Answer
$k= \pm 1$
Work Step by Step
$kx + y =0$
$x+ky=1$
Express the equations in extended matrix form:
$ [ k$ $1$ $| 0]$
$[1$ $k $| $1]$
The determinant is:
$(k \times k) - (1 \times 1)=k^2-1$
We then set this equal to zero, since that's what we want:
$k^2-1$
$k= \pm 1$
Try $k=1:$
$ [ 1$ $1$ $| 0]$
$[1$ $1$| $1]$
$R_2-R_1$ for row 2
$ [ 1$ $1$ $| 0]$
$[0$ $0$| $1]$
The last row in the matrix says $0x+0y=1$, which simplifies to $0=1$. Since this is a contradiction, $k=1$ makes the system inconsistent.
Try $k=-1:$
$ [ -1$ $1$ $| 0]$
$[1$ $-1$| $1]$
$R_2+R_1$ for row 2
$ [ 1$ $1$ $| 0]$
$[0$ $0$| $1]$
The last row in the matrix says $0x+0y=1$, which simplifies to $0=1$. Since this is also a contradiction, $k=-1$ makes the system inconsistent.
