Solving for k in a Homogenous System

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SUMMARY

The discussion focuses on solving for the variable k in a homogeneous system represented by the equations kx + 2y + 2z = 0, x + 2y + 2z = 0, and 4x + (k-1)y + 2z = 0. The matrix form of the system is presented as follows:

( k  2  2  0
  1  2  2  0
  4 (k-1) 2  0 )
. The user identifies that setting k=1 leads to infinitely many solutions, but acknowledges a mistake in the row operations, specifically the need to interchange rows to achieve a leading 1 in the first column.

PREREQUISITES
  • Understanding of homogeneous systems of equations
  • Familiarity with matrix representation of linear equations
  • Knowledge of row operations in Gaussian elimination
  • Concept of leading coefficients in matrices
NEXT STEPS
  • Study Gaussian elimination techniques for solving linear systems
  • Learn about the implications of leading 1s in row echelon form
  • Explore the concept of non-trivial solutions in homogeneous systems
  • Investigate the role of parameters in linear equations and their impact on solution sets
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Students studying linear algebra, particularly those tackling systems of equations and matrix operations, as well as educators looking for examples of solving homogeneous systems.

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Homework Statement


Find all values of k such that the following homogenous system has non-trivial solutinos:
kx + 2y + 2z = 0
x + 2y + 2z = 0
4x + (k-1)y +2z = 0

Homework Equations





The Attempt at a Solution


I put it into matrix form ( k 2 2 0
1 2 2 0
4 (k-1) 2 0 )
Then i subtracted row 2 from row 1 to get
k 2 2 0
(1-k) 0 0 0
4 (k-1) 2 0

Then i said if k=1 then there are infinite many solutions.

I don't know where to go from that.
 
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No, that's not right at all! I think you even forgot to interchange row 2 with row 1 in order to introduce a leading 1 in a11.
 

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