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Singular matrices

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Let A = [A1,...,An-1] be an (nx(n-1)) matrix. Show that B = [A1,...,An-1,Ab] is singular for every choice of b in R^n-1.

    2. Relevant equations
    Ax = 0

    3. The attempt at a solution
    I know that if B is singular that means that for the equation Bx = 0 there exists another solution another than the trivial solution (x = 0). Now if we made B have all the same columns as A except added a new column Ab, that would make B a square matrix that is (nxn). But from there, I can't figure out how to use the information I know to solve the problem...
  2. jcsd
  3. Feb 2, 2010 #2


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    If a matrix is singular, then its columns are linearly dependent. Any ideas?
  4. Feb 2, 2010 #3
    Ab=[A1b A2b ... An-1b]^T so by construction the column Ab is a linear combination of the first n-1 column vectors, regardless of what the vector b actually is. Hence detB=0
  5. Feb 2, 2010 #4
    radou isn't that only true if the matrix is two by two?

    And Matthollywood I am not sure what you are saying, could you please reword what you said.

    Thank you.
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