MAT2611 Assignment 10 2025 - Due 22 August 2025

MAT2611
ASSIGNMENT 10 2025
DUE: 22 AUGUST 2025 (MEMO)

, Instructions for the Assignments
Take care to explain all your arguments.
Only PDF …les will be accepted.
Do not type solutions, only hand written.


MAT2611
ASSIGNMENT 10
Due date: Friday, 22 August 2025




Problem 1. Find (T3 T2 T1 ) (x; y) ; where
T1 (x; y) = (4x; x y; 2x y) ;
T2 (x; y; z) = ( x; 0; x + 2y + z) ;
3 (x;
T y; z) = (2x + y z; 2y + 3z) :




1. Composition of linear transformations
, (T3 ∘ T2 ∘ T1 )(x, y) = T3 (T2 (T1 (x, y))), i.e. apply T1 first, then T2 , then T3 . We
proceed step by step:
Apply T1 . T1 (x, y) = (4x, x − y, 2x − y).

Apply T2 . Take T1 (x, y) = (x1 , x2 , x3 ) = (4x, x − y, 2x − y). Then

T2 (x1 , x2 , x3 ) = (−x1 , 0, x1 + 2x2 + x3 ) = (−4x, 0, 4x + 2(x − y) + (2x − y)).

Simplify the third component: 4x + 2(x − y) + (2x − y) = 4x + 2x − 2y + 2x − y = 8x − 3y .
So T2 (T1 (x, y)) = (−4x, 0, 8x − 3y).

Apply T3 . Now T2 (T1 (x, y)) = (y1 , y2 , y3 ) = (−4x, 0, 8x − 3y). Then

T3 (y1 , y2 , y3 ) = ( 2y1 + y2 − y3 , 2y2 + 3y3 ).

Substituting y1 = −4x, y2 = 0, y3 = 8x − 3y , we get

T3 (−4x, 0, 8x − 3y) = (2(−4x) + 0 − (8x − 3y), 2 ⋅ 0 + 3(8x − 3y)).

Simplify each component:

2(−4x) − (8x − 3y) = −8x − 8x + 3y = −16x + 3y, 3(8x − 3y) = 24x − 9y.

So T3 (T2 (T1 (x, y))) = (−16x + 3y, 24x − 9y).
Therefore,

(T3 ∘ T2 ∘ T1 )(x, y) = (−16x + 3y, 24x − 9y).