Solve Homogeneous System: Use Determinant to Check Nontrivial Solutions

Amy-Lee
Messages
27
Reaction score
0
how does one use the determinant of the coefficient matrix of a system to determine if the system has nontrivial solutions or not?
 
Physics news on Phys.org
What do you know about the coefficient matrix if its determinant equals zero? What do you know if the determinant is not equal to zero?
 
determinant = 0, homogeneous equation equals zero... therefore trivial solution
determinant not to equal 0, homogeneous equation don't equal 0... therefore nontrivial solution?
 
Amy-Lee said:
determinant = 0, homogeneous equation equals zero... therefore trivial solution
determinant not to equal 0, homogeneous equation don't equal 0... therefore nontrivial solution?

No, this is incorrect. Also, an equation is never equal to anything. For example, x + 5 = 2 is an equation, but what is it equal to? There is always an = in an equation, but that indicates that two expressions have the same value.

For a very simple example of a system of linear equations, consider this system of two equations in two unknowns:
x + y = 0
2x + 2y = 0

The determinant of the matrix of coefficients is 0, which means that the solution to this system is not unique. For this system, there is the trivial solution (x = 0, y = 0), and a whole bunch (an infinite number) of nontrivial solutions, solutions other than the trivial solution.

Here's a second example:
x + y = 0
x - y = 0
The determinant of the matrix of coefficients this time is nonzero, which means that there is exactly one solution to the system of equations, in other words, that the solution is unique. For this system, the only solution is x = 0, y = 0, the trivial solution.

For these homogenous systems of equations, the value of the determinant of the matrix of coefficients determines whether there will be a unique solution (det is nonzero), the trivial solution, or an infinite number of solutions (det = 0), including the trivial solution.
 
  • Like
Likes 1 person
what about a homogeneous system of equations with more unknowns than equations, does the above also apply?
 
Amy-Lee said:
what about a homogeneous system of equations with more unknowns than equations, does the above also apply?
No. In that case the matrix of coefficients is not square (has more columns than rows). The determinant is defined only for square matrices.

For a linear system of n equations in n variables there is a direct connection between the value of the determinant of the matrix of coefficients and whether the matrix of coefficients has an inverse. If the determinant is zero, an inverse does not exist; if the determinant is nonzero, there is an inverse.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top