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Thread: Linear Algebra: Nontrivial and trivial solutions.

  1. #1
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    Linear Algebra: Nontrivial and trivial solutions.

    I understand how to compute augmented matrices, but I am not understanding this specific answer.

    Please look here:

    Linear Algebra and Its Applications (9780321982384), Pg. 48, Ex. 4 :: Homework Help and Answers :: Slader

    I do not understand the answer as to why that is like that. (I know trivial means where the only solution to Ax=0, is where x=0, and non-trivial exists where Ax=0, and x does not = 0.

    how is x3 free?

    Free implies where there are all 0's in the row does it not?

    My buddy told me that the x3 is free because "the third row is missing." That does not make any sense to me, this problem was given 2 equations with 3 variables in each equation, denoting only 2 ROWS.
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    Re: Linear Algebra: Nontrivial and trivial solutions.

    Quote Originally Posted by math951 View Post
    I understand how to compute augmented matrices, but I am not understanding this specific answer.

    Please look here:

    Linear Algebra and Its Applications (9780321982384), Pg. 48, Ex. 4 :: Homework Help and Answers :: Slader

    I do not understand the answer as to why that is like that. (I know trivial means where the only solution to Ax=0, is where x=0, and non-trivial exists where Ax=0, and x does not = 0.

    how is x3 free?

    Free implies where there are all 0's in the row does it not?

    My buddy told me that the x3 is free because "the third row is missing." That does not make any sense to me, this problem was given 2 equations with 3 variables in each equation, denoting only 2 ROWS.
    Add a third row $0x_1+0x_2+0x_3 = 0$
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    Re: Linear Algebra: Nontrivial and trivial solutions.

    But why do we have to add a third row?
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    Re: Linear Algebra: Nontrivial and trivial solutions.

    Is it because we have 3 columns?
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    Re: Linear Algebra: Nontrivial and trivial solutions.

    So I am assuming the amount of variables we have dictates how many rows we should have. I.E. 3 variables in equation, I should assume there are 3 equations needed in total, therefore 3 rows.
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    Re: Linear Algebra: Nontrivial and trivial solutions.

    Quote Originally Posted by math951 View Post
    So I am assuming the amount of variables we have dictates how many rows we should have. I.E. 3 variables in equation, I should assume there are 3 equations needed in total, therefore 3 rows.
    We can write the same problem like this:

    $-5x_1+7x_2+9x_3 = 0$
    $x_1-2x_2+6x_3=0$

    For the second equation, we can multiply both sides by 5:
    $5x_1-10x_2+30x_3=0$

    Adding this to the first equation gives:

    $0x_1-3x_2+39x_3=0$

    Solving for $x_2$ we get: $x_2=13x_3$. Now we can plug this back into the second equation:

    $x_1-2x_2+6x_3=0$
    $x_1-2(13x_3)+6x_3=0$
    $x_1=20x_3$

    So, $x_3$ is a free variable, and we can express both $x_1$ and $x_2$ in terms of $x_3$. Namely, $x_1=20x_3$ and $x_2=13x_3$. The matrix operations simply help you do it.
    Thanks from math951
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