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Summary Linear Algebra P4
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Everything you need to know for the midterm of Linear Algebra!
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WEEK 1 : 1.1 ,1.2 , 1.3 1.4
,
§ 1.1 Systems of Linear equations
System of linear equations : Collection of one ormore linear
equations inwoning the same variable ×, .
. . .
.
✗n
2
Systems are equivalent it they have the same Solution
set
a System of linear eauation has :
4 Solution → inconsistent
no
2 1 Solution →
consistent
→
3 a Solutions
many
| } { }
× , -2×2 + ✗3=0 1 -23 0
| -23
0 2 -8 8
2×2-8×3=8 0 2 -8
5 0 -5
5×1 -
5×3=10 5 0 -5 10
=
System of linear =
coefficient =
augmented
equations matrix matrix
mxn matrix , m = # rows and n= # columns
Elementary row Operations :
•
replace one equation by the sum of itself and
a multiple of an other equation
interchange equations 2
•
multipl AN terms in an equator by a Montero
•
constant .
2 matrices are row equivalent if a sequence of
Elementary row Operations that transformatie matrix into
the other .
ij
,§ 1.2 Row reduction and echo/ on forms
leading entry is the left most nonzero entry (in a non zero row )
a matrix is in echelon form If it has the
following Properties :
1 alt nonzero rows are above rows of an zeros
any
2 each
leading entry of a row is in a column to the sight
of the leading of the
entry row above it
3 ah COIOMN
entries in a below a
leading entry are zeros
additional conditions for redliced echelon form :
4 the
leading entry in each nonzero row is 1
5 each
leading I is the only Montero entry in its column
THM 1 Each matrix and
equivalent to
only
:
is row are one
reduced echelon matrix .
1
a pinot position in a matrix A corresponds to a
leading
in the redllced echelon form .
↳ a column with a pinot position is a pivot column
stop 1-4 forward Phase
✓ step 5 backward Phase
row reduction algoritmen :
1 start with left most nonzero column ,
this a
pivot column .
The pinot position is at the top .
2
select a nonzero entry in the Pivot column
as a piloot .
A
necessary , interchange
rows to moves this entry into pinot
position .
3 use row replacement Operations to Create
2-eros in an positions below the pivot .
4 cover alle rows until there are no more nonzero
rows to modity
, 5
begin with the rightmost pinot and work/ ng
Upward and to the left ,
Create zeros
above each pinot ,
If pivot is not
a I ,
make it I
by a
scaling Operation
the variables Xi that Cor respond -
to the pivot columns
are basic variaties ,
otherWise : free variable .
↳ solution is determined Choice of the
every by a
free variable .
{ Xs is free
→
parameter,
-
c description
THM 2 : Linear is consistent if And only If
a
System
the rightmost column of the augmented matrix
is not a pivot Column .
That is ,
it and
only If
an echelon form of the augmented matrix
has no row of the form [0 0 . .
.
b]
,
with b nonzero .
If it is consistent it either has :
•
Unique Solution no free variable
•
0
many Solutions at least I free var .
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