True or false:
In a row reduction: if the original augmented matrix has no column of zeros, then the rref
cannot have a column of zeros.
True. If you look at any nonzero column, you could use row replacements to turn all
the nonzero elements in the column into Ñ' except for one. When a column is reduced to
having exactly one nonzero element, there ie no ERO that could be used to convert that
last number into a !.
Whenever a system has free variables, then the system has infinitely many solutions.
False: the system might be inconsistent. See the Existence and Uniqueness
Theorem below (discussed in Lecture 2)
Existence and Uniqueness Theorem
Let E be the augmented matrix for a system of linear equations.
ñ The system is inconsistent if and only if
a row with form Ò! ! ! ! ÞÞÞ ,Ó, where
, Á !, appears in an echelon form of E
( equivalently, if and only if the row
Ò! ! ! ÞÞÞ "Ó appears in the row reduced
echelon form of E )
ñ Otherwise, the system is consistent, and either
i) there is a unique solution (when there
are no free variables)
or
ii) there are infinitely many solutions
(when one or more variables are free)
What are the solutions of a system if the augmented matrix is
Ò" # $ % & 'Ó
The system has one equation, 5 unknowns; the matrix is already in rref
One basic variable, B" ; all other variables are free.
Solutions (infinitely many) B" œ ' #B# $B$ %B% &B&
Parametric solution, expressing basic variable in terms of free variables (the “parameters”)
What are the solutions of a system if the augmented matrix is
Ô1
0
Õ0
1
0
0
0
1
0
Ô1
Row reduce: 0
Õ0
0
1
0
1
0
0
2
0
1
0
1
0
Solutions (infinitely many)
1×
3 (an echelon fom, but not rref)
!Ø
0
1
0
2
0
1
1× Ô1
3 µ 0
!Ø Õ0
1
0
0
0
1
0
Ú
B" œ " B#
Ý
Ý
Ý
Ý B# is free
Û B$ œ $ B%
Ý
Ý
Ý
Ý B% is free
Ü B& œ !
0
1
0
0
0
1
What are the solutions of a system if the augmented matrix is
Ô1
0
Õ0
1
0
0
0
1
0
0
1
0
2
0
0
1×
3
1Ø
No solutions: 3rd row shows the system is inconsistent.
1×
3
!Ø
Properties of addition and scalar multiplication of vectors
(from textbook)
If ?ß @ß A are vectors in ‘8 and -ß . are scalars:
?@
œ@?
Ð? @Ñ A œ ? Ð@ AÑ
?!
œ!?
œ?
? Ð ?Ñ œ ? ? œ !
- Ð? @Ñ œ - ? - @
-Ð. ?Ñ
œ Ð-.Ñ?
Ð- .Ñ? œ - ? . ?
"?
œ?
___________________________________________________________________
Every vector in ‘# can be expressed as a linear combination of @" and @# . For example, it
looks (approximately) like A œ &# @" "# @#
In the figure, @" and @# can be used to create a “grid” of parallel lines: a new kind of
“graph paper” where the “grid lines” are not perpendicular. On this “graph paper”, A is
located at Ð &# ß "# Ñ
These two vectors ? and @ in ‘$ lie in a plane in ‘3 (only a limited piece of which is
pictured). Analogous to the preceding picture, ? and @ can be used to make a
“coordinate grid” (“graph paper”) on this plane.
The points in this plane are all the possible linear combinations of ? and @;
So SpanÖ?ß @× is the set of all points in this plane.
In particular, Span {u, v} contains the line in ‘$
v and 0. See the figure below.
through u and 0 and the line through
(in ‘% )
Question:
Ô "×
Ô#× Ô "×
Ö #Ù
Ö!Ù Ö $Ù
IsÖ
Ù in Span Ö Ùß Ö
Ù ?
"
"
" Ÿ
Õ %Ø
Õ#Ø Õ #Ø
does the vector equation
That is,
Ô#×
Ô " × Ô #B" B# ×
Ô "×
$B#
Ö!Ù
Ö $Ù Ö
Ù
Ö #Ù
B" Ö Ù BÖ
ÙœÖ
Ù œÖ
Ù have a solution?
"
"
B" B#
"
Õ#Ø
Õ # Ø Õ #B" #B# Ø
Õ %Ø
That is, does the system of linear equations
Ú
Ý
Ý #B" B#
$B#
Û
Ý
Ý B" B#
Ü #B" #B#
œ
"
œ
#
œ "
œ
%
have a solution?
We solve the vector equation by solving the corresponding linear system (they are
equivalent) and we do that by row reducing the augmented matrix
Augmented Matrix œ
Ô
Ö
Ö
Õ
#
!
"
#
"
$
"
#
"× Ô1
#Ù Ö!
ÙµÖ
"
2
%Ø Õ#
"
$
"
#
"× Ô1
#Ù Ö!
ÙµÖ
"
!
%Ø Õ!
Ô1
Ö!
µÖ
!
Õ!
"
"
$
%
"× Ô1
$Ù Ö!
ÙµÖ
#
!
Ø
Õ
'
!
"
"
!
!
"×
$Ù
Ù á INCONSISTENT
(
'Ø
"
$
"
%
"×
#Ù
Ù
$
'Ø
The system of linear equations has no solution, so the equivalent vector equation
has no solution.
Ô
Ö
That says: Ö
"×
Ô#× Ô "×
#Ù
Ö!Ù Ö #Ù
Ù is not in Span Ö Ùß Ö
Ù
"
"
" Ÿ
Õ %Ø
Õ#Ø Õ %Ø
In general
B" @" ÞÞÞ B: @: œ ,
(a vector equation)
has the same solutions as the linear system whose
augmented matrix is
Ò @" @#
... @: l ,Ó
As illustrated above:
@"
@#
,
Ô#×
Ô " × Ô #B" B# ×
Ô "×
$B#
Ö!Ù
Ö $Ù Ö
Ù
Ö #Ù
B" Ö Ù BÖ
ÙœÖ
Ù œÖ
Ù
"
"
B" B#
"
Õ#Ø
Õ # Ø Õ #B" #B# Ø
Õ %Ø
has the same solutions as
Ú
Ý
Ý #B" B#
$B#
Û
B
B#
Ý
Ý "
Ü #B" #B#
Ò @"
Æ
Ô
Ö
Ö
Õ
#
!
"
#
@#
Æ
"
$
"
#
œ "
œ #
œ "
œ
%
,Ó
Æ
"×
#Ù
Ù
"
%Ø
for which the augmented matrix is