MAT2611 LINEAR ALGEBRA REVISION
ASSIGNMENTS AND EXAMS
SEMESTER 1 & 2 2022
,2 2022 Semester 1: Assignment 01
ASSIGNMENT 01
Solution
Total Marks: 100
UNIQUE ASSIGNMENT NUMBER: 897513
Question 1: 10 Marks
Show that the set X with the given operations fails to be a vector space by identifying all axioms that
hold and fail to hold:
The set X = R 3 with vector addition ⊕ defined by
(a, b, c) ⊕ (x, y, z) = (1, y, c + z)
and scalar multiplication defined by k (a, b, c) = (ka, kb, kc).
In the following let u = (u 1, u2 , u3) ∈ R 3 , v = (v 1, v2, v3) ∈ R 3, w = (w 1, w2, w3 ) ∈ R 3 be arbitrary
elements of R3 and k, m ∈ R . We have
1. u ⊕ v = (1, v2, u3 + v 3 ) ∈ R3 holds. X
2. u ⊕ v = (1, v2, u3 + v 3 ).
v ⊕ u = (1, u2, v3 + u 3 ).
X
Choosing u = (1, 0, 0) and v = (0, 1, 0) we see that u ⊕ v = v ⊕ u does not hold in general.
3. u ⊕ (v ⊕ w) = u ⊕ (1, w 2 , v3 + w 3 ) = (1, w 2 , u3 + v 3 + w 3 ).
(u ⊕ v) ⊕ w = (1, v2, u3 + v 3 ) ⊕ w = (1, w2, u3 + v 3 + w 3).
Thus u ⊕ (v ⊕ w) = (u ⊕ v) ⊕ w holds.X
4. Suppose the zero vector 0 = (a, b, c) ∈3 Rexists. Then
u ⊕ 0 = (u1 , u2, u3) ⊕ (a, b, c) = (1, b, 3u+ c) = (u 1 , u2, u3) ⇔ u 1 = 1, u 2 = b, c = 0.
The zero vector property obviously does not hold, in particular the equation cannot be satisfied
for u = (0, 0, 0). X
5. Since the zero vector does not exist, the negative is undefined.
Thus the existence of negatives
does not hold. X
6. k u = (ku 1, ku2, ku3 ) ∈ R3 holds. X
7. k (u + v) = k (1, v 2, u3 + v 3 ) = (k, kv 2, ku3 + kv 3 ).
k u ⊕ k v = (1, kv 2, ku3 + kv 3).
Choosing k = 2, for example, we find k (u + v) = k u + k v does not hold in general. X
8. (k + m) u = ((k + m)u 1, (k + m)u 2, (k + m)u 3).
k u ⊕ m u = (1, mu 2, ku1 + mu 2 ).
Choosing k = m = 0, for example, we find (k + m) u = k u ⊕ m u does not hold in
general.X
4
, MAT2611/106
9. k(m u) = k (mu 1 , mu2 , mu3 ) = (kmu 1 , kmu2 , kmu3 ).
(km) u = ((km)u 1, (km)u 2 , (km)u 3).
Thus k(m u) = (km) u holds. X
10. 1 u = (1 · u 1 , 1 · u2, 1 · u3) = u holds. X
Question 2: 16 Marks
Consider the vector space M
22 of all 2 × 2 matrices.
(2.1) Show that B = {A 1 , A2, A3 , A4} is a basis for M 22 where (6)
A 1 = 3 6 ; A 2 = 0 −1 ; A 3 = 0 −8
; and A4 =
1 0 .
3 −6 −1 0 −12 −4 −1 2
Let c1 , c2 , c3, c4 ∈ R determined by
First we show linear independence of the vectors.
c1A 1 + c 2A 2 + c 3 A 3 + c 4 A 4 = 0 X
i.e.
3c1 6c1 0 −c 2 0 −8c 3 c 0
+ −c + + −c4
3c1 −6c 1 2 0 −12c 3 −4c 3 4 2c4
3c1 + c 4 6c1 − c 2 − 8c3
=
3c1 − c 2 − 12c3 − c 4 −6c 1 − 4c3 + 2c4
0 0
=
0 0
Thus we obtain the four equations
3c1 + c 4 =0
6c1 − c 2 − 8c3 =0
X
3c1 − c 2 − 12c3 − c 4 =0
−6c 1 − 4c3 + 2c4 =0
or in matrix form
3 0 0 1 c1 0
6 −1 −8 0 c2 0
= .
3 −1 −12 −1 c3 0
−6 0 −4 2 c4 0
5
, Row reduction of the augmented matrix yields
3 0 0 1 0 3 0 0 1 0
−2R1 0 0 → 0 −1 −8 −2 0
6 −1 −8
−R1 3 −1 −12 −1 0 −R2 0 −1 −12 −2 0
+2R1 −6 0 −4 2 0 0 0 −4 4 0
3 0 0 1 0
0 −1 −8 −2 0
→
0 0 −4 0 0
−R3 0 0 −4 4 0
3 0 0 1 0
0 −1 −8 −2 0
→
0 0 −4 0 0
0 0 0 4 0
Here −2R1 means subtract twice the first row from the second row (appearing right
of −2R1). Thus c4 = 0, c3 = 0, c2 = −8c 3 − 2c 4 = 0 and c 1 = −c 4/3 = 0. Since
c1 = c 2 = c 3 = c 4 = 0 X is the only solution, the matrices A1, A 2 , A 3 and A4 are linearly
independent.
Next we show that any element of M22 can be expressed as a linear combination of 1A,
A 2, A 3 and A4. Let a, b, c, d ∈ R and
a b 3a1 + a 4 6a1 − a 2 − 8a3
= a 1 A 1 + a 2A 2 + a 3A 3 + a 4A 4 =
c d 3a1 − a 2 − 12a3 − a 4 −6a 1 − 4a3 + 2a4
Thus we obtain the four equations
3a1 + a 4 =a
6a1 − a 2 − 8a 3 =b
X
3a1 − a 2 − 12a3 − c 4 =c
−6a 1 − 4a3 + 2a4 =d
or in matrix form
3 0 0 1 a1 a
6 −1 −8 0 a2 b
= .
3 −1 −12 −1 a3 c
−6 0 −4 2 a4 d
Row reduction of the augmented matrix yields
3 0 0 1 a 3 0 0 1 a
b −1 −8 −2 b − 2a
−2R1
6 −1 −8 0 → 0
−R1 3 −1 −12 −1 c −R2 0 −1 −12 −2 c − a
+2R1 −6 0 −4 2 d 0 0 −4 4 d + 2a
3 0 0 1 a
0 −1 −8 −2 b − 2a
→
0 0 −4 0 a − b + c
−R3 0 0 −4 4 d + 2a
6
