Originally Posted by

**HallsofIvy** Try to solve: Seeing "x" begining each of the equations, subtract the first from the second to get y+ 2z= 4 and the first from the third to get

$\displaystyle y+(\lambda- 3)z= \mu- 10$. Mr F says: Careless here. It's $\displaystyle {\color{red}y + (\lambda - 1)z = \mu - 6}$

Now subtract the first of those two from the second to cancel the "y"s: $\displaystyle (\lambda- 5)z= \mu- 14$. Mr F says: Wrong. A ripple effect from the above mistake. The correct result (found in a simpler way) is given in pst #6.

If [tex]\lambda- 5[/itex] is NOT 0, we can divide both sides by that to get a specific value for z and then it would be easy to solve for x and y.

Mr F says: Mixing your latex tags. Tsk tsk.

As long as $\displaystyle \lamda - 5$ is not 0, we have a unique solution. Mr F says: Darn missing b ..... "As long as $\displaystyle {\color{red}\lambda - 5}$"

If $\displaystyle \lambda- 5=0$, in other words if $\displaystyle \lambda= 5$ we cannot divide to find z but if it also happens that $\displaystyle \mu= 14$ that equations just says "0z= 0" which is true for all z.

If [tex]\lamba= 5[/itex] and $\displaystyle \m= 14$, there are an infinite number of solutions.

Mr F says: Darn missing d ..... mixing latex tags ..... darn \m

"If $\displaystyle {\color{red}\lambda= 5}$ and $\displaystyle {\color{red}m= 14}$, there are an infinite number of solutions."

If $\displaystyle \lambda- 5= 0$ and $\displaystyle \mu$ is NOT 14, then 0z would have to be equal to a non-zero number and that is impossible.

If $\displaystyle \lamba= 5$ and $\displaystyle \mu$ is not 14, there is no solution.

Mr F says: Darn missing d again ......

"If $\displaystyle {\color{red}\lambda= 5}$ and $\displaystyle {\color{red}\mu}$ is not 14, there is no solution."