(a) T: P2 → R3, T(a + bx + cx2) = a + c, c, b – c) — each case show that MDB invertible and use the fact MBD determine — standard

= [CD(1, 0, 0, 0) CD(1, 1, 0, 0) CD(1, 1, 1, 0) CD(0, 0, 0, 1)]

This is invertible and the matrix inversion algorithm (and Theorem 4) gives

MDB(T−1) = [MDB(T)]−1 =

If v = (a, b, c, d) then

CB T−1(v)] = MDB(T−1)CD(v)

Hence, we get a formula for the action of T−1:

T−1(a, b, c, d) = T−1(v) = (a − b)

=