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
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