1
LECTURE
Systems
Matrices
,
and
A
do
linear
of
,
Row
Reductions
they
do
equations
is
a
relate ?
collection
o
the form
equations of
alX
&
care/how
we
system
Ens
Forms .
Echelon
Why
Linear
of
2 ,
9 . 2x
+
,
,
X ,
+
92r x + . . .
+
+
...
G
,
nXn
AznX
+
=
b
,
b
=
:
ame X ,
where
We
can
1 )
.
2 )
.
3
We
Ax
X
,, Xz
.
)
+
ameX2 +...
, .., Yu
have
various
+
AmnYn
=
In
unknowns .
are
types of
solutions
:
consistent
Unique solution (the system
determined
Infinitely Manysolutions (the system is consistent
is
-
+
No
Solution
undetermined
(where
the system
is
inconsistent)
representa system compactly using matrices
where : A is the coefficient matrix
(mXn)
can
=
b
X
(A) (b
is
is
(nx1)
the column
rector
the com
rectorala
For example
2
Consider
This
,
becomes
[/]
↳
To
systems (systems of egations)
solve
elementary now operations
ID) We
:)
2
We
0)
3
these
snoop two
multiply
Add a
multiple
operations
are
:
nows
now
perform
we
(interchange)
by a
nonzero
of
now
one
reversible
scalar
another
to
and
preserve
set
the
soluties
.
We
talk
can
about
these
motrices
in
Echelon
Row Echelon Form (REF) if :
matrix is
1 ) all nonzero nows are above
rows
#
A
any
.
2) Each
of the
3)
A
All
leading entry of a
entry of the now
leading
entries below
a
matrix is
1)
It
now
is
in
a
above
of
all
zeros.
column to the right
.
it
leading enty are zeros
Reduced Row
REF
is in
Form :
Echelon
.
E
Form (RREF) if
is I called
leading entry in each nonzero
leading)
3 ) Each
leading I is the only nonzers in its whenn
. ) The
2
.
now
a
,
oudo
is
Example
↳
:
Given
[
Notes : We perform Gaussian elimination to get REF
Gauss-Jordan Elimination to get PREF
or
step by Step
1
.
) Make first entry of
2 )
Row 2
.
2
11
Rowl =
Pr
[
Row
Pr-2p
+
12
"
,
j 7.
31
J
=
1
↓
-
2
=
XI
1
:
=
-
0
-
7
-
2(3)
(2 ( 1))
) Make the leading entry
Row 2
already is
2(1)
-
.
3
-
(It
.
I·
.
=
-
leading 1
a
Row 2-2 Rowl
=
a
Row 2
1
-
=
=
of Row 2
1
1
+ 2=3
positive
Row 2
X2
X
[6 ?I J
This is RREF
solution
can
be
if the third
"free"
.
and the
read
directly
variable
is
Let's
discuss what it
means to have a
free variable
5
·
In RREF :
Leading vorables are variables that correspond to
E
see
Cortier
Page
-
columns
pivot
here
X ,
and X2.
Free variables
are
variables without a
T
o
has
1
in
leading
with
highlight.
column)
its
]
20 ? 1
Taking
pivot theres
-
·
(x ,
+
2x2
M
+
1x2
=
(x ,
-
+ 2x2
X
2
1xz
-
=
3xz
1xz
-
=
=
3xy
-
=
3
1
-
G
S
=
-)
Xz
=
-)
+ 3
X3
=
=>
(x ,
X,
x,
X
2(
+
2
-
X,
=
+
2
-
+
,
-
5x3
=
5
-
1
+
3xz)
6x3
+ 5x
=
5x3
-
-
xz
=
x3
=
3+2
=
3
3
3
#5-5x3
3xs
+
and
Xz = t
This
solution
our
means
form a line
solutions
t EIR ·
where
3-D
in
infinitely
has
We
can
Xy)
t
2
0
=
this
express
(X ,, X,
X
&
1
(5
5(
-1 + 3t
,
t 2
=
=
(5
=
-
when
-
+
5
started
we
1
,
,
↓ Remark
3(1)
1)
,
1)
3(2)
,
2)
2)
,
,
0)
,
. )
,
,
75
Remember
+
1
2
5(2)
-
.
5)
,
- 1 53(0)
.
