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Consider the real system of matrix A, determine all solutions.

  1. Sep 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider the real system:

    [tex]
    \begin{pmatrix}
    1 & 2 & 3 & 4\\
    4 & 3 & 2 & 1\\
    -5 & -5 & -5 & -5
    \end{pmatrix}x=\begin{pmatrix}a\\b\\c\end{pmatrix}
    [/tex]

    and denote the system by matrix A.

    1. What is the rank of A? By inspection, determine a non-zero vector in the null space of A-transpose.
    2. Under what conditions on the numbers a, b, c is the system solvable? (Use your solution of part a).
    3. Determine all solutions of the system for a = b = 1, c = -2.

    2. Relevant equations

    No relevant equations

    3. The attempt at a solution

    For part 1, I reduced to row echelon and got:

    [tex]
    \begin{pmatrix}
    1 & 0 & -1 & -2\\
    0 & 1 & 2 & 3\\
    0 & 0 & 0 & 0
    \end{pmatrix}x=\begin{pmatrix}(-c/5)+b+(4/5)c\\-b-(4/5)c\\a+b+c\end{pmatrix}
    [/tex]

    I found that rank = 2. I also found that the vector for the null space of A-transpose is:

    [tex]
    \begin{pmatrix}1\\1\\1\end{pmatrix}
    [/tex]

    For part 2, I said that the solution is solvable under the condition that a+b+c = 0 because there is only one row with all zeros.

    I am stumped on part 3. I thought that I should just plug a = b = 1, c = -2 into my reduced row echelon, giving me infinitely many solutions. I don't think this is correct though. Any advice would be helpful. Thanks.
     
  2. jcsd
  3. Sep 17, 2014 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Why don't you think there should be infinitely many solutions? You have more unknowns than equations, so it wouldn't be surprising if there were some free variables, would it?
     
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