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Homework Help: Linear Algebra Question: Determinant

  1. Jul 10, 2008 #1
    The question on my homework says:

    What can you say about the graph of the equation

    The following is in matrix form with determinant symbols around the matrix
    x y 1
    a[tex]_{1}[/tex] b[tex]_{1}[/tex] 1 = 0
    a[tex]_{2}[/tex] b[tex]_{2}[/tex] 1

    All a's and b's are constants.

    My answer was that the graph would be a line going through the origin. Like y=2x or y=3x

    Am I way off track???

    I have tried to attach the question in the document. Not sure if it will show.

    Attached Files:

  2. jcsd
  3. Jul 10, 2008 #2
    Those 1's and 2's are supposed to be subscripts.
  4. Jul 11, 2008 #3


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    No, that is not, in general, a line through the origin. What is that determinant? It not difficult to calculate.
  5. Jul 11, 2008 #4
    The determinant is b1x - a2b1

    a2b1 is a constant. So wait, I had that wrong. it is a straight line with a y-intercept which is some constant.

    How do you say that though?
  6. Jul 11, 2008 #5


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    The determinant is not that. But I don't know what this question is asking for. What can you tell from the equation of a plane/line?
  7. Jul 11, 2008 #6
    Sorry, I forgot about it being a 3x3 matrix.
  8. Jul 11, 2008 #7
    I'm not sure if the question is asking what the graph of the determinant would look like or the matrix?
  9. Jul 11, 2008 #8


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    If you expand the determinant you get an expression like f(x,y)=0. The question is what's the graph of that equation? Determinants and matrices don't have graphs. Equations have graphs.
  10. Jul 11, 2008 #9


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    Just find the determinant of the 3x3 matrix and you'll see what it describes. You have an idea of what it describes though.
  11. Jul 13, 2008 #10
    Okay. I found the determinant of this 3x3 matrix:

    It is b1x + a2y + a1b1 - a2b1 - b2x - a1y

    I simplified it: b1x - b2x + a2y - a1y - a1b1 - a2b1 = 0

    I'm not sure if this will help, but I did this next.

    b1x - b2x + a2y - a1y + C ; c = constant

    a2y-a1y = C - b1x + b2x

    y(a2 - a1) = C - x(b1 - b2)

    y = (C - x(b1 - B2)) / (a2 - a1)

    I know this is the equation of a line, but what more do I need to say?
  12. Jul 13, 2008 #11


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    Yes, it's a line. I think we are just waiting for you to retract the 'through the origin' part.
  13. Jul 14, 2008 #12


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    Could say what the gradient is and where it cuts the axes, but I think saying that it is a straight line should be sufficient.
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