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Simultaneous equation help

When writing my maths notes and doing the questions, I'm stumped on the following simulatenous equations:

a+b+c=0 a + b + c = 0

2a+c=02a + c = 0

I know that c=−2ac = -2a and a=ba = b , but I'm not sure how to find the values. Any help would be appreciated :smile:
Original post by SecretDuck
When writing my maths notes and doing the questions, I'm stumped on the following simulatenous equations:

a+b+c=0 a + b + c = 0

2a+c=02a + c = 0

I know that c=−2ac = -2a and a=ba = b , but I'm not sure how to find the values. Any help would be appreciated :smile:

What you have is a system of two equations in three variables. That means it's underdetermined: the solutions, if they exist, are not unique. You've reduced them to their simplest form, such that given aa, you can produce a solution (a,b,c)=(a,a,−2a)(a,b,c) = (a,a,-2a); that's the best you can do.

That is to say, the solution to the system is precisely b=a,c=−2ab=a, c=-2a.
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SecretDuck
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Original post by Smaug123
What you have is a system of two equations in three variables. That means it's underdetermined: the solutions, if they exist, are not unique. You've reduced them to their simplest form, such that given aa, you can produce a solution (a,b,c)=(a,a,−2a)(a,b,c) = (a,a,-2a); that's the best you can do.

That is to say, the solution to the system is precisely b=a,c=−2ab=a, c=-2a.


Oh right, it's just the lecturer came up with the answers:
a=1, b=1 and c=-2
Original post by SecretDuck
Oh right, it's just the lecturer came up with the answers:
a=1, b=1 and c=-2

(1,1,-2) is indeed a solution; indeed, in R3\mathbb{R}^3, the set of solutions is precisely those elements on the line spanned by the vector (1,1,−2)(1,1,-2).
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SecretDuck
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Original post by Smaug123
(1,1,-2) is indeed a solution; indeed, in R3\mathbb{R}^3, the set of solutions is precisely those elements on the line spanned by the vector (1,1,−2)(1,1,-2).


Yes, the answer is supposed to be an orthogonal complement to a matrix. Thank you :smile:

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