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Full rank of a matrix

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  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that the matrix ##A## is of full rank if and only if ##ad-bc \neq 0## where $$A = \begin{bmatrix}
    a & b \\
    b & c
    \end{bmatrix}$$

    2. Relevant equations


    3. The attempt at a solution
    Suppose that the matrix ##A## is of full rank. That is, rank ##2##. Then by the rank-nullity theorem, the
    dimension of the kernel is ##0##. This implies that there exists an inverse ##A^{-1}## but this will only occur if ##ad-bc \neq 0## otherwise our matrix ##A## will be singular. On the other hand, suppose ##ad-bc \neq 0##. Hence, ##A## is nonsingular and there exists an inverse ##A^{-1}## but this will occur only when the dimension of the kernel is ##0##, that is, of rank ##n = 2##.
     
  2. jcsd
  3. Sep 22, 2015 #2
    Your matrix has rank 2 iff the column vectors are linearly independent.
    Can you show that the determinant of the matrix is zero iff the column vectors are linearly dependent ?
     
  4. Sep 22, 2015 #3
    Sorry $$A=\begin{bmatrix}
    a & b \\
    c & d
    \end{bmatrix}$$ if that even matters!!!!

    How could I do this without using anything on determinants???
     
  5. Sep 22, 2015 #4
    Show that ad - bc = 0 iff ##\vec u = (a,c) ## and ##\vec v = (b,d)## are linearly dependent
     
  6. Sep 22, 2015 #5
    Cheers!!!
     
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