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Finding solutions to matrices (should be easy)

  1. Oct 17, 2011 #1

    Ush

    User Avatar

    1. The problem statement, all variables and given/known data
    Alvin is informed that the homogeneous system of equations AX = 0 has a
    one parameter family of solutions given by

    X = t[4 3 0]T

    By trial and error, he has found that

    X = [-1 5 7]T

    is a solution to the inhomogenous problem

    AX = B

    where

    B = [3 -2 -5]T

    He suspects there are more solutions

    a) Is there a solution to AX=B of the form [1 a b]T? If so, what are a and b?
    b) Is there a solution to AX=B of the form [c d 1]T? If so, what are c and d?


    2. Relevant equations

    AX = B
    X = BA-1


    3. The attempt at a solution

    attachment.php?attachmentid=40122&stc=1&d=1318907975.png

    I'm more used to simple matrices such as the following

    attachment.php?attachmentid=40123&stc=1&d=1318908297.png

    I just don't know how to apply this concept to this question,

    my test is tomorrow, any help would be appreciated =)
     

    Attached Files:

  2. jcsd
  3. Oct 17, 2011 #2

    Mark44

    Staff: Mentor

    The second equation above is true if and only if A is invertible.
    From the given information, we can say that
    [tex]A t\begin{bmatrix}4\\3\\0\end{bmatrix} + A \begin{bmatrix}-1\\5\\7\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix} + \begin{bmatrix}3\\-2\\-5\end{bmatrix}[/tex]
    for any value of t.

    For a), Is there some particular value of t for which the two vectors on the right add up to <1, a, b>T?
    The b part is similar.
     
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