Augmented matrix linear algebra

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SUMMARY

The discussion centers on determining conditions for the augmented matrix of a linear system to have no solution, many solutions, or a unique solution. The correct conditions are identified as: (a) no solution when b=1/2 and a≠1, (b) many solutions when b=1/2 and a=1, and (c) a unique solution when b≠1/2. Participants shared their attempts at row reduction, confirming that proper echelon form leads to these conclusions.

PREREQUISITES
  • Understanding of augmented matrices in linear algebra
  • Familiarity with row reduction techniques
  • Knowledge of pivot positions and free variables
  • Basic concepts of linear systems and their solutions
NEXT STEPS
  • Study the process of row reduction in detail
  • Learn about the implications of pivot positions in matrices
  • Explore the concept of free variables in linear systems
  • Investigate the conditions for consistency in linear equations
USEFUL FOR

Students and educators in linear algebra, particularly those focusing on solving systems of equations using augmented matrices and row reduction techniques.

devilaudy
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Homework Statement



Given the system whose augmented matrix is
 1 1 1 1 
 1 −1 0 a 
 0 1 b 0 
Determine (if possible) conditions on a and b such that this system has (a) no solution (b) many solutions (c) a unique solution.


Homework Equations



-Row reduction
-No solution: row of zero
-Many solutions: one or more free variables
-Unique solution: pivot in every column

The Attempt at a Solution



I tried to do it, but my answers were wrong. The good answers are a) b=1/2, a≠1 b) b=1/2, a=1 c) b≠1/2.

I tried to reduce the matrix and I had:

 0 0 1-2b 1-3a 
 1 0 b 2a 
 0 1 b a 

Can someone explain how to reduce the matrix properly or what am I doing wrong?
 
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devilaudy said:

Homework Statement



Given the system whose augmented matrix is
 1 1 1 1 
 1 −1 0 a 
 0 1 b 0 
Determine (if possible) conditions on a and b such that this system has (a) no solution (b) many solutions (c) a unique solution.


Homework Equations



-Row reduction
-No solution: row of zero
-Many solutions: one or more free variables
-Unique solution: pivot in every column

The Attempt at a Solution



I tried to do it, but my answers were wrong. The good answers are a) b=1/2, a≠1 b) b=1/2, a=1 c) b≠1/2.

I tried to reduce the matrix and I had:

 0 0 1-2b 1-3a 
 1 0 b 2a 
 0 1 b a 

Can someone explain how to reduce the matrix properly or what am I doing wrong?

Right off the bat I swapped the 2nd and 3rd rows. After that, I row reduced to get a matrix in echelon form. I get the same answers as your "good" ones.
 

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