Test for compatibility of equations - Determinant |A b|

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    Determinant Test
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Discussion Overview

The discussion revolves around the compatibility of equations in a system represented by the matrix equation $Ax=b$, particularly focusing on the use of the determinant of the augmented matrix $|A \ b|$ as a criterion for solvability. Participants explore the implications of the dimensions of the matrix and the conditions under which solutions may exist.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants question the validity of using the determinant $|A \ b|$ for a non-square matrix, suggesting it may refer to Matlab notation $A \ b$ instead.
  • There is uncertainty about the definition of the augmented matrix $(A \mid b)$ and whether it can be considered square given the system's dimensions.
  • Some participants propose that if the system has a solution, then $b$ must be in the column space of $A$, implying a relationship between $b$ and the columns of $A$.
  • One participant suggests that if the equations are independent, there may be no solution, while dependent equations could lead to a solution with a determinant of zero.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the determinant in the context of the augmented matrix and the conditions for compatibility of the equations. No consensus is reached regarding the correct application of the determinant criterion.

Contextual Notes

There are unresolved questions regarding the definitions and properties of the matrices involved, particularly concerning the dimensions and the implications of the determinant in this context.

mathmari
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Hey! :o

Let $Ax=b$ be a system of linear equations, where the number of equations is by one larger than the number of unknown variables, so the matrix $A$ is of full column rank.

Why can the test for combatibility of equations use the criterion of the determinant $|A \ b|$ ? (Wondering)
 
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mathmari said:
Hey! :o

Let $Ax=b$ be a system of linear equations, where the number of equations is by one larger than the number of unknown variables, so the matrix $A$ is of full column rank.

Why can the test for combatibility of equations use the criterion of the determinant $|A \ b|$ ?

Hey mathmari!

What is this "test for compatibility of equations"? (Wondering)

How is $|A \ b|$ a determinant? It's not a square matrix is it? (Wondering)
Did you perhaps mean the Matlab notation [M]A \ b[/M], which means $A^+b$?
 
Klaas van Aarsen said:
What is this "test for compatibility of equations"? (Wondering)

The test if the system is solvable.
Klaas van Aarsen said:
How is $|A \ b|$ a determinant? It's not a square matrix is it? (Wondering)
Did you perhaps mean the Matlab notation [M]A \ b[/M], which means $A^+b$?

But isn't the augmented matrix $(A\mid b)$ a square matrix, since the number of equations is by one larger than the number of unknown variables and that means that $A$ is a $n\times (n-1)$ matrix, or not? (Wondering)
 
mathmari said:
The test if the system is solvable.

But isn't the augmented matrix $(A\mid b)$ a square matrix, since the number of equations is by one larger than the number of unknown variables and that means that $A$ is a $n\times (n-1)$ matrix, or not?

Ah okay.

If the system has a solution, then $b$ must be in the column space of $A$ yes? (Wondering)
That is, we can write $b$ as a linear combination of the column vectors in $A$.

Doesn't that imply that the determinant of the augmented matrix $(A\mid b)$ is zero? (Thinking)
 
I suspect you mean "compatibility", not "combatibiity"!

No one wants equations to fight!

If the number of equations is greater than the number of unknowns and the equations are "independent" there is no solution. There may be a solution if the equations are "dependent" which means the determinant must be 0.
 
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