Respuesta :

Answer:

Yes , S is a basis for [tex]P_3[/tex].

Step-by-step explanation:

Given

S=[tex]\left\{4t-12,5+t^3,5+3t,-3t^2+\frac{2}{3}\righ\}[/tex].

We can make a matrix

Let A=[tex]\begin{bmatrix}-12&4&0&0\\5&0&0&1\\5&3&0&0\\\frac{2}{3}&0&-3&0\end{bmatrix}[/tex]

All rows and columns are linearly indepedent and S span [tex]P_3[/tex].Hence, S is a basis of [tex]P_3[/tex]

Linearly independent means any row or any column should not combination of any rows or columns.

Because  a subset of V with n elements is a basis if and only if it is linearly independent.

Basis:- If B is a subset  of a vector space V over a field F .B is basis of V if satisfied the following conditions:

1.The elements of B are linearly independent.

2.Every element of vector V spanned by the elements of B.