Sketch of Lecture 5
Example 33. If v1 =
2
1
Tue, 9/1/2015
1
1
and v2 =
, then c1v1 + c2v2 =
2c1 c2
c1 + c2
.
Example 34. (Geometric description of R2) A vector x1 represents the point (x1, x2) in
x2
the plane.
Better: an arrow from the origin to (x1, x2)
Sketch of Lecture 6
Thu, 9/3/2015
Working with spans
Review 41. spancfw_v1, v2,
, vm is the set of all linear combinations
x1v1 + x2v2 +
where x1, x2,
Example 42.
If so,
, xm can be any real numbers.
1
Is 1 in
0
1
1 as
write
0
Solution.
+ xm v m ,
span
Sketch of Lecture 8
Thu, 9/10/2015
The transpose of a matrix
Denition 62. Interchanging the rows and columns of A produces its transpose AT .
Example
1
0
(a)
2
(b)
1 2
3 0
(c)
63.
2 T
1 =
4
1 3
3 0
1 0 2
2 1 4
T
=
T
=
1 3
3 0
A matrix A such that AT = A
Sketch of Lecture 12
Thu, 9/24/2015
Review. Linear independence, matrix inverses and such
Example 89. Suppose that A is a 3 3 matrix, and that Ax = 0 has the solution x =
[ 2 1 3 ]T . What does that tell us about the columns of A?
col 1
col 2
col 3
0
o
Sketch of Lecture 13
Tue, 9/29/2015
What we have learned so far
Linear systems
Systems of equations can be written as Ax = b.
x1 2x2 = 1
x1 + 3x2 = 3
1 2
x=
1 3
Sometimes, we represent the system by its augmented matrix
Thirdly, we can write the system
Sketch of Lecture 14
Tue, 10/6/2015
Example 95. Find the general solution to
Solution. We eliminate
1 2 1 4 3
1 0 1 0 1
>
1 2 1 4
x=
1 0 1 0
RREF
(do it!)
3
.
1
1 0 1 0 1
.
0 1 1 2 1
We set x3 = s1 and x4 = s2 because these are our free variables. Then, x
Sketch of Lecture 10
Thu, 9/17/2015
1
1
1
Review 76. Are the vectors 1 , 2 , 1 linearly independent?
1
3
3
Solution. The vectors are linearly independent if and only if the system
1
1
1
1
2
3
1
0
1 x = 0
3
0
has only the trivial solution x = 0. By t
Sketch of Lecture 11
Review.
3 1
2 1
Never write
A
B
1 1
2 3
Tue, 9/22/2015
1 0
. Hence,
0 1
=
1
3 1
2 1
1 1
.
2 3
=
for matrices! Why?
Because it is unclear whether you mean AB 1 or B 1A. (And order matters a lot with matrices!)
Example 82. Solve
3 1
x=
Sketch of Lecture 16
Thu, 10/15/2015
Let us begin with recalling the denition of a basis and rephrasing Theorem 105.
Review. Three equivalent ways to determine whether a set of vectors (from V ) is a basis of
a vector space V :
if the vectors span V and
Sketch of Lecture 4
Review 24. The augmented matrix
Thu, 8/27/2015
1
0
0
2
0
0
0
1
0
2 1
3
1
1
1
is in echelon form but not in (row-)reduced echelon form (RREF). To put it in RREF, we do
R1 2R3 R1
1
2 0
0 3
R2 3R3 R2
>
0
0 1
0 2 .
0
0 0
1
1
We can now r
Sketch of Lecture 3
Tue, 8/25/2015
The row-reduced echelon form
Review. Let us solve the following system by Gaussian elimination.
x1 2x2 + x3 = 0
3x1 4x2 5x3 = 8
4x1 + 5x2 + 9x3 = 9
First, we write down the augmented matrix, then we perform elementary ro
Sketch of Lecture 2
Thu, 8/20/2015
Matrix notation
2 4 2
1 3
3
2x1 4x2 = 2
x1 + 3x2 = 3
(augmented matrix)
Example 11. What is the augmented matrix for the following linear system?
x1 + 2x2 x3 = 1
x 2 + x3 = 2
2x1 + 4x2 + x3 = 5
Example 12. Let us solve t
TEMA DE PROIECTARE PEISAGISTIC - MASTER ANUL I
SPAII TERASATE DE RELAXARE I ODIHN NTR-UN PARC
Pe un teren denivelat, n cadrul unei dilatri spaiale dintr-un parc, se va crea o succesiune de spaii amenajate
compoziional terasat, pentru odihn i relaxare. Dil
Sketch of Lecture 1
Tue, 8/18/2015
Introduction to systems of linear equations
Denition 1. A linear equation in the variables x1, ., xn is an equation that can be written as
a1x1 + a2x2 +
for some numbers b, a1, a2,
+ a n xn = b
, an .
Example 2. Which of
Sketch of Lecture 15
Tue, 10/13/2015
Review. spaces, basis, dimension, col(A), row(A), null(A)
Example 104.
1 3
2 4
(a) Consider V = col
Since the vectors
1
2
,
3
4
. By denition, V = span
1
,
2
3
4
.
are linearly independent (it is only two, and they are