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System of homogeneous equations

  1. Mar 17, 2015 #1
    I got three equations:
    In my textbook, its written "eliminating l, m, n we get:"
    1& -c& -b\\
    -c& 1& -a\\
    -b& -a& 1\\
    but if I take l, m, n as variables and since ##l=\frac{\Delta_1}{\Delta}## (Cramer's rule) and ##\Delta_1=0##, then if ##\Delta=0##,you get an indeterminate form.
    Is the expression given in my textbook correct?
  2. jcsd
  3. Mar 17, 2015 #2


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    Staff Emeritus
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    It's not clear what is going on in your text book. To me, it looks like l, m, and n are the unknown variables for this system. If elimination were correctly carried out on the matrix of coefficients, then you would be left with only 1's on the main diagonal and only zeros to the lower left of the main diagonal.

    In any event, the solution of a system homogeneous linear equations requires special consideration. If the determinant of the matrix of coefficients is not equal to zero, then l = m = n = 0 is the only solution to the system. If the determinant of the matrix of coefficients equals zero, there is an infinite number of solutions.


    This article at least gets the basics right.
  4. Mar 17, 2015 #3
    Its taken from a question:
    "If the planes x=cy+bz, y=az+cx and z=bx+ay pass through a line, then the value of ##a^2+b^2+c^2+2abc=##?
    They took the direction cosines of the lines to be l, m, n and since the line is perpendicular to the normals of all three planes, you get the above 3 equations.
  5. Mar 17, 2015 #4
    If three planes going through a line, then the homogeneous system of linear equation you wrote has a non-zero solution. The coefficient matrix is square, so this can happen if and only if its determinant is non-zero. And the coefficient matrix is exactly the matrix you wrote, so its determinant must be zero. That what they meant by "eliminating l, m, and n".
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