If B is the standard basis of the space set of prime numbers P 3ℙ3 of polynomials, then let Bequals={1,t, t2,t3}. Use coordinate vectors to test the linear independence of the set of polynomials below. Explain your work. 1 plus 6 t squared minus t cubed1+6t2−t3, t plus 2 t cubedt+2t3, 1 plus t plus 6 t squared

The set of polynomial is linearly independent

Explanations:

For clarity and easiness of expression, the complete solution is handwritten in the file attached. However, the following are the steps taken:

1) The coordinate vectors of the polynomials are formed as:

(1,0,6,-1), (0,1,02) and (1,1,6,0)

2) A matrix is formed using the coordinates of the polynomials as the columns of the matrix

3) The polynomial formed is row reduced to form an echelon matrix.

4) As seen from the reduced echelon form of the matrix, the rank of the matrix = 3

5) since the rank = number of polynomials = 3, the set of polynomials is linearly independent.

1; 35 is the ratio

step-by-step explanation:

a 270° rotation about point p says on e d g e n u it y (had to space it out because it says it's a misspelling so i couldn't submit it