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Test for compatibility of equations - Determinant |A b|

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,028
Hey!! :eek:

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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,710
Hey!! :eek:

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 $A^+b$?
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,028
What is this "test for compatibility of equations"? (Wondering)
The test if the system is solvable.


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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,710
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)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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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