How to find the number of solutions to a system of equations with one parameter which cannot be reduced?

Question

I have the system of linear equations where $k$ is the parameter: $$ \begin{cases} x+ky+z = 0 \\ kx+y+kz=0 \\ (k+1)x-y+z=0 \end{cases} $$

For which the matrix is: $$ \begin{bmatrix} 1 & k & 1 \\ k & 1 & k \\ k+1 & -1 & 1 \end{bmatrix} $$

I managed to reduce the matrix to: $$ \begin{bmatrix} 1 & k & 1 \\ 0 & 1-k^2 & 0 \\ 0 & 2+k & k \end{bmatrix} $$

However, I don't see how this can be reduced further. As far as I know I need the echelon form to be able to determine how many solutions exist. Should I just plug in $k=-2$ in order to reduce it further? If yes then what can be said about the number of solutions if $k \neq =2$?

EDIT: I'm only beginning linear math course so I can't use any advanced techniques (like determinants etc.)

Answer 1

Case 1: Assume that $1-k^2 \not=0$ (i.e., $k \not=1, k \not=-1$). Then we can add $\frac{-(2+k)}{1-k^2}$ times the second row to the third row. This leads to: $$ \begin{bmatrix} 1 & k & 1 \\ 0 & 1-k^2 & 0 \\ 0 & 0 & k \end{bmatrix} $$

Case 2: If $1-k^2=0$ (i.e. $k= \pm 1$) then what you have above is equal to $$ \begin{bmatrix} 1 & k & 1 \\ 0 & 0 & 0 \\ 0 & 2+k & k \end{bmatrix} \to \begin{bmatrix} 1 & k & 1 \\ 0 & 2+k & k \\ 0 & 0 & 0 \end{bmatrix} $$

Can you take it from here in both cases?

Answer 2

Let's call the matrix you've obtained $A$, and let $v = (x,y,z)$. You want solutions for

$$ A\cdot v = \vec{0} $$

Now, if $A$ is invertible (i.e $det A \neq 0$), then left-multiplying by $A^{-1}$ gets us

$$ v = \vec{0} $$

You can check that, since $det A = k(1-k^2)$, this only happens if $k$ is $0, 1$ or $-1$. In any other case, $A$ is non invertible, so $v \to A \cdot v$ is not injective and therefore there are $v, w$ such that

$$ A\cdot v = A\cdot w \iff A\cdot (v -w) = \vec{0} $$

and any scalar multiplication of $(v-w)$ is a solution as well. So as a very general analysis, the system has no non-zero solutions for $k = 0,1,-1$ and an infinite amount for any other value of $k$.

Answer 3

Your system equation is parameterized. Instead of calculating you can calculate the determinant of the said system matrix, and equate it to zero, to obtain the singular points.

For your system of equation,

$$k(k^2 - 1) = 0 \implies k = 0, \pm 1$$

So, your system of equation will be singular at $0, \pm 1$.

The characteristic equation for the set of linear equation,

$A(A^2 - I) = 0$

Where $A$ is your system matrix.