I've row reduced my system of equations to:
x + 2y -z = -3, y -(k-3)z = -5, (k^2 - 2k)z = 5k+11
I have solved for no solutions (system is inconsistent for k=0 or k=2)
Have mostly solved for all unique solutions (I believe): (system is consistent for all other values than k =0 or k=2.)
However when solving for infinitely many solutions, I've solved the last equation {equation 3 - (k^2 - 2k) = 5k+11}
To find that there are infinitely many solutions at k= 3 1/2 +/- (sqrt(93))/2
My question is regarding what happens when k=9,
This would make the second equation:
y - [(9)-3]z = -5, 1 - 6 = -5, -5=-5
As this is a redundant statement, does this result in a case of infinitely many solutions when k=9? It would remove the need for the inclusion of this statement in the system of equations would it not, does that mean it doesn't impact the other two at all?
I've attempted to research this and figure it out so I apologise if it's a simple answer but I haven't been able to find one.
At this stage I am leaning towards the fact that it wouldn't impact on the other two, but not sure how this makes sense as it still exists as a point(?),line(?),plane(?) (not sure on this either sorry, am having a hard time rationalising it in my head) within the confines of the question.
So TL;DR could someone please help me with what to do in the case of if k=9, does this lead to infinitely many solutions, does it not affect it at all, does it mean there's another unique solution? etc.
Thank you in advance :)
\begin{align} x + 2y -z &= -3, \\ y -(k-3)z &= -5,\\ (k^2 - 2k)z &= 5k+11 \end{align}
Let $k=9$:
\begin{align} x + 2y -z &= -3, \\ y -6z &= -5,\\ (9^2 - 18)z &= 5(9)+11 \end{align}
Let's write the LHS in terms of a matrix and we obtain:
$$\begin{bmatrix} 1 & 2 & -1 \\ 0 & 1 & -6 \\ 0 & 0 & 9^2-18 \end{bmatrix}$$
The determinant of this matrix is non-zero and hence it is invertible. Hence there is a unique solution.
Remark: The second equation is not redundant. We can't just drop $y$ and $z$ from the second equation.