The system x + 2y + z = 0, 2x + ky + 3z = 0 and x + y + 2z = 0 has solutions other than x = y = z =…

Step 1. The determinant. $

= 1(2k - 3) - 2(4 - 3) + 1(2 - k) = k - 3$. Setting it to$0$gives$k = 3$.

Step 2. Solve for $k = 3$. The first equation minus the third gives $y - z = 0$, so $y = z$.

Step 3. Find $x$. The first equation gives $x = -2y - z = -3z$.

Step 4. Put $z = 1$. $x = -3$ and $y = 1$.