I've learned to find the solutions to linear systems using Gaussian Elimination. Moving on, I've found a new kind of exercise I hadn't done before:
Find the values for $k$ that make this system:
- Inconsistent (no solutions)
- With unique solution
- With infinite solutions $$\begin {cases} kx + y + z = 1\\ x + ky + z = 1\\ x + y + kz = 1 \end {cases}$$
Let's see…
For inconsistency
If I wanted this system to be inconsistent, I would need a row to be of the form $0 + 0 + 0 = 1$… But even if $k = 0$, I would also need two of the three $x,y,z$ variables to also be $0$ to have such a row, which is not necessarily true. So what then?
For unique solution
For unique solution, I would need tha rank of the matrix to be $3$. I think that in this case, it is enough to say $k = 1$, because it would cause the diagonal to be a bunch of $1$s, and I think there was a theorem saying that when this happens the matrix can always be reduced to the identity, thus having an unique solution. I'm not sure how to word this though.
For infinite solutions
For infinite solutions, the rank of the matrix has to be less than $3$… and I'm not even sure how to proceed.
Best Answer
Hints: The determinant is:
$$k^3-3 k+2 = (k-1)^2(k+2) = 0 \implies k = 1, -2$$
We want the determinant to be nonzero for unique solutions.
If we substitute $k = 1$ in the system, and do a RREF we have:
$$\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$
If we substitute $k = -2$ in the system, and do a RREF we have:
$$\left( \begin{array}{cccc} 1 & 0 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$
Can you now derive conclusions for inconsistent, unique solution and infinite solutions?