Linear Algebra Exercises-n-Answers.pdf

70 Linear Algebra, by Hefferon
2 For both of these structures, these checks are all routine. As with the prior question, most of the
checks consist only of remarking that property is so familiar that it does not need to be proved.
3 There is no multiplicative inverse for 2 so the integers do not satisfy condition (5).
4 These checks can be done by listing all of the possibilities. For instance, to verify the commutativity
of addition, that a + b = b + a, we can easily check it for all possible pairs a, b, because there are
only four such pairs. Similarly, for associativity, there are only eight triples a, b, c, and so the check
is not too long. (There are other ways to do the checks, in particular, a reader may recognize these
operations as arithmetic ‘mod 2’.)
5 These will do.
+ 0 1 2
0 0 1 2
1 1 2 0
2 2 0 1
· 0 1 2
0 0 0 0
1 0 1 2
2 0 2 1
As in the prior item, the check that they satisfy the conditions can be done by listing all of the cases,
although this way of checking is somewhat long (making use of commutativity is helpful in shortening
the work).
Topic: Crystals
1 Each fundamental unit is 3.34 × 10 −10 cm, so there are about 0.1/(3.34 × 10 −10 ) such units. That
gives 2.99 × 10 8 , so there are something like 300, 000, 000 (three hundred million) units.
2 (a) We solve
( ) ( )
1.42 1.23
c 1 + c
0 2 =
0.71
( ) 5.67
3.14
=⇒ 1.42c 1 + 1.23c 2 = 5.67
0.71c 2 = 3.14
to get c 2 =≈ 4.42 and c 1 ≈ 0.16.
(b) Here is the point located in the lattice. In the picture on the left, superimposed on the unit cell
are the two basis vectors ⃗ β 1 and ⃗ β 2 , and a box showing the offset of 0.16 ⃗ β 1 + 4.42 ⃗ β 2 . The picture on
the right shows where that appears inside of the crystal lattice, taking as the origin the lower left
corner of the hexagon in the lower left.
So this point is in the next column of hexagons over, and either one hexagon up or two hexagons
up, depending on how you count them.
(c) This second basis
makes the computation easier
( ) ( )
1.42 0
c 1 + c
0 2
1.42
( 1.42
〈
0
=
)
,
( )
5.67
3.14
( ) 0
〉
1.42
=⇒ 1.42c 1 = 5.67
1.42c 2 = 3.14
(we get c 2 ≈ 2.21 and c 1 ≈ 3.99), but it doesn’t seem to have to do much with the physical structure
that we are studying.
3 In terms of the basis the locations of the corner atoms are (0, 0, 0), (1, 0, 0), . . . , (1, 1, 1). The locations
of the face atoms are (0.5, 0.5, 1), (1, 0.5, 0.5), (0.5, 1, 0.5), (0, 0.5, 0.5), (0.5, 0, 0.5), and (0.5, 0.5, 0). The
locations of the atoms a quarter of the way down from the top are (0.75, 0.75, 0.75) and (0.25, 0.25, 0.25).
The atoms a quarter of the way up from the bottom are at (0.75, 0.25, 0.25) and (0.25, 0.75, 0.25).
Converting to Ångstroms is easy.