Tutorial Letter 203/2/2017: Linear Algebra
Tutorial Letter 203/2/2017: Linear Algebra
Tutorial Letter 203/2/2017: Linear Algebra
LINEAR ALGEBRA
MAT1503
Semester 2
BARCODE
university
Define tomorrow. of south africa
SEMESTER 2
ASSIGNMENT 03
Fixed closing date: 11 September 2017
Unique Assignment: 789872
QUESTION 1
x−6 y−4
(a) Note that t = = =z−2
5 3
∴ 3x − 18 = 5y − 20
⇒ 3x − 5y = −2 .......................................................... (1)
Also x − 6 = 5z − 10
∴ x − 5z = −4 .......................................................... (2)
Next y − 4 = 3z − 6
∴ y − 3z = −2 .......................................................... (3)
4x − 5y − 5z = −6..........................................................(*)
x + y − 8z = −6..........................................................(**)
So the system is
4x − 5y − 5z = −6
x + y − 8z = −6 [check]
2
MAT1503/203/2/2017
x + 4y + z = 0
1
⇒ x + 4 − 10 =0
5 1
∴ x = 10 = 2
x4 =0
x3 + 2x4 = 2 ⇒ x3 = 2
x2 + 3x3 + 7x4 =5
⇒ x2 + 3 · 2 + 0 =5
∴ x2 = 5 − 6 =1
x1 + 2x2 − x3 + x4 = −3
⇒ x1 + 2 (1) − 2 + 0 =3
∴ x1 =3
11 19 −2
1 3 3
| 3
−11 1
→ 0 1 2 | 2
R
3 1
− R1 → R1
1
0 0 0 | 81 R
2 2
− R2 → R2
QUESTION 2
(a) 2x + y = c
3y + z = c
3
x + 4z = c
The augemented matric is
2 1 0 | c
0 3 1 | c
1 0 0 | c
1 0 4 | c
→ 0 3 1 | c R1 ↔ R3 → R1
2 1 0 | c R3 ↔ R1 → R3
1 0 4 | c
0 3 1 | c
0 1 −8 | c R3 − 2R1 → R3
1 0 4 | c
→ 0 1 −8 | −c R2 ↔ R3 → R2
0 3 1 | c R3 ↔ R2 → R3
1 0 4 | c
→ 0 1 −8 | −c
0 0 25 | 4c R3 − 3R2 → R3
∴ 25z = 4c
25
∴ c= z
4
to make c a positive integer z must be a positive multiple of 4. The smallest value is z = 4
This makes 25 the smallest value of c. Row 2 gives y − 8z = −c whence c = 8z − y.
To make c a positive integer here
y
8z > y. So z >
8
Thus y must be a multiple of 8. That is the only control of y on c.
Similarly x + 4z = c only dictates that x > −4z and that x must be an integer.
The smallest value of c is 25.
(b) The coefficient determinant of the sustem is
1 2 3
1 k 4
1 2 (k + 2)
k 4 − 2 1 4
1 k
= 1 + 3
2 k+2 1 k+2 1 2
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MAT1503/203/2/2017
= k (k + 2) − 8 − 2 (k + 2 − 4) + 3 (2 − k)
= k 2 + 2k − 8 − 2k + 4 + 6 − 3k
= k 2 − 3k + 2
k 2 − 2k + 2 =
6 0
⇒ (k − 1) (k − 2) =6 0
⇒ k 6= 1 and k 6= 2
So solutions exist.
(iii) When k = 2, thee are infinitely many solution.
a b c
→
−
(c) (i) A e = d →
−
Ae2= e , Ae3= f
→
−
g h i
by matrix multiplication.
5
(ii) Be1 = →
−
v 1, B→
−
e 2, = →
−
v 2, B→
−
e 3, = →
−
v 3, B→
−
e i , 1 ≤ i ≤ 3. There was a typo here.
QUESTION 3
u = (0, 3, 0) , v = (1, 0, 4) and w = (2, 4, 0)
(a)
2v − 2u = 2 (1, 0, 4) − 2 (0, 3, 0)
= (2, 0, 8) − (0, 6, 0)
= (2, −6, 8)
(b)
||2u + 3v − w|| = ||(0, 6, 0) + (3, 0, 12) − (2, 4, 0)||
= ||(1, 2, 12)||
√
= 12 + 22 + 122
√
= 149
(c) This is
||−3u − (w − 4v)|| = ||−3u − w + 4v||
= ||(v1 − 3, 0) − (2, 4, 0) + (4, 0, 16)||
q
= ||(2, −7, 16)|| = 22 + (−7)2 + 162
√ √
= 4 + 49 + 256 == 53 + 256
√
= 309 units.
(d)
(v · w) v
Projv w =
||v||2
2
Projv w = (1, 0, 4)
17
(e) Area = ||v × w||
i j k
v × w = j 0 4
2 4 0
= (−16, 8, 4)
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MAT1503/203/2/2017
q
∴ ||v × w|| = (−16)2 + 82 + 42
√
= 256 + 64 + 16
√
= 336 square units
(−16, 8, 4) · (x, y − 3, z) = 0
⇒ −16x + 8y − 24 + 4z = 0
⇒ −16x + 8y + 4z − 24 = 0
= 7 [cos 0 + i sin 0]
=7
√
|z1 |
(iv) zz21 = = √7 =1
|z2 | 7
7
(b) Let z be a 4th root of 8 so that z 4 = 8. Suppose z = r [cos θ + i sin θ] is the polar form of z.
Now 8 = 8 + 0i.
Let
Thus
1
r4 = 8 and r = 8 4
cos 4θ = and sin 4θ = sin π
So 4θ = π + 2kπ, k = 0, 1, 2, 3
π + 2kπ
Thus θ = .
4
1 π + 2kπ π + 2k
This gives zk = z (k) = 8 4 cos + i sin
4 4