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Matrix Linear Systems Equation

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Question in the image.


    2. Relevant equations
    I got a but can seem to get the answer for b which is $$\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$$

    where for any scalar c
    3. The attempt at a solution
    I managed to get it to $$\vec{x}(A-1)=0$$

    $$= \begin{pmatrix} 0 3 \\ -1 0\end{pmatrix}=\begin{pmatrix} 1 0 \\ 0 1\end{pmatrix}$$
    Then I reduced the matrix to reduced row echolon form
    but after that I seem not to get the answer.

    Thank you
     

    Attached Files:

    Last edited by a moderator: Mar 12, 2012
  2. jcsd
  3. Mar 12, 2012 #2

    Mark44

    Staff: Mentor

    You made a mistake in how you calculated A - I. (It's not A - 1. IOW, you don't subtract 1 from each entry of A.)
     
  4. Mar 12, 2012 #3
    I get (A-I),
    which would then be
    $$\begin{pmatrix} 0 3 \\ 0 0\end{pmatrix}$$
    but this would mean x1 and x2 = 0
     
    Last edited by a moderator: Mar 12, 2012
  5. Mar 12, 2012 #4

    Mark44

    Staff: Mentor

    No, you don't get A - 1. Here's the way it goes:

    Ax = x
    => Ax - x = 0
    => (A - I)x = 0

    In the last equation, that is A - I, the 2 x 2 identity matrix, NOT the matrix whose entries are all 1's.

    Calculate A - I, and row reduce it to find the solution set.
     
  6. Mar 12, 2012 #5
    Thanks, I understand that part,careless mistake but when I reduce it I get
    x1 and x2 = 0
     
  7. Mar 12, 2012 #6

    Mark44

    Staff: Mentor

    No. The top row says 0x1 + 3x2 = 0, so x1 is arbitrary.
     
  8. Mar 12, 2012 #7
    then why is the answer $$\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$$



    $$\vec {x} = x1\begin{pmatrix} 1 \\ 0\end{pmatrix}$$
     
    Last edited by a moderator: Mar 12, 2012
  9. Mar 12, 2012 #8

    Mark44

    Staff: Mentor

    Fixed your LaTeX. You need to put the [noparse]$$[/noparse] thing at the end of your expression, too.
    Those are pretty much the same thing. In the first, c is an arbitrary number; in the second, x1 is an arbitrary number.
     
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