b) Since we have a linear relationship between the columns, column 3 can not be a pivot
column, so the rank is at most 2. The problem also tell us that the three a, vectors point in-
different directi
3 (30 pts.) Let A be any matrix and R it's row reiduee'd eehelon form. Answer True
or False to. the statements below and briey explain. (Note, if there are
any eountereXamples to a statement below you
4 (28 13135.) A Sudoko puzzle solution such as the example on the last page is a 9:59.
matrix A that among other properties has the numbers 1 through 9 once .in-
every row and in every column.
Hint 1:
MATH 3000 (Azoff)
Linear Algebra
Fall 2009
Last updated December 18.
Grades have been reported to the registrar. The median on the final exam was 153 out
of 200.
Exams are available for pick-up.
email
MATH 3000 (Azoff) Fall 2009
Study Guide for Third Hour Test (Section 3.6 and Chapters
4, 5, and 6)
Section 3.6:
Abstract Vector Spaces (up through Example 9, i.e., not
responsible for inner products)
MATH 3000 (Azoff) Fall 2009
Study Guide for Chapters 1 and 2 (First Hour Test)
Definitions
Rn
unit vector
line in Rn (parametric form)
hyperplane in Rn (cartesian or normal form)
linear combination of
MATH 3000 (Azoff) Fall 2009
Study Guide for Chapters 1 and 2
Definitions
Rn
unit vector
line in Rn (parametric form)
hyperplane in Rn (cartesian or normal form)
linear combination of vectors
dot produ
MATH 3000 (Azo)
Fall 2009
Practice First Exam with Answers (100 points)
1 (16 points). Find the general solution to the following system of three equations
in four unknowns.
x1 + 3 x 2 x 3 3 x 4 = 0
2
MATH 3000 (Azo)
Fall 2009
Practice First Hour Exam (100 points)
1 (16 points). Find the general solution to the following system of three equations
in four unknowns.
x1 + 3 x 2 x 3 3 x 4 = 0
2 x1 + 6
MATH 3000 (Azo)
Fall 2009
Notes to Problems from Section 3.6
3.6.2. (Parts a), c), and e) were assigned.) Decide which sets of vectors are linearly independent.
10
01
11
a) The set of two-by two matri
MATH 3000, FALL 2011
INSTRUCTOR: DANIEL KRASHEN
Basic
Course Information
Lectures: MWF 10:10-11:00, Boyd 222
Oce Hours: Wednesdays 11:00-12:00, Boyd 437 or by appointment
Oce Phone: 542-2555
Email: dk
3 (30 pts.) Suppose the nonzero vectors a1. :12, :13 point in different. directions in R3 but
3:11 + 2:12 | cfw_223 2 zero vector .
The matrix A has those vectors 11.11 [12, ea in its columns.
(a) Des
3. [Fr-om fall 2006 exam 1, problem 4.
(a) If A is a 3*byn5 matrix, what information do you have about the nullspace of A?
(b) In the vector space M of all 3 X 3 matrices? what subspace is spanned by
QUIZ l ANSWERS
l 1 D
a). cfw_12 points We reduce: 4 1 bg r G 1 b2 4b1 4.- D 1 g 4391
2 D D
So the equation becomes:
cfw_3, ?\.(x1)_cfw_ $4, \_
K0 o 3- Ks3+s3 ms.)
b (6 points) ICllnly when b3 +451 bl
4 (28 13135.) Start with this 2 by 4 matrix:
A: [2 3 1 1l
P [6932]
(a) Find all special Solutions to Act : D and describe the. nullsijace ofA.
(b) Find the complete solutionmeaning all solutions (3:11
Solutions
The solutions for all problems from previous exams are posted on the ISo web page. Solutions to the rst four
problems are:
1. The differences between the solutions must be in the nullspace.
18.06 Professor Strang Quiz 1 March 3, 2008'
Grading
1
Your PRINTED name is:
2
3
4
Please circle your recitation:
1 M 2 2-131 A. Ritter 2-0285 2-1192 afr
2 M 2 4-149 A. Tievsky 2-492 3-4093 tievsky
3
18.06 Quiz 1 Solution
Hold on Monday, 2 hiarch 2009 at 11am in Walker Gym.
Total: 60 points.
Problem 1: Your classmate, Nyarlathotep, performed the usual elimination steps
to convert A to echelon form
1 (18 13135.) Start with an invertible 3 by 3 matrix A. Construct B by subtracting 4
times row 1 of A from row 3. How do you nd 51 from. Al? You
can anst in matrix notation, but you must also answer i
Some practice problems
The 18.96 web site has exams from previous terms that you can download, with solutions. Ive listed a few practice
exam problems that I like below, but there are plenty more to c
b) If we did it using Gauss-Jordan elimination:
[1111001 111100
1 3 3 0 1 0. W '0 2 2 1 1 '0
1 3 7 0 0 1 '0 2 6 1 0 1
W '0 2 2 1 1 0
004011
(This is L1 on the right hand side, so we can multiply and s
18.06 Spring 2009 Exam 1 Practice
General comments
Exam 1 covers the rst'S lectures of 18.06:
1. The Geometry of Linear Equations
2. Elimination and Matrix lOperations (elimination, pivots, etcetera;
The nullspace is .all linear cemb'inatiens of these twe vecters; it will be a plane in R4.
b) One way to strive fer a particular selutien is just te leek at the set-up: em in is the negative
ef the 4t
and (from the rst row) 1:1 2 1 4:132 = 5; setting 33; = 0,174 = 1, we get (from the
second row) 1:2 2 3 and (from the rst row):t1 = 3 4:52 2 15. Hence, N(A) is
spanned by two special solutions as foll
Elimination on A leads to U:
1 1 11M M
AI=133 552:0
2 (24 pts.)
137133 1
leads to
[111 I] 01
cfw_Ia-"2022 153:0
0'04 1:3 1
(a) Factor the rst matrix A into A = LU and also into A = LDLT.
1:) Find the