3-Variable Simultaneous Equations Question
Tried this old FP4 question but unsure how to solve part b? I couldn't see that any of the equations were multiples or parallel (such as in an inconsistent system)
Looked at the MS (below) and not sure where they derived 5b - 25 = 0 from? Feel like I'm probably forgetting something basic but not sure - is anyone able to help?
Looked at the MS (below) and not sure where they derived 5b - 25 = 0 from? Feel like I'm probably forgetting something basic but not sure - is anyone able to help?
(edited 2 years ago)
For 3 variables youve got to be a bit more systematic as it will be a combination of the equations. So the third plane will be parallel to a combination of the first two. What did you do for the part a)?
When you reduce it, you'll find, as per the mark scheme,
5b - 25 = 0
This will be inconsistent (no solution) for b != 5.
When you reduce it, you'll find, as per the mark scheme,
5b - 25 = 0
This will be inconsistent (no solution) for b != 5.
(edited 2 years ago)
Original post by mqb2766
For 3 variables youve got to be a bit more systematic as it will be a combination of the equations. So the third plane will be parallel to a combination of the first two. What did you do for the part a)?
When you reduce it, you'll find, as per the mark scheme,
5b - 25 = 0
This will be inconsistent (no solution) for b != 5.
When you reduce it, you'll find, as per the mark scheme,
5b - 25 = 0
This will be inconsistent (no solution) for b != 5.
Just to say that I don't see that you *have* to end up with the equation 5b-25 = 0; it will depend on how you solve it. (I got 3 = 8 - b, for example).
Original post by DFranklin
Just to say that I don't see that you *have* to end up with the equation 5b-25 = 0; it will depend on how you solve it. (I got 3 = 8 - b, for example).
sure, thats why I asked about how they did a).
Original post by mqb2766
For 3 variables youve got to be a bit more systematic as it will be a combination of the equations. So the third plane will be parallel to a combination of the first two. What did you do for the part a)?
When you reduce it, you'll find, as per the mark scheme,
5b - 25 = 0
This will be inconsistent (no solution) for b != 5.
When you reduce it, you'll find, as per the mark scheme,
5b - 25 = 0
This will be inconsistent (no solution) for b != 5.
Original post by DFranklin
Just to say that I don't see that you *have* to end up with the equation 5b-25 = 0; it will depend on how you solve it. (I got 3 = 8 - b, for example).
For part a) I calculated the determinant of the 3x3 matrix as below?
Ok you could do an equivalent answer using row reduction methods on the augmented matrix (extra column corresponding to the (3,4,b) vector).
Or spot that with a=-3, the third row is a linear combination of the first two? Then apply the same linear combination to the right hand side (3,4,b)?
To get the linear combination a*row1 + b*row2 = row3
x: 2a + b = 2
y: -a + 2b = 1
Edit - as Dfranklin below notes, this can be used to answer both parts by considering the z coefficients (part a) and then the right hand side (part b).
Or spot that with a=-3, the third row is a linear combination of the first two? Then apply the same linear combination to the right hand side (3,4,b)?
To get the linear combination a*row1 + b*row2 = row3
x: 2a + b = 2
y: -a + 2b = 1
Edit - as Dfranklin below notes, this can be used to answer both parts by considering the z coefficients (part a) and then the right hand side (part b).
(edited 2 years ago)
Original post by beachpanda
For part a) I calculated the determinant of the 3x3 matrix as below?
To answer the 2nd part, you basically have to try to solve the equations when a = -3 and see what happens. (You'll find you can write the (LHS of the) 3rd equation as a combination of the first 2 and therefore the value of the RHS is forced to equal 5). (And so the equations are inconsistent unless b = 5).
The disadvantage of doing (a) using a determinant is that it doesn't help with part (b) at all. In a question like this, it's arguably "better" to effectively do parts (a) and (b) "simultaneously"; try to solve the equations, you'll end up realising that you "can't" solve them in a standard way for a particular value of a (which gives you your answer to (a)) and then continue to solve for that particular value of (a).
That said: in this case I don't think it's actually less work. [I'm used to when Cambridge used to set fairly standard Tripos/Entrance Exam questions that were similar but a little more involved and it was pretty uniformly the case that it wasn't a good idea to try to find the determinant].
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