SOLUTIONS TO PROBLEMS 9 (ODD NUMBERS)
9.1
The cross product is
i
j
k
a×b = 1
2
3 = 7i + 10j − 9k .
4 −1 2
The triple product is
4 −1 2
b ⋅ a ×c = 1
2 3 = 4(6 + 18) + 1(3 −15) + 2(−6 −10) = 52.
5 −6 3
9.3
This may be obtained in several ways using the rules for manipulating determinants.
For example, adding multiples of row 1 to rows 2, 3, and 4, reduces column 3 to 1, 0,
0, 0:
3
7
1
2
9
18
0
3
2
4
0 −1
−14 −35
0 −6
Δ=
9
18
3
2
4
−1
−14 −35
−6
=
,
where we have expanded about column 3. Replacing c2 → c2 − 2c1 , then gives
Δ=
9
0
3
2
0 −1 = 7
9
2 −1
−14 −7 −6
9.5
3
= −105.
Expanding along the first row gives
1
1
0
0 
0
0 −2
1
0 
0
1 −2
1 
0 = −2Δn−1 − Δn−2 .






0
0
0
0
1
−2
Δn = −2Δn−1 − 0
Hence Sn = Δn + Δn−1 = −(Δn−1 + Δn−2 ) = −Sn−1 for n > 2 , so that
Sn = −Sn−1 = (−1)2 Sn−2 = (−1)n−2 S 2 = (−1)n
on evaluating S 2 explicitly. Hence
S9.1
Δn = (−1)n − Δn−1 = (−1)n −[(−1)n−1 − Δn−2 ]
n
= 2(−1)n + Δn−2 = 3(−1) − Δn−3 etc.
= m(−1)m + (−1)m Δn−m .
Putting m = n − 2 , and evaluating Δ2 = 3 explicitly, gives
Δn = (n − 2)(−1)n−2 + 3(−1)n−2 = (n + 1)(−1)n .
9.7
This is equivalent to finding a solution of the homogeneous equations
αx + 3y − 2z = 0,
−3x + αy + (α + 4)z = 0,
− x + 3y + 4z = 0,
with z = 1 . Such a solution exists if
α 3 −2
α 3 −5
=
−3 α α + 4
−3 α 4
−1 3 1
−1 3
4
= α2 −17α + 42 = (α − 3)(α −14) = 0,
i.e. if α = 3 or 14. For the latter, (1) becomes
14x + 3y − 2z = 0
−3x + 14y + 18z = 0
− x + 3y + 4z = 0
with solutions
x :y :z =
14 18
−3 18
−3 14
:−
:
3 4
−1 4
−1 3
= 2 : −6 : 5
by (9.13a). Since z = 1 , the values of x and y are x = 2 5, y = −6 5 .
9.9
(a)
⎛ 1 −2 0 ⎞⎟ ⎛ 9 3 −6 ⎞⎟ ⎛ −8 −5
6 ⎞⎟⎟
⎜⎜
⎟⎟ ⎜⎜
⎟⎟ ⎜⎜
⎟⎟
⎜
⎟ ⎜
⎟ ⎜
A − 3B = ⎜⎜ 3
2 5 ⎟⎟ − ⎜⎜ 3 0
6 ⎟⎟ = ⎜⎜ 0
2 −1 ⎟⎟ ,
⎟
⎟
⎟⎟
⎜⎜
⎟ ⎜
⎟ ⎜⎜
⎜⎜⎝ −1 3 1 ⎟⎟⎠ ⎜⎜⎜⎝ −6 12
9 ⎟⎟⎠ ⎜⎜⎝ 5 −9 −8 ⎟⎟⎠
⎛ 1
1 −6 ⎞⎟⎟
⎜⎜
⎟
⎜
AB = ⎜⎜ 1 23 13 ⎟⎟⎟,
⎟⎟
⎜⎜
⎜⎜⎝ −2 3 11 ⎟⎟⎠
⎛ 8 −10
3 ⎞⎟⎟
⎜⎜
⎟⎟
⎜
BA = ⎜⎜ −1
4
2 ⎟⎟.
