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i just need someone to check my awnsers and working XD thanks
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
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