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Math Help - How do I prove if a vector is a solution to a linear system?

  1. #1
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    How do I prove if a vector is a solution to a linear system?

    (I do not study in America so forgive me if I don't translate the terms correctly)

    I am having trouble solving this question:

    Vectors "u" (-2, 4, 4, 2) and "v" (4, -2, -2, 4) are solutions to a linear system (M) with 4 unknowns. And it is known that (2, 2, 2, 2) is not a solution.

    a) Prove that this system is not homogenous (that the b vector does not equal 0. I don't know if this is the correct translation)

    b) Prove that (0, 2, 2, 0) is a solution.

    "a" I was able to answer: there is a rule that says, in a homogenous matrix, with a1,...,ak answers, that every linear combination (again, don't know if this is correct translation) of a1,...,ak is also a solution of the system:
    Meaning, if I prove (2, 2, 2, 2) is a linear combination of "u" and "v" then that proves that this system is not homogenous. (Since, otherwise, it should be homogenous, according to the law)

    -2 4 2 1 -2 -1 1 0 1
    4 -2 2 = 2 -1 1 = 0 1 1
    4 -2 2 2 -1 1 0 0 0
    -2 4 2 1 -2 -1 0 0 0

    1 ( -2, 4, 4, -2) + 1 (4, -2, -2, 4) = (2, 2, 2, 2)

    This proves that (2, 2, 2, 2) is a linear combination of "u" and "v". And since it is a given that (2, 2, 2, 2) is not a solution, then (M) is not homogenous.

    b) I believe i answered "a" correctly, but I do not know how to answer this second part. I have tried many things. If someone can help me that would be great! thank you so much.
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  2. #2
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    Quote Originally Posted by jayshizwiz View Post
    (I do not study in America so forgive me if I don't translate the terms correctly)

    I am having trouble solving this question:

    Vectors "u" (-2, 4, 4, 2) and "v" (4, -2, -2, 4) are solutions to a linear system (M) with 4 unknowns. And it is known that (2, 2, 2, 2) is not a solution.

    a) Prove that this system is not homogenous (that the b vector does not equal 0. I don't know if this is the correct translation)

    b) Prove that (0, 2, 2, 0) is a solution.

    "a" I was able to answer: there is a rule that says, in a homogenous matrix, with a1,...,ak answers, that every linear combination (again, don't know if this is correct translation) of a1,...,ak is also a solution of the system:
    Meaning, if I prove (2, 2, 2, 2) is a linear combination of "u" and "v" then that proves that this system is not homogenous. (Since, otherwise, it should be homogenous, according to the law)

    -2 4 2 1 -2 -1 1 0 1
    4 -2 2 = 2 -1 1 = 0 1 1
    4 -2 2 2 -1 1 0 0 0
    -2 4 2 1 -2 -1 0 0 0

    1 ( -2, 4, 4, -2) + 1 (4, -2, -2, 4) = (2, 2, 2, 2)

    This proves that (2, 2, 2, 2) is a linear combination of "u" and "v". And since it is a given that (2, 2, 2, 2) is not a solution, then (M) is not homogenous.

    b) I believe i answered "a" correctly, but I do not know how to answer this second part. I have tried many things. If someone can help me that would be great! thank you so much.

    Let ax_1+bx+2+cx_3+dx_4=K\neq 0 be one of the equations in the (non-homogeneous, you proved this correctly) system, then

    I\,:\, -2a_1+4a_2+4a_3-2a_4=K

    II\,:\,4a_1-2a_2-2a_3+4a_4=K .

    Now, multiply eq. I by 2, add this to eq. II and then divide by 3...

    Tonio
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  3. #3
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    Let <br /> <br />
ax_1+bx_2+cx_3+dx_4=K\neq 0<br />
be one of the equations in the (non-homogeneous, you proved this correctly) system, then

    <br /> <br />
I\,:\, -2a_1+4a_2+4a_3-2a_4=K<br />

    <br /> <br />
II\,:\,4a_1-2a_2-2a_3+4a_4=K<br />

    Now, multiply eq. I by 2, add this to eq. II and then divide by 3...

    Tonio
    Thanks. I am still pretty new at this stuff so I have a few last questions. Since both equations = K, can't I also write

    -2a_1+4a_2+4a_3-2a_4=4a_1-2a_2-2a_3+4a_4

    which would make

    6a_1-6a_2-6a_3+6a_4=0

    So what does that mean exactly? Is it legal to do that?

    Also, is it possible to get the result

    2a_1+2a_2+2a_3+2a_4=K using your method of combining equations??
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