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Math Help - Matrix equation proof

  1. #1
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    Matrix equation proof

    If a matrix equation AX=B has two solutions then why does it have infinitely many solutions?

    Clue: if X_1 and X_2 are solutions then look into X_3=sX_1+tX_2  (s,t\in R)

    Now, I really dont know where to start on this one. I would appreciate some help just to get started.
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  2. #2
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    Take s and t such that s+t=1. Let X_1, X_2 be such that AX_1 = AX_2 = B. Now, what can you say about X_3 = s \cdot X_1 + t \cdot X_2 ?
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  3. #3
    Super Member girdav's Avatar
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    I guess the unknown is X.
    If X_1 and X_2 are solutions, we will have A\left(sX_1+tX_2\right) = sB+tS =\left(s+t\right)B and if s+t=1 then sX_1+tX_2 will be a solution, that can be different from X_1 and X_2 for some s and t well chosen.
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  4. #4
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    Heres what I got so far

    This line <br /> <br />
X_3=sX_1+tX_2    (s,t\in R)<br />

    means that X_3 represents the multitude of solutions.Right?
    And if i rewrite the original equation to <br /> <br />
A\left(sX_1+tX_2\right)=B<br />
and prove that its true for any (s,t\in R) then im done. Right?


    I tried this:
    <br /> <br />
A\left(sX_1+tX_2\right)=B<br />

    AsX_1+AtX_2=B

    sAX_1+tAX_2=B

    Now I can do that right? Because s and t are scalars.

    And since AX_1=AX_2

    I get

    sAX_1+tAX_1=B

    (s+t)AX_1=B

    I can rewrite the equation <br /> <br />
sAX_1+tAX_2=B<br />
to <br /> <br />
sB+tB=(s+t)B=B<br />

    because <br /> <br />
AX_1=AX_2=B<br />

    The final equation looks like this

    <br /> <br />
(s+t)AX_1=(s+t)B<br />
which is true for any (s,t\in R) since (s+t) is a scalar and can be divided from the equation.


    Honestly, if this is correct Im still unsure so any enlightment is still appreciated.
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  5. #5
    Super Member girdav's Avatar
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    You could get this equation more directly by multiplying by s+t because X_1 is a solution.
    You can try to find s and t such as sX_1+tX_2 is a solution given that X_1 and X_2 are solutions.
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  6. #6
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    Im not quite sure if I understand. I see that i could cut to the chase a lot faster but I guess its the "proof" part that I tried to elaborate from the clue-part. Can I consider your first reply an answer to the problem?

    Im having a hard time grasping the s and t part.

    And is there something wrong about my attempt to an answer besides it being unnecessarily long?
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  7. #7
    Super Member girdav's Avatar
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    If X_1 and X_2 are solutions of AX=B then what can you say about t X_1+\left(1-t\right)X_2 for t\in \mathbb R?
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  8. #8
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    Still confused

    I should probably give up since I still do not understand. Im assuming that X_1 and X_2 are two different matrixes that satisfy the equation. Whats the purpose of adding them together? Let alone the purpose of adding the for some well chosen values of s and t?

    So what can I say about

    <br /> <br />
t X_1+\left(1-t\right)X_2<br />

    ..nothing really!
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  9. #9
    Super Member girdav's Avatar
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    You can multiplate it by A.
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