Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can I assume that an equation (with 3 variables) is unsolvable if its determinant equals zero.Sure, it not true if all it's other determinants are zero as well, but it seems to be unlikely

it is true for a determinant such as (1,0,0),(0,1,0),(0,0,1) but thats not a real equation.

if this is the case is there an easy way to prof that? otherwise could anyone provide an example with for which this rule doesn't work.

share|improve this question

closed as not constructive by Belgi, no identity, Thomas, Austin Mohr, J. M. Oct 30 '12 at 22:52

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

3  
"All it's other determinants"? A matrix only has one determinant. –  EuYu Oct 29 '12 at 18:31
2  
Also, what is the determinant of an equation ? –  Belgi Oct 29 '12 at 18:33
    
Here is an example of what I mean when I say 'solve equation' with a determinant cliffsnotes.com/study_guide/… –  kirill Oct 29 '12 at 18:35
    
Oh, you're talking about evaluating the determinant. If your determinant is found to be $0$, then you've "solved" for the determinant already (you found that it's $0$). So what do you mean by unsolvable exactly? –  EuYu Oct 29 '12 at 18:37
    
as Manos said, I'm referring to Cramer's Rule –  kirill Oct 29 '12 at 20:53

1 Answer 1

up vote 1 down vote accepted

What you are referring to is known as "Cramer's Rule". When we have an $n \times n$ linear system of equations $Ax=b$ and the determinant of the coefficient matrix $A$ is nonzero (equivalently $A$) is invertible, then we can apply Cramer's rule, which involves divisions with $det(A)$ to obtain the unique solution $A^{-1}b$. It might be possible though that $det(A)=0$, in which case Cramer's Rule does not apply, and still the system has a solution. The correct criterion to check is whether $b \in \mathcal{R}(A)$, i.e. the right hand side vector must lie in the range space of $A$, i.e. we can construct $b$ as a linear combination of columns of $A$. A simple example is the system $\left[\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right] x = \left[\begin{array}{c} 3\\ 0 \end{array} \right]$. Then every vector of the form $\left[\begin{array}{c} 3\\ \alpha \end{array} \right], \, \alpha \in \mathbb{R}$, is a solution.

share|improve this answer
    
Could you explain it in a simpler way? is there a need to verify that one (at least) of the other determinant is non zero to state that the equation has no solution? while the solution is defined to be in the Real number rage? –  kirill Oct 29 '12 at 20:53
1  
In my example i am using the real number field $\mathbb{R}$. I am not aware of the significance of the other determinants appearing in Cramer's rule in the nonexistence of a solution, but since determinants are hard to compute, it is better to use the criterion $rank[A |b]=rank[A]$. A solution exists if and only if the above equality is true. Checking this is very easy, e.g. if you are using MATLAB via the "rank" function. Do you know what rank is? –  Manos Oct 29 '12 at 22:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.