Question 33
(MON 4) In what case does the system
$$\begin{aligned}x+y+mz&=a,\\x+my+z&=b,\\mx+y+z&=c\end{aligned}$$
have a solution? Find the conditions under which the unique solution of the above system is an arithmetic progression.
$$\begin{aligned}x+y+mz&=a,\\x+my+z&=b,\\mx+y+z&=c\end{aligned}$$
have a solution? Find the conditions under which the unique solution of the above system is an arithmetic progression.