EXERCISE SET 1.1
1.
(b) Not linear because of the term x1x3.
2
(d) Not linear because of the term x1 .
(e) Not linear because of the term x3/5.
1
7.
Since each of the three given points must satisfy the equation of the curve, we have the
system of equatio
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Introductory Sociology CLEP Quick Prep Sheet
Welcome to the Quick Prep Sheet for the Introductory Sociology CLEP!
Sociology is the scientific study of social structure, or patter
Mohammed Ali
Omar Faruk
Mrs. Bibbs 4th hr.
02/22/16
Malcolm X
On the 21st of February 1965 in Omaha, Nebraska, a little boy
that went by the name of Malcolm X was born. Little did the
people know, he was soon to be a person that gave electrifying
speeches
Sitting in my 4th grade English class listening to a Malcom xs speech was the first
time I heard of Malcom x. I remember my teacher telling us about how influential he
was to the African American people. Hearing about the great man, I decided to learn
mor
Omar Faruk
10/23/15
5th hour
If only one sport could exist in the world, what would it be?
If only one sport could exist in the world, it would be wild. To support that claim, the
obvious answer would be boxing. Official boxing athletes are always legenda
McMaster University
Department of Mathematics & Statistics
MATH 1B03 Linear Algebra I
Fall 2010
COURSE DESCRIPTION.
Vector spaces given by solutions to linear systems. Linear independence, dimension. Determinants. Eigenvalues, eigenvectors and diagonalisa
EXERCISE SET 11.1
1.
(a) Substituting the coordinates of the points into Eq. (4) yields
x
1
2
1
1 =0
1
y
1
2
which, upon cofactor expansion along the rst row, yields 3x + y + 4 = 0; that is,
y = 3x 4.
(b) As in (a),
x
0
1
1
1 =0
1
y
1
1
yields 2x + y 1 =
EXERCISE SET 10.1
3.
(b) Since two complex numbers are equal if and only if both their real and imaginary parts
are equal, we have
x+y=3
and
xy=1
Thus x = 2 and y = 1.
5.
(a) Since complex numbers obey all the usual rules of algebra, we have
z = 3 + 2i (1
EXERCISE SET 7.1
1.
(a) Since
3
det( I A) = det
8
0
= ( 3)( + 1)
+ 1
the characteristic equation is 2 2 3 = 0.
(e) Since
det( I A) = det
0
0
2
=
the characteristic equation is 2 = 0.
3.
(a) The equation (I A)x = 0 becomes
3
8
0 x1 0
=
+ 1 x2 0
EXERCISE SET 5.1
11.
This is a vector space. We shall check only four of the axioms because the others follow
easily from various properties of the real numbers.
(1)
(4)
(5)
(6)
If f and g are real-valued functions dened everywhere, then so is f + g. We m
EXERCISE SET 4.1
3.
We must nd numbers c1, c2, c3, and c4 such that
c1(1, 3, 2, 0) + c2(2, 0, 4, 1) + c3(7, 1, 1, 4) + c4(6, 3, 1, 2) = (0, 5, 6, 3)
If we equate vector components, we obtain the following system of equations:
c1 + 2c2 + 7c3 + 6c4 = 0
3c1
EXERCISE SET 3.1
1.
z
(a)
z
(c)
(3, 4, 5)
(3, 4, 5)
y
y
x
(e)
x
( 3, 4, 5)
( j)
z
z
(3, 0, 3)
y
y
x
x
3.
(a)
(e)
5.
P2
P
1
P P2 = (3 4, 7 8) = (1, 1)
1
= (2 3, 5 + 7, 4 2) = (5, 12, 6)
(a) Let P = (x, y, z) be the initial point of the desired vector and a
Mill refers to greatest happiness principle as the moral
foundation of his theory of utilitarianism. In his words, pleasure
from pain is the only things desirable as ends. Utilitarianism was
accused of being a swinish doctrine because it only appeals to
p