G5. (5 points) next, let’s turn our attention to the vector space p2, which is the set of polynomial with degree at most 2, together with polynomial addition and scalar multiplication. let p1 = 1−x+2x2 p2 = 3+x p3 =5−x+4x2 p4 =−2−2x+2x2. is the span{p1,p2,p3,p4}={k1p1 +k2p2 +k3p3 +k4p4 : k1,k2,k3,k4 ∈r}=p2? that is, can every polynomial of degree at most two be written as a linear com- bination of p1, p2, p3 and p4? explain.

No, those vectors do not span P₂

Step-by-step explanation:

At first, notice we have 4 vectors from a vector space of dimension 3 ({1,x,x²} is the standard base of P₂) so we are sure they are linearly dependent (because there are more than 3 vectors) i.e. it´s possible to write one of them in terms of other in the same set:

p4=p1-p2

Now, we conclude the span {p1,p2,p3,p4} = {p1,p2,p3}

If we want to know the subspace that this span gets, we work with the matrix of the vectors made as follows:

The row-reduction of this matrix show us the basis of the same row-subspace even in the case that some of them are linearly dependent:

This shows that the span {p1,p2,p3}={(1+0.5x²),(x-1.5x²)}

Looking in the previous equality, the span {p1,p2,p3} is generated by two linearly independent vectors, but P₂ needs 3 linearly independent vectors to be generated.

The answer is that {p1,p2,p3,p4}⊂P₂.

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !