Question

If vectors (1,-2, K) in R3 is a linear combination of the vector (2,-1,-5) and (3,0,-2) then the value of K is
(a) K= - 5
(b) K= - 1
(c) K= - 2
(d) K= - 8​

Answer 1

Answer:

k= -8

Step-by-step explanation:

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Answer 2

Answer :

d) k = -8

Solution :

• Given : Vector (1 , -2 , k) ∈ R³ is a linear combination of the vector (2 , -1 , -5) and (3 , 0 , -2)
• To find : k = ?

It is given that , the vector (1 , -2 , k) ∈ R³ is a linear combination of the vector (2 , -1 , -5) and (3 , 0 , -2) .

Thus , let (1 , -2 , k) = a(2 , -1 , -5) + b(3 , 0 , -2)

→ (1 , -2 , k) = (2a , -a , -5a) + (3b , 0 , -2b)

→ (1 , -2 , k) = (2a + 3b , -a , -5a - 2b)

→ 1 = 2a + 3b ........(1)

-2 = -a .........(2)

k = -5a - 2b ........(3)

From eq-(2) , we have ,

a = 2

Now ,

Putting a = 2 in eq-(1) , we get

→ 1 = 2×2 + 3b

→ 1 = 4 + 3b

→ 3b = 1 - 4

→ 3b = -3

→ b = -3/3

→ b = -1

Now ,

Putting a = 2 and b = -1 in eq-(3) , we get

→ k = -5×2 - 2×(-1)

→ k = -10 + 2

→ k = -8

Hence , k = -8 .

Some important information :

Vector space :

(V , +) be an algebraic structure and (F , + , •) be a field , then V is called a vector space over the field F if the following conditions hold :

1. (V , +) is an abelian group .
2. ku ∈ V ∀ u ∈ V and k ∈ F
3. k(u + v) = ku + kv ∀ u , v ∈ V and k ∈ F .
4. (a + b)u = au + bu ∀ u ∈ V and a , b ∈ F .
5. (ab)u = a(bu) ∀ u ∈ V and a , b ∈ F .
6. 1u = u ∀ u ∈ V where 1 ∈ F is the unity .

♦ Elements of V are called vectors and the lements of F are called scalars .

♦ If V is a vector space over the field F then it is denoted by V(F) .

Linear combination :

A vector v in a vector space V is called a linear combination of the vectors v₁ , v₂ , v₃ , . . . , vₖ if v can be expressed in the form :

v = c₁v₁ + c₂v₂ + c₃v₃ + . . . + cₖvₖ

where c₁ , c₂ , c₃ , . . . , cₖ are scalars and are called weights of linear combination .