For which values of k does the system x + 2y - 2z = 0, 2x - y + kz = 0, 3x + y - z = 0 have a nontrivial solution? Give all solutions in each case.

Step-by-step solution

1. 1

   Note what kind of system this is. All three right-hand sides are $0$, so the system is homogeneous and $(0,0,0)$ is always a solution. The real question is whether there are others, which happens exactly when the coefficient determinant vanishes.

2. 2

   Eliminate using the two equations that do not contain $k$. Equations 1 and 3 involve only known coefficients, so work with them first. From equation 3,

   $y=z-3x,$

   and substituting into equation 1:

   $x+2(z-3x)-2z=x+2z-6x-2z=-5x=0\;\Longrightarrow\;x=0.$

   So $x=0$ for every value of $k$ — the parameter cannot rescue the first component.

3. 3

   Feed $x=0$ back through the system. With $x=0$, equation 3 gives $y=z$, and equation 2 becomes

   $-y+kz=0\;\Longrightarrow\;-z+kz=0\;\Longrightarrow\;(k-1)z=0.$

   Everything now hinges on whether the factor $k-1$ is zero.

4. 4

   Split into the two cases. If $k\neq 1$ then $k-1\neq 0$ forces $z=0$, hence $y=0$ and the only solution is the trivial one:

   $(x,y,z)=(0,0,0).$

   If $k=1$ the equation $(k-1)z=0$ is automatically satisfied, so $z=t$ is free and $y=t$, $x=0$:

   $(x,y,z)=(0,t,t),\qquad t\in\mathbb{R}.$

5. 5

   Confirm with the determinant. Expanding the coefficient determinant along the first row,

   $\begin{vmatrix}1&2&-2\\2&-1&k\\3&1&-1\end{vmatrix}=1(1-k)-2(-2-3k)-2(2+3)=1-k+4+6k-10=5k-5=5(k-1).$

   This is zero precisely when $k=1$, matching the case split exactly. For $k=1$ a quick check: $(0,1,1)$ gives $0+2-2=0$ ✓, $0-1+1=0$ ✓, $0+1-1=0$ ✓.