Unformatted text preview: ISM: Linear Algebra Section 1.2 1'1 —2 3
3:2 _ —1 2
13 — 5 o + t 1
1'4 1 U h. Yes, from our solution in part [a], if we plug in 1 for 3:1 and 13 for 1'4, we obtain
Sit—25:1 ands=13, which leadstot=Q, and 32:5,$3=Q. So we have the solution: 31 = 1, 3:2 = 5, 1'3 = Q and $4 = 13, which is an arithmetic
progression. 43. It is required that 1",, = ﬂing—1 +$k+1], or 23;, = Ik_1 + $5,“, or $1, — 31,4 = 1:5,“ —
1.1,. This means that the diﬂ'erence of an},r two consecutive terms must he the same;
we are looking at the ﬁnite arithmetic sequences. Thus the solutions are of the form
[31,32,33, ... ,3“) = [t,t+1", t+2r, .. .,t+[n—1]r], where t and 1" are arbitrary constants. 2 1 c5
49. We begin by solving the system. Our augmented matrix begins as: [l 3 15
1 c 45 C‘JC‘JC‘J 1 c :13 59,0
and is reduced to U 1 IJE £50 . In order for 2:, y and z to be integers, C must he a c c 13 go
multiple of 25. We want the smallest positive choice, so C = 25. 511.;{1}=a+e+c12+ar3 andwelearnthat ﬂu} = 11:3, 111) =o+b+c+d=2,
f{2]=n+2b+4c+8ul=ﬂ.a1so, 2
ff[t}dt=ct+lbt2+ éct3+ idtﬂﬁ = 2o+ﬂb+§c+ 411:4.
I] 2
1 D ﬂ [1 3
Now, we set up our matrix, 1 1 1 1 2 . However, when we reduce this, the
1 2 4 8 D
2 2 g 4 4
last line becomes I] = 1, meaning that the system is inconsistent. 2? ...
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 Spring '09
 Derksen
 Algebra

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