System of homogeneous equations

Click For Summary

Discussion Overview

The discussion revolves around a system of homogeneous equations derived from the intersection of three planes in three-dimensional space. Participants explore the implications of the determinant of the coefficient matrix and the conditions under which the system has non-trivial solutions.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant presents a system of equations and questions the correctness of a textbook statement regarding the elimination of variables leading to a determinant condition.
  • Another participant suggests that if elimination is performed correctly, the resulting matrix should have specific properties, indicating confusion about the textbook's explanation.
  • A third participant provides context by referencing a question about planes and their intersection, explaining how the equations relate to direction cosines.
  • A later reply asserts that for the system to have a non-zero solution, the determinant of the coefficient matrix must be zero, reinforcing the earlier claims about the determinant's role.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the textbook's statement and the implications of the determinant of the coefficient matrix. There is no consensus on the correctness of the textbook's explanation or the implications of the determinant condition.

Contextual Notes

Participants note that the solution of a homogeneous system requires careful consideration of the determinant, with implications for the existence of non-trivial solutions. The discussion highlights the complexity of the relationships between the variables and the conditions under which solutions exist.

AdityaDev
Messages
527
Reaction score
33
I got three equations:
l-cm-bn=0
-cl+m-an=0
-bl-am+n=0
In my textbook, its written "eliminating l, m, n we get:"
$$
\begin{vmatrix}
1& -c& -b\\
-c& 1& -a\\
-b& -a& 1\\
\end{vmatrix}=0
$$
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?
 
Physics news on Phys.org
AdityaDev said:
I got three equations:
l-cm-bn=0
-cl+m-an=0
-bl-am+n=0
In my textbook, its written "eliminating l, m, n we get:"
$$
\begin{vmatrix}
1& -c& -b\\
-c& 1& -a\\
-b& -a& 1\\
\end{vmatrix}=0
$$
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?

It's not clear what is going on in your textbook. 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.

http://en.wikipedia.org/wiki/System_of_linear_equations

This article at least gets the basics right.
 
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.
 
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".
 
  • Like
Likes   Reactions: AdityaDev

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K