Discussion of a system as k varies

Problem

For which values of $k$ is the system $\begin{cases} kx+y=1 \\ x+ky=1\end{cases}$ (a) determinate; (b) inconsistent; (c) indeterminate? (Study $\det A = k^2-1$.)

Solution

The determinant of the coefficient matrix is $\det A = \begin{vmatrix} k & 1 \\ 1 & k\end{vmatrix} = k^2 - 1 = (k-1)(k+1).$ (a) Determinate — when $\det A \ne 0$, that is for $\boxed{k \ne \pm 1}$: the system has a unique solution (by symmetry $x=y=\tfrac{1}{k+1}$). (b) Inconsistent — for $k=-1$ the system becomes $\begin{cases} -x+y=1 \\ x-y=1\end{cases}$; adding the two equations gives $0=2$, absurd: no solution. (c) Indeterminate — for $k=1$ the two equations coincide ($x+y=1$): infinitely many solutions depending on a parameter.

Links

Topics: Linear systems
Concepts: Determinant · Inconsistent system · Indeterminate system
Skills: Case-based reasoning · Solve systems
Exercise type: System solving