Question

Given:
\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{Bmatrix} x_1 \\ x_2 \\ x_3 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} \]
The above system of equations represents a

• A. plane
• B. line
• C. volume
• D. point

Hint:

Find the rank of the coefficient matrix; the solution set has dimension (number of unknowns minus rank).

Answer 1

The Correct Option is B

Think about the two equations as two planes in 3D space and reason about how planes intersect, instead of using the rank-nullity formula.

1. Each equation is one plane: $x_1+x_2+x_3=0$ is a flat plane through the origin, since it is one linear equation in three variables with zero on the right side. Likewise $x_1+2x_3=0$ is a second plane through the origin.
2. Two distinct, non-parallel planes meet in a line: the normal vector of the first plane is $(1,1,1)$ and of the second is $(1,0,2)$. These normal vectors are not scalar multiples of each other, so the planes are not parallel and are not the same plane. Two distinct, non-parallel planes in 3D space always intersect in exactly one straight line.
3. Rule out the other shapes: the system cannot describe a volume, since that would need the equations to place no real restriction on the variables, which is impossible here since each equation is a genuine constraint. It cannot be a point either, since a point needs three independent constraints (planes) to pin down a unique $(x_1,x_2,x_3)$, but only two planes are given here, leaving one free direction.

Solving directly confirms this: from $x_1+2x_3=0$, $x_1=-2x_3$; putting this into the first equation gives $x_2=x_3$. So every solution is $(x_1,x_2,x_3)=(-2t,t,t)$ for any real $t$, tracing out a straight line through the origin as $t$ varies.

The system of equations represents a line, matching option (B).