1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Please Help. Matrix with variables

  1. Aug 7, 2008 #1
    This is a pre-calculus problem. I don't know if I got the answer right, and I don't quite understand what I'm doing. I read the section several times and understand regular augmented matrices but this problem is confusing me a bit. There is no mention in the text about matrices with variables in the matrices. There is no answer in the back of the book.

    Find this matrix in rref form and for what values of x is this valid.

    matrix[1 x -1] My work so far:[x 1 -1] Switched order of rows
    [x 1 -1] [1 x -1]
    [-1 1 x] [-1 1 x] (continues below and to the left)



    [0 1-x^2 x-1] added multiples of middle row to other rows
    [1 x -1]
    [0 x+1 x-1]

    Added multiple of top row to middle row, x can't equal 1 or -1 (multiple is x/[(x +1)(x-1)])

    [0 1-x^2 x-1]
    [1 0 (x/(x+1)) -1]
    [0 x+1 x-1]


    [0 -(x^2-1) x-1]
    [1 0 -1/(x+1)] simplify
    [0 x+1 x-1]

    [1 0 -1/(x+1)] Simplify to rref. Again, x can’t be 1 or -1
    [0 1 -1/(x+1)]
    [0 1 (x-1)/(x+1)]


    the three systems at this point:

    a = -1/(x+1)
    b = -1/(x+1)
    b = (x-1)/(x+1)

    Setting the last to = to each other i get

    (1/(x-1)) times (-1-(x-1)) = 0

    and x=0, x doesn't equal 1 again


    a= -1 and b=-1 plugging in

    so i get

    [1 0 -1]
    [0 1 -1]
    [-1 1 0]

    and through addition of the first row to the third row

    [1 0 -1]
    [0 1 -1]
    [0 1 -1] for the rref equivalent matrix


    I FIND IT WEIRD THAT THE ONLY NUMBER X COULD BE IS 0. AM I WRONG??

    Why did the creator of this problem put "x" in the matrix. If you have three variables and three equations, why not make a 3 by 4 augmented matrix?

    If I write this out

    a + bx = -1
    ax + b = -1
    -a + b = x

    IT DOESN'T SEEM LIKE A LINEAR EQUATION? WHY IS IT BEING SOLVED WITH "LINEAR ALGEBRA." Is X only allowed to be a single number (an unknown coefficient)?

    Thank you for your help
     
  2. jcsd
  3. Aug 7, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper


    Your initial matrix is

    [1 x -1]
    [x 1 -1]
    [-1 1 x]

    ?
     
  4. Aug 7, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    So the original matrix is
    [tex]\left[\begin{array}{ccc} 1 & x & -1 \\ x & 1 & -1 \\ -1 & 1 & x\end{array}\right]


    Why would you want them to "equal each other"? There is nothing like that in the problem. Clearly the rref form for this is
    [tex]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & \frac{x-1}{x+1}\end{array}\right][/tex]
    or simply to
    [tex]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{array}\right][/tex]

    as long as x is NOT either 1 or -1.

    x is 0 in order that what happens? The problem, as you stated it, was simply to row reduce the matrix. That can be done for any x except 1 or -1.

    Nothing was said about equations. A matrix in general does not have to be the "augmented" matrix of any set of equations.
     
  5. Aug 22, 2008 #4
    Thank you that was very helpful.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Please Help. Matrix with variables
  1. Help with a Matrix (Replies: 6)

  2. Matrix help (Replies: 1)

Loading...