1a. The linear transformation with matrix B given by
[5 6]
[2 4]
maps the unit square onto the parallelogram in question. Therefore, the area of the parallelogram is the area of the square (1) times the scaling factor for this transformation, which is det(A) = 20 - 12 = 8. Thus the parallelogram has area 8.
1b. The desired matrix is just the inverse matrix of B, which can be computed using the determinant above and the adjugate matrix as
[ 1/2 -3/4]
[-1/4 5/8]
2a. This set of polynomials is a subspace. To see this, check the three properties of subspaces:
i) The zero vector (p(t) = 0) is in the set, since it is equal to 0*t^2.
ii) If p(t) = at^2 and q(t) = bt^2, then
(p+q)(t) = p(t) + q(t) = at^2 + bt^2 = (a+b)t^2 is an element of the set.
iii) If p(t) = at^2 and c is a real number, then
(cp)(t) = c*p(t) = cat^2 is an element of the set.
2b. This set of polynomials is not a subspace. One reason for this is that the zero polynomial, p(t) = 0, does not satisfy p(0) = 1, so it is not contained in the set.