MAT2611 Assignment 4 2026 - Distinction | Due 27 July 2026

,(b) < w, v + 2u >
Solution:

First compute v + 2u:

v + 2u = (1, 2) + 2(1, 3) = (1, 2) + (2, 6) = (3, 4)

Now compute < w, v + 2u >:

< w, v + 2u >= wT A(v + 2u)

Given w = (−3, 1) and v + 2u = (3, 4):

0 1 3
< w, v + 2u >= [−3 1] [ ][ ]
1 2 4
​ ​ ​




First compute A(v + 2u):

0 1 3 0(3) + 1(4) 4
A(v + 2u) = [ ][ ] = [ ]=[ ]
1 2 4 1(3) + 2(4) 11
​ ​ ​ ​




Then:

4
< w, v + 2u >= [−3 1] [ ] = 3(4) + 1(11) = 12 + 11 = 1
11
​




Answer: < w, v + 2u >= 1



(c) ∥w∥2
Solution:

∥w∥2 =< w, w >= wT Aw (Williams, 2019: 352)

0 1 3
< w, w >= [ 3 1] [ ][ ]
1 2 1
​ ​ ​




First compute Aw :

0 1 −3 0(−3) + 1(1) 1
Aw = [ ][ ] = [ ]=[ ]
1 2 1 1(−3) + 2(1) −1
​ ​ ​ ​ ​

,