Comprehensive Solutions: Linear Algebra and Number Systems Review

This assignment solution covers a range of problems in linear algebra and number systems. It includes solving equations and inequalities, determining the truth value of set theory statements, expressing vectors in terms of other vectors within a triangle, and performing vector operations such as finding dot products, unit vectors, and angles between vectors. The solution demonstrates step-by-step calculations and explanations for each problem, providing a comprehensive guide for understanding the underlying concepts and methodologies. It also involves resolving vectors into parallel and perpendicular components. The detailed solutions are suitable for students seeking to enhance their problem-solving skills in linear algebra and number systems.

Linear Algebra and Number System
1. Solving
a) Solving for x in:
−x
3 + 2
x =2
The least common multiplier of 3, x is 3x: Multiplying the above equation by 3x gives
−x
3 ( 3 x ) + 2
x ( 3 x ) =2 ( 3 x )−x2+ 6=6 x ⇔ x2 +6 x−6
Using quadratic equation:x=−b ± √ b2−4 ac
2 a x=−6 ± √36+ 24
2 ¿−3 ± √15
∴ x=−3+ √15 x=−3− √15
b) Solving the inequality −x
3 + 2
x ≥2
i) Needs to consider two cases separately, i.e. x > 0 and x < 0, in order to partition
the set of all possible values of x into disjoint subsets, and considering each of
them separately.
ii) Solving
−x
3 + 2
x −2 ≥ 0−x2+ 6−6 x ≥ 0
Using symbols ∨ and ∧ to denote the disjunction and conjunction of predicates
−x2+ 6−6 x ≥ 0 ⇔( x +3+ √ 15 ) ( x+3− √ 15 ) ≥ 0 ⇔
[ ( x +3+ √ 15>0 ) ∧ ( x +3− √ 15>0 ) ] ∨ [ ( x +3+ √ 15<0 ) ∧ ( x+3− √ 15<0 ) ] ⇔
[ ( x ≥−3− √15 ) ∧ ( x ≥−3+ √15 ) ] ∨ [ ( x ≤−3− √ 15 ) ∧ ( x ≤−3+ √15 ) ] ⇔
( x ≤−3− √15 ) ∨ ( x ≤−3+ √15 )
Thus the solution is:

(−∞,−3− √15 ] ∪ ( 0 ,−3+ √15 ]
2. True or false with reason why.
Statement True/False Reason
a) { 1 } ∈ { 1,2 , x } True Each set contains an element the other
b) { 0,0,0,0 }={0 } False
c) π ∈ Z False π=3.142 which is not an element of the set of
real numbers
d) ∅ ∈ {0 } False Empty set is not 0
e) 2 ∈{2 } True 2 is an element of set 2
f) 2 ⊆{2 } False 2 is not a subset of set 2
g) [ 2,2 1
2 ] ⊆ [1,3 ] True The first set is a suset of the second set
h) [ 1,3 ] ⊆ (1,3) True {1,3} is a subet of {1,2,3}
i) (−1,1 )={−1,0 ,1 } True The first set is equal to the second set.
j) ( 0,4 ) ∩ {0,4 }=[0,4] True The intersection of the two sets are [0,4]
k) ( 0,4 ) ∪ {0,4 }=[0,4 ] False The union of the two sets are {0,1,2,3,4}
l) [ 0,2 ] ∪ [2,4]=[0,4 ] False The intersection of the two sets is [4]
3. Arbitrarily triangle
a) The diagram showing the information given is shown below (note: Figure not drawn to
scale):

b) Expressing ⃗ PRin terms of a∧b⃗ PR=⃗ PQ+⃗ QR=a+b
c)⃗ Ppin terms of a∧b⃗ Pp=⃗ PQ +⃗ Qp⃗ Qp= 1
2 b ∴⃗ Pp=a+ 1
2 b
d)⃗ Qqin terms of a∧b⃗ Qq=⃗ QR−1
2⃗ PR¿ b−1
2 ( a+b )¿ 1
2 ( −a+ b )⃗
Rrin terms of a∧b⃗ Rr=⃗ RQ +⃗ Qr ¿−⃗ QR− 1
2⃗ QP¿−b−1
2 a
e) Vector ⃗ Pp+⃗ Qq+⃗ Rr⃗ Pp+⃗ Qq+⃗ Rr=a+ 1
2 b+ 1
2 ( −a+ b ) −b− 1
2 a ¿ 0
4. Considering the given vectors c=i−2 j+ k∧d=2 j−2 k, and points
A= ( 1,4,1 ) , B= ( 2 ,−3,1 )∧C= ( 2 ,−1,5 )
a) Find the following:
i) Vector⃗
AB=B−A= (2 ,−3,1 )− ( 1,4,1 )= (1 ,−7,0 )⃗ BC =C−B= ( 2 ,−1,5 )− ( 2 ,−3,1 ) = ( 0,2,4 )⃗
AC=C− A= ( 2 ,−1,5 )− (1,4,1 ) ,=(1 ,−5,4 )
ii) ||⃗ AC||||⃗ AC||= √ 12 +52 +42= √42
iii) The exact distance from A to B

|⃗ AB|= √12 +72 +02=5 √2
iv) The unit vector in the direction of ⃗ BC⃗
BCu =⃗ BC
|⃗BC|= 1
√02+22 +42 ( 0,2,4 )¿ (0 , √5
5 , 2 √5
5 )
v) The point on the negative y-axis which is the same distance from the origin as A
is ( 1 ,−4,1 )
vi) Dot product
c . d=1 ( 0 )−2 ( 2 ) +1 (−2 )=−6⃗ BC .⃗ AC =0 ( 1 ) +2 ( −5 ) +4 ( 4 ) =6
vii) Vector c point is in a direction closer to −d
viii) exact angle between – dand the direction of the positive y-axis
θ=cos−1 ¿ ¿ ¿ 45 °
ix) the exact angle between ⃗ BC ∧⃗ ACθ=cos−1
(⃗
BC .⃗ AC
|⃗BC||⃗ AC| )¿ cos−1
( 6
2 √ 5 √ 42 )
¿ 78.0522°
b) Resolving vectors
i) From the diagram

ii) Vector c into two vectors
c∥= d . c
c . c c¿− 6
4 ( 1,−2,1 )¿−3
2 i+ 3 j− 3
2 kc⊥=d− d . c
c . c c ¿ ( 0 , 2 ,−2 ) − ( −3
2 ,3 ,−3
2 )
¿ 3
2 i− j− 1
2 k