Respuesta :
Answer with explanation:
Let, A=[1,1,2]
B=[1,2,1]
C=[2,1,5]
⇒Now, Writing vector , A in terms of Linear combination of C and B
A=x B +y C
⇒[1,1,2]=x× [1,2,1] + y×[2,1,5]
1.→1 = x +2 y
2.→ 1=2 x +y
3.→ 2= x+ 5 y
Equation 3 - Equation 1
→3 y=1
[tex]y=\frac{1}{3}[/tex]
[tex]1=x+\frac{2}{3}\\\\x=1 -\frac{2}{3}\\\\x=\frac{1}{3}[/tex]
So, Vector A , can be written as Linear Combination of B and C.
⇒Now, Writing vector , B in terms of Linear combination of A and C
Now, let, B = p A+q C
→[1,2,1]=p× [1,1,2] +q ×[2,1,5]
4.→1= p +2 q
5.→2=p +q
6.→1=2 p +5 q
Equation 5 - Equation 4
-q =1
q= -1
→2= p -1
→p=2+1
→p=3
So, Vector B , can be written as Linear Combination of A and C.
⇒Now, Writing vector , C in terms of Linear combination of A and B
C=m A + n B
[2,1,5] = m×[1,1,2] + n× [1,2,1]
7.→2= m+n
8.→1=m +2 n
9.→5=2 m + n
Equation 8 - Equation 7
n= -1
→m+ (-1)=2
→m=2+1
→m=3
So, Vector C , can be written as Linear Combination of A and B.
So, All the three vectors , A=[1,1,2],B=[1,2,1],C=[2,1,5] can be written as Linear combination of each other.
⇒≡But , the two vectors, (0.4,3.7,-1.5) (0.2,0),can't be written as Linear combination of each other as first vector is of order, 1×3, and second is of order, 1×2.
None of the selections is correct.
How to apply linear combinations and linear independence to determine the existence of a relationship with a given vector
In this case, we must check the existence of a set of real coefficients such that the following two linear combinations exist:
[tex]\alpha_{1}\cdot (1, 1, 2)+\alpha_{2}\cdot (1, 2,1)+\alpha_{3}\cdot (2, 1, 5) = (0.4, 3.7, -1.5)[/tex] (1)
[tex]\alpha_{4}\cdot (1,1,2)+\alpha_{5}\cdot (1,2,1) + \alpha_{6}\cdot (2,1,5) = (0, 2, 0)[/tex] (2)
Now we proceed to solve each linear combination:
First system
[tex]\alpha_{1}+\alpha_{2}+2\cdot \alpha_{3} = 0.4[/tex]
[tex]\alpha_{1}+2\cdot \alpha_{2}+\alpha_{3} = 3.7[/tex]
[tex]2\cdot \alpha_{1}+\alpha_{2}+5\cdot \alpha_{3} = -1.5[/tex]
The system has no solution, since the third equation is a linear combination of the first and second ones.
Second system
[tex]\alpha_{4}+\alpha_{5}+2\cdot \alpha_{6} = 0[/tex]
[tex]\alpha_{4}+2\cdot \alpha_{5}+\alpha_{6} = 2[/tex]
[tex]2\cdot \alpha_{4}+\alpha_{5}+5\cdot \alpha_{6} = 0[/tex]
The system has no solution, since the third equation is a linear combination of the first and second ones.
None of the selections is correct. [tex]\blacksquare[/tex]
To learn more on linear combinations, we kindly invite to check this verified question: https://brainly.com/question/9672435
