MAT2611
ASSIGNMENT 09
Due Thursday,8 August 2025
, Problem 1
Given a basis S = {v1 , v2 } for R2 , where
v1 = (2, 3), v2 = (1, 0),
and a linear transformation T : R2 → R3 defined by
T (v1 ) = (2, 3, 1), T (v2 ) = (1, 2, 4),
determine a formula for T (x, y) and evaluate T (2, −3).
Any vector (x, y) ∈ R2 can be expressed as a linear combination of v1 and v2 :
(x, y) = a(2, 3) + b(1, 0) = (2a + b, 3a).
Solving for a and b:
y 2y
3a = y ⇒ a = , 2a + b = x ⇒ b = x − .
3 3
Using linearity:
y 2y
T (x, y) = aT (v1 ) + bT (v2 ) = (2, 3, 1) + x − (1, 2, 4).
3 3
Compute:
2y y 2y 4y 8y
T (x, y) = + x − , 2x − , 4x −
, y,
3 3 3 3 3
y 7y
= x, 2x − , 4x − .
3 3
Evaluate at (x, y) = (2, −3):
−3 7 · (−3)
T (2, −3) = 2, 2 · 2 − ,4 · 2 − = (2, 5, 15).
3 3
1
ASSIGNMENT 09
Due Thursday,8 August 2025
, Problem 1
Given a basis S = {v1 , v2 } for R2 , where
v1 = (2, 3), v2 = (1, 0),
and a linear transformation T : R2 → R3 defined by
T (v1 ) = (2, 3, 1), T (v2 ) = (1, 2, 4),
determine a formula for T (x, y) and evaluate T (2, −3).
Any vector (x, y) ∈ R2 can be expressed as a linear combination of v1 and v2 :
(x, y) = a(2, 3) + b(1, 0) = (2a + b, 3a).
Solving for a and b:
y 2y
3a = y ⇒ a = , 2a + b = x ⇒ b = x − .
3 3
Using linearity:
y 2y
T (x, y) = aT (v1 ) + bT (v2 ) = (2, 3, 1) + x − (1, 2, 4).
3 3
Compute:
2y y 2y 4y 8y
T (x, y) = + x − , 2x − , 4x −
, y,
3 3 3 3 3
y 7y
= x, 2x − , 4x − .
3 3
Evaluate at (x, y) = (2, −3):
−3 7 · (−3)
T (2, −3) = 2, 2 · 2 − ,4 · 2 − = (2, 5, 15).
3 3
1
