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Non singular matrix proof

  1. Apr 28, 2010 #1
    a matrix is non singular only if its det does not equal zero. Calculate its inverse.

    How do I go about proving this? I can only think of a counter example where matrix is singular given identical rows or columns or multiples of each other, which will generate a det of 0.

    What do you think?
  2. jcsd
  3. Apr 28, 2010 #2


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    Prove what? That a matrix is non-singular if and only if its determinant is non-zero? What is your definition of "non-singular". I suspect, from that addtional "Calculate its inverse" that "non-singular" is defined as "has an inverse" (or, more precisely, that "singular" is defined as "does not have an inverse" and "non-singular" is the reverse of that. Okay, how would you find the inverse of a matrix? Does the determinant come into that?
  4. Apr 28, 2010 #3
    By non singular I mean a inverse exists. I believe the inverse is the adjoint/ det of the matrix. So the det cant be 0.
  5. Apr 29, 2010 #4


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    That's half way. You also need to show that if the determinant is non-zero then the matrix is invertible. Since if the determinant is non-zero, 1/det exists, all you need to do is show that "adjoint" always exists.
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