10
J
Dece
173
-
-
-
-
solved
ordered trope
an
as
-
=
=
-
5(0)
25 ,
-
-
Xz zt
,
(5
5 5t
=
,
x2
(5-5t
=
EEIR
t
E
since
solutions. The
parameterized by to
↓
space ,
↓
To elaborate,
many
we
had :
any "reductions
before
37
[2
·
(x ,
+ 2x2
2x ,
+
-
Xz
=
3x2 + X
77
3
=
↑
When
Integration
we
a
che
started .
1(5) + 2(1)
0
-
=
3
After
Since
have
we
it
i
[ ]
[]
+
O
the
solution
Interjection
We
#Now
what
ordered triple
an
as
↓
is
set
.
All
I
Basic
do
we
mean
think
idea ,
rectors of
this
form.
INotation
triples.
ordered
use
-
,
e
completed,
we
vectors !
by
geometry.
I rector is an object that has magnitude (length) and
direction. For an example
the
on
arrow from
o (3 4)
-
origin
,
in the
plane
,
.
anything that is
a
: Fo is i
rector
.
has an
errow
Usually expressed above it
.
Mathematically a
,
(h)
it
.
has a
So
to
(3 4)
or
·
example ,
the system
describing
This is
is
list of numbers like
2-Dim rector because
ordered
an
called
a
components
this
in
vector
and
(5 , -1 , 0)
75 , 3 ,
how the solutions
1)
is
particular solution
direction Vector
is a
ast Varies.
a
,
change
In Linear
Algebra
terms : The solution set
equation
of
line ,
a
.
is
a
vector
solution set
=
+t
solution
particular
direction vector
↑
The solution
that passes
extends
CARE
Now
I
,
is
through and
,
a
a
line
IRS
in
Defined
:
by a point
direction rector along which it
.
END
We
set here
OF
APPLICATIONS +
ABOUT
REF
how
must
know
let's
practice
WE
REF and PREF
solve
(into
WHY
.
REF and
The
Consider
Matrix
OF
PREF
AND
to
CASES
USE
simplifications .
RREF
REF)
C
=
( 3
H
108
1 )
Start
.
original
R2
=
12
OI
3
108
R3-R ,
PEMDAS
(
22 - 1
5- 2
E
-
.
=
E
0
2 =
203
=
/
-
3
Ot O
R3
=
Rz + 2 R2
I
(
S
=>
colorl)
,
original
original original
=
Crowl
R2- 2 R
(
Ry
first pivot
with
123
01
-
00
3
-1
22
+
-
5+
(I
(6)
This
is
we
have
We
REF
can
.
.
2
.
this
matrix
to
3
-
00-1
last
1233
(1
(
=
-
,
,
the
right of
to
RREF
point.
continue
I
0
0
positive
pivot
0
to
(
0
I
S
Clear above first
R, = R
Rz
each
01
C
=>
pivot
below
123
) Make
each
zeros
use
S
1
because
pivot
3 Rs
2
,
3
-
3)
=
(1
,
2
,
0)
! 2)
⑳of
R2 + 3 Rs
120
(889 (
Row 2
becomes
=
10 , 1 , 0
previous
and
.)
3
Clear
Ri
Ri
=
(1
-
2R2
-
0
,
2
I
=>
Final
RREF
-
0
2
,
100
0
01
001
-0)
Matrix
G
=
(
,
0
I
man
%)
,
achieved
here
This
.
-
the
pivot
second
above
identity matrix meaning
,
by
the
chance
system
,
has
is
a
also
unique
-
solution
-
.
Extra
(x ,
+
0x + 0xz
0x ,
+
1x2
+
Oxy
0x + 0xz +
(xz
#
=
=
=
E
#
Points : REF
Hey
=
Each
pivot
Zeros
the
is to
right
of
the
below each pivot
need not be 1
.
Pivots
REF
Points :
y
-
All
pivot positions are
Zeros
above and
below
each
pilot
above
.
pivot
Examples
B
REF
Each
123
(
=
REF
of
014
002
because
above
Only
:
(
all
B
nonzero
(
=
rows
are
above
any
zera
in
column
nows.
leading entry pivot is to the
right of leading entry
below each Pilot
Zeros
.
it
RREF
c
(6
=
all
each
%8
.
oo 1
pivots
pivot
C
only
the
is
nonio
entry
PREF
D
=
(
Leading
Teres
01
-
03
is
land ?
nows
in
Edea: REF
,
3
O
abow
RREF
(
102
and
,
below
pivots
pivots are
1
each
.
pint
move right
plus
O
below
above each
each
pico
point.