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