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Homework Help: Prove: A 2x2 matrix is nonsingular if and only if the determinant != 0

  1. Aug 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove: A 2x2 matrix is nonsingular if and only if the determinant != 0

    3. The attempt at a solution

    I need to prove this, using logic and maybe the theorem that a n x n matrix is nonsingular if and only if it is row equivalent to I_n.

    I could use a push in the right direction. I need to show a proof for both ways, as it is an if and only if statement.
  2. jcsd
  3. Aug 6, 2010 #2
    I'm no linear algebra expert, but I would start by writing that if the matrix is written:
    a b
    c d

    Then it's singular if a = x*b and c=x*d (or a=x*c and b=x*d), and work my way to the formula for the determinant.
  4. Aug 6, 2010 #3
    Ok, but where does this come from?
  5. Aug 6, 2010 #4
    It would come from the fact that if the matrix is singular, then column one a multiple of the other (or one row is a multiple of the other). Maybe this isn't how it's supposed to be done though, so you might be better off waiting for somebody else's ideas than going off mine :/
  6. Aug 6, 2010 #5
    For one direction, assume matrix A is invertible. Then there is matrix B such that AB=1, where 1 is identity matrix. Take the determinant of both sides of AB=1.

    For the other direction, assume det A is invertible. Using the determinant and cofactor expansions, you can find a nice form for the inverse of A
  7. Aug 7, 2010 #6
    I probably should have mentioned. I don't have the definition of the determinant to work with.

    the problem actually says Show that the 2 x 2 matrix A is nonsingular if and only if ad-bc != 0.

    I've figured out the If matrix A is nonsingular, then ab-bc != 0 side. I just need the if ad - bc != 0, Matrix A is nonsingular side.
    Last edited: Aug 7, 2010
  8. Aug 8, 2010 #7
    Given matrices of the form
    a b = A
    c d


    d -b = B
    -c a

    How can you ensure AB=1?
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