Answer:
Yes , S is a basis for
.
Explanation:
Given
S=
.
We can make a matrix
Let A=
![\begin{bmatrix}-12&4&0&0\\5&0&0&1\\5&3&0&0\\(2)/(3)&0&-3&0\end{bmatrix}](https://img.qammunity.org/2020/formulas/mathematics/college/w208zga57lmc1scg4iukuwo6721wtpic97.png)
All rows and columns are linearly indepedent and S span
.Hence, S is a basis of
![P_3](https://img.qammunity.org/2020/formulas/mathematics/college/iamokljzz49c9x5sayh79ae0dmotcy0pdh.png)
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.