cos 90° = 0, sin 90° = 1 → 12i [1 mark]

Scoring Breakdown

Common Mark Deductions

• Multiplying angles instead of adding them
• Using the wrong trig values for 90° (some write cos 90° = 1)

Key Phrases To Include

• multiply magnitudes
• add angles
• cos 90° = 0
• sin 90° = 1
• 12∠90°
• 12i

Marks

3

Topic

Complex Numbers — n-th Roots (De Moivre Roots)

Difficulty

hard

Template Id

T10

Examiner Tip

Always write 'k = 0, 1, ..., n−1' in your formula line. This shows the examiner you know there are n distinct roots, and you earn the method mark even if arithmetic on one root is wrong.

Model Answer

Given: z = 4∠120°, n = 2 (square root) Formula (n-th roots of a complex number): z^(1/n) = r^(1/n) ∠[(θ + 360°k)/n], k = 0, 1, ..., n−1 Step 1 — Compute r^(1/2): r = 4 → r^(1/2) = √4 = 2 Step 2 — Root for k = 0: θ₀ = (120° + 360°·0)/2 = 120°/2 = 60° z₀ = 2∠60° Step 3 — Root for k = 1: θ₁ = (120° + 360°·1)/2 = 480°/2 = 240° z₁ = 2∠240° ∴ The two square roots are: 2∠60° and 2∠240°

Question Type

numerical

Answer Structure

• Line 1: State n-th root formula explicitly [setup, partial credit if formula is correct]
• Line 2: Compute r^(1/2) = 2 [0.5 mark embedded in step]
• Line 3: k = 0 root → 2∠60° [1 mark]
• Line 4: k = 1 root → 2∠240° [1 mark]
• Line 5: State that exactly n = 2 roots exist [completes the 3-mark answer]

Scoring Breakdown

Common Mark Deductions

• Finding only one root (k = 0) and stopping
• Using degrees/n without adding 360°k for subsequent roots
• Computing r^(1/n) incorrectly (e.g., 4^(1/2) = 2 is correct, but some write 4/2 = 2 by coincidence and cannot generalize)

Key Phrases To Include

• n-th roots formula
• r^(1/n)
• (θ + 360°k)/n
• k = 0 and k = 1
• two distinct roots

Marks

3

Topic

Vectors — Angle Between Vectors

Difficulty

medium

Template Id

T11

Examiner Tip

Write the formula cos θ = (u · v)/(|u||v|) before substituting — this is a separate mark in some marking schemes and demonstrates you are using the correct approach.

Model Answer

Given: u = (1, 2, 2), v = (2, 0, 1) Step 1 — Compute the dot product u · v: u · v = (1)(2) + (2)(0) + (2)(1) = 2 + 0 + 2 = 4 Step 2 — Compute magnitudes |u| and |v|: |u| = √(1² + 2² + 2²) = √(1 + 4 + 4) = √9 = 3 |v| = √(2² + 0² + 1²) = √(4 + 0 + 1) = √5 Step 3 — Apply the angle formula: cos θ = (u · v) / (|u| · |v|) = 4 / (3 × √5) = 4 / (3√5) = 4 / 6.7082 ≈ 0.5963 θ = cos⁻¹(0.5963) ≈ 53.4° ∴ The angle between u and v is approximately 53.4°

Question Type

numerical

Answer Structure

• Block 1: Compute u · v = 4 [1 mark]
• Block 2: Compute |u| = 3 and |v| = √5 [1 mark]
• Block 3: Apply cos θ formula and evaluate θ ≈ 53.4° [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the cross product magnitude formula |u × v| = |u||v|sinθ instead of the dot product
• Computing |u| = 1 + 2 + 2 = 5 instead of √(1 + 4 + 4) = 3
• Giving the answer in radians when degrees are requested

Key Phrases To Include

• cos θ = (u · v) / (|u||v|)
• dot product
• magnitude
• inverse cosine
• 53.4°

Marks

5

Topic

Cramer's Rule — 3×3 System

Difficulty

hard

Template Id

T12

Examiner Tip

In a 5-mark problem, never skip the verification step. If your final answers are wrong but your determinant setup is correct, you still earn 3–4 marks. Verification also shows the examiner exactly where any error occurred, enabling maximum partial credit.

