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Math Help - Linear Algebra - Linear Equations

  1. #1
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    Linear Algebra - Linear Equations

    I am having trouble answering a multiple choice question, in this question there can be multiple answers and there are 5 choices.

    3a+b+9c-2d=20
    7b -8d=2
    3a +9c =24

    I found b=-26 and d=-23. I am pretty sure that is right, however if you insert that into the first equation you get 3a+9c = 0. However that may be a part of the answer, heres the possible answers.

    A. 'c' can be determined uniquely
    B. If a=-1, then c=4
    C. The solution has three paramaters.
    D. The solution has two parameters.
    E. 'd' can be expressed in terms of 'a' and 'b' alone.

    Remember you can select more than one option. Any insight would be much appreciated.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by the_philosipher View Post
    I am having trouble answering a multiple choice question, in this question there can be multiple answers and there are 5 choices.

    3a+b+9c-2d=20
    7b -8d=2
    3a +9c =24

    I found b=-26 and d=-23. I am pretty sure that is right, however if you insert that into the first equation you get 3a+9c = 0. However that may be a part of the answer, heres the possible answers.

    A. 'c' can be determined uniquely
    B. If a=-1, then c=4
    C. The solution has three paramaters.
    D. The solution has two parameters.
    E. 'd' can be expressed in terms of 'a' and 'b' alone.

    Remember you can select more than one option. Any insight would be much appreciated.
    a similar question was asked here

    see if you understand what was done
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  3. #3
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    odd

    I actually had that question in my homework as well, weird. I understand that question, I got as far as removing row 3, but for some reason I removed row 1 as well, the rest I understand. In this question though I am still stuck and can't seem to find the right help from that thread. Thanks for the idea though.

    My problem stems from using a matrix to isolate the third row as [0 0 0 0 / 4]. Does that not mean that its no solution, or non-consistent?

    And just to check, 'b' and 'd' can be determined uniquely because of EQN 2 and 3.

    I also find that if i substitute the values I found for 'b' and 'd' into EQN 1 I end up with 3a+9c=0 which doesn't help much when EQN 2 is 3a+9c=24.

    So I'm still quite lost.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by the_philosipher View Post
    I actually had that question in my homework as well, weird. I understand that question, I got as far as removing row 3, but for some reason I removed row 1 as well, the rest I understand. In this question though I am still stuck and can't seem to find the right help from that thread. Thanks for the idea though.

    My problem stems from using a matrix to isolate the third row as [0 0 0 0 / 4]. Does that not mean that its no solution, or non-consistent?
    yes, that would mean it is inconsistent, but you did something wrong, i did not have that problem
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  5. #5
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    Okay, so I have b=-14/3 and d=-13/3.

    I have checked and it fits the EQN 1. But where do I go from here? I would still have to make 1 parameter because I am left with both the 'a' and 'c' variables.

    I don't see how there could be two parameters or how you could determine 'a'. When I select the first three options it says I am wrong ( one parameter, can determine b uniquely, can determine d uniquely). What am I missing?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by the_philosipher View Post
    I am having trouble answering a multiple choice question, in this question there can be multiple answers and there are 5 choices.

    3a+b+9c-2d=20
    7b -8d=2
    3a +9c =24

    I found b=-26 and d=-23. I am pretty sure that is right, however if you insert that into the first equation you get 3a+9c = 0. However that may be a part of the answer, heres the possible answers.

    A. 'c' can be determined uniquely
    B. If a=-1, then c=4
    C. The solution has three paramaters.
    D. The solution has two parameters.
    E. 'd' can be expressed in terms of 'a' and 'b' alone.

    Remember you can select more than one option. Any insight would be much appreciated.
    Subtract the third equation from the first, that gives you two equations in
    b and d which you can solve to find that b=6 and d=5.

    That leaves you with one equation in a and c: a+3c=8. Choose one of these
    as a parameter to solve for the other, so your solution depends on one parameter.

    So lets see:

    A is true
    B is false as when a=-1, b=3.
    C is false the solution has 1 parameter
    D is false the solution has 1 parameter
    E is true d=(0)a+(5/6)b

    RonL
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  7. #7
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    Wow I'm not entirely sure why that was so difficult. For some reason I thought i could manipulate EQN 1 so that b-2d=4 was left, which you cannot and that was the reason for my confusion. Thanks for your help.
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