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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).

Consider the equations : [tex] A \vec{x} = \vec{b} [/tex]. where

Referring to Cramers Rule, the determinate on the bottom is just the determinate of

As for the top however, yes it is definitely possible for this to be zero. All that this requires is that

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statdad

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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,

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uart

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statdad

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You are correct - I missed his qualifying statement.isa 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.

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uart

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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

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.

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statdad

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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

* elimination

* substitution

* guass/jordan

algorithms.

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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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