Math 568
Row Echelon Form and Number of Solutions
1. Row Echelon Form In these notes we will define one of the most important forms of a matrix. It is one of the "easier" forms of a system to solve, in particular, only back-substitution is needed to compl
Mathematics 568 Au, Wi, Sp, Su (1st Term) Prerequisite:
3 cr.
Introductory Linear Algebra I
Mathematics 254. Not open to students with credit for 571. Catalog Description: The n-dimensional Euclidean space and its subspaces; matrices as mappings; matrix a
Math568 Homework II Solution
Lee, Gangyong
1
Exercises 2.1
1. Solution: x y +
3
5z = 0 is a linear equation.
11. Solution: We have 3x 6y = 0, so x = 2y . Hence, the solution set is
x = 2t
y=t
19. Solution: We have
x 2y = 1
y=3
So y = 3 and x = 2y + 1 = 2
1
Exercise 1.1
17. Solution:
xa
xa
( x a) 2 x
a x
(a x) + a
x
x
2
=
=
=
=
=
=
=
2(x 2a)
2x 4a
(2x 4a) 2x
4a
4a + a
3a
3a
Exercise 1.2
1
2
3. Solution: u = 2 and v = 3, so we have
3
1
uv = 12+23+31
= 2 + 6 + 3 = 11
1
2, so we have
9. Solution: u =
3
u
Henc
Quiz 12
Autumn 2013
Name:
Math 2568
Problem 1 (6 points)
1
1
1 2
0 2 1
2
. Find the eigenvalues of A, and for each eigenvalue of A, nd the algebraic
Let A =
1 1 3
2
0
0 1
0
multiplicity, the geometric multiplicity, and a basis for the associated eigen
Quiz 6
Autumn 2013
Name:
Math 2568
Problem 1 (6 points)
Are the vectors v1 = (0, 1, 1, 1) , v2 = (0, 0, 1, 1) , v3 = (1, 1, 1, 0) linearly independent? For full credit, be sure to
justify your answer with an explanation.
Solution
.
Problem 2 (6 points)
Co
Quiz 5
Name:
Autumn 2013
Math 2568
There are two problems on this quiz; make sure to look at the back.
Problem 1 (6 points)
Find the angle between the vectors a = (2, 1) and b = 3
1
2,
1
2
.
Solution
.
Problem 2 (6 points)
Let R be a real number, and set
Quiz 8
Name:
Autumn 2013
Math 2568
Problem 1 (6 points)
Let P be the vector space of degree 3 polynomials. Consider the subset
Q = cfw_p P : 2 p (1) + p (3) = 0.
Is Q a subspace of P ? If so, explain why. If not, explain why not.
Solution
.
Problem 2 (6 p
Quiz 7
Name:
Autumn 2013
Math 2568
Problem 1 (6 points)
1
0 2
1
10
2
1
1 6 7 which is row equivalent to 0 1 12 4 .
Consider the matrix A = 3
4 1
4
8
00
0
0
For full credit, nd a basis for (and dimension of) the null space of A, a basis for (and dimension
Quiz 11
Name:
Autumn 2013
Math 2568
Problem 1 (6 points)
Compute the eigenvalues and eigenvectors of the matrix
A=
1
1
6
6
.
Solution
.
Problem 2 (6 points)
43
4
26
38
7 4 4 6
is cfw_3, 4, 3, 5 , with associated eigenvectors
The set of eigenvalues of A
Math 568
Systems of Linear Equations and Matrices
In these notes, we define a linear system and their associated matrices. We also indicate the algebra which can be preformed on these objects.
1. Definitions and Notation A linear equation in n variables i
Math 568 (05/27/11)
Lecturer : Gangyong Lee
5.3 The Gram-Schmidt Process and
the QR Factorization
The Gram-Schmidt Process
1
2
1 and x2 = 0
Example 5.12 Let W = span(x1 , x2 ), where x1 =
0
1
Construct an orthogonal basis for W .
1. Theorem 5.15 The Gram
Math 568 (05/25/11)
Lecturer : Gangyong Lee
5.2 Orthogonal Complements and
Orthogonal Projections
Orthogonal Complements
Denition Let W be a subspace of Rn . We say that a vector v in Rn is
orthogonal to W if v is orthogonal to every vector in W . The set
Math 568 (04/01/11)
Lecturer : Gangyong Lee
1.3 Lines and Planes
Lines in R2 and R3
1. Denition The normal form of the equation of a line l in R2 is
n (x p) = 0 or n x = n p
where p is a specic point on l and n = 0 is a normal vector for l.
The general fo
Math 568 (03/30/11)
Lecturer : Gangyong Lee
1.2 Length and Angle: The Dot Product
The Dot Product
1. Denition If u =
u1
u2
.
.
.
and v =
un
v1
v2
.
.
.
,
vn
then the dot product u v of u and v is dened by
u v = u1 v1 + u2 v2 + + un vn
1
3
Example 1.15
Math 568 (03/28/11)
Lecturer : Gangyong Lee
1.1 The Geometry and Algebra of Vectors
Vectors in the Plane
1. Denition of Vector A vector is a direct line segment that corresponds to
a displacement from one point A to another point B .
The vector from A to
Math568 Homework VIII Solution
Lee, Gangyong
1
Exercises 4.3
13
.
2 6
(a) characteristic polynomial
|A I | = 2 7 + 12
(b) eigenvalues
2 7 + 12 = 0, and we have 1 = 3 and 2 = 4.
(c) a basis for each eigenspace
2 3 0
3
3
1 = 3: [A 1 I |0] =
. So we have sol
Math568 Homework VII Solution
Lee, Gangyong
1
Exercises 4.2
7. Solution: By using cofactor expansion along the third row, we have
5 22
52
22
52
1 1 2 = 3
0
+0
= 3 (2 2 2 1) = 6
12
1 2
1 1
3 00
9. Solution: By using cofactor expansion along the third row,
Math568 Homework VI Solution
Lee,Gangyong
1
Exercises 3.6
1. Solution: A =
2 1
34
1
2
,u =
2 1
34
2 1
34
TA (u) = Au =
TA (v ) = Av =
1
. So the standard matrix is
1
1
1
and T (e3 ) =
00+1
0+03
. So we have
1
0
=
2
11
3
8
=
1
2
1+0
=
10
11
.
1 1
11. Solut
Math568 Homework V Solution
Lee, Gangyong
1
Exercises 3.3
11. Solution: The coecient matrix is A =
The inverse of A is A1 =
1
2315
x = A1 b =
3 1
5 2
3 1
5 2
21
.
53
3 1
=
. So the solution is
5 2
1
2
=
5
9
23. Solution:
ABXA1 B 1
ABXA1
ABX
BX
X
=
=
=
=
=
Math 2568 Homework #1 due: 20 Jan 2017 (at beginning of class or beforehand)
All these problems must be done by hand no software package or computer language
can be used.
Include the vertical line in augmented matrices, and put in the header information t