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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- Nov 30th 2009, 01:54 PMprincessamme@hotmail.comlinear 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 - Nov 30th 2009, 03:35 PMpickslides
Solve this

$\displaystyle \left[ \begin{array}{c}

x\\

y\\

z\end{array}\right]=\left[ \begin{array}{ccc}

1 & 2 & m \\

2 & 3 & -2 \\

m & 1 & m^2 \end{array} \right]^{-1} \left[ \begin{array}{c}

0\\

m\\

3\end{array}\right]$ - Dec 1st 2009, 04:51 AMHallsofIvy
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 $\displaystyle 1- 2m\ne 0$, $\displaystyle y= \frac{3}{1- 2m}$. 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.

- Dec 1st 2009, 05:11 AMRaoh
hi(Happy)

$\displaystyle \left\{\begin{matrix}

x+2y+mz=0\\

2x+3y-2z=m\\

mx+y+m^2z=3

\end{matrix}\right.$

Using Gauss method,

$\displaystyle \Rightarrow \left\{\begin{matrix}

x+2y+mz=0 \\

-y-2z(1+m)=m\\

y(1-2m)=3

\end{matrix}\right.$

in case $\displaystyle m=\frac{1}{2}$,there will be no solution.

in case $\displaystyle m \neq \frac{1}{2}$,you'll have $\displaystyle y=\frac{3}{1-2m}$, substitute and get the other values . - Dec 1st 2009, 05:23 AMRaoh
- Dec 2nd 2009, 06:34 AMprincessamme@hotmail.com
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 (Happy)