Solve the homogeneous system x + 2y - 2z = 0, 2x - y + z = 0, 3x + y - z = 0.

Step-by-step solution

1. 1

   Note that the trivial solution always exists. Every right-hand side is $0$, so $(0,0,0)$ solves the system automatically. The interesting question is whether there are others — which happens exactly when the three equations are linearly dependent.

2. 2

   Add equations 2 and 3 to eliminate two variables at once. The $-y$ and $+y$ cancel, and so do $+z$ and $-z$:

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

   This pairing is worth spotting: two variables disappear in a single addition because equations 2 and 3 have exactly opposite $y$ and $z$ coefficients.

3. 3

   Substitute $x=0$ into the remaining equations. Equation 2 becomes $-y+z=0$, so $z=y$. Equation 1 becomes $2y-2z=0$, so $y=z$ again — the same condition, not a new one. That repetition is the algebraic fingerprint of a dependent system: three equations impose only two independent constraints.

4. 4

   Parametrise the solution set. With $x=0$ and $y=z$ free to slide together, set $y=z=t$:

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

   Geometrically the three planes all contain the line spanned by $(0,1,1)$, so they meet in a line rather than at a single point.

5. 5

   Verify with the determinant and by substitution. Expanding along the first row,

   $\begin{vmatrix}1&2&-2\\2&-1&1\\3&1&-1\end{vmatrix}=1(1-1)-2(-2-3)-2(2+3)=0+10-10=0,$

   confirming a nontrivial solution must exist. Checking $(0,1,1)$: $0+2-2=0$ ✓, $0-1+1=0$ ✓, $0+1-1=0$ ✓.