MA 3280 Lecture 16 - Spans and Linear Independence
Wednesday, March 5, 2014.
Objectives: Material from 4.6 and 4.7: The span of a collection of vectors, linear dependence, and linear
independence.
The
MA 3280 Practice Test I
Name
Monday, February 17, 2014
Note: Test I is Wednesday (2/19/14).
1.
Solve
2x1
x1
+
+
3x2
2x2
=
=
4
.
1
2a.
The set of solutions to a linear equation in two unknowns, a1 x1 +
MA 3280 Practice Test I Answers in Red
Name
Monday, February 17, 2014
Note: Test I is Wednesday (2/19/14).
Solve
1.
2x1
x1
+
+
3x2
2x2
=
=
4
. ( 5, 2 )
1
2a.
The set of solutions to a linear equation
MA 3280 Lecture 22 - Eigenvectors and Eigenvalues
Monday, March 31, 2014.
Objectives: Material from 9.2: Eigenvectors and eigenvalues
Existence of a unique solution
We kind of already know this, but l
MA 3280 Lecture 23 - More on Eigenvectors
Wednesday, April 2, 2014.
Objectives: Discuss the diagonalization of a matrix.
There is something called a diagonalization for a matrix. This comes easily fro
MA 3280 Lecture 21 - Properties of Determinants
Friday, March 28, 2014.
Objectives: Material from 8.6: Basic properties of determinants.
Lets start by stating some of the basic properties of determina
MA 3280 Lecture 20 - Minors and Cofactors
Wednesday, March 26, 2014.
Objectives: Material from 8.7.
Another way of describing a determinant is in terms of minors and cofactors. Lets look at our deniti
MA 3280 Lecture 19 - Determinants
Monday, March 24, 2014.
Objectives: Material from 8.2 and 8.3: Order 1, 2, and 3 determinants.
At the beginning of the semester, and when we talked about nding invers
MA 3280 Lecture 18 - Bases
Friday, March 14, 2014.
Objectives: Material from 4.8 and 4.9: Bases, dimension, and nding bases from spanning sets.
Basis. Suppose we have a collection of vectors cfw_ u1 ,
MA 3280 Lecture 17 - Linear Independence
Friday, March 7, 2014.
Objectives: Material from 4.7: linear dependence and linear independence.
Consider the vectors u1 = 1, 2, 1 , u2 = 0, 2, 1 , u3 = 1, 0,
MA 3280 Practice Test II
Name
Monday, March 10, 2014
Note: Test II is Wednesday (3/12/14).
1.
Solve the equation
1
0
3
1
x1 1 + x2 1 + x3 5 = 2 .
1
2
8
2
(1)
2.
Let u = 1, 2, 3, 4 and v = 1, 4, 3, 2