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Linear Algebra Homework Help

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the matrix A = 1 1 2 1
    1 2 3 4
    2 4 7 2t-6
    2 2 6-t t
    Find the values of t for which the matrix A is invertible


    2. Relevant equations



    3. The attempt at a solution

    For A to be invertible, the determinant should not be 0.

    So what I did is find the determinant of the 4x4 matrix.
    Since this 4x4 matrix doesn't have any zero.
    I find 4 3x3 determinants which leads to 3 2x2 determinants. And a total of 12 2x2 determinants.
    After I sum all of them, my equation lead to 4t^2-7t+50 is not equals to zero.
    And I can't solve for t.
    This is weird because when I use online calculator for matrix, for t is 0,1,3, or any random number I put in, there is an inverse matrix.
     
  2. jcsd
  3. Feb 13, 2012 #2

    vela

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    The determinant of that matrix is 2t2-17t+26. I would suggest row-reducing the matrix a bit to make finding the determinant less tedious.
     
  4. Feb 13, 2012 #3

    Deveno

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    you needn't even row-reduce that much. clearing out the first column reduces the problem to finding a 3x3 determinant, and one can use the rule of sarrus:

    [tex]\begin{vmatrix}a&b&c\\d&e&f\\g&h&k\end{vmatrix} = aek + bfg + cdh - ceg - afh - bdk[/tex]
     
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