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For what value(s) of x does A^-1 exist.

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Let A =
    (x+1 2 )
    (x+5 x+1)

    For what value(s) of x, if any, does A-1 exist?

    2. Relevant equations



    3. The attempt at a solution
    So let me see if I get the gist of this. An n x n matrix is invertible if and only if rank (A) = n. So unless there is a value of x which would make rank A < 2, then it is invertible. Since I can see no way to reduce both x+5 and x+1 to zero using one value, then A is invertible for all real numbers?
     
  2. jcsd
  3. Feb 4, 2013 #2
    I don't know if you have learned the concept of determinants, but a matrix A is invertible if and only if its determinant is nonzero. The determinant for 2x2 matrices of the form
    (a b)
    (c d)
    is given by det A = ad - bc

    Computing the determinant and factoring the quadratic polynomial you'll get, will give you the answer.
     
  4. Feb 4, 2013 #3
    Aha. So, det(A) = ((x+1)(x+1))-(2(x+5)) = x2-9 = (x+3)(x-3)
    So A-1 exists for all values ≠ 3, -3.
     
    Last edited: Feb 4, 2013
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