Mastering Linear Algebra and Number System: Equations, Sets, Vectors, and More

​In this report we will discuss about linear algebra & number system and below are the summaries point:- Solving equations involving x by using the quadratic equation and considering different cases for inequalities. Assessing the truth or falsehood of various statements about sets and their elements. Analyzing vectors, distances, angles, and resolving vectors into components in a given scenario.

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