Consistency of a set of equations

[shrabastee]

Homework Statement

(q-1)x+py=0
(1-q)x-py=0
x+y=1

In this linear equation system:
a. if p and q lie between 0 and 1 how many solutions does this system have and why?
b. Ignoring the above condition, find the values of p and q so that the system has NO solution.

Homework Equations

Umm..

The Attempt at a Solution

The equation sets are consistent and yield a non-trivial solution if det A is non-zero, and the equations are linearly independent, right? Or do I need to look at ranks and augmented matrices and all those things?

I was thinking of equating the determinant of the coefficient matrix to 0, to see when they would be inconsistent and hence unsolvable, and i got p=0 that way, but I don't think it's right. I get very confused with this unsolvable/undefined/multiple solutions business, can someone please give me a step-by-step method to check whether a given set of equations is solvable? And also help me with thi sum? Thanks. :)

 

[vela]

I suggest you form the augmented matrix and go from there.

 

[shrabastee]

Okay, thanks :)

 

[SammyS]

Look at the sum of the first two equations.

 

[AJKing]

What is the "Augmented matrix"?

 

[vela]

http://en.wikipedia.org/wiki/Augmented_matrix

 

[shrabastee]

So the system has infinite solutions, right? Since x+y=1 is the only equation we are left with, once we add the first two?

But what about the second part? When will the system have no solution?

 

[vela]

You didn't use the matrix, did you?

 

[ehild]

So the system has infinite solutions, right? Since x+y=1 is the only equation we are left with, once we add the first two?

No, you have two independent equations. The second one is just the same as the first one, multiplied by -1.

ehild

 

[SammyS]

No, you have two independent equations. The second one is just the same as the first one, multiplied by -1.

ehild

Yes, that's what I hoped OP (shrabastee) would recognize after following my hint.

Now, shrabastee, take one of the first two equations along with the third, and solve simultaneously.

 

[shrabastee]

So how many solutions does the system have? If p and q lie between 0 and 1 it has one solution, does it? Which I can find by solving them simultaneously. And, they'll have no solution if the ratio of the coefficients of the variables is equal. Is that it?

 

[shrabastee]

You didn't use the matrix, did you?

No I didn't :( Augmented matrices confuse me too much. :(

 

[ehild]

So how many solutions does the system have? If p and q lie between 0 and 1 it has one solution, does it? Which I can find by solving them simultaneously. And, they'll have no solution if the ratio of the coefficients of the variables is equal. Is that it?

Yes, there is one solution if p and q are different. There would not be any solution if the denominator was p+1-q=0. Is it possible or not in the range of p and q?
Yes, there is no solution if p=q-1, except in case p=0, q=1. As both p and q are between 0 and 1, it can happen if the interval was meant closed. How many solutions are there if 0 and 1 values are allowed for p, q?ehild

 

[vela]

No I didn't :( Augmented matrices confuse me too much. :(

I asked because if you had used one, you would have avoided the mistake of thinking you get rid of both equations when you added the first to the second and got 0=0. Using a row operation on the augmented matrix would have left you with one row of all zeros, but the other two row (i.e., the other equations) would have remained.

 

[shrabastee]

Yes, there is one solution if p and q are different. There would not be any solution if the denominator was p+1-q=0. Is it possible or not in the range of p and q?
Yes, there is no solution if p=q-1, except in case p=0, q=1. As both p and q are between 0 and 1, it can happen if the interval was meant closed. How many solutions are there if 0 and 1 values are allowed for p, q?


ehild

Can I find specific values for p and q? Or does any value of p and q that satisfies the equations q-p=1 qualify as a solution? p and q have no restrictions in this case, they've asked us to ignore the "lies between 0 and 1" condition.

 

[SammyS]

Can I find specific values for p and q? Or does any value of p and q that satisfies the equations q-p=1 qualify as a solution? p and q have no restrictions in this case, they've asked us to ignore the "lies between 0 and 1" condition.

No. As long as q-p ≠ 1 you have a solution. If 0 < p,q < 1 , all values of p,q in that interval give a solution !

 

[Ray Vickson]

Homework Statement

(q-1)x+py=0
(1-q)x-py=0
x+y=1

In this linear equation system:
a. if p and q lie between 0 and 1 how many solutions does this system have and why?
b. Ignoring the above condition, find the values of p and q so that the system has NO solution.

Homework Equations

Umm..

The Attempt at a Solution

The equation sets are consistent and yield a non-trivial solution if det A is non-zero, and the equations are linearly independent, right? Or do I need to look at ranks and augmented matrices and all those things?

I was thinking of equating the determinant of the coefficient matrix to 0, to see when they would be inconsistent and hence unsolvable, and i got p=0 that way, but I don't think it's right. I get very confused with this unsolvable/undefined/multiple solutions business, can someone please give me a step-by-step method to check whether a given set of equations is solvable? And also help me with thi sum? Thanks. :)

Ignore one of the first two equations (since one is just the negative of the other). Then, if we have, say (q-1)x + py = 0, and x + y = 1, the system has a unique solution if the determinant of the coefficient matrix [[q-1,p][1,1]] is nonzero, and has either no solution or infinitely many solutions if the determinant = 0; that is a theorem, that holds for general linear equation systems. For 0 <= q,p <= 1 there is just one value of (q,p) that makes the determinant equal to zero, and you have found it: q=1, p=0. In that case the system reads as 0*x+0*y=0, x+y=1. You should be able to see whether we have infinitely many solutions or no solution.

RGV