exam 2 question

[kimk2kr]

consider the simultaneous linear equations
            px+3y+z=5
            2x-y+2pz=3
            x+4y+pz=6
Which is true:

a) there is no unique solution when p^2=1 and infinitely many solutions when p=-1
b) there is no unique solution when p^2=1 and no solution when p=1
c) there is no unique solution when p^2≠1 and no solution when p≠-1
d) there is a unique solution when p^2≠1 and infinitely many solutions when p≠-1
e) there are infinitely many solutions when p=1 and no solution when p=-1

i need a proper working out for this questions
help please

[Special At Specialist]

Hmm... I know that parallel lines have no solution and all of the equations you gave me were lines, but I've never done three dimensional problems before, so I'm afraid I can't help you much.

If it was two dimensions, I would solve it like this:
1. Parallel lines have no solution, so look for parallel lines.
2. The best way to do this is to look for equations which have the same gradient but a different translation.
3. If you can't do that, a different method is to turn it into matrix form and if the discriminant equals 0, then it has no solutions.
4. Find all algebraic values that fit these conditions.

[Vincezor]

Haha this is from Kilbaha 2011 exam 2!

Yeah I had trouble with this question. By putting the information in a matrix and finding the determinant, we can then solve that to find which values will yield either infinite solutions or no solutions (det = 0)

So anyway you get p = -1 or 1 (or p^2=1)

Now what else you could do is setup 3 similtaneous equations, to solve for x,y,z.

When you do so, you get

As you can see when p = -1, there is no solution (as x=z=undefined?)

Looking at the options, E is the only answer.

[kimk2kr]

Haha this is from Kilbaha 2011 exam 2!

Yeah I had trouble with this question. By putting the information in a matrix and finding the determinant, we can then solve that to find which values will yield either infinite solutions or no solutions (det = 0)

So anyway you get p = -1 or 1 (or p^2=1)

Now what else you could do is setup 3 similtaneous equations, to solve for x,y,z.

When you do so, you get

As you can see when p = -1, there is no solution (as x=z=undefined?)

Looking at the options, E is the only answer.

tbh i dont really understand why i have to solve for x,y and z...
when i subbed  1 into the equations none of them were equal to each other
whereas the answer tells you that its infite solution when p=1

[dc302]

I don't suppose you learn row reduction in Methods do you now...

[kimk2kr]

I don't suppose you learn row reduction in Methods do you now...

row reduction......?

[madoscar65]

For that question, you can also use calculator and find when determinant=0

[Greatness]

I don't suppose you learn row reduction in Methods do you now...

Is that the rref( ____ ) command? (reduced row - echleon form)

Well i used that, and i got 
Then you can do what vincezor said

[Special At Specialist]

Bump this topic because it is an important question which has not yet been resolved.

Using a calculator, we get the solution:
x = 2 / (p + 1)
y = 1
z = 2 / (p + 1)

From these solutions, we can see that p ≠ -1, so there will be no unique solution when p = -1.

If we substitute p = 1, we get: x = y = z = 1. However, this is still a unique solution. All this means is that if p = 1 then (1, 1, 1) is a solution. So this doesn't tell us anything.

From what it seems: if p = -1 then there is no unique solution and if p ≠ -1 then there is a unique solution (as shown above with various values of p).

None of the multiple choice options seem to be valid.

[paulsterio]

None of the multiple choice options seem to be valid.

yeah...nah!








When


      (1)

     (2)

      (3)


(3) - (1)





Substitute into (1), (2) and (3)









Hence, there are infinite solutions


When


  (4)

    (5)

     (6)


(6) + (4)













Transpose and make positive:








Hence, there are no solutions when

The answer is E.

[kamil9876]

Bump this topic because it is an important question which has not yet been resolved.

Using a calculator, we get the solution:
x = 2 / (p + 1)
y = 1
z = 2 / (p + 1)

The problem is that the calculator probably does this symbolically and assumes p is not -1 (obviously), but also somewhere along the way assumes p is not 1. So you had no way of telling whether p=1, p=2 or whatever was problematic. The most general way to solve this class of problems is to do the row reduction by hand (but I'm guessing this isn't in vce?) or calculate the determinant and see where it is 0, then do some casework as paulsterio did.