Linear Algebra Chapter 1 Solutions to Exercises Exercise 1.1. Let A ...

Linear Algebra
Chapter 1
Solutions to Exercises
Exercise 1.1. Let
A =
[ ]
1 2
, B =
3 4
[ ]
4 3
2 1
Compute AB, BA. You should get different answers.
Solution: [ ] [ ] [ ]
1 2 4 3 8 5
= .
3 4 2 1 20 13
[ ] [ ] [ ]
4 3 1 2 13 20
= .
2 1 3 4 5 8
[ ]
[ ]
1 2
a b
Exercise 1.2. Let A = as in 1.1, and suppose B = , where a+c =
3 4
c d
d, 2c = 3b. Show that AB = BA.
Solution:
while
[ ] [ ] [ ]
1 2 a b a + 2c b + 2d
AB =
=
,
3 4 c d 3a + 4c 3b + 4d
[ ] [ ]
a b 1 2
BA =
=
c d 3 4
[
a + 3b 2a + 4b
c + 3d 2c + 4d
Using the two given equations a + c = d, 2c = 3b, one sees that AB = BA.
Exercise 1.3. Compute [ ] [ ]
2 1 1 −1
1 1 −1 2
and [
1 −1
−1 2
You should get the same answer both times.
Solution: Both products are I.
Exercise 1.4. Using the trig identities
] [ ]
2 1
.
1 1
cos(θ + φ) = cos θ cos φ − sin θ sin φ, sin(θ + φ) = sin θ cos φ + sin φ cos θ,
]
.
show that [
cos θ
sin θ
] [ ]
− sin θ cos φ − sin φ
=
cos θ sin φ cos φ
[ ]
cos(θ + φ) − sin(θ + φ)
.
sin(θ + φ) cos(θ + φ)
Solution:
[ ] [ ] [ ]
cos θ − sin θ cos φ − sin φ cos θ cos φ − sin θ sin φ − cos θ sin φ − sin θ cos φ
=
sin θ cos θ sin φ cos φ sin θ cos φ + sin φ cos θ cos θ cos φ − sin θ sin φ
[ ]
cos(θ + φ) − sin(θ + φ)
=
,
sin(θ + φ) cos(θ + φ)
using the trig identities.
1

2
Exercise 1.5. [ The powers ] of a matrix are computed as A 2 = AA, A 3 = AAA,
0 −1
etc. Let A = . Compute A
1 −1
2 , A 3 , A 4 , A 100 .
Solution:
A 2 =
[ ]
−1 1
, A 3 = I, A 4 = A, A 100 = AA 99 = A(A 3 ) 33 = A(I) 33 = A.
−1 0
Exercise 1.6. Find the inverses of
A =
[ ]
2 1
, B =
1 1
[ ]
0 −1
, C =
1 −1
[ ]
1 b
.
0 1
Solution:
A −1 =
[ ]
1 −1
, B −1 =
−1 2
[
−1
]
1
−1 0
C −1 =
[ ]
1 −b
.
0 1
Exercise 1.7. Let
Show that
[
cos θ − sin θ
A =
sin θ cos θ
]
.
[ ]
A −1 cos(−θ) − sin(−θ)
=
.
sin(−θ) cos(−θ)
Solution:
First note that det A = sin 2 θ + cos 2 θ = 1, so
A −1 = 1 [
cos θ sin θ
1 − sin θ cos θ
]
.
To get the desired expression for A −1 , recall that cos θ is an even function and sin θ
is an odd function.
Exercise 1.8. There are only two numbers that are their own inverses, namely 1
and -1. Find five matrices that are their own inverses.
Solution:
There are infinitely many such matrices, besides the obvious ones like
[ ]
±1 0
.
0 ±1
For example, take any nonzero number b, and consider the matrices
[ ]
−1 b
,
0 1
[
0 b
1
b
0
]
.

3
Exercise 1.9. Find a matrix A having at least one nonzero entry, such that
A 2 =
[ ]
0 0
.
0 0
[ ]
1 0
Solution: Any matrix of the form works. The most general such matrix
0 1
(you only have to find one of them) is
[ ]
a b
, where a 2 + bc = 0.
c −a
Exercise 1.10. Let
[ ] [ ] [ ]
a b
e f
i j
A = , B = , C = .
c d
g h
k l
Show that (AB)C = A(BC).
Solution:
We compute
[ ] [ ]
ae + bg af + bh i j
(AB)C =
ce + dg cf + dh k l
[ ]
(ae + bg)i + (af + bh)k (ae + bg)j + (af + bh)l
=
(ce + dg)i + (cf + dh)k (ce + dg)j + (cf + dh)l
[ ]
a(ei + fk) + b(gi + hk) a(ej + fl) + b(gj + hl)
=
c(ei + fk) + d(gi + hk) c(ej + fl) + d(gj + hl)
[ ] [ ]
a b ei + fk ej + fl
=
c d gi + hk gj + hl
= A(BC).