2.2: 6, 15
6. Let A be a nonsingular matrix. Show that
det(A−1 ) =
1
.
det(A)
Answer. Recall that for n × n matrices A and B, det(A) det(B) = det(AB) by
Theorem 2.2.3. Thus
det(A−1 ) det(A) = det(A−1 A) = det(I) = 1.
But det(A−1 ) and det(A) are scalars, so we have
det(A−1 ) =
1
,
det(A)
as required.
15. Let A and B be n × n matrices. Prove that if AB = I then BA = I.
What is the significance of this result in terms of the definition of a nonsingular
matrix?
Answer. Suppose AB = I. By Exercise 14 in this section, A and B are both
nonsingular. Then by Theorem 1.3.3, (AB)−1 = B −1 A−1 . But AB = I, so
(AB)−1 = I −1 = I. Thus
BA =
=
=
=
=
=
(BI)A
B(AB)−1 A
B(B −1 A−1 )A
(BB −1 )(A−1 A)
I2
I.
Therefore, if AB = I then BA = I.
This result is significant because it means that we only need to check that
AB = I (or BA = I) in order to verify that B is the inverse of A.
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