Download ECON3120/4120 Mathematics 2, autumn 2005 Problem

Department of Economics
October 2005
Arne Strøm
ECON3120/4120 Mathematics 2, autumn 2005
Problem solutions for seminar no. 6, 10–14 October 2005
(For practical reasons some of the solutions may include problem parts that are
not on the problem list for this seminar.)
EMEA, 15.7.3 (= LA, 2.1.5)
Using the definitions of vector addition and multiplication of a vector by a real
number, we get
3(x, y, z) + 5(−1, 2, 3) = (4, 1, 3) ⇐⇒ (3x − 5, 3y + 10, 3z + 15) = (4, 1, 3)
Since two vectors are equal if and only if they are component-wise equal, this
vector equation is equivalent to the equation system
3x − 5 = 4,
3y + 10 = 1,
3z + 15
with the obvious solution
x = 3,
y = −3,
z = −4.
EMEA, 15.7.4 (= LA, 2.1.6)
(a) For every vector x we have x + 0 = x, so if x + 0 = 0, then x = 0, which
means that all components of x are 0.
(b) Since 0x = 0 for every vector x, this equation gives us no information about
the components of x.
EMEA, 15.7.6 (= LA, 2.1.9)
(Since we do not know the components of a and b, this is an exercise in “abstract”
vector manipulation.)
4x − 7a = 2x + 8b − a ⇐⇒ 4x − 2x = 8b − a + 7a
⇐⇒ 2x = 6a + 8b ⇐⇒ x = 3a + 4b.
EMEA, 15.7.8 (= LA, 2.2.4)
The dot product (inner product, scalar product, Norwegian: prikkproduktet,
indreproduktet, skalarproduktet) is
(x, x − 1, 3) · (x, x, 3x) = x2 + (x − 1)x + 9x = 2x2 + 8x = 2x(x + 4).
The two vectors are mutually orthogonal when their dot product equals 0, that is,
for x = 0 and for x = −4.
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EMEA, 15.8.2 (= LA, 2.3.2)
(a) Note: In the EMEA version of this problem, x = (1 − λ)a + λb, whereas in
LA, x = λa + (1 − λ)b. For convenience (= because of laziness), the illustration
uses the definition from LA. EMEA readers please read λ as 1 − λ.
y
λ=0
λ = 1/4
λ = 1/2
λ = 3/4
b
1
−1
λ=1
a
1
2
3
x
Problem 15.8.2
(b) As λ runs from 0 to 1, x will trace out the line segment from a to b in EMEA
(from b to a in LA). By the point-point formula for a straight line (“topunktsformelen for en rett linje”), the straight line through a = (3, 1) and b = (−1, 2)
has the equation
y−1=
1
1
7
2−1
(x − 3) = − (x − 3) ⇐⇒ y = − x +
⇐⇒ x + 4y = 7. (∗)
−1 − 3
4
4
4
For every λ in R the point (x, y) = (1 − λ)a + λb = (3 − 4λ, 1 + λ) satisfies (∗), that
is, the point lies on L. Conversely, if (x0 , y0 ) is a point on L, then x0 = 7−4y0 , and
if we let λ = y0 − 1, then (1 − λ)a + λb = (3 − 4λ, 1 + λ) = (7 − 4y0 , y0 ) = (x0 , y0 ).
EMEA, 15.9.2 ((a),(b),(c) = LA, 2.6.2(a),(c),(d))
(a) The line L has the parameter representation
x1 = −t + 2,
x2 = 2t − 1,
x3 = t + 3.
If we let t = 0, we get precisely the point a = (2, −1, 3) on L.
If the point (1, 1, 1) were to lie on L, there must exist a t such that
−t + 2 = 1,
2t − 1 = 1,
t + 3 = 1.
But the first of these equations gives t = 1, which is not a solution of the last
equation. Thus (1, 1, 1) cannot lie on L.
(b) To an arbitrary t there corresponds the point
x = (2, −1, 3) + (−t, 2t, t) = a + tv,
where v = (−1, 2, 1). Thus, the vector v is a direction vector for the line L.
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(c) The plane P that passes through a and is perpendicular to L, has v as a
normal vector. (Every direction vector for L is a normal vector for P and vice
versa.) Therefore P is given by the equation
v · x = v · a,
which gives
−x1 + 2x2 + x3 = −1.
