MAT3701 EXAM PACK |COMPLETE SOLUTIONS & PAST PAPERS| 2025.

MAT3701
EXAM
PACK

2025

,MAT3701 Linear Algebra III - Oct/Nov 2024 Exam Solutions

QUESTION 1 [20 marks]

(1.1) Cancellation Law and Zero Vector

(a) What is the zero vector of V? [2 marks]

The zero vector of a vector space V over field F is the unique element 0 ∈ V such that
for every vector x ∈ V: x + 0 = 0 + x = x

The zero vector serves as the additive identity in the vector space.

(b) Prove the zero vector is unique [4 marks]

Proof by contradiction:

Suppose there exist two zero vectors 0₁ and 0₂ in V.

Step 1: Since 0₁ is a zero vector: 0₂ + 0₁ = 0₂

Step 2: Since 0₂ is a zero vector: 0₁ + 0₂ = 0₁

Step 3: By commutativity of vector addition: 0₂ + 0₁ = 0₁ + 0₂

Step 4: From steps 1, 2, and 3: 0₂ = 0₂ + 0₁ = 0₁ + 0₂ = 0₁

Therefore 0₁ = 0₂, proving the zero vector is unique.

(1.2) Prove W is a subspace [6 marks]

Given: V = ℝ², α ∈ ℝ, and W = {(x, y) ∈ ℝ² : y = αx}

To prove W is a subspace, we must show:

1. 0 ∈ W

, 2. Closure under addition
3. Closure under scalar multiplication

Step 1: Zero vector The zero vector in ℝ² is (0, 0). For (0, 0): y = 0 and αx = α(0) = 0
Since 0 = α(0), we have (0, 0) ∈ W ✓


Step 2: Closure under addition Let (x₁, y₁), (x₂, y₂) ∈ W. Then y₁ = αx₁ and y₂ = αx₂. (x₁,
y₁) + (x₂, y₂) = (x₁ + x₂, y₁ + y₂) = (x₁ + x₂, αx₁ + αx₂) = (x₁ + x₂, α(x₁ + x₂)) Since y₁ + y₂
= α(x₁ + x₂), we have (x₁ + x₂, y₁ + y₂) ∈ W ✓


Step 3: Closure under scalar multiplication Let (x, y) ∈ W and c ∈ ℝ. Then y = αx. c(x, y)
= (cx, cy) = (cx, c(αx)) = (cx, α(cx)) Since cy = α(cx), we have c(x, y) ∈ W ✓


Therefore W is a subspace of V.

(1.3) Lagrange interpolation [8 marks]

Find polynomial through points (1, 2), (2, 1), (3, 2).

Step 1: Set up Lagrange basis polynomials For points (x₀, y₀) = (1, 2), (x₁, y₁) = (2, 1),
(x₂, y₂) = (3, 2):

L₀(x) = ((x-x₁)(x-x₂))/((x₀-x₁)(x₀-x₂)) = ((x-2)(x-3))/((1-2)(1-3)) = ((x-2)(x-3))/((-1)(-2)) =
((x-2)(x-3))/2

L₁(x) = ((x-x₀)(x-x₂))/((x₁-x₀)(x₁-x₂)) = ((x-1)(x-3))/((2-1)(2-3)) = ((x-1)(x-3))/(1(-1)) = -((x-
1)(x-3))

L₂(x) = ((x-x₀)(x-x₁))/((x₂-x₀)(x₂-x₁)) = ((x-1)(x-2))/((3-1)(3-2)) = ((x-1)(x-2))/(2(1)) = ((x-
1)(x-2))/2

Step 2: Construct interpolating polynomial P(x) = y₀L₀(x) + y₁L₁(x) + y₂L₂(x) P(x) = 2 ·
((x-2)(x-3))/2 + 1 · (-((x-1)(x-3))) + 2 · ((x-1)(x-2))/2 P(x) = (x-2)(x-3) - (x-1)(x-3) + (x-1)(x-
2)

, Step 3: Expand and simplify (x-2)(x-3) = x² - 5x + 6 -(x-1)(x-3) = -(x² - 4x + 3) = -x² + 4x
- 3 (x-1)(x-2) = x² - 3x + 2

P(x) = (x² - 5x + 6) + (-x² + 4x - 3) + (x² - 3x + 2) P(x) = x² - 4x + 5

Polynomial: P(x) = x² - 4x + 5

Verification:


• P(1) = 1 - 4 + 5 = 2 ✓
• P(2) = 4 - 8 + 5 = 1 ✓
• P(3) = 9 - 12 + 5 = 2 ✓


QUESTION 2 [36 marks]

(2.1) Linear map analysis

(a) Prove T is linear [4 marks]

T : ℝ³ → ℝ³ defined by T(x, y, z) = (x, y, 0)

Step 1: Prove additivity Let u = (x₁, y₁, z₁) and v = (x₂, y₂, z₂) be vectors in ℝ³. T(u + v)
= T((x₁, y₁, z₁) + (x₂, y₂, z₂)) = T(x₁ + x₂, y₁ + y₂, z₁ + z₂) = (x₁ + x₂, y₁ + y₂, 0) = (x₁, y₁,
0) + (x₂, y₂, 0) = T(u) + T(v) ✓


Step 2: Prove homogeneity Let u = (x, y, z) ∈ ℝ³ and c ∈ ℝ. T(cu) = T(c(x, y, z)) = T(cx,
cy, cz) = (cx, cy, 0) = c(x, y, 0) = cT(u) ✓


Therefore T is linear.

(b) Find N(T) [2 marks]

N(T) = {(x, y, z) ∈ ℝ³ : T(x, y, z) = (0, 0, 0)} T(x, y, z) = (x, y, 0) = (0, 0, 0) This requires x
= 0, y = 0, and z can be any real number.