Systems solving with no solution, unique, and infinite solutions?

[dylanhouse]

Homework Statement

Find the values of k in the following system of linear equations such that, the system has no solution, the system has a unique solution, and the system has infinitely many solutions.

x+y+2kz = 0
−2x−y+6z = −3k
−x+2y+(k² −3k)z = 9

Homework Equations

The Attempt at a Solution

I'm not really sure how to do this. I tried putting it into an augmented matrix and solving, but this is terribly messy and I can't get anywhere with it :$

 

[Vorde]

Well can you show us your work? It is a messy system, but if you lead us through your work we might be able to help you/give you tips! :)

 

[dylanhouse]

I think this is right so far? :$

 

[dylanhouse]

The matrix should actually be augmented. The last column to the right.

 

[Vorde]

The first thing I'd suggest is when moving from matrix 1 in your picture to matrix 2, add twice the first row to the second instead of (negative)twice the third row to the second. It made it a much easier matrix, now can you see where to go?

 

[HallsofIvy]

Once you have reduced the augmented matrix, any value of k that makes the first three entries in any row, but not the last entry in that row, 0, gives a system that has no solution. Do you see why?
Consider a system of equations that reduces to
\begin{bmatrix}1 & 0 & 0 & A \\ 0 & 1 & 0 & B \\ 0 & 0 & 0 & C\end{bmatrix}
where C is not 0. That is equivalent to the system of equation x= A, y= B, 0z= 0= C.
No value of z makes that true.

A value of k that does NOT give the situation above makes all four entries in a row 0 gives a system that has an infinite number of solutions. Do you see why?
Consider a system of equations that reduces to
\begin{bmatrix}1 & 0 & 0 & A \\ 0 & 1 & 0 & B \\ 0 & 0 & 0 & 0\end{bmatrix}
That is equivalent to the system of equation x= A, y= B, 0z= 0= 0 which is true for any value of z.