Can someone check my work? EASY PEASY MATRICES

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The forum discussion revolves around solving systems of linear equations using augmented matrices and Gauss reduction. The user presents multiple problems, including writing systems of equations from given matrices and determining reduced row echelon forms. Key solutions include the correct identification of augmented matrices for given equations and the application of row operations to achieve reduced echelon forms, confirming that the user successfully solved for variables x, y, and z in various scenarios.

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  • Understanding of linear equations and systems
  • Familiarity with augmented matrices
  • Knowledge of Gauss elimination and row operations
  • Basic proficiency in matrix algebra
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  • Learn about matrix operations and their applications in solving systems of equations
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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
 
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You're more likely to get a response if you don't throw a whole laundry list of problems at us.
 
1 a and b are correct
 
Mark44 said:
You're more likely to get a response if you don't throw a whole laundry list of problems at us.

sorry,
 

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