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Lecture 7 – Appendix B: Some sample problems from Boas
Here are some solutions to the sample problems assigned for Chapter 3 (Sections 3, 6,
7 and 9 ).
§3.3: 5
Solution: We want to find the determinant of a 5x5 matrix by first simplifying the
matrix, using the general invariances of determinants ( e.g ., adding and subtracting
rows), and then use the Laplace development. We have
3 3 4 4 4 2 1
5 5 5 1
7 0 1 3 5 7 0 1 3 5
2 1 0 1 4 2 1 0 1 4
7 3 2 1 4 15 3 0 8 8
8 6 2 7 4 8 6 2 7 4
1 3 5 7 5 1 3 5 7 5
7 0 1 3 5 7 0 1 3 5
2 1 0 1 4 2 1 0 1 4
15 3 0 8 8 15 3 0 8 8
22 6 0 13 14 22 6 0 13 14
1 3 5 7 5 36 3 0 8 30
2 1 1 4
15 3 8 8
1
22 6 13 14
36
R R R R R R
R R R
^
2 2 8 1
3 3 13 1
4 4 8 1
3 3 1
2 1 1 4
31 5 0 40
48 7 0 66
3 8 30 52 5 0 62
31 5 40 31 5 40
48 66 31 40
1 48 7 66 48 7 66 5 7
21 22 21 22
52 5 62 21 0 22
R R R
R R R
R R R
R R R
§3.3: 17
Solution: Here we apply Cramer’s rule to the equations described the Lorentz
transformation of Special Relativity to find the inverse transformation. We have
2 2
2 2 2
2
2 2
2
2 2 2
2
x x vt v x x
t t vx c^ v^ c^ t^ t
x v
t^ x^ vt
x x vt
v v c
v c
x
t vx c v c t
t t vx c
v v c
v c
^
^
^
It should no surprise that the form of the transformation is what we would obtain by
boosting in the opposite direction, v v.
§3.6: 2
Solution: Here we practice finding various combinations of 2 matrices. We find
2
A B AB
BA
A B
A B
A
^ ^ ^ ^ ^ ^
^ ^ ^
^
^ ^ ^
2
2
det 5 150 125 25 5det 5,
det 5 5 det 25 1 25;
det 3 18 0 18 3det 6
det 3 3 det 9 2 18.
A A
A A
B B
B B
Thus n = 2 as expected for 2x2 matrices. The Mathematica version of most of these
steps can be found on the web page.
§3.6: 22
Solution: Now we use matrices to solve a set of simultaneous equations. We have in
standard matrix notation
x y z x
x y z y
x y z z
So now we work out the inverse of the coefficient matrix using our new technology.
The pieces are
1
1 1 1
det 2 1 1 1 2 2 1 4 3 1 4 3 4 7 1 12,
3 2 2
4 7 1 4 4 0 4 4 0
1
4 1 5 , 7 1 3 7 1 3.
12
0 3 3 1 5 3 1 5 3
T
M
C C M
It is informative to check this result via
1
MM
^ ^ ^ ^ ^
So finally we have
4 4 0 4 16 4 0 1
1 1
7 1 3 1 28 1 15 1.
12 12
1 5 3 5 4 5 15 2
x
y
z
^
It is easy to confirm the validity of this result by plugging back into the original
equations.
§3.7: 35
Solution: We want to think about the following matrix as a combination of rotations
and reflections. First we check the determinant to find
So we conclude that this transformation contains a reflection. We also note that the
2x2 sub matrix in the upper vector left corner corresponds to an active 90 degree
rotation about the z axis, x y y , x (see Eq. 7.12), while the lower right corner
means a reflection through the xy plane, z z ,
x y
y x
z z
^
1
1
i i i
AA i i i i
i
i
A A i i i i
i i i
^ ^ ^
§3.9: 8
Solution: We want to verify some properties of the Hermitian conjugate. (a) From
Eq. (9.10) in Boas we have
T T T
AB B A. Here we go through the same argument
including the conjugation. We have
^
†
† † † † † † †
.
T T T
kl lj jk jk kj lk jl
l l
jl lk jk l
AB A B A B A B A B
B A B A AB B A
(b) Now we apply the transpose (or the Hermitian conjugate) to a product of more
that 2 matrices (Eq. 9.11 in Boas) by using the associative property of matrix
multiplication (see Eq. 9.8 in Boas). We have
T T^ T T T T T
T T T T
ABCD A BCD B CD A CD B A
D C B A
