Question

Determine the values of k such that the system of linear equations does not have a unique solution. (Enter your answers as a comma-separated list.)x + y + kz = 9x + ky + z = 2kx + y + z = 5

Answer 1

$\begin{cases}x+y+kz=9\\x+ky+z=2\\kx+y+z=5\end{cases}$

Consider the system in matrix form, with coefficient matrix

$A=\begin{bmatrix}1&1&k\\1&k&1\\k&1&1\end{bmatrix}$

Then the system

$A\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\2\\5\end{bmatrix}$

has a unique solution if $A$ is invertible, which requires that $\det A\neq0$. Compute the determinant: expanding along the first row gives

$\det A=\begin{vmatrix}k&1\\1&1\end{vmatrix}-\begin{vmatrix}1&1\\k&1\end{vmatrix}+k\begin{vmatrix}1&k\\k&1\end{vmatrix}$

$\det A=(k-1)-(1-k)+(1-k^2)=-1+2k-k^2=-(1-k)^2$

This is zero if $k=1$, and non-zero otherwise, hence there is no unique solution if $k=1$.