# Linear algebra

Watch
Announcements
#1

Hello, I'm currently doing part b) and getting a little stuck.

My instinct says that I need to write 5 equations, such as x(1) = 2 + s + t and then use the 5 equations to find s, t and then x(1) to x(5)?

However, I'm having real problems eliminating anything... so perhaps I'm not doing what the question is asking?

Or am I meant to be setting x(1) to be 18 (15+2+1) = 2 + s + t?

Also I don't know how I'm to know that there is a unique solution via the inequalities, but I know there is one via the RREF as rk(A) = rk(A|b).

Thanks!!
0
2 years ago
#2
(Original post by Bameron)

Hello, I'm currently doing part b) and getting a little stuck.

My instinct says that I need to write 5 equations, such as x(1) = 2 + s + t and then use the 5 equations to find s, t and then x(1) to x(5)?
Well, the question explictly tells you to form 5 inequalties, so I'd say your instincts are wrong. (I also have no idea where 2 + s + t came from!).

Your first inequality should be: "Since , and ".

For how to proceed after you have the 5 inequalities, you just have to look for ways to find conditions on s and t that eventually allow you to find them.

0
#3
Ahh I don't know what I'm doing... I've followed your example and completed the other 4 inequalities, but I don't really know what you mean by finding conditions for s and t other than that they are both greater than or equal to 0 and I don't know what using that fact does for me.

Unfortunately your hint of x(2) + x(3) >= 0 hasn't helped me much... x(2) + x(3) = -10t... so -10t >= 0.

(Original post by DFranklin)
Well, the question explictly tells you to form 5 inequalties, so I'd say your instincts are wrong. (I also have no idea where 2 + s + t came from!).

Your first inequality should be: "Since , and ".

For how to proceed after you have the 5 inequalities, you just have to look for ways to find conditions on s and t that eventually allow you to find them.

Ahh I don't know what I'm doing... I've followed your example and completed the other 4 inequalities, but I don't really know what you mean by finding conditions for s and t other than that they are both greater than or equal to 0 and I don't know what using that fact does for me.

Unfortunately your hint of x(2) + x(3) >= 0 hasn't helped me much... x(2) + x(3) = -10t... so -10t >= 0.
0
2 years ago
#4
(Original post by Bameron)
Ahh I don't know what I'm doing... I've followed your example and completed the other 4 inequalities, but I don't really know what you mean by finding conditions for s and t other than that they are both greater than or equal to 0 and I don't know what using that fact does for me.

Unfortunately your hint of x(2) + x(3) >= 0 hasn't helped me much... x(2) + x(3) = -10t... so -10t >= 0.
If -10t >=0, what can we say about t? And what else do we know about t? So...
0
#5
(Original post by DFranklin)
If -10t >=0, what can we say about t? And what else do we know about t? So...
If -10t>= 0 then t has to be negative to be greater than 0, but we also know t is greater than 0, so we have a contradiction?
0
2 years ago
#6
(Original post by Bameron)
If -10t>= 0 then t has to be negative to be greater than 0, but we also know t is greater than 0, so we have a contradiction?
No, we don't have a contradiction because the inequalities are not strict. is the only value for which and both hold.
0
#7
Ok so that leaves me with x(5) = 0, x(2) + x(3) = 0

from x(1), 2+s >= 0 so s >= -2, but from x(4), 4s >= 0, so s>=0, which I already knew from the question.
Last edited by Bameron; 2 years ago
0
#8
Having s also = 0 give me the correct answer, but I don't know how to prove s = 0 from the other side, just that I know s>=0
0
2 years ago
#9
(Original post by Bameron)
Having s also = 0 give me the correct answer, but I don't know how to prove s = 0 from the other side, just that I know s>=0
Not sure about in addition to ... that would imply which is negative.

You can show that must take on a different specific value though. Notice that the equations for are off by a multiple of -1 from each other (after imposing the condition ). For them both to be greater than (or equal to) 0, must take on a specific value which you can determine.
Last edited by RDKGames; 2 years ago
0
#10
Ok s=4/7 and this give me the correct portion sizes. Thank you both for being so patient with me, I'm going to run through the question again to make sure I understand everything.
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Are you tempted to change your firm university choice on A-level results day?

Yes, I'll try and go to a uni higher up the league tables (5)
55.56%
Yes, there is a uni that I prefer and I'll fit in better (0)
0%
No I am happy with my choice (4)
44.44%
I'm using Clearing when I have my exam results (0)
0%