ASSIGNMENT 09
Due Thursday,8 August 2025
, Problem 1
Given a basis S = {v1 , v2 } for R2 with v1 = (2, 3) and v2 = (1, 0), and a linear transfor-
mation T : R2 → R3 such that
T (v1 ) = (2, 3, 1), T (v2 ) = (1, 2, 4),
find a formula for T (x, y) and compute T (2, −3).
Any (x, y) ∈ R2 can be expressed as:
(x, y) = av1 + bv2 = 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
Then,
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
T (x, y) = x, 2x − , 4x − .
3 3
Substitute (x, y) = (2, −3):
−3 7 · (−3)
T (2, −3) = 2, 2 · 2 − ,4 · 2 − = (2, 5, 15).
3 3
Answer:
y 7y
T (x, y) = x, 2x − , 4x − , T (2, −3) = (2, 5, 15).
3 3
1