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Singular matrix

  1. Apr 27, 2012 #1
    I have to proove that a singular (2 x 2)-matrix can be written as

    a b
    ta tb

    or

    ta tb
    a b

    My attempt is not a real proof, and as I'm very inexperienced with writing proofs, maybe someone could write it, so that I will understand it in the future.

    Attempt.

    Let B =
    a b
    c d

    From the definition, det(B)= ad-bc.

    For a singular matrix, det(B) = 0.

    Hence ad-bc=0 <=> ad=bc <=> a/c=b/d.

    We have that one row is a multiple of the other.

    If A=
    a b
    ta tb
    then we have a/ta=b/tb <=> 1/t=1/t, and that's true for all real t > 0.

    And if
    A =
    ta tb
    a b
    Then we have ta/a=tb/b <=> t=t,
    which is true for all t,a,c.
     
  2. jcsd
  3. Apr 27, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The final portion of your proof is backwards.

    Your are saying that if [itex]A= \begin{bmatrix}a & b \\at & bt\end{bmatrix}[/itex] then |A|= 0.

    What you want to prove is the other way: if |A|= 0 then [itex]A= \begin{bmatrix}a & b \\at & bt\end{bmatrix}[/itex].

    You are correct that if ad- bc= 0, then a/c= b/d. Let t= a/c= b/d.
     
  4. Apr 27, 2012 #3
    How about assuming the matrix is on the form:
    [itex]\left(\begin{array}{cc}
    a & b \\
    ta & tb\end{array}\right)
    [/itex]
    Then the determinant is:
    [itex]atb - tab = 0,\forall t,a,b\in\mathbb{R}[/itex]
    q.e.d. Can you work the other one out then?
     
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