When to Parametrise: Explaining b(iii)

[Gregg]

Homework Statement

I've done the first parts. b(iii) I don't understand, the mark scheme uses a parameter to solve it. (x=t). I don't understand why it's necessary, actually I can get a solution? The textbook is rubbish that's why I'm asking so much. An explanation would be good.

 

[rock.freak667]

Because when p=1,q=8 and r=0, 24p-3q-r=0 right?


So in the augmented matrix, you will have an entire row of zeroes. Which means you will have an infinite number of solutions, which will all depend on a parameter,t.

 

[Gregg]

How exactly do you do it? Does matter which you parametrise? the text doesn't give a clear definition and explanation.

Here's the example

solve
x-3y+4z=4
2x-y+3z=8
3x+y+2x=12

Solution:

To eliminate y (3)x3 + (1)

10x+10z=40

(2)+(3)

5x+5z=40

Setting z=t, a parameter gives x=4-1. Substituting in equation 3, 12-3t + y + 2t=12

y=t

{x,y,z} = {4,0,0} + t{-1,1,1} A common line.

 

[rock.freak667]

No it doesn't really matter which one you set as t.

Just once you eliminate one variable from the three equations and then make one equation from the remaining two, you can set either as the parameter and get the same answer.

 

[HallsofIvy]

Back in (bi) you showed that the given system of equations never has a unique solution, presumably by showing that the determinant of the coefficient matrix was 0. (a= b so (a-b)= 0)

So for given p, q, r, the system either has no solution or has an infinite number of solutions. In (bii) you showed that if 24p- 3q- r= 0, the system is "consistent", so has at least one solution and so has and infinite number of solutions.

Now, with p= 1, q= 8, and r= 0 24p- 3q- r= 24(1)- 3(8)- 0= 0 so this system has an infinite number of solutions. You cannot expect to get a unique solution. You need a parameter to get all solutions. Basically, you cannot solve for specific values of x, y, and z but you can solve for two of them in terms of the other one. Use that one as parameter.

 

[Gregg]

Back in (bi) you showed that the given system of equations never has a unique solution, presumably by showing that the determinant of the coefficient matrix was 0. (a= b so (a-b)= 0)

So for given p, q, r, the system either has no solution or has an infinite number of solutions. In (bii) you showed that if 24p- 3q- r= 0, the system is "consistent", so has at least one solution and so has and infinite number of solutions.

Now, with p= 1, q= 8, and r= 0 24p- 3q- r= 24(1)- 3(8)- 0= 0 so this system has an infinite number of solutions. You cannot expect to get a unique solution. You need a parameter to get all solutions. Basically, you cannot solve for specific values of x, y, and z but you can solve for two of them in terms of the other one. Use that one as parameter.

It doesn't seem to work when I try to do it.