⎟⎟
⎜⎜
⎜⎜⎝ 7 21 23 ⎟⎟⎠
S9.2
(1)
(b) XY is defined only if the number of columns in X is the same as the number of
columns in Y. So AC and DA are not defined, but
⎛ 1 −7 −2 ⎞⎟
⎜
⎟⎟,
CA = ⎜⎜
⎜⎜ 21 4 28 ⎟⎟⎟
⎝
⎠
⎛ 25
25 ⎞⎟⎟
⎜
⎟,
CD = ⎜⎜
⎜⎜ 17 −16 ⎟⎟⎟
⎝
⎠
⎛ 3
6
⎜⎜
AD = ⎜⎜⎜ 2 17
⎜⎜
⎜⎝ −5 −5
⎛ 37
7
⎜⎜
⎜⎜
DC = ⎜
5 −13
⎜⎜
⎜⎜⎝ −18
21
⎞⎟
⎟⎟
⎟⎟,
⎟⎟
⎟⎟
⎠
11 ⎞⎟⎟
⎟⎟
7 ⎟⎟,
⎟⎟
−15 ⎟⎟⎠
are defined.
9.11
The rotation does not change z, while x and y transform as in (9.48) and Figure 9.1,
so that
⎛ cos θ −sin θ 0 ⎞⎟
⎜⎜
⎟
R(θ) = ⎜⎜ sin θ cos θ 0 ⎟⎟⎟.
⎜⎜
⎟⎟
⎜⎝ 0
0
1 ⎟⎠
Then
⎛ cos θ −sin θ 0 ⎞⎟⎛ cos θ −sin θ 0
⎜⎜
⎟⎟⎜⎜
1
1
2
2
⎜⎜
⎟⎟⎜⎜
R(θ1 )R(θ2 ) = ⎜ sin θ1
cos
θ
0
sin
θ
cos
θ
0
⎜
⎟
1
2
2
⎟⎟⎜⎜
⎜⎜
⎟
⎜⎜ 0
0
1 ⎟⎟⎠⎜⎜⎝ 0
0
1
⎝
⎞⎟
⎟⎟
⎟⎟
⎟⎟ .
⎟⎟
⎟⎟⎠
Multiplying out, and using the trigonometric identities for cos(θ1 + θ2 ) and
sin(θ1 + θ2 ) , gives
R(θ1 )R(θ2 ) = R(θ1 + θ2 ) = R(θ2 + θ1 ) = R(θ2 )R(θ1 ) .
Similarly, evaluating the inverse gives
⎛ cos θ sin θ 0 ⎞⎟ ⎛⎜ cos(−θ) −sin(−θ) 0 ⎞⎟
⎜⎜
⎟⎟
⎟⎟ ⎜⎜
−1
⎟
R (θ) = ⎜⎜⎜ −sin θ cos θ 0 ⎟⎟⎟ = ⎜⎜ sin(−θ)
cos(θ) 0 ⎟⎟ = R(−θ)
⎟⎟
⎟⎟ ⎜⎜
⎜⎜
0
0
1 ⎟⎠ ⎜⎜⎝
⎜⎝
0
0
1 ⎟⎟⎠
and also R−1(θ) = RT (θ) as required.
9.13
(a)
(A + B)3 = (A + B)(A + B)(A + B)
= (A + B)(A2 + AB + BA + B2 )
= (A3 + A2B + ABA + AB2 + BA2 + BAB + B2A + B3 )
(b)
S9.3
e Ae B = (I + A + 12 A2 +)(I + B + 12 B2 +)
= I + A + B + 12 (A2 + 2AB + B2 ) +
e (A+B) = I + (A + B) + 12 (A + B)2 
= I + (A + B) + 12 (A2 + AB + BA + B2 ) +
so that
e Ae B = e (A+B)
(1)
requires that A and B commute, i.e. AB = BA . If this is satisfied, then A and B
behave algebraically like ordinary numbers and (1) is satisfied to all orders in A and
B.
