Question

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The set where p1(t) = t, p2(t) = t2, p3(t) = 2t + 3t2 C: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t + t2

Answer 1

Sets A and B are linearly independent because no vector within these sets can be written as a linear combination of the others. Set C is linearly dependent because one of the vectors can be written as a linear combination of the other two.

Step-by-step explanation:

To determine which of the sets of vectors is linearly independent, we can examine if any of the vectors can be written as a linear combination of the others. If not, the set is linearly independent.

Set A: {p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t}. These vectors are linearly independent because there is no way to write one of these vectors as a linear combination of the other two.

Set B: {p1(t) = t, p2(t) = t2, p3(t) = 2t + 3t2}. These vectors are linearly independent for the same reason as Set A.

Set C: {p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t + t2}. These vectors are linearly dependent because p3(t) can be written as a linear combination of p1(t) and p2(t) (p3(t) = 3*p1(t) + 1*p2(t)).

Therefore, Set C is not linearly independent while Sets A and B are linearly independent.

Answer 2

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

`p_(1)(t) = 1`,
`p_(2)(t)= t^(2)` and
`p_(3)(t) = 3 + 3\cdot t`:

`\alpha_(1)\cdot p_(1)(t) + \alpha_(2)\cdot p_(2)(t) + \alpha_(3)\cdot p_(3)(t) = 0`

`\alpha_(1)\cdot 1 + \alpha_(2)\cdot t^(2) + \alpha_(3)\cdot (3 +3\cdot t) = 0`

`(\alpha_(1)+3\cdot \alpha_(3))\cdot 1 + \alpha_(2)\cdot t^(2) + \alpha_(3)\cdot t = 0`

The following system of linear equations is obtained:

`\alpha_(1) + 3\cdot \alpha_(3) = 0`

`\alpha_(2) = 0`

`\alpha_(3) = 0`

Whose solution is
`\alpha_(1) = \alpha_(2) = \alpha_(3) = 0`, which means that the set of vectors is linearly independent.

`p_(1)(t) = t`,
`p_(2)(t) = t^(2)` and
`p_(3)(t) = 2\cdot t + 3\cdot t^(2)`

`\alpha_(1)\cdot p_(1)(t) + \alpha_(2)\cdot p_(2)(t) + \alpha_(3)\cdot p_(3)(t) = 0`

`\alpha_(1)\cdot t + \alpha_(2)\cdot t^(2) + \alpha_(3)\cdot (2\cdot t + 3\cdot t^(2))=0`

`(\alpha_(1)+2\cdot \alpha_(3))\cdot t + (\alpha_(2)+3\cdot \alpha_(3))\cdot t^(2) = 0`

The following system of linear equations is obtained:

`\alpha_(1)+2\cdot \alpha_(3) = 0`

`\alpha_(2)+3\cdot \alpha_(3) = 0`

Since the number of variables is greater than the number of equations, let suppose that
`\alpha_(3) = k`, where
`k\in\mathbb{R}`. Then, the following relationships are consequently found:

`\alpha_(1) = -2\cdot \alpha_(3)`

`\alpha_(1) = -2\cdot k`

`\alpha_(2)= -2\cdot \alpha_(3)`

`\alpha_(2) = -3\cdot k`

It is evident that
`\alpha_(1)` and
`\alpha_(2)` are multiples of
`\alpha_(3)`, which means that the set of vector are linearly dependent.

`p_(1)(t) = 1`,
`p_(2)(t)=t^(2)` and
`p_(3)(t) = 3+3\cdot t +t^(2)`

`\alpha_(1)\cdot p_(1)(t) + \alpha_(2)\cdot p_(2)(t) + \alpha_(3)\cdot p_(3)(t) = 0`

`\alpha_(1)\cdot 1 + \alpha_(2)\cdot t^(2)+ \alpha_(3)\cdot (3+3\cdot t+t^(2)) = 0`

`(\alpha_(1)+3\cdot \alpha_(3))\cdot 1+(\alpha_(2)+\alpha_(3))\cdot t^(2)+3\cdot \alpha_(3)\cdot t = 0`

The following system of linear equations is obtained:

`\alpha_(1)+3\cdot \alpha_(3) = 0`

`\alpha_(2) + \alpha_(3) = 0`

`3\cdot \alpha_(3) = 0`

Whose solution is
`\alpha_(1) = \alpha_(2) = \alpha_(3) = 0`, which means that the set of vectors is linearly independent.

The set of vectors A and C are linearly independent.