Question

Choose solution set $S$ corresponding to the system of two equations \[ x-2y+z=0,\qquad x-z=0 \] (Note: $\mathbb{R}$ denotes the set of real numbers).

• A. $S=\left\{\alpha\!\begin{pmatrix}1\\[2pt]1\\[2pt]1\end{pmatrix}\;\middle|\;\alpha\in\mathbb{R}\right\}$
• B. $S=\left\{\alpha\!\begin{pmatrix}1\\[2pt]1\\[2pt]1\end{pmatrix}+\beta\!\begin{pmatrix}1\\[2pt]0\\[2pt]1\end{pmatrix}\;\middle|\;\alpha,\beta\in\mathbb{R}\right\}$
• C. $S=\left\{\alpha\!\begin{pmatrix}1\\[2pt]1
  [2pt]1\end{pmatrix}+\beta\!\begin{pmatrix}2\\[2pt]1
  [2pt]2\end{pmatrix}\;\middle|\;\alpha,\beta\in\mathbb{R}\right\}$
• D. $S=\left\{\alpha\!\begin{pmatrix}1\\[2pt]0\\[2pt]1\end{pmatrix}\;\middle|\;\alpha\in\mathbb{R}\right\}$

Hint:

Use one equation to eliminate a variable, then check how many free parameters remain: two planes intersecting in $\mathbb{R}^3$ typically give a line $\Rightarrow$ a single direction vector.

Answer 1

The Correct Option is A

From $x-z=0\Rightarrow z=x$. Substitute in $x-2y+z=0$ to get $x-2y+x=0\Rightarrow 2x-2y=0\Rightarrow x=y$.
Thus $x=y=z=t$ for some $t\in\mathbb{R}$. Hence \[ (x,y,z)^\top=t\begin{pmatrix}1\\ [2pt]1\\[2pt]1\end{pmatrix}, \] so the solution set is one–dimensional and equals $S=\left\{\,\alpha\!\begin{pmatrix}1\\[2pt]1\\[2pt]1\end{pmatrix}\middle|\alpha\in\mathbb{R}\right\}$, which matches \fbox{(A)}.