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Math Help - Explanation of a matrix, linear system

  1. #1
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    Explanation of a matrix, linear system

    The matrix is:

    [-1 0 -3] [k1] [0]
    [1 0 3] [k2] = [0]
    [1 0 3] [k3] [0]

    Now I understand that if you let k1=3 and k3=-1, then k2=0:

    [3 ]
    [0 ]
    [-1]

    Another possible solution, according to the back of the book is

    [0]
    [1]
    [0]

    Is it correct for me to think that 1 is an aribtrary number and that it really could be anything? Thanks,

    Kim
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  2. #2
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    Hello, Kim Nu!

    I'm not sure what your question is.
    The matrix equation is rather silly, isn't it?


     \begin{bmatrix}\text{-}1 & 0 & \text{-}3 \\ 1 & 0 & 3 \\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} \:=\:\begin{bmatrix}0\\0\\0\end{bmatrix}
    Now I understand that if you let: . \begin{bmatrix}x\\y\\z\end{bmatrix}\:=\:\begin{bma  trix}3 \\ 0 \\ \text{-}1 \end{bmatrix}
    Another possible solution, according to the book is: . \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}

    The matrix equation gives this system: . \begin{array}{ccc}\text{-}x-3z &=&0 \\ x+3z &=&0 \\ x+3z &=& 0 \end{array}

    . . which is just one equation: . x+3z\:=\:0 \quad\Rightarrow\quad x \:=\:-3z


    A solution would be: . \begin{array}{ccc}z &=& \text{any number} \\ x &=& -\text{3 times z} \\ y &=& \text{any number} \end{array}


    Therefore: . \begin{bmatrix}x\\y\\z\end{bmatrix} \;=\;\begin{bmatrix} 3s \\ t \\ s\end{bmatrix}\quad\hdots . for any values of s\text{ and }t

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  3. #3
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    Soroban,

    Thanks for the response!

    So under the system you set up, when y=1, does that mean that z has to be 0? I'm guessing not as x and z are the only ones that are dependent on each other.
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    Quote Originally Posted by Kim Nu View Post
    Soroban,

    Thanks for the response!

    So under the system you set up, when y=1, does that mean that z has to be 0? I'm guessing not as x and z are the only ones that are dependent on each other.
    That's correct: y and z are both free variables and are independent from each other. Also note that Soroban's final answer should be \left[\begin{matrix}x\\y\\z\end{matrix}\right]<br />
= \left[\begin{matrix}-3s\\t\\s\end{matrix}\right].
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