MAT2611
ASSIGNMENT 5
2025
, PROBLEM 1
B = {p1 , p2 } and B′ = {q1 , q 2 }
p1 = 1 + 2x and p2 = 3 − x
q1 = 2 − 2x and q 2 = 4 + 3x
a).
q1 = a1 p1 + a2 p2
2 − 2x = a1 (1 + 2x) + a2 (3 − x)
2 − 2x = a1 + 2a1 x + 3a2 − a2 x
2 − 2x = (a1 + 3a2 ) + (2a1 − a2 )x
Therefore
a1 + 3a2 = 2 1
2a1 − a2 = −2 2
From 1 a1 = 2 − 3a2 and substitute into 2
2a1 − a2 = −2 2
2(2 − 3a2 ) − a2 = −2
4 − 6a2 − a2 = −2
−7a2 = −2 − 4
−7a2 = −6
6
a2 =
7
ASSIGNMENT 5
2025
, PROBLEM 1
B = {p1 , p2 } and B′ = {q1 , q 2 }
p1 = 1 + 2x and p2 = 3 − x
q1 = 2 − 2x and q 2 = 4 + 3x
a).
q1 = a1 p1 + a2 p2
2 − 2x = a1 (1 + 2x) + a2 (3 − x)
2 − 2x = a1 + 2a1 x + 3a2 − a2 x
2 − 2x = (a1 + 3a2 ) + (2a1 − a2 )x
Therefore
a1 + 3a2 = 2 1
2a1 − a2 = −2 2
From 1 a1 = 2 − 3a2 and substitute into 2
2a1 − a2 = −2 2
2(2 − 3a2 ) − a2 = −2
4 − 6a2 − a2 = −2
−7a2 = −2 − 4
−7a2 = −6
6
a2 =
7
