This document discusses linear dependence and independence of vectors. It defines linear dependence as a set of vectors that can be written as a linear combination of scalars times the vectors equaling the zero vector. Linear independence is defined as scalars equaling zero if the linear combination equals the zero vector. Examples are provided to demonstrate determining if a set of vectors is linearly dependent or independent by setting up a system of equations and calculating the determinant.

1. VECTOR CALCULUS AND LINEAR
ALGEBRA (2110015)
Made By :-
DIVYANGSINH RAJ (150990119004)
NAVED FRUITWALA (150990119006)
UTKARSH GANDHI (150990119007)
Branch :- MECHANICAL ENGINEERING
Semester :- 2ᴺᴰ

2. TOPIC :-
LINEAR DEPENDENCE(LD)
&
LINEAR
INDEPENDENCE(LI)

3. Linear Dependence and
Independence :-
• A finite set of vector of a vector space is said to be
Linearly Dependent(LD) if there exists a set of
scalars k1 , k2 , ……, kn such that,
k1u1 + k2u2 +……+knun= ō
• A finite set of vector of a vector space is said to be
Linearly Independent(LI) if there exists scalars k1 ,
k2 , ……, kn such that,
k1u1 + k2u2 +……+knun= ō => k1= k2= ……= kn=0

4. • Properties For LI - LD
• Property 1: Any subset of a vector space is either
L.D. or L.I.
• Property 2: A set of containing only ō vector that is
{ō} is L.D.
• Property 3: A set is containing the single non zero
vector is L.I.
• Property 4: A set having one of the vector as zero
vector is L.D.

5. EXAMPLES
• 1.] Consider the set of vectors to check LI or LD
{(1,0,0),(0,1,0),(0,0,1)} in R³.
• Solution :-
Let k₁,k₂,k₃ belongs to R such that,
k₁(1,0,0)+k₂(0,1,0)+k₃(0,0,1)=(0,0,0)
(k₁,k₂,k₃)=(0,0,0)
k₁=0,k₂=0,k₃=0
Therefore, the set {i,j,k} is LI.

6. • 2.] Determine whether the vectors are LI in R³ (1-
2,1),(2,1,-1),(7,-4,1).
• Solution :-
Let k₁,k₂,k₃ belongs to R such that,
k₁(1,-2,1)+k₂(2,1,-1)+k₃(7,-4,1)=(0,0,0)
k₁+2k₂+7k₃=0
-2k₁+1k₂-4k₃=0
k₁-k₂+k₃=0

7. |A|=
|A|= 1[(1)(1)-(-4)(-1)]
-2[(-2)(1)-(-4)(1)]
+7[(-2)(-1)-(1)(1)]
|A|= -3-4+7
|A|=0
Since the determinant of the system is zero, the
system of these equations has a nontrivial
solution. That is at least one of k₁,k₂,k₃ is nonzero.
Thus the vectors are LD.
1 2 7
−2 1 −4
1 −1 1

9. Solution:-
i.) 2-x+4𝑥2
, 3+6x+2𝑥2
, 2+10x-4𝑥2
Let,
𝑘1 𝑝1+𝑘2 𝑝2+𝑘3 𝑝3=0
=>𝑘1(2-x+4𝑥2
)+𝑘2(3+6x+2𝑥2
)+𝑘3(2+10x−4𝑥2
)=0
2𝑘1+3𝑘2+2𝑘3=0
-𝑘1+6𝑘2+10𝑘3=0
4𝑘1+2𝑘2 −4 𝑘3=0

10. ~ 𝐴 =
~ A=
= 2(-24-20)-3(4-40)+2(-2-24)
= -88+108-52
2 3 2
−1 6 10
4 2 −4
2 3 2
−1 6 10
4 2 −4

11. 𝐴 = -32 ǂ 0
Therefore the system has unique solution
The given vectors are not L.D(i.e they are L.I).

12. Solution:-
ii.) A= 2+x+𝑥2
, x+2𝑥2
, 2+2x+3𝑥2
Let,
𝑘1 𝑝1+𝑘2 𝑝2+𝑘3 𝑝3=0
𝑘1(2+x+𝑥2
)+𝑘2(x+2𝑥2
)+𝑘3(2+2x+3𝑥2
)=0
2𝑘1+0𝑘2+2𝑘3=0
𝑘1+𝑘2+2𝑘3=0
𝑘1+2𝑘2 +3 𝑘3=0

13. ~ A=
~ 𝐴 =
= 2(3-4) + 2(2-1)
=2(-1) + 2(1)
=0
2 0 2
1 1 2
1 2 3
2 0 2
1 1 2
1 2 3

14. 𝐴 = 0
Therefore the system has infinitely many solution
The given vectors are L.D(i.e they are not L.I).