Linear Algebra Exercises-n-Answers.pdf

74 Linear Algebra, by Hefferon
3 (a) Setting
gives this
(L 2 M 1 T −2 ) p1 (L 0 M 0 T −1 ) p2 (L 3 M 0 T 0 ) p3 = (L 0 M 0 T 0 )
2p 1 + 3p 3 = 0
p 1 = 0
−2p 1 − p 2 = 0
which implies that p 1 = p 2 = p 3 = 0. That is, among quantities with these dimensional formulas,
the only dimensionless product is the trivial one.
(b) Setting
(L 2 M 1 T −2 ) p 1
(L 0 M 0 T −1 ) p 2
(L 3 M 0 T 0 ) p 3
(L −3 M 1 T 0 ) p 4
= (L 0 M 0 T 0 )
gives this.
2p 1 + 3p 3 − 3p 4 = 0
p 1 + p 4 = 0
−2p 1 − p 2 = 0
(−1/2)ρ 1+ρ 2 ρ 2↔ρ 3
−→
ρ 1+ρ 3
2p 1 + 3p 3 − 3p 4 = 0
−→ −p 2 + 3p 3 − 3p 4 = 0
(−3/2)p 3 + (5/2)p 4 = 0
Taking p 1 as parameter to express the torque gives this description of the solution set.
⎛ ⎞
1
{ ⎜ −2
⎟
⎝−5/3⎠ p ∣
1 p 1 ∈ R}
−1
Denoting the torque by τ, the rotation rate by r, the volume of air by V , and the density of air by
d we have that Π 1 = τr −2 V −5/3 d −1 , and so the torque is r 2 V 5/3 d times a constant.
4 (a) These are the dimensional formulas.
dimensional
quantity formula
speed of the wave v L 1 M 0 T −1
separation of the dominoes d L 1 M 0 T 0
height of the dominoes h L 1 M 0 T 0
acceleration due to gravity g L 1 M 0 T −2
(b) The relationship
(L 1 M 0 T −1 ) p 1
(L 1 M 0 T 0 ) p 2
(L 1 M 0 T 0 ) p 3
(L 1 M 0 T −2 ) p 4
= (L 0 M 0 T 0 )
gives this linear system.
p 1 + p 2 + p 3 + p 4 = 0
0 = 0
−p 1 − 2p 4 = 0
ρ 1+ρ 4
−→
p 1 + p 2 + p 3 + p 4 = 0
p 2 + p 3 − p 4 = 0
Taking p 3 and p 4 as parameters, the solution set is described in this way.
⎛ ⎞ ⎛ ⎞
0 −2
{ ⎜−1
⎟
⎝ 1 ⎠ p 3 + ⎜ 1
⎟
⎝ 0 ⎠ p ∣
4 p 3 , p 4 ∈ R}
0 1
That gives {Π 1 = h/d, Π 2 = dg/v 2 } as a complete set.
(c) Buckingham’s Theorem says that v 2 = dg · ˆf(h/d), and so, since g is a constant, if h/d is fixed
then v is proportional to √ d .
5 Checking the conditions in the definition of a vector space is routine.
6 (a) The dimensional formula of the circumference is L, that is, L 1 M 0 T 0 . The dimensional formula
of the area is L 2 .
(b) One is C + A = 2πr + πr 2 .
(c) One example is this formula relating the the length of arc subtended by an angle to the radius
and the angle measure in radians: l − rθ = 0. Both terms in that formula have dimensional formula
L 1 . The relationship holds for some unit systems (inches and radians, for instance) but not for all
unit systems (inches and degrees, for instance).