1. If A is invertible then B = IB = (A−1A)B = A−10 = 0, contrary to assumption.
2. a. By the definition of matrix multiplication column k of AB is exactly ABk .
3.AB =
I2 but
.
5. and
are both invertible (in fact
A−1 =
A and
B−1 =
B), but
is not invertible by Theorem 3 because det (
A +
B) = 0. For a simpler example, take
A =
U and
B = −
U for any invertible matrix
U .
7.(A + B)(A − B) = A2 + AB − BA − B2, so (A − B)(A + B) = A2 − B2 if and only if AB − BA = 0.
9. It is routine that A3 = I, so AA2 = I = A2A . This implies that A−1 = A2.
11. a. AA−1 = I = A−1A shows A is the inverse of A−1, that is (A−1)−1 = A . Next, (AB)(B−1A−1) = AIA−1 = AA−1 = I, and similarly (B−1A−1)(AB) = I. Hence AB is invertible and (AB)−1 = B−1A−1.
13. a.If AB = I then det A det B = det (AB) = det I = 1 . Hence det A is a unit so, by Theorem 7, A is invertible. But then A−1 = A−1I = A−1(AB) = B, whence BA = A−1A = I as required.
15. a. The only nonzero entry in EikEmj must come from entry k of row i of Eik and column m of Emj . The result follows.
c. For each i and j, ...