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Math 102 - Homework 6 (selected problems)
David Lipshutz
Problem 1. (Strang, 4.4: #14)
Use Cramerās Rule to solve for y.
(a)
ax + by = 1
cx + dy = 0
(b)
ax + by + cz = 1
dx + ey + f z = 0
gx + hy + iz = 0
Proof.
(a)
A =
[
a b
c d
]
b =
[
]
B 2 =
[
a 1
c 0
]
y =
det B 2
det A
c
ad ā bc
(b)
A =
a b c
d e f
g h i
b =
B 2 =
a 1 c
d 0 f
g 0 i
y =
det B 2
det A
di ā f g
det A
Problem 2. (Strang, 4.4: #18)
Find A ā 1 from the cofactor formula C T / det A. Use symmetry in part (b):
(a) A =
(b) A =
Proof.
(a)
C =
A
ā 1
C
T
det A
(b)
C =
A
ā 1
C
T
det A
Problem 3. (Strang, 4.4: #36)
The triangle with corners (0, 0), (1, 0), (0, 1) has area
1 2
. The pyramid with four corners
(0, 0 , 0), (1, 0 , 0), (0, 1 , 0), (0, 0 , 1) has what volume? The pyramid with five corners at (0, 0 , 0 , 0)
and the rows of I has what volume?
Proof. Take the case with n corners, then the volume is the integral over all positive x 1 ,... , xn
such that
ān
i= xi ā¤ 1, i.e.
V =
{xi:x 1 +Ā·Ā·Ā·+xnā¤ 1 }
dx 1 Ā· Ā· Ā· dxn
Now make a substitution, ui = x 1 + Ā· Ā· Ā· + xi, so the integral becomes,
V =
{ui:u 1 ā¤Ā·Ā·Ā·ā¤unā¤ 1 }
du 1 Ā· Ā· Ā· dun
There are n! ways to arrange the ui and integrating over each permutation gives the same
volume by summing over every permutation, we are integrating over every point in the unit
cube, ā
P
{ui:uP 1 ā¤Ā·Ā·Ā·ā¤uP nā¤ 1 }
du 1 Ā· Ā· Ā· dun =
P
V = n!V = 1
Hence V =
1 n!
. So the volume of the pyramid is
1 3!
1 6
and the volume over the four
dimensional pyramid is
1 4!
1 24
Problem 4. (Strang, 5.1: #14)
Find the rank and all four eigenvalues for both the matrix of ones and the checkerboard
matrix:
A =
C =
Which eigenvectors correspond to nonzero eigenvalues?
Proof. The matrix A has rank 1, so it has one nonzero eigenvalue and three zero eigenvalues.
If Ī» 1 = Ī» 2 = Ī» 3 = 0, then Ī» 4 =
i= Ī»i = Tr(A) = 4 which from inspection has associated
eigenvector [1, 1 , 1 , 1]
T
. Then matrix C has rank 2, so it has two nonzero eigenvalues and
two zero eigenvalues. Note that if x is an eigenvector of C, since the first and third row of
C are the same as well as the second and fourth row, x must have the pattern [a, b, a, b]
T .
Plugging in we get:
Cx = Ī»x ā
2 b
2 a
2 b
2 a
Ī»a
Ī»b
Ī»a
Ī»b
ā a =
Ī»b
, b =
Ī»a
ā a =
Ī»
2 a
ā Ī» = Ā± 2
