Math 222 - Selected Homework Solutions from
Sections 2.2 and 2.3
Instructor - Al Boggess
Fall 1998
Page 98 - Section 2.2
5 We are to show det(A) = n det A. The key is to write
A = (I )A
Now take determinants:
det(A) = det(IA) = det(I ) det A
Now, I has n factors of down the diagonal and so det(I ) = n .
Thus det(A) = n det A, as desired.
6 Suppose A is a nonsingular matrix (so A,1 exists). We are to show
det(A,1 ) = det1 A
The key equation is
I = A,1 A
Taking determinants, we get
1 = det(AA,1 ) = (det A)(det A,1 )
Solving, we get
as desired.
1 = det(A,1 )
det A
1
9 If A and B are row equivalent by row operations of type I (switch
rows) and of type III (add a multiple of one row to another), then the
determinants of A and B dier only by a minus sign, at most. This
is because, type III operations do not change the determinant at all
and type I operations change the sign. If only type III operations are
involved, then det A = det B .
11 We assume that A = LU where L is lower triangular (with ones
on the diagonal) and U is upper triangular (with diagonal elements
u11; u22 ; u33 . Clearly, det L = 1 and det U = u11 u22u33 . Therefore
det A = det L det U = u11 u22 u33 .
12 We are to prove that AB is nonsingular if and only if A and B are
nonsingular. We rst assume that A and B are nonsingular and then
prove that AB is nonsingular. The key equation is
det(AB ) = det A det B
If A and B are nonsingular, then det A and det B are nonzero. So the
product, det A det B is nonzero and hence det(AB ) 6= 0. Thus AB is
nonsingular.
Now assume that AB is nonsingular and we will show that both A
and B are nonsingular. Again we use the equation
det(AB ) = det A det B
If AB is nonsingular, then det(AB ) is not zero. This implies that
BOTH det A and det B are nonzero. Thus both A and B are nonsingular, as desired.
Page 104 - Section 2.3
8 We are to show
The key equation is
det(adjA) = (det A)n,1
A,1 = det1 A adjA
Using problem 5 on page 98 with = det1 A and problem 6, we get
1 = det A,1 = 1 det(adjA)
det A
(det A)n
2
Solving for det(adjA) we get
det(adjA) = (det A)n,1
as desired.
12 We are to show that if det A = 1, then adj(adjA) = A. The key
equation is
A,1 = det1 A adjA If det A = 1, then adjA = A,1 or A = (adjA),1 . Now apply ** with
A replaced by adjA. We obtain
1 adj(adjA)
(adjA),1 = det(adj
A)
Now as already mentioned, the left side is A. On the right side, since
adjA = A,1 , we have det(adjA) = det(A,1 ) = 1= det A = 1. Therefore from the previous equation
A = adj(adjA)
as desired.
3