Is it possable for Cramer's rule to fail?

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In a linear system of equations where there is a solution, is it possible for Cramer's rule to fail? By fail, as in end up with a zero determinate on either the top or the bottom or both.
 

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  • #2
uart
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Well lets look at would would lead to a zero determinate on either the top or the bottom.

Consider the equations : [tex] A \vec{x} = \vec{b} [/tex]. where A is a matrix and both x and b column vectors.

Referring to Cramers Rule, the determinate on the bottom is just the determinate of A. So if the equations do indeed have a unique solution then this cannot be zero.

As for the top however, yes it is definitely possible for this to be zero. All that this requires is that b be a linear combination of the remaining columns. But does this mean that Cramers rule has failed? Not at all. There is no problems in a fraction having a numerator of zero, it just means that the answer is zero (for the particular element of vector x under consideration).
 
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  • #3
statdad
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There is a little more to the possibilities of zeros in the numerator, denominator, or both. The easiest way to demonstrate this is with a [tex] 2 \times 2 [/tex] system, but the ideas are the same for larger systems.

First

[tex]
\begin{align*}
2x + 3y & = 10 \\
10x + 15y & = 50
\end{align*}
[/tex]

Clearly this has infinitely many solutions. When you try to solve this with Cramer's rule every determinant is zero, even though the system has (infinitely many) solutions.

Now consider this slightly different system.

[tex]
\begin{align*}
2x + 3y & = 10 \\
10x + 15y & = 51
\end{align*}
[/tex]

this system does not have any solutions, but the determinant of the coefficient matrix is zero once again, but neither of the determinants you would use in the numerator is zero
 
  • #4
uart
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Statdad, when the OP stated "In a linear system of equations where there is a solution" I took that to mean that there is a unique solution to the equations. In this case then all that I said above is true. Clearly Cramers rule will fail if there is not a unique solution.
 
  • #5
statdad
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Statdad, when the OP stated "In a linear system of equations where there is a solution" I took that to mean that there is a unique solution to the equations. In this case then all that I said above is true. Clearly Cramers rule will fail if there is not a unique solution.
You are correct - I missed his qualifying statement.
 
  • #6
uart
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Faken to sumarize the important points of the above.

If the equations have just one solution then Cramers rule will not fail.

As statdad's examples correctly point out however, if Cramers rule does fail then you cannot assume that there are no solutions, as this will also happen in the case where there are multiple solutions.

The bottom line is that there much are better methods than Cramers rule, especially when dealing with sets of equations that don't necessarily have a unique solution.
 
  • #7
statdad
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"The bottom line is that there much are better methods than Cramers rule, especially when dealing with sets of equations that don't necessarily have a unique solution."

True - Cramer's rule is of interest primarily for historical and theoretical purposes.
I would say (regarding the items I posted) that I don't tell students that Cramer's rule has failed in situations like those: I refer to those results as flags about whether the system has infinitely many or zero solutions. Minor language, but it fits with the identification of infinitely many vs zero solutions from
* elimination
* substitution
* guass/jordan
algorithms.
 
  • #8
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Saying that Cramer's rule fails is like saying that the MVT fails because for f(x)=|x|, there's no point in (-1,1) with f'(0)=0. A theorem never fails, but people may fail if they don't pay attention to the hypotheses.
 
  • #9
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Wow, thank you everyone!

You gave me a very nice solid answer.

Your right, I apparently wasn't thinking when i said that the numerator could no be zero, it obviously can.
 

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