linear equation help

**princessamme@hotmail.com** · November 30th 2009, 02:54 PM

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

**pickslides** · November 30th 2009, 04:35 PM

Solve this

$\left[ \begin{array}{c} x\\ y\\ z\end{array}\right]=\left[ \begin{array}{ccc} 1 & 2 & m \\ 2 & 3 & -2 \\ m & 1 & m^2 \end{array} \right]^{-1} \left[ \begin{array}{c} 0\\ m\\ 3\end{array}\right]$

**HallsofIvy** · December 1st 2009, 05:51 AM

Originally Posted by princessamme@hotmail.com

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.

**Raoh** · December 1st 2009, 06:11 AM

hi

$\left\{\begin{matrix} x+2y+mz=0\\ 2x+3y-2z=m\\ mx+y+m^2z=3 \end{matrix}\right.$
Using Gauss method,
$\Rightarrow \left\{\begin{matrix} x+2y+mz=0 \\ -y-2z(1+m)=m\\ y(1-2m)=3 \end{matrix}\right.$
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 .

**Raoh** · December 1st 2009, 06:23 AM

Originally Posted by Raoh

hi

$\left\{\begin{matrix} x+2y+mz=0\\ 2x+3y-2z=m\\ mx+y+m^2z=3 \end{matrix}\right.$
Using Gauss method,
$\Rightarrow \left\{\begin{matrix} x+2y+mz=0 \\ -y-2z(1+m)=m\\ y(1-2m)=3 \end{matrix}\right.$
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 $\left [ x=u-2,z=u \right ]$ .

**princessamme@hotmail.com** · December 2nd 2009, 07:34 AM

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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