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they get harder btw ;)

1:

a) Write out the system of linear equations in variables, x,y,z represented by the augmneted matrix

( 1 2 0 5)

( 3 1 -1 0)

( 0 -2 1 2)

my attempt at 1a:

x + 2y = 5

3x + y -z = 0

-2y + z =2

1b:

write down the augmented matrix corresponding to this system of linear equations:

4a + 2c - d = 8

4a + b +3c +2d = 0

b + c + d = 4

-3a + 8b = 12

(4 0 2 -1 8)

(4 1 3 2 0)

(0 1 1 1 4)

(-3 8 0 0 12)

i think that was it?

2a:

state whether each of the following matrices is in reduced echelon form. If it is not, then give a reason and say what single row operation is needed to bring it to reduced echelon form.

I)

1 0 0 1

0 1 0 -2

0 0 2 -5

not in reduced echelon form

divide p3 by 2

II)

1 1 0 3 0

0 0 1 2 0

0 0 1 2 1

not in reduced echelon form ( don't know how to do this one)

NOT SURE WHAT SINGLE OPERATION will bring to reduced echelon form

III)

1 0 -2 0

0 1 0 0

0 0 0 1

not in reduced echelon form(not sure how to remove the -2)

NOT SURE WHAT SINGLE OPERATION will bring to reduced echelon form

IV)

1 0 2 0 1

0 0 0 1 2

0 1 0 0 -3

not in reduced echelon form

Swap row three with row two

V)

1 0 1 0

0 1 0 1

0 0 1 1

not in reduced echelon form

row one - row three

VI)

0 1 5 0 1

0 0 0 1 2

0 0 0 0 0

not in reduced echelon form (there is a 5 in the way)

NOT SURE WHAT SINGLE OPERATION will bring to reduced echelon form

VII)

1 4 0 1

0 0 0 0

0 0 1 3

not in reduced echelon form(there is a 4 in the way)

NOT SURE WHAT SINGLE OPERATION will bring to reduced echelon form

VIII)

1 0 2 0 3

0 0 1 0 0

0 0 0 1 5

not in reduced echelon form

row one - (2*row2)

For the following systems of equations, write down the Augmented matrix and use Gauss reduction as defined in lectures to find it's reduced row echelon form. hence find the solution of the System

3a:

x + 3y + z = 4

2x + 6y -z = -1

3x +11y +5z =20

matrix/my solution,

(1 3 1 4)

(2 6 -1 -1)p2-2p1

(3 11 5 20)p3-3p1

=

(1 3 1 4)

(0 0 -3 -9)p2 swap p3

(0 2 2 8)

=

(1 3 1 4)

(0 2 2 8)p2/2

(0 0 -3 -9)

=

(1 3 1 4)p1-3p2

(0 1 1 4)

(0 0 -3 -9)

=

(1 0 -2 -8)

(0 1 1 4)

(0 0 -3 -9)p3/-3

=

(1 0 -2 -8)p1+2p3

(0 1 1 4)p2-p3

(0 0 1 3)

=

(1 0 0 -2)

(0 1 0 1)

(0 0 1 3)

that is the reduced echelon form

x = -2

y = 1

z = 3

3b:

3y+2z = 25

3x + y + 5z = 29

x + 2y + 5z = 28

my solution/matrix

(0 3 2 25)

(3 1 5 29)

(1 2 5 28)p3 swap with p1

=

(1 2 5 28)

(3 1 5 29)p2-3p1

(0 3 2 25)

=

(1 2 5 28)

(0 -5 -10 -55)p2/-5

(0 3 2 25)

=

(1 2 5 28)p1-2p2

(0 1 2 11)

(0 3 2 25)p3-3p2

=

(1 0 1 6)

(0 1 2 11)

(0 0 -4 -8)p3/-4

=

(1 0 1 6)p1-p3

(0 1 2 11)p2-2p3

(0 0 1 2)

=

(1 0 0 4)

(0 1 0 7)

(0 0 1 2)

this is the reduced echelon form

x = 4

y = 7

z = 2

4a LAST QUESTION

while on holiday in europe, bills' daily exspenses were:

in england, 20 on accomodation, 40 on food 30 on travel

in france, 60 on accomodation, 30 on food, 20 on travel

in spain, 40 on accomodation, 35 on food, 35 on travel

in total he spent 580 on accomodation, 530 on food and 420 on travel

using e,f and s write down a system of linear equations and solve using gauss reduction to find how many days bill spen tin each country, show all working with clearly labelled reduction steps.

MY ATTEMPT

england = x, france = y, spain = z (just to make it simpler)

20x + 60y + 40z = 580

40x + 30y + 35z = 530

30x + 20y + 35z = 420

matrix

(20 60 40 580)p1/20

(40 30 35 530)

(30 20 35 420)

=

(1 3 2 29)

(40 30 35 530)p2 - 40p1

(30 20 35 420)p3 - 30p1

=

(1 3 2 29)

(0 -90 -45 -630)p2/-90

(0 -70 -25 -450)

=

(1 3 2 29) p1-3p2

(0 1 0.5 7)

(0 -70 -25 -450)p3 + 70p2

=

(1 0 0.5 8)

(0 1 0.5 7)

(0 0 10 40)p3/10

=

(1 0 0.5 8)p1+(.5*p3)

(0 1 0.5 7)p2+(.5*p3)

(0 0 1 4)

=

(1 0 0 10)

(0 1 0 9)

(0 0 1 4)

this is the reduced echelon form

x = 10

y = 9

z = 4

10 days in england, 9 in france, 4 in spain

please check for me and thank you very much XD