(d) The point of intersection between L and the plane 3x1 + 5x2 − x3 = 6 is given
by that parameter value t for which the point (−t + 2, 2t − 1, t + 3) on L satisfies
the equation for the plane, that is
3(−t + 2) + 5(2t − 1) − (t + 3) = 6.
This equation gives t = 4/3, and we get the point (2/3, 5/3, 13/3).
EMEA, 15.3.1 (= LA, 3.2.4)
See the answers in the book. Note that in (d), the product AB is not defined,
since A is 2 × 2 and B is 3 × 2.
EMEA, 15.3.3 (= LA, 3.2.5)
See the answer in the back of the book. If you prefer to calculate A(BC) directly
instead of using the associative law, A(BC) = (AB)C, then you will need the
product
⎞
⎛
14 −4 10
BC = ⎝ 21
0 27 ⎠
11 −4 13
EMEA, 15.3.5 (= LA, 3.2.6)
(a) We know that A is an m × n matrix. Let B be a p × q matrix. The matrix
product AB is defined if and only if n = p, and BA is defined if and only if q = m.
So for both AB and BA to be defined, it is necessary and sufficient that B is an
n × m matrix.
(b) (In LA, the matrix B is called X.) We know from part(a) that if BA and
x y
AB are defined, then B must be a 2 × 2 matrix. So let B =
. Then
z w
x y
1 2
x + 2y 2x + 3y
BA =
=
z + 2w 2z + 3w
z w
2 3
and
1 2
x y
x + 2z
y + 2w
AB =
=
.
2 3
z w
2x + 3z 2y + 3w
Hence,
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BA = AB
⎧
x + 2y = x + 2z
⎪
⎪
⎪
⎪
⎪
⎨ 2x + 3y = y + 2w
⇐⇒
⎪
z + 2w = 2x + 3z
⎪
⎪
⎪
⎪
⎩
2z + 3w = 2y + 3w
The first and last of these four equations are true if and only if y = z, and if y = z,
then the second and third are true if and only if x = w − y. Hence, the matrices
B that commute with A are precisely the matrices of the form
w−y y
1 0
−1 1
B=
=w
+y
,
y
w
0 1
1 0
where y and w can be any real numbers.
Exam problem 120
(a) Formula (5.4.3) in FMEA ((1.4.3) in MA II) gives x = Ce−2t + 1.
(b) z = ẇ satisfies ż + 2z = 2, so z = Ce−2t + 1 for a constant C. Hence,
1
w = z dt = (Ce−2t + 1) dt = − Ce−2t + t + D,
2
with D as a new constant. w(0) = 0 gives D = 12 C, and w(− 12 ) =
C = 2. Hence D = 1. Thus the function we seek is
1
2
− e gives
w(t) = −e−2t + t + 1.
Exam problem 127
(a) Let us first find the indefinite integral, using integration by parts:
1
ln x3
3 1
1
−
· − dx
dx
=
3
ln
x
·
dx
=
3
ln
x
·
−
x2
x2
x
x
x
3
3 ln x
3 ln x 3
+
− + C.
=−
dx = −
2
x
x
x
x
(We could have tried the substitution u = ln x instead. That would give x = eu ,
dx = eu du and
ln x3
−2u u
dx = 3ue
e du = 3ue−u du,
x2
and then we would have to use integration by parts here, too.)
The definite integral is then
e
e
3(1 + 1) 3(0 + 1)
ln x3
3(ln x + 1)
=−
+
= 3 − 6e−1 .
dx = −
2
x
x
e
1
1
1
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(b) The equation can be written as
(∗∗)
x2 − 25
dx
= (−1) ·
,
dt
2x
x > 5.
This is obviously a separable equation, and the standard procedure gives
2x dx
= (−1) dt,
x2 − 25
ln(x2 − 25) = −t + C,
(1)
(C1 = eC )
x2 − 25 = C1 e−t ,
x = 25 + C1 e−t .
Note that we have used the assumption x > 5. We√want the solution that passes
through P = (t0 , x0 ) = (0, 10). This leads to 10 = 25 + C1 , so C1 = 75 and
√
√
x = 25 + 75e−t = 5 1 + 3e−t .
The slope dx/dt of the curve at P is given by the differential equation (∗∗):
75
15
dx
=−
=− .
dt
20
4
It is clear from (∗∗) that dx/dt is always negative when x > 5, so all solutions are
decreasing.
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