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Math Help - Solving a system of linear equations.

  1. #1
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    Solving a system of linear equations.

    Hi Everyone,

    I've started university lately and i'm currently taking a math class in linear algebra. I'm going well so far but i'm having some issues with solving a system of linear equations when there is a unknown value in the system of equation. I must be missing a point somewhere. but here i go.


    x - 2y + 3z = 1
    -3x + y - 3z = 2
    -2x - y + (kČ -2k)z = k +3


    I have to find k in order for the system to have at least one solution.
    Any help would be appreciated as the more i try to solve it the more i get confused.
    I think i'm getting confused by the (kČ-2k) section of the equation.

    thanks.
    Last edited by mr fantastic; September 21st 2010 at 02:45 PM. Reason: Edited title.
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Hyrules View Post
    Hi Everyone,

    I've started university lately and i'm currently taking a math class in linear algebra. I'm going well so far but i'm having some issues with solving a system of linear equations when there is a unknown value in the system of equation. I must be missing a point somewhere. but here i go.


    x - 2y + 3z = 1
    -3x + y - 3z = 2
    -2x - y + (kČ -2k)z = k +3


    I have to find k in order for the system to have at least one solution.
    Any help would be appreciated as the more i try to solve it the more i get confused.
    I think i'm getting confused by the (kČ-2k) section of the equation.

    thanks.
    find determinant of the system ... and on depending of the value of "k" you will have one, none or infinity of solutions ...
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  3. #3
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    ouch... Thanks for the answer unfortunatly the determinant is the next chapter in the math book. I have to do it some other way or should i say the long way.
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Hyrules View Post
    ouch... Thanks for the answer unfortunatly the determinant is the next chapter in the math book. I have to do it some other way or should i say the long way.
    in your case ...

    D= \begin{vmatrix}<br />
1&-2&3\\<br />
-3&1&-3\\<br />
-2&1&(k^2-2k)<br />
\end{vmatrix} = 1\cdot 1\cdot (k^2-2k) + (-2)\cdot (-3) \cdot  (-2) + 3\cdot (-3) \cdot  1 - \big{[} 3\cdot 1\cdot (-2) + (-2)\cdot (-3)\cdot  (k^2-2k) + 1 \cdot (-3) \cdot  1\big{]}  = . . . . solve this one

    if determinant of that is  D \neq 0 than sys have unique solution ....




    1° when  D\neq 0 and if at least one of  D_x\vee D_y\vee D_z\vee  \neq 0 than solutions are unique and given

    \displaystyle x = \frac {D_x}{D} , y= \frac {D_y}{D} , z = \frac {D_z}{D}


    2° if  D= 0 and if at least one of  D_x\vee D_y\vee D_z\vee  \neq 0 than system of equations have no solutions


    3° if  D= 0 and if  D_x= D_y= D_z= 0 than there can be :

    a) all sub determinants of D are zeros, and coefficients with the unknowns are proportional there are 2 cases :

    - all independent members of system are proportional with coefficient with unknowns than there are infinity of solutions
    - all independent members of system are not proportional with coefficient with unknowns than there are no solutions

    b) if at least one sub determinant of D are different from zero than there are infinity of solution




    Edit: ah sorry ... I assumed that you are familiar with this, because you post question on university sub forum .... sorry
    Last edited by yeKciM; September 21st 2010 at 01:36 PM.
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  5. #5
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    Maybe my post should be in pre-u section of this forum. I believe that the american and canadian school system is different. University for us i believe is college in the states though i'm not sure. Here we have the equivalent of elementary->highschool->college->University.
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  6. #6
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    Express x from the first equation and insert it to 2 and 3.
    You have two equations with y and z.
    Express y from one equation and insert it to the third.
    Finally you have
    a(k)z + b(k) =c(k).
    Now examine it.
    1) if a(k)<>0 there is only one solution.
    2) if a(k)=0 and b(k)=c(k) there are many solutions.
    3) if a(k)=0 and b(k)<>c(k) there is no solution.
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  7. #7
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    great that was the answer i was looking for ! Thanks.
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