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Math Help - very confused matrices

  1. #1
    Member i_zz_y_ill's Avatar
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    very confused matrices

    STUCK ON PART (III) hellp

    (i) find the inverse of 4 1 k , k doesnt =5
    3 2 5
    8 5 13
    DONE
    (ii) solve the smultaneous equation

    4x + y + 7z = 12
    3x + 2y + 5z = m
    8x + 5y +13z = 0
    giving x,y,z in terms of m,,,DONE i got x= -6-11m
    y= -6+2m
    z=6+6m correct i think
    (iii) find the value of p for which the simultaneous equations have solutions, and fn general solution in this case

    4x+y+5z=12
    3x+2y+5z=p
    8x+5y+13z=0


    how does it have solutions??,ithought this was just det doesnt = 0,,confused as to how im supposed to do this part when k doesnt = 5 either,,,,,,help pls
    Last edited by i_zz_y_ill; May 3rd 2008 at 08:32 AM. Reason: i got it thnx
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  2. #2
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    Hello, i_zz_y_ill!

    (iii) Find the value of p for which the system has solutions.

    . . \begin{array}{ccc}4x+y+5z&=& 12 \\<br />
3x+2y+5z&=&p \\ 8x+5y+13z& =& 0 \end{array}

    We have: . \left| \begin{array}{ccc|c} 4 & 1 & 5 & 12 \\ 3 & 2 & 5 & p \\ 8 & 5 & 13 & 0 \end{array}\right|

    \begin{array}{c}R_1-R_2 \\ \\ R_3-2R_1\end{array}\left|\begin{array}{ccc|c}1 & \text{-}1 & 0 & 12-p \\ 3 & 2 & 5 & p \\ 0 & 3 & 3 & \text{-}24 \end{array}\right|

    \begin{array}{c}\\ R_2 - 3R_1 \\ R_3\div3 \end{array}\left|\begin{array}{ccc|c}1 & \text{-}1 & 0 & 12-p \\ 0 & 5 & 5 & 4p-36 \\ 0 & 1 & 1 & \text{-}8 \end{array}\right|

    \begin{array}{c}\text{Switch} \\R_2\text{ and }R_3\\ \end{array}<br />
\left|\begin{array}{ccc|c}1 & \text{-}1 & 0 & 12-p \\ 0 & 1 & 1 & \text{-}8 \\ 0 & 5 & 5 & 4p-36 \end{array}\right|

    \begin{array}{c}R_1+R_2 \\ \\ R_3-5R_2\end{array}<br />
\left|\begin{array}{ccc|c}1 & 0 & 1 & 4-p \\ 0 & 1 & 1 & \text{-}8 \\<br />
0 & 0 & 0 & 4p+4 \end{array}\right|


    The last row of the matrix is all zeros.
    To have a solution, the last term must also be zero. **

    . . 4p + 4 \:=\:0\quad\Rightarrow\quad\boxed{ p \:=\:-1}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We have: . \begin{array}{ccc}x + z &=& 5 \\ y + z &=&\text{-}8 \end{array} \quad\Rightarrow\quad \begin{array}{ccc} x &=& 5 - z \\ y &=& \text{-}8-z \\ z&=&z \end{array}


    On the right, replace z with the parameter t.


    And we have: . \begin{Bmatrix}x &=& 5-t \\ y &=&\text{-}8-t \\ z &=& t \end{Bmatrix}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    If the last row were something like: . 0\;\;0\;\;0\;\;|\;\;3

    . . the system would have no solution . . . remember?

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