Reducing the system $x_1+kx_2-x_3=0$, $-x_1+x_2+k^2x_3=0$, $x_1+kx_2+(k-2)x_3=0$

Question

I can't seem to figure out how to reduce this system of equations:

$$\begin{align} x_1+kx_2-\phantom{(k-2)}x_3 &=0 \\ -x_1+\phantom{k}x_2+\phantom{(k-)}k^2x_3 &=0 \\ x_1+kx_2+(k-2)x_3 &=0 \end{align}$$

I put them into a matrix and solved it down to:

$$\begin{align} x_1+kx_2-\phantom{2}x_3 &=0 \\ -10x_1-8x_2-2x_3 &=0 \\ (k-3)x_3 &=0 \end{align}$$

but I'm not sure where to go from there. I'm trying to figure out if there is a unique solution or infinitely many.

Answer 1

Reduce the matrix to the triangular form

$$\left(\begin{array}{ccc}1 & k & -1 \\ 0 & k+1 & k^2-1\\0& 0 & k-1\end{array}\right).$$

If $k\ne \pm 1$ then the RREF has three pivots so there is unique solution, $(0,0,0)$. If $k=1$ or $k=-1$ then there are only two pivots, so the system has infinitely many solutions.

Answer 2

An homogeneous system has non trivial solutions if the determinant of the matrix $A$ of the coefficients is $0$. In this case $$k^2-1=0$$ Two cases:

• $k=-1$, $$A=\left( \begin{array}{ccc} 1 & -1 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & -3 \\ \end{array} \right)$$ The system can be reduced as $$\left\{ \begin{array}{lll} x_1-x_2-x_3=0 \\ x_1-x_2-3x_3=0 \\ \end{array} \right.$$ Solutions are $k\,(1,1,0),\;\forall k\in\mathbb{R}$

• $k=1$

$$A=\left( \begin{array}{ccc} 1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & -1 \\ \end{array} \right)$$ The system becomes $$\left\{ \begin{array}{lll} x_1+x_2-x_3=0 \\ x_1-x_2-x_3=0 \\ \end{array} \right.$$ Solutions are $k\,(1,0,1),\;\forall k\in\mathbb{R}$.