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Consider the following set of $3n$ linear equations in $3n$ variables:

$\begin{matrix} x_1-x_2=0 & x_4-x_5 =0 & \dots & x_{3n-2}-x_{3n-1}=0 \\ x_2-x_3=0 & x_5-x_6=0 & & x_{3n-1}-x_{3n}=0 \\ x_1-x_3=0 & x_4-x_6 =0 & & x_{3n-2}=x_{3n}=0 \end{matrix}$

Let $S \subseteq \mathbb{R}^{3n}$ be the set of solutions to this set of equations. Then,

- $S$ is empty
- $S$ is a subspace of $\mathbb{R}^{3n}$ of dimension 1
- $S$ is a subspace of $\mathbb{R}^{3n}$ of dimension n
- $S$ is a subspace of $\mathbb{R}^{3n}$ of dimension $n-1$
- $S$ has exactly $n$ elements

0 votes

$\text{$x_1=x_2=x_3$}$

$\text{$x_4=x_5=x_6$}$

$…...$

$…...$

$x_{3n-2} = x_{3n-1} = x_{3n}$

$\text{So, solution space is $\left \{\begin{pmatrix} x_1\\ x_1\\ x_1\\ x_4\\ x_4\\ x_4\\ ...\\ x_{3n-2}\\ x_{3n-2}\\ x_{3n-2}\\ \end{pmatrix}\right\}$}$

$\text{$= \left\{ x_1\begin{pmatrix} 1\\ 1\\ 1\\ 0\\ 0\\ 0\\ ...\\ 0\\ 0\\ 0\\ \end{pmatrix} + x_4\begin{pmatrix} 0\\ 0\\ 0\\ 1\\ 1\\ 1\\ ...\\ 0\\ 0\\ 0\\ \end{pmatrix} + …… +x_{3n}\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ ...\\ 1\\ 1\\ 1\\ \end{pmatrix} \right\}$} \rightarrow{(1)}$

$\text{So, solution space contains $3n/3 = n$ basis vectors which are:}$

$\text{$ \left\{\begin{pmatrix} 1\\ 1\\ 1\\ 0\\ 0\\ 0\\ ...\\ 0\\ 0\\ 0\\ \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 0\\ 1\\ 1\\ 1\\ ...\\ 0\\ 0\\ 0\\ \end{pmatrix} , …… ,\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ ...\\ 1\\ 1\\ 1\\ \end{pmatrix} \right\}$}$

$\text{These vectors are orthogonal and so they form a linearly independent set and }$

$\text{from (1), it is clear that they span the whole solution space.}$

$\text{Hence, subspace $S$ has dimension $n.$}$

$\text{$x_4=x_5=x_6$}$

$…...$

$…...$

$x_{3n-2} = x_{3n-1} = x_{3n}$

$\text{So, solution space is $\left \{\begin{pmatrix} x_1\\ x_1\\ x_1\\ x_4\\ x_4\\ x_4\\ ...\\ x_{3n-2}\\ x_{3n-2}\\ x_{3n-2}\\ \end{pmatrix}\right\}$}$

$\text{$= \left\{ x_1\begin{pmatrix} 1\\ 1\\ 1\\ 0\\ 0\\ 0\\ ...\\ 0\\ 0\\ 0\\ \end{pmatrix} + x_4\begin{pmatrix} 0\\ 0\\ 0\\ 1\\ 1\\ 1\\ ...\\ 0\\ 0\\ 0\\ \end{pmatrix} + …… +x_{3n}\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ ...\\ 1\\ 1\\ 1\\ \end{pmatrix} \right\}$} \rightarrow{(1)}$

$\text{So, solution space contains $3n/3 = n$ basis vectors which are:}$

$\text{$ \left\{\begin{pmatrix} 1\\ 1\\ 1\\ 0\\ 0\\ 0\\ ...\\ 0\\ 0\\ 0\\ \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 0\\ 1\\ 1\\ 1\\ ...\\ 0\\ 0\\ 0\\ \end{pmatrix} , …… ,\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ ...\\ 1\\ 1\\ 1\\ \end{pmatrix} \right\}$}$

$\text{These vectors are orthogonal and so they form a linearly independent set and }$

$\text{from (1), it is clear that they span the whole solution space.}$

$\text{Hence, subspace $S$ has dimension $n.$}$