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For which a is the mtrix nonsingular

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data
    For which a [tex]\in[/tex] C is the 3x3 matrix:

    [tex]\[\left[ \begin{matrix}
    a & 2 & 0 \\
    0 & -1 & 2 \\
    1 & 0 & a \\
    \end{matrix} \right]\]
    [/tex]
    nonsingular ?

    2. Relevant equations
    Ax = 0


    3. The attempt at a solution
    Well, the problem is that I think I know what to do, but I can't remember how it's done :S
    If Ax = 0, then it's nonsingular, so that's all I have to show.
    And then I multiply the matrix with x ([x1, x2, x3]) I just get the same matrix as I have above, but with zeros in the fourth column.
    So if I put in on REF I get that x1 = -a, x2 = a2/2.
    And then what ? In my head in seems so easy, but I just can't figure it out :S
    What is the a's that's satisfy that the matrix is nonsingular ?
    If I use RREF on AB = BA = I, I get that a [tex]\neq[/tex] (+-)2. But I can only do that if I have assumed that the matrix, A, is nonsingular, which I have not shown yet.
    So, what to do ? :)


    Regards
     
  2. jcsd
  3. Feb 1, 2010 #2

    Mark44

    Staff: Mentor

    That's not the definition of nonsingularity. The equation above is true for x = 0 for any square matrix. If the matrix is nonsingular, the equation will be true only for x = 0.

    Another approach that I think you have forgotten, is that the determinant of a nonsingular (AKA invertible) matix is nonzero. Try taking the determinant and see what value(s) a needs to be so that the matrix is nonsingular.
     
  4. Feb 2, 2010 #3
    Hmmm, but I think I'm not allowed to use determinants yet. We haven't "learned" it yet, so...
     
  5. Feb 2, 2010 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, then, go back to your first idea but keep Mark44's caution in mind: for what values of a does
    [tex] \begin{bmatrix} a & 2 & 0 \\ 0 & -1 & 2 \\ 1 & 0 & a \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}ax+ 2y \\ -y+ 2z \\x+ az\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\\end{bmatrix}[/tex]
    for x, y, and z NOT all 0?
     
  6. Feb 2, 2010 #5
    Well, if I'm allowed to just solve the equations but substituting some of the equations into some of the others, I get that: a = 2 or a = -2.
     
  7. Feb 2, 2010 #6

    Mark44

    Staff: Mentor

    Yes, those are correct. And you can check these by substituting into your original matrix to show that Ax = 0.
     
  8. Feb 2, 2010 #7
    Ok, great, thank you :)
     
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