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
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 =
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
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
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.
(Existe
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
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 ,
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
P
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,
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
.
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
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 ch
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 +