Math 222 - Selected Homework Solutions for
Assignment 2
Instructor - Al Boggess
Spring 1998
Page 76 - 1.5
14 We are given that A is an n n matrix with Ax = 0 for all vectors x.
We are to show that A = 0 which means that we must show the ith,
jth entry, a , is zero for all 1 i; j n.
Now since Ax = 0 for all x, we can let x = e , for any j , and get
ij
j
0
0 = Ae = B
@
j
a11 : : : a1 : : : a1
..
.
...
j
n
a 1 ::: a
n
nj
:::a
nn
001
1 BB .. CC 0 a1 1
BB . CC B . C
.. C
. AB
BB 1. CCC = @ .. A
a
@ .. A
j
nj
0
Therefore, we conclude that the jth column of A is zero for each 1 j n. Hence all the entries of A are zero, which means that A is the
zero matrix.
15 This problem follows easily from problem 14. We assume that Bx =
Cx for all x. We want to show that B = C . These statements are
equivalent to the following: 1) assume (B , C )x = 0 for all x and 2)
show that B , C = 0. From problem 14, if (B , C )x = 0 for all x,
then B , C = 0, as desired.
Page 91 - Section 2.1
10 We assume that A is an n + 1 n + 1 matrix with two identical rows.
We will show that det A = 0. The proof is by mathematical induction.
When n = 1, A is a 2 2 matrix with both rows the same:
1
A = aa11 aa12
11 12
!
whose determinant is a11 a12 , a11 a12 = 0. This proves the case n = 1.
For the inductive step, we assume that the determinant of any n n
matrix with two identical rows is zero, and we will show the same is
true for n + 1 n + 1 matrices. Let A be the n + 1 n + 1 matrix
0 a11
A=B
@ ...
: : : a1 +1 1
..
.
;n
...
a +1 1 : : : a +1 +1
n
;
n
CA
;n
Two of the rows of A are identical by hypothesis. By switching rows,
we can assume that neither of these two rows are the LAST one (and
switching rows only changes the determinant by a + or - sign). By
expanding by minors along the last row:
det A =
+1
X
n
k
=1
(,1) det A +1
k
n
;k
where A +1 is the minor with the last row and the kth column
deleted. Now, A +1 is an nn matrix with two identical rows (A +1
inherits the same two identical rows from A). Therefore det A +1 = 0
by the induction hypotheseis, for 1 k n + 1. Thus, det A = 0, as
desired.
11 a) No; for example let A = B = I , the identity matrix. A + B = 2I
whose determinant is 4. However, det I + det I = 2.
b) Yes; you can either write out a detailed computation for the 2 2
case or wait for the proof in the next section for the general case.
c) Yes; from part b), det(AB ) = det A det B which equals det B det A,
which in turn from part b) equals det(BA).
n
;k
n
;k
n
n
2
;k
;k