9.15
(a) A matrix A is unitary if it satisfies the relation AA† = A†A = I . Now A† is
⎛
⎞
⎜⎜ 0 (−1 −i) 6 2 6 ⎟⎟
⎟⎟
⎜
⎟⎟ ,
A† = ⎜⎜ 1
0
0
⎟⎟
⎜⎜
⎜⎜ 0 (1 + i) 3 1 3 ⎟⎟⎟
⎝
⎠
so
⎛
⎞⎟ ⎛
⎞
0
1
0
⎜⎜
⎟⎟ ⎜⎜ 0 (−1 −i) 6 2 6 ⎟⎟⎟
⎜
⎟⎜
⎟
AA† = ⎜⎜ (−1 + i) 6 0 (1 −i) 3 ⎟⎟ ⎜⎜ 1
0
0 ⎟⎟⎟
⎟⎟ ⎜⎜
⎜⎜
⎟⎟
⎟⎟ ⎜
⎜⎜
2 6
0
1 3
⎟⎠ ⎜⎝ 0 (1 + i) 3 1 3 ⎟⎟⎠
⎝
⎛ 1
0
1
⎜⎜
⎜
= ⎜⎜ 0 (1 −i)(1 + i) 6 + (1 −i)(1 + i) 3 (−1 + i) 3 + (1 −i) 3
⎜⎜
−(1 + i) 3 + (1 + i) 3
2 3+1 3
⎜⎝ 0
⎞⎟ ⎛
⎟⎟ ⎜⎜ 1 0 0
⎟⎟ = ⎜
⎟⎟ ⎜⎜ 0 1 0
⎟⎟ ⎜⎜⎝ 0 0 1
⎠
Likewise, A†A = I .
(b) By construction,
⎛
A = ⎜⎜⎜ 1 2
⎜⎝ 3 2
T
T
⎞⎟
⎛
⎞
⎟⎟ = ⎜⎜ 1 3 ⎟⎟⎟,
⎟⎠
⎜⎜⎝ 2 2 ⎟⎠
and AS = 12 (A + AT ) is a symmetric and AAS = 12 (A − AT ) is an anti-symmetric
matrix.
So,
⎞
1⎛
AS = ⎜⎜⎜ 2 5 ⎟⎟⎟,
2 ⎜⎝ 5 4 ⎟⎠
and
⎞
1⎛
AAS = ⎜⎜⎜ 0 −1 ⎟⎟⎟,
2 ⎜⎝ 1 0 ⎟⎠
and
1⎛
A = ⎜⎜⎜ 2 5
2 ⎜⎝ 5 4
9.17
⎞⎟ 1 ⎛ 0 −1 ⎞⎟
⎟⎟ + ⎜⎜
⎟.
⎟⎠ 2 ⎜⎜⎝ 1 0 ⎟⎟⎠
(a) From the definition, S = ST and A = −AT , so that
S9.4
⎞⎟
⎟⎟
⎟⎟.
⎟⎟
⎟⎠
Tr(SA) = ∑ (SA)ii = ∑ ∑ Sik Aki = ∑ ∑ Ski Aik
i
i
k
k
i
= Tr(ST AT ) = −Tr(SA),
which implies that Tr(SA) = 0 .
(b) Let the two diagonal matrices be Dij = dii δij and Fij = fii δij . Then,
(DF)ij = ∑ Dik Fkj = ∑ dii δik f jj δkj = dii f jj ∑ δik δkj
k
k
k
= dii f jj δij = dii fii .
Similarly, by interchanging F and D we can show that (FD)ij = fii dii so that
FD = DF and F and D commute.
9.19
The determinant of A is
−2 1
2
−2 1
A = −1 1
1
= −1 1
0
−2 1 = 1,
0 =−
−1 1
0 2 −1
0 2 −1
so A is non-singular. The co-factors are:
A11 = −3, A12 = −1, A13 = −2,
A21 = 5,
A22 = 2,
A31 = −1, A32 = 0,
A23 = 4,
A33 = 1.
