Results 1 to 10 of 10

Math Help - Algebra: find all the solutions of each system of equations?

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    46

    Algebra: find all the solutions of each system of equations?

    Algebra: find all the solutions of each system of equations?

    X^1 - 2X^2 + X^3 = 1
    X^2 - 2X^3 + X^4 = 0
    X^3 - 2X^4 + X^5 = 2
    X^4 - 2X^5 + X^6 = 0
    X^5 - 2X^6 + X^7 = 1

    im not sure how to start such a big system of equations ....do you have to isolate all the different x's(x^1,x^2....)?
    any help much appreciated!!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Question: does x^2 mean (as it usually does) x^{2}? Or does it mean (as I suspect it does) x_{2}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    46
    ya your right its the second one!!! you said sorry i dont know how to do that!!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Interesting. Your system looks like a discretization of a DE. Anyway. Here's your system in matrix form:

    \begin{bmatrix}<br />
1 &-2 &1 &0 &0 &0 &0\\<br />
0 &1 &-2 &0 &0 &0 &0\\<br />
0 &0 &1 &-2 &1 &0 &0\\<br />
0 &0 &0 &1 &-2 &1 &0\\<br />
0 &0 &0 &0 &1 &-2 &1<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\\ x_{6}\\ x_{7}<br />
\end{bmatrix}=\begin{bmatrix}<br />
1 \\0 \\2 \\0 \\1<br />
\end{bmatrix}.

    Correct? I would start doing ERO's on the augmented matrix

    \left[\begin{matrix}<br />
1 &-2 &1 &0 &0 &0 &0\\<br />
0 &1 &-2 &0 &0 &0 &0\\<br />
0 &0 &1 &-2 &1 &0 &0\\<br />
0 &0 &0 &1 &-2 &1 &0\\<br />
0 &0 &0 &0 &1 &-2 &1<br />
\end{matrix}\left|\begin{matrix}<br />
1 \\0 \\2 \\0 \\1<br />
\end{matrix}\right].

    What do you get?

    [EDIT] Actually, you can do the back substitution immediately, because your matrix is upper triangular. I think you'll need two parameters to describe all of the solutions.
    Last edited by Ackbeet; September 22nd 2010 at 11:14 AM. Reason: Back Substitution.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2010
    Posts
    46
    im confused how you could do back substitution here i thought you had to have a solution for one value of x immediately to do this??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Sorry. See the edit in Post # 4.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2010
    Posts
    46
    im confused how you could do back substitution here i thought you had to have a solution for one value of x immediately to do this??
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Well, you don't have enough equations to get a unique solution. You're going to have a two-parameter family of solutions (I get two from the fact that you have 7 unknowns and 5 equations. 7 - 5 = 2.) What I would do is use two parameters s=x_{6} and t=x_{7}. Then you take the last equation there and solve for x_{5} as follows:

    x_{5}-2x_{6}+x_{7}=1, so

    x_{5}=1+2x_{6}-x_{7}=1+2s-t. Now that you have x_{5}, you can proceed to obtain x_{4}, and so on. Make sense? Your final answer should look like this:

    \begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\\  x_{6}\\x_{7}\end{bmatrix}=\begin{bmatrix}a\\b\\c\\  d\\e\\f\\g\end{bmatrix}+s\begin{bmatrix}h\\i\\j\\k  \\l\\m\\n\end{bmatrix}+t\begin{bmatrix}o\\p\\q\\r\  \u\\v\\w\end{bmatrix},

    where a through r are known, and u through w are known. You've already got the following:

    \begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\\  x_{6}\\x_{7}\end{bmatrix}=\begin{bmatrix}a\\b\\c\\  d\\1\\0\\0\end{bmatrix}+s\begin{bmatrix}h\\i\\j\\k  \\2\\1\\0\end{bmatrix}+t\begin{bmatrix}o\\p\\q\\r\  \-1\\0\\1\end{bmatrix}.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Mar 2010
    Posts
    46
    Sorry about the delay I have all the values of x in terms of s and t
    X1=12+6s-5t
    X2=8+5s-4t
    X3=5+4s-3t
    X4=2+3s-2t
    X5=2s-t+1
    x6=s
    X7=t

    But I'm not sure how you got your final answer with this or is it ok to leave it in the form I have it
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Your answer is equivalent in form to the form that I gave. Which one is preferable depends on your teacher. If I were teaching, I'd accept either. To convert from one to the other, just read off the coefficients. For example:

    \begin{bmatrix}a\\b\\c\\d\\e\\f\\g\end{bmatrix}=\b  egin{bmatrix}12\\8\\5\\2\\1\\0\\0\end{bmatrix},

    which is the constant vector. Make sense?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Analytical solutions for system of diff equations?
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: April 11th 2011, 03:10 PM
  2. solutions to system of equations
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: March 17th 2011, 05:26 AM
  3. System of linear equations - types of solutions.
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: August 28th 2010, 11:47 AM
  4. Find solutions of linear system
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 19th 2010, 02:58 PM
  5. Replies: 1
    Last Post: September 1st 2009, 07:22 PM

Search Tags


/mathhelpforum @mathhelpforum