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.