Model Answer

Given system: 2x − y + z = 3 x + y − z = 0 3x + 2y + z = 7 Step 1 — Write coefficient matrix A and vector b: A = | 2 −1 1| b = |3| | 1 1 −1| |0| | 3 2 1| |7| Step 2 — Compute det A (expand along Row 1): det A = 2|1 −1| − (−1)|1 −1| + 1|1 1| |2 1| |3 1| |3 2| = 2[(1)(1)−(−1)(2)] + 1[(1)(1)−(−1)(3)] + 1[(1)(2)−(1)(3)] = 2[1+2] + 1[1+3] + 1[2−3] = 2(3) + 1(4) + 1(−1) = 6 + 4 − 1 = 9 Step 3 — Compute det Aₓ (replace column 1 with b): Aₓ = | 3 −1 1| | 0 1 −1| | 7 2 1| det Aₓ = 3[(1)(1)−(−1)(2)] − (−1)[(0)(1)−(−1)(7)] + 1[(0)(2)−(1)(7)] = 3[1+2] + 1[0+7] + 1[0−7] = 3(3) + 7 − 7 = 9 x = det Aₓ / det A = 9/9 = 1 Step 4 — Compute det Aᵧ (replace column 2 with b): Aᵧ = | 2 3 1| | 1 0 −1| | 3 7 1| det Aᵧ = 2[(0)(1)−(−1)(7)] − 3[(1)(1)−(−1)(3)] + 1[(1)(7)−(0)(3)] = 2[0+7] − 3[1+3] + 1[7−0] = 14 − 12 + 7 = 9 y = det Aᵧ / det A = 9/9 = 1 Step 5 — Compute det A_z (replace column 3 with b): A_z = | 2 −1 3| | 1 1 0| | 3 2 7| det A_z = 2[(1)(7)−(0)(2)] − (−1)[(1)(7)−(0)(3)] + 3[(1)(2)−(1)(3)] = 2[7−0] + 1[7−0] + 3[2−3] = 14 + 7 − 3 = 18 z = det A_z / det A = 18/9 = 2 Verification: Eq.1: 2(1) − 1 + 2 = 3 ✓ Eq.2: 1 + 1 − 2 = 0 ✓ Eq.3: 3 + 2 + 2 = 7 ✓ ∴ x = 1, y = 1, z = 2

Question Type

numerical

Answer Structure

• Block 1: Write A matrix and b vector, compute det A = 9 [1 mark]
• Block 2: Form Aₓ, compute det Aₓ = 9, find x = 1 [1 mark]
• Block 3: Form Aᵧ, compute det Aᵧ = 9, find y = 1 [1 mark]
• Block 4: Form A_z, compute det A_z = 18, find z = 2 [1 mark]
• Block 5: Verification in all three equations [1 mark — shows rigor and catches errors]

Scoring Breakdown

Common Mark Deductions

• Replacing wrong column for each unknown
• Arithmetic error in 3×3 cofactor expansion (most common — show each 2×2 sub-determinant clearly)
• Not checking det A ≠ 0 before proceeding
• Omitting verification — 5-mark problems are expected to have a check

Key Phrases To Include

• Cramer's Rule
• det A ≠ 0
• replace i-th column with b
• xᵢ = det Aᵢ / det A
• verification

Marks

2

Topic

Matrices — Matrix Multiplication

Difficulty

easy

Template Id

T13

Examiner Tip

Write the dimension check '(2×2)(2×2) → 2×2 result' as your first line. For larger matrices, this step prevents wasted effort on non-conformable multiplication.

