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Thread: Reduce Augmented Matrix to find value to consistent equations.

  1. #1
    Member alexgeek's Avatar
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    Reduce Augmented Matrix to find value to consistent equations.

    I'm not sure how to do this one, I row reduced a bit but don't know where to go with it.
    $\displaystyle
    \begin{bmatrix}
    1 & 2 & 2 & | & 1 \\
    2 & 1 & 3 & | & 3 \\
    4 & 5 & 7 & | & \lambda \\
    \end{bmatrix}
    $

    How do I find the value of $\displaystyle \lambda$ for which the equations are consistent?

    Thanks
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  2. #2
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    Quote Originally Posted by alexgeek View Post
    I'm not sure how to do this one, I row reduced a bit but don't know where to go with it.
    $\displaystyle
    \begin{bmatrix}
    1 & 2 & 2 & | & 1 \\
    2 & 1 & 3 & | & 3 \\
    4 & 5 & 7 & | & \lambda \\
    \end{bmatrix}
    $

    How do I find the value of $\displaystyle \lambda$ for which the equations are consistent?

    Thanks
    Reduce it to upper triangular and post your results.

    I just did that myself and I don't think it can be consistent.
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  3. #3
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    What did you get when you "row reduced a bit"? Yes, there is a single value of $\displaystyle \lambda$ that makes this system consistent. What it is should be clear after you have "done" the first column.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    What did you get when you "row reduced a bit"? Yes, there is a single value of $\displaystyle \lambda$ that makes this system consistent. What it is should be clear after you have "done" the first column.
    I was lazy so I let my calculator do the ref. By doing it by hand, I see the answer is 1.
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  5. #5
    Member alexgeek's Avatar
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    Oh I reduced it a bit last night then didn't know where to go, tried it this morning and got 5 but if you did it with a calculator I doubt I'm right.

    $\displaystyle R_2 - 2R_1 $
    $\displaystyle
    \begin{bmatrix}
    1 & 2 & 2 & | & 1 \\
    0 & -3 & -1 & | & 1 \\
    4 & 5 & 7 & | \lambda \\
    \end{bmatrix}
    $

    $\displaystyle R_1 + 2R_2 $
    $\displaystyle
    \begin{bmatrix}
    1 & -4 & 0 & | & 3 \\
    0 & -3 & -1 & | & 1 \\
    4 & 5 & 7 & | \lambda \\
    \end{bmatrix}
    $

    $\displaystyle R_3 -4 R_1 $
    $\displaystyle
    \begin{bmatrix}
    1 & -4 & 0 & | & 3 \\
    0 & -3 & -1 & | & 1 \\
    0 & 21 & 7 & | \lambda -12 \\
    \end{bmatrix}
    $

    $\displaystyle R_3 + 7 R_2 $
    $\displaystyle
    \begin{bmatrix}
    1 & -4 & 0 & | & 3 \\
    0 & -3 & -1 & | & 1 \\
    0 & 0 & 0 & | \lambda -5 \\
    \end{bmatrix}
    $

    So I think if $\displaystyle \lambda = 5 $ then the equations are linearly dependent as there's a row if zeros yes?

    Thanks
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  6. #6
    A Plied Mathematician
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    After row reduction, an inconsistent system looks like a row of zeros before the augmented matrix sign, but a nonzero entry after it. Such an equation can't hold, can it? So you have two choices here: either the system is consistent and has infinitely many solutions (but only one choice for $\displaystyle \lambda$), or the system is inconsistent and has no solutions (but any other choice for $\displaystyle \lambda$ is fine).
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  7. #7
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    Yes, that is correct. But you are doing too much work. Your second step, adding a multiple of the second row to the first, is unecessary. Always work by columns, from left to right. Since there is already a "1" in the upper left, you don't need to do anything to the first row. Subtract twice the first row from the second and 4 times the first row from the third to get
    $\displaystyle \left[\begin{array}{ccc}1 & 2 & 2 \\ 0 & -3 & -1 \\ 0 & -3 & -1\end{array}\left|\begin{array}{c}1 \\ 1\\ \lambda- 4\end{array}\right]$

    Now since the first three numbers in the last two rows are the same, subracting the second row from the third will give "0 0 0" so the matrix is not "invertible"- there is not a unique solution, which is what dwsmith really meant to say. If $\displaystyle \lambda= 5$, then $\displaystyle \lambda- 4= 1$ so that the two last rows are identical. In that case, there are an infinite number of solutions so the system certainly is "consistent". If $\displaystyle \lambda$ is any other number, there is NO solution which is what "inconsistent" means.
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