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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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March 21, 2022
Number of pages
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Written in
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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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