So, by (9.65),
A
−1
⎛ −3 5 −1
⎜⎜
= ⎜⎜ −1 2
0
⎜⎜
⎜⎝ −2 4 −1
⎞⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟⎠
and
⎛ 6 −2 ⎞⎟ ⎛
⎜⎜
⎟⎟ ⎜⎜ −3 5 −1
⎜
⎟⎟ = ⎜⎜ −1 2
4
0
⎜
X=A ⎜
0
⎟⎟ ⎜
⎜⎜
⎜⎜
⎟
⎟
3 ⎠ ⎝ −2 4 −1
⎜⎝ 1
−1
9.21
⎞⎟⎛⎜ 6 −2 ⎞⎟ ⎛
⎞
⎟⎟ ⎜⎜ 1 3 ⎟⎟
⎟⎟⎜⎜
⎟⎟⎜ 4
0 ⎟⎟⎟ = ⎜⎜ 2 2 ⎟⎟⎟ .
⎟⎟⎜⎜
⎟⎟
⎟ ⎜
⎟⎟⎠⎜⎜⎝ 1
3 ⎟⎟⎠ ⎜⎜⎝ 3 1 ⎟⎠
Using the notations of equation (9.72), we have
⎛ 2
3 −1 ⎞⎟⎟
⎜⎜
⎟⎟
⎜
A = ⎜⎜ 1 −1
1 ⎟⎟ with det A ≡ Δ = −15.
⎟⎟
⎜⎜
⎜⎜⎝ −1
1
2 ⎟⎟⎠
Also
S9.5
⎛ 0
3 −1 ⎞⎟⎟
⎜⎜
⎟⎟
⎜⎜
Δx = det ⎜ 1 −1
1 ⎟⎟ = −3,
⎟⎟
⎜⎜
⎜⎜⎝ 2
1
2 ⎟⎟⎠
⎛ 2 0 −1 ⎞⎟
⎜⎜
⎟⎟
⎜
⎟
Δy = det ⎜⎜ 1 −1
1 ⎟⎟ = −3
⎟⎟
⎜⎜
⎜⎜⎝ −1
1
2 ⎟⎟⎠
and
⎛ 2
3 0 ⎞⎟⎟
⎜⎜
⎟
⎜⎜
Δz = det ⎜ 1 −1 1 ⎟⎟⎟ = −15.
⎟⎟
⎜⎜
⎜⎜⎝ −1
1 2 ⎟⎟⎠
so that
x=
9.23
Δ
Δ
1
1
= , y = y = , z = x = 1.
Δ
5
Δ
5
Δ
Δx
(a) the condition for a unique solution is
4 2 α
det A = 7 3 4 ≠ 0
1 1 2
i.e. 4a −12 ≠ 0 . Thus, unique solutions exist for α ≠ 3 , independent of the value of
β.
(b) For α = 2 det A = −4 and
A−1
⎛
2
⎜
1 ⎜⎜⎜
= − ⎜ −10
4 ⎜⎜
⎜⎜⎝
4
−2
6
−2
2 ⎞⎟⎟
⎟⎟
− 2 ⎟⎟ ,
⎟⎟
− 2 ⎟⎟⎠
where the matrix is the transposed matrix of the co-factors of A. Thus,
⎛ x ⎞⎟
⎛ 3 ⎞⎟ ⎛⎜
12
⎜⎜
⎜
⎜
⎟⎟
⎜⎜ y ⎟⎟ = A−1 ⎜⎜ 8 ⎟⎟⎟ = ⎜⎜ −5 2
⎜⎜
⎟ ⎜
⎜⎜
⎟⎟
⎜⎜⎝ 4 ⎟⎟⎟⎠ ⎜⎜⎜
⎟
⎜⎝ z ⎠
3
⎜⎝
⎞⎟
⎟⎟
⎟⎟ .
⎟⎟
⎟⎟
⎟⎠
(c) For α = 3 , multiplying the first equation by 2 and comparing to the sum of the
second and third equations, gives
8x + 4y + 6z = 2β
and
8x + 4y + 6z = 12 ,
respectively. So in case (i) β = 6 , a solution exists, but is not unique, while in case
(ii) β = 2 , no solutions exist.
S9.6