Results 1 to 13 of 13

Math Help - Show that the system of equation is consistent

  1. #1
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Show that the system of equation is consistent

    Show that the system of equation <br />
x - 3y -8z = -10;<br />
<br />
3x+y-4z=0;<br />
and <br />
2x+5y+6z =13<br />
is consistent and find the solution .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zorro View Post
    Show that the system of equation <br />
x - 3y -8z = -10;<br />
<br />
3x+y-4z=0;<br />
and <br />
2x+5y+6z =13<br />
is consistent and find the solution .
    One way of doing this:

    Express the sytem in matrix form: A X = B.

    Construct the coefficient matrix A and show that the \det(A) \neq 0.

    Solve A X = B: A X = B \Rightarrow X = A^{-1} B.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    I am still not getting the anw=swer

    Quote Originally Posted by mr fantastic View Post
    One way of doing this:

    Express the sytem in matrix form: A X = B.

    Construct the coefficient matrix A and show that the \det(A) \neq 0.

    Solve A X = B: A X = B \Rightarrow X = A^{-1} B.

    I getting the det (A) = 0

    I dont know what to do??????
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zorro View Post
    I getting the det (A) = 0

    I dont know what to do??????
    OK, the determinamt is equal to zero (I should have checked). This means that either there's no solution or an infinite number of solutions. So my suggestion is no good after all.

    In fact, since x = 3y + 8z from the first equation, the second and third equations become:

    y + 2z = 0 and y + 2z = 13/2.

    Therefore the equations are INconsistent and there is no solution to be found.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Do u have a links for this

    Quote Originally Posted by mr fantastic View Post
    OK, the determinamt is equal to zero (I should have checked). This means that either there's no solution or an infinite number of solutions. So my suggestion is no good after all.

    In fact, since x = 3y + 8z from the first equation, the second and third equations become:

    y + 2z = 0 and y + 2z = 13/2.

    Therefore the equations are INconsistent and there is no solution to be found.
    Mr fantastic could u please provide me with a link for solving such type of question's.....
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by mr fantastic View Post
    OK, the determinamt is equal to zero (I should have checked). This means that either there's no solution or an infinite number of solutions. So my suggestion is no good after all.

    In fact, since x = 3y + 8z from the first equation, the second and third equations become:

    y + 2z = 0 and y + 2z = 13/2.

    Therefore the equations are INconsistent and there is no solution to be found.
    Unfortunately x=1, y=1, z=1 IS a solution (So the system is consistent as at least one solution exists. The given solution is not unique!).

    CB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by zorro View Post
    Show that the system of equation <br />
x - 3y -8z = -10;<br />
<br />
3x+y-4z=0;<br />
and <br />
2x+5y+6z =13<br />
is consistent and find the solution .
    HI

    x-3y-8z=-10 ---- 1

    3x+y-4z=0 -----2

    2x+5y+6z=13 ---- 3

    1x2 -3 y+2z=3\Rightarrow z=\frac{3-y}{2} ---4

    Substitute 4 into equation 3 .

    2x+5y+6(\frac{3-y}{2})=13

    x=2-y

    so (x,y,z)=(2-y , y , (3-y)/2)

    (x,y,z)=y(-1,1,-1/2)+(2,0,3/2) where y\in R

    and if y=1 , (x,y,z)=(1,1,1) which agrees with CB .

    Not sure if this is right though .
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    I couldnt understand

    Quote Originally Posted by mathaddict View Post
    HI

    x-3y-8z=-10 ---- 1

    3x+y-4z=0 -----2

    2x+5y+6z=13 ---- 3

    1x2 -3 y+2z=3\Rightarrow z=\frac{3-y}{2} ---4

    Substitute 4 into equation 3 .

    2x+5y+6(\frac{3-y}{2})=13

    x=2-y

    so (x,y,z)=(2-y , y , (3-y)/2)

    (x,y,z)=y(-1,1,-1/2)+(2,0,3/2) where y\in R

    and if y=1 , (x,y,z)=(1,1,1) which agrees with CB .

    Not sure if this is right though .

    How did u get y+2z=3\Rightarrow z=\frac{3-y}{2} ---4
    could u please explain
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Nov 2009
    Posts
    263

    Smile

    Quote Originally Posted by zorro View Post
    How did u get y+2z=3\Rightarrow z=\frac{3-y}{2} ---4
    could u please explain
    ((multiply equation 1 by 2) - equation 3) divide by -11

    Quote Originally Posted by mathaddict View Post
    HI

    x-3y-8z=-10 ---- 1

    3x+y-4z=0 -----2

    2x+5y+6z=13 ---- 3

    1x2 -3 y+2z=3\Rightarrow z=\frac{3-y}{2} ---4

    Substitute 4 into equation 3 .

    2x+5y+6(\frac{3-y}{2})=13

    x=2-y

    so (x,y,z)=(2-y , y , (3-y)/2)

    (x,y,z)=y(-1,1,-1/2)+(2,0,3/2) where y\in R

    and if y=1 , (x,y,z)=(1,1,1) which agrees with CB .

    Not sure if this is right though .
    the solutions are the set \{(x,y,z)|(x,y,z)=t(-1,1,-1/2)+(2,0,3/2), t \in \mathbb{R}\}
    there's an infinite number of solutions
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by dedust View Post
    ((multiply equation 1 by 2) - equation 3) divide by -11



    the solutions are the set \{(x,y,z)|(x,y,z)=t(-1,1,-1/2)+(2,0,3/2), t \in \mathbb{R}\}
    there's an infinite number of solutions
    All of which is just of just academic interest here, to show consistency it is sufficient to show that there is at least one solution.

    CB
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    OK, the determinamt is equal to zero (I should have checked). This means that either there's no solution or an infinite number of solutions. So my suggestion is no good after all.

    In fact, since x = 3y + 8z -10 from the first equation, Mr F says: An embarrassing correction in red. So what follows is wrong.

    the second and third equations become:

    y + 2z = 0 and y + 2z = 13/2.

    Therefore the equations are INconsistent and there is no solution to be found.
    Therefore the second and third equations become y + 2z = 3 and y + 2z = 3. Therefore there are an infinite number of solutions (the one given by CB is once such). These solutions can be expressed in parametric form. The system is consistent.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523
    sorry, but i am still unable to understand this ????????????
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    I'm not sure if this is a right approach to this problem due to teaching myself the material but could you not just use gauss-jordan elimination.

    The last row would be

    0x +0y + 0z = 0

    therefore showing that there is infinite number of solutions
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need to show that Ax=w is consistent
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: September 18th 2011, 04:20 PM
  2. Replies: 3
    Last Post: September 13th 2011, 06:08 AM
  3. how do i show this is a consistent estimator?
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: December 3rd 2010, 04:27 PM
  4. Show that Equations are Consistent
    Posted in the Advanced Algebra Forum
    Replies: 16
    Last Post: September 22nd 2010, 03:59 PM
  5. Show that a system is controllable
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: May 4th 2009, 05:26 PM

Search Tags


/mathhelpforum @mathhelpforum