Undetermined system with no solutions?

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The discussion centers on demonstrating that the given system of equations has no solutions. It is noted that the system is undetermined due to having more unknowns than equations. The determinant of the matrix formed by the coefficients is initially stated to be non-zero, indicating linear independence, but later corrections reveal it is actually zero. The key point is that a linear combination of the equations leads to a contradiction, confirming the absence of solutions. Thus, the conclusion is that the system indeed has no solutions.
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So I have the system { 2x+3y+z-3v=2 ;x-y+2z+v=0 ;3x+2y+3z-2v=-2} . I have to show that the system has no solutions.. I notice that it has more unknowns than equations so it is an undetermined system.If I form the matrix ? |(2 1 3 ) (3 -1 2) (1 2 3)| I notice that the determinant is diff from zero so this three vectors are linearly indipendent. Now how do I form a matrix using ( 2 0 -2) and these three vectors with the condition that they have a determinant diff from zero? Because if they have a determinant diff from zero,the system has no solutions.
 
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Elaia06 said:
So I have the system { 2x+3y+z-3v=2 ;x-y+2z+v=0 ;3x+2y+3z-2v=-2} . I have to show that the system has no solutions.. I notice that it has more unknowns than equations so it is an undetermined system.If I form the matrix ? |(2 1 3 ) (3 -1 2) (1 2 3)|
I notice that the determinant is diff from zero so this three vectors are linearly independent.
Right, but what are you deducing from that?
If there are no solutions there must be a linear combination of the equations which eliminates all variables but not all constants. It's fairly easy to spot.
 
I wrote the question in the wrong way..the answer on my textbook is " The system has no solutions"..I just have to show if the system has or not solutions...
 
But if the system has no solutions the vectors forming the matrix including 2 0 -2 should be linearly indipendent /...
 
Elaia06 said:
But if the system has no solutions the vectors forming the matrix including 2 0 -2 should be linearly indipendent /...
Not sure what you mean by 'including 2 0 -2'. Do you mean the 5x3 matrix formed by moving the constants across to the left? The non-existence of solutions does not require all the equations to be linearly independent. Trivial example:
x - 1 = 0
x - 2 = 0
x - 2 = 0
In the OP, I can see a linear combination of the equations which eliminates all the variables but leaves a contradictory statement regarding the constants. That implies there are no solutions.
The det of |(2 1 3 ) (3 -1 2) (1 2 3)| is 0.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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