Matrices
We have already defined what we mean by a matrix.
In this section,
we introduce algebraic operations into the set of matrices.
Definition. If A
k by
we define A
n,
t
and B are two matrices of the same size, say
B to be the k by
the corresponding
MODEL ANSWERS TO HWK #12
(18.022 FALL 2010)
(1) (i) F = (2by 2y) + (2y 2ay)k = 0. Hence a = b = 1.
2
2
(ii) fx = y , so f = xy + g(y, z). fy = 2xy + gy = 2xy + 2yz, so g = y 2 z + h(z). Now,
3
3
fz = y 2 + h = y 2 + z 2 , and h = z3 . Therefore f = xy 2 +
18.02 Practice Exam 2 A Solutions
Problem 1.
a) f = (y 4x3 ) + x; at P , f = 3, 1.
b) w 3 x + y.
Problem 2.
dh
h
.2.
ds u
s
b) Q is the northernmost point on the curve h = 2200; the vertical distance between consecutive
h
h
100
.3.
level curves is about
18.02 Practice Exam 1B
Problem 1.
Let P , Q and R be the points at 1 on the x-axis, 2 on the y-axis and 3 on the z-axis, respectively.
a) (6) Express QP and QR in terms of j and k.
i,
b) (9) Find the cosine of the angle P QR.
Problem 2. Let P = (1, 1, 1)
Linear Spaces
we have seen (12.1-12.3 of Apostol) that n-tuple space
V
,
has the following properties :
Addition:
1.
(Commutativity) A + B = B + A.
2.
(Associativity) A + (B+c) = (A+B) + C.
3.
(Existence of zero) There is an element
such 'that A + - = A
0
18.02 Practice Exam 1 A
z
Problem 1. (15 points)
A unit cube lies in the rst octant, with a vertex at the origin (see gure).
O
a) Express the vectors OQ (a diagonal of the cube) and OR
, ,
(joining O to the center of a face) in terms of k.
x
b) Find the
18.02 Problem Set 2, Part II Solutions
1. a)
(b) See the diagram. Because the lines are parallel, we can cut through them
with a single plane which is orthogonal to each line. The normal vectors
n1 , n2 , n3 lie in this plane.
(c) One way to do this part
18.02 Practice Exam 1B Solutions
Problem 1.
a) P = (1, 0, 0), Q = (0, 2, 0) and R = (0, 0, 3). Therefore QP = 2 and QR = 2 + 3k.
1, 2, 0 0, 2, 3
4
QP QR
b) cos =
= 12 + 22 22 + 32 = 65
QP QP
Problem 2.
a) P Q = 1, 2, 0, P R = 1, 0, 3.
k
2 0 = 6
18.02 Problem Set 1, Part II Solutions
1. There are several ways to set up the tetrahedron for this problem. The
simplest way is to inscribe it in the unit cube, so that it has vertices (0, 0, 0),
(1, 1, 0), (0, 1, 1), (1, 0, 1). If you didnt think of thi
MODEL ANSWERS TO HWK
(18.022 FALL 2010)
(1) We want to show that
that
Z 1
2
I =
e
R1
e
1
x
dx
0
x
Z
dx =
1
e
y
p
10
. We show that I =
dy =
0
Z
1
0
Z
1
x +y )
e
R1
0
e
dxdy =
0
x
dx =
p
2
by proving
.
4
(i) The square of side a de ned by 0 < x < a and 0 <
18.02 Practice Exam 1 A Solutions
Problem 1.
1
a) OQ = + + k; OR = + +
2
1, 1, 1
OQ OR
b) cos = =
|OQ | |OR |
3
1
k.
2
1
1
2 , 1, 2
3
2
2 2
=
.
3
Problem 2.
dR
Velocity: V =
= 3 sin t, 3 cos t, 1.
dt
Problem 3.
Speed: |V | =
9 sin2 t + 9 cos2 t +
18.02 Practice Exam 2 A
Problem 1. (10 points: 5, 5)
Let f (x, y) = xy x4 .
a) Find the gradient of f at P : (1, 1).
b) Give an approximate formula telling how small changes x and y produce a small change
w in the value of w = f (x, y) at the point (x, y)
MODEL ANSWERS TO HWK #11
(18.022 FALL 2010)
(1) (6.1.1)
(a) x0 = ( 3, 4) so |x0 | = 5, hence
Z
Z 2
f ds = 5
(2
x
3t + 8t
2)dt = 50 .
0
(b) x0 = ( sin t, cos t) so |x0 | = 1, hence
Z
Z
f ds =
(cos t + 2 sin t) = 4 .
x
p
0
0
0
(2) (6.1.3) x = (1, 1, 3 t/2)