Ax=b1 and Ax=b2 are consistent. Is the system Ax=b1+b2 necessarily consistent?


by iamzzz
Tags: axb1, axb2, consistent, necessarily
iamzzz
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#1
Nov20-12, 09:30 PM
P: 22
1. The problem statement, all variables and given/known data
Suppose A is m*n matrix b1 and b2 are m*1 vector and the systems Ax=b1 and Ax=b2 are consistent. Is the system Ax=b1+b2 necessarily consistent?


2. Relevant equations



3. The attempt at a solution

I think Ax = b1 + b2 should be consistent but i dont know how to prove..
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Dick
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#2
Nov20-12, 09:37 PM
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Quote Quote by iamzzz View Post
1. The problem statement, all variables and given/known data
Suppose A is m*n matrix b1 and b2 are m*1 vector and the systems Ax=b1 and Ax=b2 are consistent. Is the system Ax=b1+b2 necessarily consistent?


2. Relevant equations



3. The attempt at a solution

I think Ax = b1 + b2 should be consistent but i dont know how to prove..
No thoughts about how to prove at all? A(x1+x2)=Ax1+Ax2. Think about it some more.
iamzzz
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#3
Nov20-12, 09:59 PM
P: 22
Quote Quote by Dick View Post
No thoughts about how to prove at all? A(x1+x2)=Ax1+Ax2. Think about it some more.
let x1 be the solution of Ax1=b1 and x2 be the solution of Ax2=b2

So Ax1+Ax2=b1+b2=A(x1+x2)
so (x1+x2)could be x
prove ?
Is this correct ?

Dick
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#4
Nov20-12, 10:03 PM
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Ax=b1 and Ax=b2 are consistent. Is the system Ax=b1+b2 necessarily consistent?


Quote Quote by iamzzz View Post
let x1 be the solution of Ax1=b1 and x2 be the solution of Ax2=b2

So Ax1+Ax2=b1+b2=A(x1+x2)
so (x1+x2)could be x
prove ?
Is this correct ?
I would say let x1 be ANY solution to Ax=b. There may be more than one. But why are you asking "Is this correct?". What part of it are you worried about?
iamzzz
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#5
Nov20-12, 10:19 PM
P: 22
Quote Quote by Dick View Post
I would say let x1 be ANY solution to Ax=b. There may be more than one. But why are you asking "Is this correct?". What part of it are you worried about?
I mean does that prove the problem ?

Anyway thanks for the help
Dick
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#6
Nov20-12, 10:22 PM
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Quote Quote by iamzzz View Post
I mean does that prove the problem ?

Anyway thanks for the help
I'm just saying I would feel better if you KNEW it solved the problem instead of having to ask. Yes, it's fine.


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