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 $\displaystyle 1- 2m\ne 0$, $\displaystyle y= \frac{3}{1- 2m}$. 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.
hi
$\displaystyle \left\{\begin{matrix}
x+2y+mz=0\\
2x+3y-2z=m\\
mx+y+m^2z=3
\end{matrix}\right.$
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 .