ALL 13 CHAPTERS COVERED
SOLUTIONS MANUAL
,Chapter 1
Problem 1.1 (a) Ab is not defined.
(b) Since C is a 3 × 3 matrix and A is 3 × 2, the product CA is defined and is the
3 × 2 matrix
1 2 1 2 1 4 6
CA = 3 0 −1 1 1 = 6 0
4 1 1 0 3 9 8
(c) A + Cb is not defined, since Cb is 3 × 1 and A is 3 × 2.
2 2
(d) A + D = 3 6
6 6
0 1
(e) bT D = ( 1 1 −1 ) 2 5 = ( −4 3)
6 3
(f) The product DAT is a 3 × 3 matrix, as is C, so this expression is defined:
0 1 ( ) 1 1 3
2 1 0
DAT = 2 5 = 9 7 15 ,
1 1 3
6 3 15 9 9
5
,6 CHAPTER 1.
1 1 3 1 2 1 2 3 4
T
DA + C = 9 7
15 + 3 0
−1 = 12 7 14
15 9 9 4 1 1 19 10 10
1
(g) bT b = ( 1 1 −1 ) 1 = (3). As mentioned in Section 1.8.2, we
−1
normally just identify a 1 × 1 matrix by its entry. So we may write (3) as 3.
1 1 1 −1
(h) bbT = 1 ( 1 1 −1 ) = 1 1 −1
−1 −1 −1 1
1 2 1 1 2
(i) Cb = 3 0 −1 1 = 4
4 1 1 −1 4
Problem 1.2 The matrix aT b is the product of a 1 × n matrix and an n × 1
matrix, so it is a 1 × 1 matrix (which can be identified with the scalar it
represents; that is, the scalar given by the inner product ⟨a, b⟩ ). The product bT a
is also a 1 × 1 matrix. Since the product aT b is a 1 × 1 matrix, it is symmetric,
so that
aT b = (aT b)T = bT a.
Problem 1.3 We have
( )( ) ( ) ( )
3 7 x y 3x + 7z 3y + 7w 1 0
= =
0 −1 z w −z −w 0 1
Hence,
3x + 7z = 1
3y + 7w = 0 3x + 0 = 1 1 7
=⇒ =⇒ x = , y = , z = 0, w = −1
−z = 0
3y − 7 = 0 3 3
−w = 1
(1 7
)
Therefore B −1 = 3 3 .
0 −1
, 7
Using the method in Activity 1.24, we have
( )
−1 1 −1 −7
|B| = −3, so that B =− ,
3 0 3
which, of course, agrees with the previous result for B −1 .
You can now check that this is indeed B −1 by multiplying out BB −1 and B −1 B
to obtain the identity matrix.
Problem 1.4 To obtain A, you can first take the transpose of both sides of the
equation ( )
−1 T 3 5
(A ) =
1 2
to obtain ( )
−1 3 1
A = .
5 2
Now A = (A−1 )−1 , so we need the inverse of the above matrix. This can be
found using the method given in Activity 1.24. You should obtain
( )
2 −1
A= .
−5 3
Problem 1.5 Since A is symmetric, we have aij = aji , and since B is
skew-symmetric we must have bij = −bji . Then the matrices are
1 −4 7 0 −3 5
A = −4 6 0 B= 3 0 2.
7 0 2 −5 −2 0
Problem 1.6 Since (A + AT )T = AT + A = A + AT , the matrix A + AT is
symmetric. Similarly, (A − AT )T = AT − A = −(A − AT ), so this matrix is
skew symmetric.
Adding the matrices A + AT and A − AT we obtain 2A, so we can write
1 1
A = (A + AT ) + (A − AT ),
2 2
which is the sum of a symmetric and a skew-symmetric matrix.
SOLUTIONS MANUAL
,Chapter 1
Problem 1.1 (a) Ab is not defined.
(b) Since C is a 3 × 3 matrix and A is 3 × 2, the product CA is defined and is the
3 × 2 matrix
1 2 1 2 1 4 6
CA = 3 0 −1 1 1 = 6 0
4 1 1 0 3 9 8
(c) A + Cb is not defined, since Cb is 3 × 1 and A is 3 × 2.
2 2
(d) A + D = 3 6
6 6
0 1
(e) bT D = ( 1 1 −1 ) 2 5 = ( −4 3)
6 3
(f) The product DAT is a 3 × 3 matrix, as is C, so this expression is defined:
0 1 ( ) 1 1 3
2 1 0
DAT = 2 5 = 9 7 15 ,
1 1 3
6 3 15 9 9
5
,6 CHAPTER 1.
1 1 3 1 2 1 2 3 4
T
DA + C = 9 7
15 + 3 0
−1 = 12 7 14
15 9 9 4 1 1 19 10 10
1
(g) bT b = ( 1 1 −1 ) 1 = (3). As mentioned in Section 1.8.2, we
−1
normally just identify a 1 × 1 matrix by its entry. So we may write (3) as 3.
1 1 1 −1
(h) bbT = 1 ( 1 1 −1 ) = 1 1 −1
−1 −1 −1 1
1 2 1 1 2
(i) Cb = 3 0 −1 1 = 4
4 1 1 −1 4
Problem 1.2 The matrix aT b is the product of a 1 × n matrix and an n × 1
matrix, so it is a 1 × 1 matrix (which can be identified with the scalar it
represents; that is, the scalar given by the inner product ⟨a, b⟩ ). The product bT a
is also a 1 × 1 matrix. Since the product aT b is a 1 × 1 matrix, it is symmetric,
so that
aT b = (aT b)T = bT a.
Problem 1.3 We have
( )( ) ( ) ( )
3 7 x y 3x + 7z 3y + 7w 1 0
= =
0 −1 z w −z −w 0 1
Hence,
3x + 7z = 1
3y + 7w = 0 3x + 0 = 1 1 7
=⇒ =⇒ x = , y = , z = 0, w = −1
−z = 0
3y − 7 = 0 3 3
−w = 1
(1 7
)
Therefore B −1 = 3 3 .
0 −1
, 7
Using the method in Activity 1.24, we have
( )
−1 1 −1 −7
|B| = −3, so that B =− ,
3 0 3
which, of course, agrees with the previous result for B −1 .
You can now check that this is indeed B −1 by multiplying out BB −1 and B −1 B
to obtain the identity matrix.
Problem 1.4 To obtain A, you can first take the transpose of both sides of the
equation ( )
−1 T 3 5
(A ) =
1 2
to obtain ( )
−1 3 1
A = .
5 2
Now A = (A−1 )−1 , so we need the inverse of the above matrix. This can be
found using the method given in Activity 1.24. You should obtain
( )
2 −1
A= .
−5 3
Problem 1.5 Since A is symmetric, we have aij = aji , and since B is
skew-symmetric we must have bij = −bji . Then the matrices are
1 −4 7 0 −3 5
A = −4 6 0 B= 3 0 2.
7 0 2 −5 −2 0
Problem 1.6 Since (A + AT )T = AT + A = A + AT , the matrix A + AT is
symmetric. Similarly, (A − AT )T = AT − A = −(A − AT ), so this matrix is
skew symmetric.
Adding the matrices A + AT and A − AT we obtain 2A, so we can write
1 1
A = (A + AT ) + (A − AT ),
2 2
which is the sum of a symmetric and a skew-symmetric matrix.