Model Answer

Given: A = |1 2|, B = |5 6| |3 4| |7 8| Matrix dimensions: A is 2×2, B is 2×2 → AB is 2×2 (AB)₁₁ = (1)(5) + (2)(7) = 5 + 14 = 19 (AB)₁₂ = (1)(6) + (2)(8) = 6 + 16 = 22 (AB)₂₁ = (3)(5) + (4)(7) = 15 + 28 = 43 (AB)₂₂ = (3)(6) + (4)(8) = 18 + 32 = 50 ∴ AB = |19 22| |43 50|

Question Type

numerical

Answer Structure

• Line 1: State dimension check: (2×2)(2×2) → 2×2 result [setup]
• Lines 2–5: Compute all four entries using row × column rule [1 mark for method]
• Line 6: Display final 2×2 matrix AB [1 mark for correct answer]

Scoring Breakdown

Common Mark Deductions

• Multiplying element-by-element (Hadamard product) instead of row × column
• Computing BA instead of AB (order matters)
• Dimension error — not checking inner dimensions match

Key Phrases To Include

• row × column
• dimension check
• (m×n)(n×p) = m×p
• AB ≠ BA in general

Marks

1

Topic

Complex Numbers — Polar to Rectangular Conversion

Difficulty

easy

Template Id

T14

Examiner Tip

Quadrant III means both a and b are negative. A quick sketch of the unit circle in the margin takes 3 seconds and prevents sign errors that cost the mark.

Model Answer

cos 210° = −√3/2, sin 210° = −1/2 z = 2(−√3/2 + i(−1/2)) = −√3 − i ∴ z = −√3 − i

Question Type

very_short_answer

Answer Structure

• Line 1: Evaluate cos 210° and sin 210° from unit circle [key step]
• Line 2: Multiply by r = 2 to get rectangular form [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using cos 210° = +√3/2 (forgetting Quadrant III sign)
• Multiplying only the real part by r and leaving the imaginary coefficient unscaled

Key Phrases To Include

• cos 210° = −√3/2
• sin 210° = −1/2
• a = r cosθ
• b = r sinθ

Marks

5

Topic

Vectors — Dot Product and Cross Product Applications

Difficulty

hard

Template Id

T15

Examiner Tip

In a combined 5-mark problem covering both dot and cross product, explicitly label each part (a) and (b) and draw a separation line. Examiners mark each part independently — a wrong answer in part (a) does not forfeit marks in part (b).

Model Answer

Given: p = (2, 3, −1), q = (1, −2, 2) Part (a) — Dot Product (scalar): p · q = (2)(1) + (3)(−2) + (−1)(2) = 2 − 6 − 2 = −6 ∴ p · q = −6 Since p · q ≠ 0, the vectors are NOT perpendicular. Part (b) — Area of Parallelogram = |p × q| Step 1 — Compute p × q: p × q = | i j k | | 2 3 −1 | | 1 −2 2 | i: (3)(2) − (−1)(−2) = 6 − 2 = 4 j: −[(2)(2) − (−1)(1)] = −[4 + 1] = −5 k: (2)(−2) − (3)(1) = −4 − 3 = −7 p × q = (4, −5, −7) = 4i − 5j − 7k Step 2 — Compute |p × q|: |p × q| = √(4² + (−5)² + (−7)²) = √(16 + 25 + 49) = √90 = 3√10 ≈ 9.49 ∴ Area of parallelogram = |p × q| = 3√10 ≈ 9.49 square units

Question Type

numerical

Answer Structure

• Block 1: Compute p · q = −6 and note non-perpendicularity [1 mark]
• Block 2: Set up 3×3 determinant for p × q [1 mark]
• Block 3: Compute all three components of p × q = (4, −5, −7) [1 mark]
• Block 4: Compute |p × q| = √90 [1 mark]
• Block 5: Simplify to 3√10 and state the geometric meaning (area of parallelogram) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing which operation gives area (it is |cross product|, not dot product)
• Forgetting negative sign on j-component of cross product
• Leaving area as √90 without simplifying (minor deduction in some marking schemes)

Key Phrases To Include

• dot product = scalar
• cross product = vector
• area of parallelogram = |p × q|
• |p × q| = |p||q|sinθ
• 3√10