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Solving for x in a determinant

  1. Oct 25, 2013 #1
    Solve for x
    x -6 -1
    2 -3x x-3 = 0
    -3 2x x+2


    Attempt-
    x-2 3(x-2) -(x-2)
    2 -3x x-3
    -3 2x x+2

    1 3 -1
    2 -3x x-3
    -3 2x x+2

    1 3 -1
    0 -3x-6 x-1 = 0
    0 -x+3 2(x-1)
    x =-3

    By doing this i'm getting just one value of x. How do i get the values of x like 2 and 1?
     
  2. jcsd
  3. Oct 25, 2013 #2

    jedishrfu

    Staff: Mentor

    Its very hard to read your determinant or even that it is a determinant.

    Code (Text):


    [ (x)   (-6)   (-1)  ]
    [ (2)   (-3x)  (x-3) ]
    [ (-3)  (2x)   (x+2) ]

     
    Is this right?

    You need to evaluate the determinant to get a polynomial in x and then factor it.
     
    Last edited: Oct 25, 2013
  4. Oct 25, 2013 #3
    Its 2x in (column 2,row 3).
     
    Last edited: Oct 25, 2013
  5. Oct 25, 2013 #4

    jedishrfu

    Staff: Mentor

    Noted and fixed in my prior post.

    Did my answer help?
     
  6. Oct 25, 2013 #5

    Mark44

    Staff: Mentor

    Why don't you just expand the determinant directly?

    It would be helpful to us if you told us what you are doing in each step above. Apparently in the 2nd step of your attempt, you have divided the first row by x - 2. In doing this, you are ignoring the possibility that x - 2 = 0, thereby losing x = 2 as a solution. Instead of dividing by x - 2, bring x - 2 out as a factor.

    The basic idea is this:
    $$ \begin{vmatrix} ka & kb \\ c & d \end{vmatrix} = k\begin{vmatrix} a & b \\ c & d\end{vmatrix}$$
     
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