Answer:
Vectors [tex]p_{1} = 2+5\cdot x + 3\cdot x^{2}[/tex], [tex]p_{2} = 4+11\cdot x +6\cdot x^{2}[/tex] and [tex]p_{3} = 1+2\cdot x +x^{2}[/tex] are linearly independent.
Step-by-step explanation:
From Linear Algebra, we must remember that a set of vectors is linearly independent if and only if coefficients of its linear combinations are all zeroes. That is:
[tex]\Sigma\limits_{i=1}^{n} \alpha_{i}\cdot p_{i}= 0[/tex], where [tex]\alpha_{i} = 0[/tex]. (Eq. 1)
Where:
[tex]p_{i}[/tex] - i-th Polynomial of the set of vectors, dimensionless.
[tex]\alpha_{i}[/tex] - i-th coefficient associated with the i-ith polynomial of the set of vectors, dimensionless.
Let [tex]p_{1} = 2+5\cdot x + 3\cdot x^{2}[/tex], [tex]p_{2} = 4+11\cdot x +6\cdot x^{2}[/tex] and [tex]p_{3} = 1+2\cdot x +x^{2}[/tex] elements of the set of vectors, whose linear combination is:
[tex]\alpha_{1}\cdot p_{1}+\alpha_{2}\cdot p_{2}+\alpha_{3}\cdot p_{3}= 0[/tex]
[tex]\alpha_{1}\cdot (2+5\cdot x+3\cdot x^{2})+\alpha_{2}\cdot (4+11\cdot x +6\cdot x^{2})+\alpha_{3}\cdot (1+2\cdot x+x^{2}) = 0[/tex]
[tex](2\cdot \alpha_{1}+4\cdot \alpha_{2}+\alpha_{3})+(5\cdot \alpha_{1}+11\cdot \alpha_{2}+2\cdot \alpha_{3})\cdot x+(3\cdot \alpha_{1}+6\cdot \alpha_{2}+\alpha_{3})\cdot x^{2} = 0[/tex]
Values of coefficient are contained in the following homogeneous system of linear equations:
[tex]2\cdot \alpha_{1}+4\cdot \alpha_{2}+\alpha_{3} = 0[/tex] (Eq. 2)
[tex]5\cdot \alpha_{1}+11\cdot \alpha_{2}+2\cdot \alpha_{3} = 0[/tex] (Eq. 3)
[tex]3\cdot \alpha_{1}+6\cdot \alpha_{2}+\alpha_{3} = 0[/tex] (Eq. 4)
The solution of this system is:
[tex]\alpha_{1} = 0[/tex], [tex]\alpha_{2} = 0[/tex], [tex]\alpha_{3} = 0[/tex]
Which means that given set of vectors are linearly independent.
