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07.05.2021 •

Mathematics

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Ответ:

nekobaby75

26.09.2020

Mathematics

Vectors $p_{1} = 2+5\cdot x + 3\cdot x^{2}$ , $p_{2} = 4+11\cdot x +6\cdot x^{2}$ and $p_{3} = 1+2\cdot x +x^{2}$ 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:

$\Sigma\limits_{i=1}^{n} \alpha_{i}\cdot p_{i}= 0$ , where $\alpha_{i} = 0$ . (Eq. 1)

Where:

$p_{i}$ - i-th Polynomial of the set of vectors, dimensionless.

$\alpha_{i}$ - i-th coefficient associated with the i-ith polynomial of the set of vectors, dimensionless.

Let $p_{1} = 2+5\cdot x + 3\cdot x^{2}$ , $p_{2} = 4+11\cdot x +6\cdot x^{2}$ and $p_{3} = 1+2\cdot x +x^{2}$ elements of the set of vectors, whose linear combination is:

$\alpha_{1}\cdot p_{1}+\alpha_{2}\cdot p_{2}+\alpha_{3}\cdot p_{3}= 0$

$\alpha_{1}\cdot (2+5\cdot x+3\cdot x^{2})+\alpha_{2}\cdot (4+11\cdot x +6\cdot x^{2})+\alpha_{3}\cdot (1+2\cdot x+x^{2}) = 0$

$(2\cdot \alpha_{1}+4\cdot \alpha_{2}+\alpha_{3})+(5\cdot \alpha_{1}+11\cdot \alpha_{2}+2\cdot \alpha_{3})\cdot x+(3\cdot \alpha_{1}+6\cdot \alpha_{2}+\alpha_{3})\cdot x^{2} = 0$

Values of coefficient are contained in the following homogeneous system of linear equations:

$2\cdot \alpha_{1}+4\cdot \alpha_{2}+\alpha_{3} = 0$ (Eq. 2)

$5\cdot \alpha_{1}+11\cdot \alpha_{2}+2\cdot \alpha_{3} = 0$ (Eq. 3)

$3\cdot \alpha_{1}+6\cdot \alpha_{2}+\alpha_{3} = 0$ (Eq. 4)

The solution of this system is:

$\alpha_{1} = 0$ , $\alpha_{2} = 0$ , $\alpha_{3} = 0$

Which means that given set of vectors are linearly independent.

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-8x +40 > -16 Please give steps! And please do be quick, this is timed.