Answer: since [B]^-1[B] = I.
Explanation:
To find the matrix for T relative to B, we need to find the coordinates of the vectors T(b), T(b^2), and T(b^3) with respect to the basis B.
We have:
T(b) = 6T(b^2) + 1T(b^3) = b, -5b^2
T(b^2) = 1T(b^2) + 0T(b^3) = 7b, -3b^2
T(b^3) = 0T(b^2) - 2T(b^3) = 0, 4b^2
To find the matrix [T], we write the coordinates of T(b), T(b^2), and T(b^3) as columns:
[T] = [b, 7b, 0; -5b^2, -3b^2, 4b^2]
To check this matrix, we can apply it to the basis vectors and see if we get the same coordinates as the vectors T(b), T(b^2), and T(b^3):
[T][b] = [b, 7b, 0][1; 0; 0] = [b; -5b^2]
[T][b^2] = [b, 7b, 0][0; 1; 0] = [7b; -3b^2]
[T][b^3] = [b, 7b, 0][0; 0; 1] = [0; 4b^2]
These are the same as the coordinates we found for T(b), T(b^2), and T(b^3), so our matrix [T] is correct.
To find ITIB, we first need to find the inverse of the matrix [B] whose columns are the basis vectors b, b^2, and b^3. We can do this by row reducing the augmented matrix [B | I]:
[1 0 0 | 1 0 0]
[0 1 0 | 0 1 0]
[0 0 1 | 0 0 1]
So [B] is already in reduced row echelon form, and its inverse is just I:
[B]^-1 = [1 0 0; 0 1 0; 0 0 1]
Therefore,
ITIB = [B]^-1[T][B] = [T]
since [B]^-1[B] = I.