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

1. Feb 27, 2013

### dylanhouse

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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 :$2. Feb 27, 2013 ### Vorde Well can you show us your work? It is a messy system, but if you lead us through your work we might be able to help you/give you tips! :) 3. Feb 27, 2013 ### dylanhouse I think this is right so far? :$

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4. Feb 27, 2013

### dylanhouse

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

5. Feb 27, 2013

### 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?

6. Feb 28, 2013

### 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.