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Math Help - linear equation help

  1. #1
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    linear equation help

    can anyone help me with this question

    solve completely the following system

    x+2y+mz=0
    2x+3y-2z=m
    mx+y+m^2z=3

    Im a bit stumped on what to do with the m? any help would be appreciated
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  2. #2
    Master Of Puppets
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    Solve this

     \left[ \begin{array}{c}<br />
x\\<br />
y\\<br />
z\end{array}\right]=\left[ \begin{array}{ccc}<br />
1 & 2 & m \\<br />
2 & 3 & -2 \\<br />
m & 1 & m^2 \end{array} \right]^{-1} \left[ \begin{array}{c}<br />
0\\<br />
m\\<br />
3\end{array}\right]
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  3. #3
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    Quote Originally Posted by princessamme@hotmail.com View Post
    can anyone help me with this question

    solve completely the following system

    x+2y+mz=0
    2x+3y-2z=m
    mx+y+m^2z=3

    Im a bit stumped on what to do with the m? any help would be appreciated
    If you don't want to use matrices, you can just solve it as you would in basic algebra: Multiplying the first equation by m and subtracting from the third equation elimnates both x and z: (m- m)x+ (1- 2m)y+ (m^2- m^2)z= 3- 0 so (1- 2m)y= 3. If 1- 2m\ne 0, y= \frac{3}{1- 2m}. Put that value of y into the first and second equations to get two equations for x and z. if m= 1/2, 1- 2m= 0 so there is no solution.
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  4. #4
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    Smile

    hi
    \left\{\begin{matrix}<br />
x+2y+mz=0\\ <br />
2x+3y-2z=m\\ <br />
mx+y+m^2z=3<br />
\end{matrix}\right.
    Using Gauss method,
    \Rightarrow \left\{\begin{matrix}<br />
x+2y+mz=0 \\ <br />
-y-2z(1+m)=m\\ <br />
y(1-2m)=3<br />
\end{matrix}\right.
    in case m=\frac{1}{2},there will be no solution.
    in case m \neq \frac{1}{2},you'll have y=\frac{3}{1-2m}, substitute and get the other values .
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  5. #5
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    Quote Originally Posted by Raoh View Post
    hi
    \left\{\begin{matrix}<br />
x+2y+mz=0\\ <br />
2x+3y-2z=m\\ <br />
mx+y+m^2z=3<br />
\end{matrix}\right.
    Using Gauss method,
    \Rightarrow \left\{\begin{matrix}<br />
x+2y+mz=0 \\ <br />
-y-2z(1+m)=m\\ <br />
y(1-2m)=3<br />
\end{matrix}\right.
    in case m=\frac{1}{2},there will be no solution.
    in case m \neq \frac{1}{2},you'll have y=\frac{3}{1-2m}, substitute and get the other values .
    There is probably another case,
    when m=-1, y will equal 1 and we will have the solution \left [ x=u-2,z=u \right ] .
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  6. #6
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    Thank you for all your help, I originally used gaussian elimination, however i could get a nice solution so i assumed what i was doing was wrong
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