Solutions For Calculus Quiz #1
1. (a) u and w are linearly independent; v and w are linearly independent.
(b) Since u and v are linearly dependent, the graph of {c1u + c2v|c1, c2 ∈ R} is the line y = 2x on R2.
(c) Since u and w are linearly independent, the graph of {c1u+c2w|c1, c2 ∈ R} is the whole R2 plane.
2.
½ x − y + 2z = 3
−5x + 2y − 7z = 6 Let z = t.
Then we have
x = −t − 4 y = t − 7 z = t
3. (a)
0 0 4.6 3.7
0.7 0 0 0
0 0.5 0 0 0 0 0.1 0
(b)
0 0 4.6 3.7
0.7 0 0 0
0 0.5 0 0 0 0 0.1 0
1500
500 250 50
=
1355 1050 250
25
.
0 0 4.6 3.7
0.7 0 0 0
0 0.5 0 0 0 0 0.1 0
1355 1050 250
25
=
1243
935 525 25
.
4. (a) N(0) =
µ 100 100
¶ . N(1) =
µ 1.3 2 0.07 0
¶ µ 100 100
¶
=
µ 330 7
¶ . N(2) =
µ 1.3 2 0.07 0
¶ µ 330 7
¶
=
µ 443 23
¶ . N(3) =
µ 1.3 2 0.07 0
¶ µ 443 23
¶
=
µ 622 31
¶ . N(4) =
µ 1.3 2 0.07 0
¶ µ 622 31
¶
=
µ 871 44
¶ . N(5) =
µ 1.3 2 0.07 0
¶ µ 871 44
¶
=
µ 1220 61
¶ . N(6) =
µ 1.3 2 0.07 0
¶ µ 1220 61
¶
=
µ 1708 85
¶ . N(7) =
µ 1.3 2 0.07 0
¶ µ 1708 85
¶
=
µ 2390 120
¶ .
(b)
t 1 2 3 4 5 6 7
q0(t) 3.3 1.342 1.404 1.4 1.4 1.4 1.399 q1(t) 0.07 3.286 1.348 1.42 1.386 1.393 1.412
1
(c) Both ratios q0(t) and q1(t) seem to approach 1.4.
(d) det
µ 1.3 − λ 2 0.07 −λ
¶
= λ(λ − 1.3) − 0.14 = λ2− 1.3λ − 0.14 = (λ − 1.4)(λ + 0.1) = 0
∴ λ1 = 1.4, λ2 = −0.1.
µ 1.3 2 0.07 0
¶ µ x1 x2
¶
=1.4 µ x1
x2
¶ .
½ 1.3x1+ 2x2 = 1.4x1
0.07x1 = 1.4x2 ∴ x1 = 20x2, v1 = µ 20
1
¶ . µ 1.3 2
0.07 0
¶ µ x1 x2
¶
=-0.1 µ x1
x2
¶ .
½ 1.3x1+ 2x2 = −0.1x1
0.07x1 = −0.1x2 ∴ x2 = −0.7x1, v2 = µ 10
−7
¶ .
(e) The growth parameter equals the largest eigenvalue, that is, 1.4.
And
µ 2000 100
¶
is a stable age distribution.
5. (a)
0.24x + 0.21y + 0.17z = 500 × 1.3 0.04x + 0.07y + 0z = 100 × 1.3 0.08x + 0.12y + 0z = 180 × 1.3
24x + 21y + 17z = 65000 4x + 7y + 0z = 13000 8x + 12y + 0z = 23400
(b)
24 21 17 65000 4 7 0 13000 8 12 0 23400
(R1) (R2) (R3) (R2)
(R1) (R3)
4 7 0 13000 24 21 17 65000 8 12 0 23400
(R4) (R5) (R6) (R4)
(R5) − 6(R4) (R6) − 2(R4)
4 7 0 13000 0 −21 17 −13000 0 −2 0 −2600
(R7) (R8) (R9)
1 4(R7)
−12(R9)
1 17(R8)
1 74 0 3250 0 1 0 1300 0 −2117 1 −130017
(R10) (R11) (R12) (R10) − 74(R11)
(R11) (R12) + 2117(R11)
1 0 0 975 0 1 0 1300 0 0 1 1430017
∴ x = 975, y = 1300, z = 1430017 .
2
