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Math Help - Row operations on a matrix

  1. #1
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    Row operations on a matrix

    A = [1 2 5 3 8]
    [3 6 11 6 20]
    [1 3 6 0 16]
    [6 12 23 12 40]
    [5 11 22 9 40]


    Hi there i'm really struggling to find the determinant of this matrix using row operations
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by michael11 View Post
    A = [1 2 5 3 8]
    [3 6 11 6 20]
    [1 3 6 0 16]
    [6 12 23 12 40]
    [5 11 22 9 40]


    Hi there i'm really struggling to find the determinant of this matrix using row operations
    Notice that the fourth row is almost the second one multiplied by two : Let's do \mathrm{row}4=\mathrm{row}4-2\times \mathrm{row} 2 (it'll make appear "a lot" of 0) :

    D=\left|<br />
\begin{tabular}{ccccc}<br />
1&2&5&3&8\\<br />
3&6&11&6&20\\<br />
1&3&6&0&16\\<br />
0&0&1&0&0\\<br />
5&11&22&9&40\\<br />
\end{tabular}<br />
\right|=-1\times\left|<br />
\begin{tabular}{ccccc}<br />
1&2&3&8\\<br />
3&6&6&20\\<br />
1&3&0&16\\<br />
5&11&9&40\\<br />
\end{tabular}<br />
\right|

    Now let's do \mathrm{row}4=\mathrm{row}4-\mathrm{row}3-\mathrm{row}2-\mathrm{row}1 (same idea as before, it makes appear several 0) :

    <br />
D=-\left|<br />
\begin{tabular}{cccc}<br />
1&2&3&8\\<br />
3&6&6&20\\<br />
1&3&0&16\\<br />
0&0&0&-4\\<br />
\end{tabular}<br />
\right|=4\times\left|<br />
 \begin{tabular}{ccc}<br />
 1&2&3\\<br />
 3&6&6\\<br />
 1&3&0\\<br />
 \end{tabular}<br />
 \right|

    Can you go on ?
    Last edited by flyingsquirrel; June 16th 2008 at 01:45 PM.
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  3. #3
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    hi,

    i tried to finish it and i got detA = -3408

    don't think thats right some how
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Wooo, that's a big number.

    How did you expand D=4\times\left|<br />
 \begin{tabular}{ccc}<br />
 1&2&3\\<br />
 3&6&6\\<br />
 1&3&0\\<br />
 \end{tabular}<br />
 \right| ?
    Last edited by flyingsquirrel; June 16th 2008 at 11:23 PM.
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  5. #5
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    I multiplied the 4 through, then found the determinant using the lattice method?

    but something tells me that is wrong
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  6. #6
    Super Member flyingsquirrel's Avatar
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    I assume that the lattice method is Sarrus' rule ?

    Repeat the first two columns :

    <br />
 \begin{tabular}{|ccc|cc}<br />
{\color{red}  1}&{\color{green}2}&{\color{blue}3}&1&2\\<br />
 3&{\color{red}6}&{\color{green}6}&{\color{blue}3}&  6\\<br />
 1&3&{\color{red}0}&{\color{green}1}&{\color{blue}3  }\\<br />
 \end{tabular}<br />

    Now multiply the red numbers together, multiply the green numbers together, multiply the blue numbers together and add these three products. Let's call the result D_1.

    <br />
 \begin{tabular}{|ccc|cc}<br />
 1&2&{\color{magenta}3}&{\color{yellow}1}&{\color{c  yan}2}\\<br />
 3&{\color{magenta}6}&{\color{yellow}6}&{\color{cya  n}3}&6\\<br />
 {\color{magenta}1}&{\color{yellow}3}&{\color{cyan}  0}&1&3\\<br />
 \end{tabular}<br />

    Same thing : multiply together the numbers of the same color (except black ) and add the three products. Again, let's call the result D_2.

    As D=4\times\left|<br />
 \begin{tabular}{ccc}<br />
 1&2&3\\<br />
 3&6&6\\<br />
 1&3&0\\<br />
 \end{tabular}<br />
 \right|, the determinant is given by D=4(D_1-D_2).

    Hope that helps.
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  7. #7
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    Hello, michael11!

    I assumed that they wanted Row Operations all the way . . .

    flyingsquirrel had: . \left|\begin{array}{ccc} 1 & 2 & 3 \\ 3 & 6 & 6 \\ 1 & 3 & 0 \end{array}\right| \times 4


    \begin{array}{c}\\ R_2 - 3R_1 \\ R_3-R_1\end{array}\left|\begin{array}{ccc}1 & 2 & 3 \\ 0 & 0 & \text{-}3 \\ 0 & 1 & \text{-}3 \end{array}\right| \times 4 \quad = \quad \left|\begin{array}{cc}1 & 2 \\ 0 & 1 \end{array} \right| \times 12


    \begin{array}{c}R_1-2R_2 \\ \\ \end{array}\left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| \times 12 \quad=\quad\boxed{ 12}



    Terrific job with the colors, flyingsquirrel!
    .
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