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Exam question

  1. Jun 22, 2010 #1
    Q:Prove that a square matrix A is invertible iff A is non singular.
    My Ans: Since the inverse of a square matrix is given by,
    A^ = (1/|A|)adj.A (Where A^ is A inverse)
    If |A|=0, A^ is not defined.
    i.e, A^ exist only if A is non singular. In other words, a square matrix A is invertible iff A is non singular.
    Conversly, Let |A|=0, i.e., let A be singular.
    then, A^=(1/|A|)adj.A = (1/0)adj.A (not defined)
    Hence we conclude that A^ exist only if A is non singular.

    I got only 1 out of 5 marks for this answer. What is missing in my answer?
     
  2. jcsd
  3. Jun 22, 2010 #2
    Let A be a square invertible matrix.
    Then there exist a finite number of elementary operations on A that will transform A to I. That is, E1E2...EnA=I. Hence, there exists an inverse of A, namely E1E2...En. So A is nonsingular.
    Thus, if A is invertible, then A is nonsingular.

    Conversely, let A be nonsingular.
    Then there exists an inverse of A.
    Hence there exist a finite number of elementary operations on A such that A is transformed to I. That is A^-1 = E1E2...En and (A^-1)A=I. So A is invertible.
    Thus, if A is nonsingular, then A is invertible.

    Therefore, A is invertible iff A is nonsingular.
     
  4. Jun 22, 2010 #3
    Why my answer is incorrect?
     
  5. Jun 22, 2010 #4
    Strictly speaking, "invertible" means that a matrix can be transformed into the identity matrix by a finite series of elementary operations. Your proof did not use the definition.

    One problem with constructing the proof is that invertible and nonsigular are often used interchangeably. But this is precisely because of the bi-implication.

    Are you taking linear algebra this summer or are you rehashing an old exam?
     
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