Answer:
The set is a basis for P2
Step-by-step explanation:
Remember, a set B is a basis for a vector space V if B is a linear independent set and gen(B)=V. Also, remember that the polynomials [tex]x^2+2x-3, 4x^2+ x-1, 5x^2+4x-4[/tex] are linearly independent if the unique scalars [tex]a,b,c[/tex] such that
[tex]a(x^2+2x-3)+b(4x^2+ x-1)+c(5x^2+4x-4)=0[/tex]
are [tex]a=b=c=0[/tex]
Expanding the above equation we have that
[tex]ax^2+2ax-3a+4bx^2+ bx-b+5cx^2+4cx-4c=0\\(a+4b+5c)x^2+(2a+b+4c)x+(-3a-b-4c)=0[/tex]
Then we need values of a, b and c such that
[tex]a+4b+5c=0\\2a+b+4c=0\\-3a-b-4c=0[/tex]
So, we have a homogeneous system with augmented matrix
[tex]A= \left[\begin{array}{ccc}a&4b&5c\\2a&b&4c\\-3a&-b&-4c\end{array}\right][/tex]
Now, we apply row operations for solve the system
1. To the second row of A we subtract the first row twice. And to the third row we add the first row three times. We get the matrix:
[tex]\left[\begin{array}{ccc}a&4b&5c\\0&-7b&-6c\\0&11b&11c\end{array}\right][/tex]
2. We multiply the third row of the previous matrix by 1/11
[tex]\left[\begin{array}{ccc}a&4b&5c\\0&-7b&-6c\\0&b&c\end{array}\right][/tex]
3. To the third row of the previous matrix we add 7 times the second row.[tex]\left[\begin{array}{ccc}a&4b&5c\\0&-7b&-6c\\0&0&42c\end{array}\right][/tex]
Since the rank of the matrix is 3 then the system has unique solution. But the system is homogeneous, therefore the solution is
[tex](a,b,c)=(0,0,0)[/tex]
This show that the polynomials are linear independent and therefore the set is a basis for P2
