# Test for compatibility of equations - Determinant |A b|

#### mathmari

##### Well-known member
MHB Site Helper
Hey!! 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 $|A \ b|$ ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey!! 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"? How is $|A \ b|$ a determinant? It's not a square matrix is it? Did you perhaps mean the Matlab notation [M]A \ b[/M], which means $A^+b$?

#### mathmari

##### Well-known member
MHB Site Helper
What is this "test for compatibility of equations"? The test if the system is solvable.

How is $|A \ b|$ a determinant? It's not a square matrix is it? 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? #### Klaas van Aarsen

##### MHB Seeker
Staff member
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? 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? #### HallsofIvy

##### Well-known member
MHB Math Helper
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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