Linear Algebra/Comparing Set Descriptions

From Wikibooks, the open-content textbooks collection

Current revision (unreviewed)

This subsection is optional. Later material will not require the work here.

[edit] Comparing Set Descriptions

A set can be described in many different ways. Here are two different descriptions of a single set:

$\{\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}z\,\big|\, z\in\mathbb{R}\} \quad\text{and}\quad \{\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}w\,\big|\, w\in\mathbb{R}\}.$

For instance, this set contains

$\begin{pmatrix} 5 \\ 10 \\ 15 \end{pmatrix}$

(take $z = 5$ and $w = 5 / 2$ ) but does not contain

$\begin{pmatrix} 4 \\ 8 \\ 11 \end{pmatrix}$

(the first component gives $z = 4$ but that clashes with the third component, similarly the first component gives $w = 4 / 5$ but the third component gives something different). Here is a third description of the same set:

$\{\begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix}+\begin{pmatrix} -1 \\ -2 \\ -3 \end{pmatrix}y\,\big|\, y\in\mathbb{R}\}.$

We need to decide when two descriptions are describing the same set. More pragmatically stated, how can a person tell when an answer to a homework question describes the same set as the one described in the back of the book?

[edit] Set Equality

Sets are equal if and only if they have the same members. A common way to show that two sets, $S 1$ and $S 2$ , are equal is to show mutual inclusion: any member of $S 1$ is also in $S 2$ , and any member of $S 2$ is also in $S 1$ .^[1]

Example 4.1

To show that

$S_1= \{\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}c+\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}d\,\big|\, c,d\in\mathbb{R}\}$

equals

$S_2= \{\begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix}m+\begin{pmatrix} -1 \\ -3 \\ 0 \end{pmatrix}n\,\big|\, m,n\in\mathbb{R}\}$

we show first that $S_1\subseteq S_2$ and then that $S_2\subseteq S_1$ .

For the first half we must check that any vector from $S 1$ is also in $S 2$ . We first consider two examples to use them as models for the general argument. If we make up a member of $S 1$ by trying $c = 1$ and $d = 1$ , then to show that it is in $S 2$ we need $m$ and $n$ such that

$\begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix}m +\begin{pmatrix} -1 \\ -3 \\ 0 \end{pmatrix}n =\begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}$

that is, this relation holds between $m$ and $n$ .

$\begin{array}{*{2}{rc}r} 4m &- &n &= &2 \\ 1m &- &3n &= &0 \\ & &0 &= &0 \end{array}$

Similarly, if we try $c = 2$ and $d = - 1$ , then to show that the resulting member of $S 1$ is in $S 2$ we need $m$ and $n$ such that such that

$\begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix}m +\begin{pmatrix} -1 \\ -3 \\ 0 \end{pmatrix}n =\begin{pmatrix} 3 \\ -3 \\ 0 \end{pmatrix}$

that is, this holds.

$\begin{array}{*{2}{rc}r} 4m &- &n &= &3 \\ 1m &- &3n &= &-3 \\ & &0 &= &0 \end{array}$

In the general case, to show that any vector from $S 1$ is a member of $S 2$ we must show that for any $c$ and $d$ there are appropriate $m$ and $n$ . We follow the pattern of the examples; fix

$\begin{pmatrix} c+d \\ -c+d \\ 0 \end{pmatrix}\in S_1$

and look for $m$ and $n$ such that

$\begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix}m +\begin{pmatrix} -1 \\ -3 \\ 0 \end{pmatrix}n =\begin{pmatrix} c+d \\ -c+d \\ 0 \end{pmatrix}$

that is, this is true.

$\begin{array}{*{2}{rc}r} 4m &- &n &= &c+d \\ m &- &3n &= &-c+d \\ & &0 &= &0 \end{array}$

Applying Gauss' method

$\begin{array}{rcl} \begin{array}{*{2}{rc}r} 4m &- &n &= &c+d \\ m &- &3n &= &-c+d \end{array} &\xrightarrow[]{-(1/4)\rho_1+\rho_2} &\begin{array}{*{2}{rc}r} 4m &- &n &= &c+d \\ & &-(11/4)n &= &-(5/4)c+(3/4)d \end{array} \end{array}$

gives $n = (5 / 11) c - (3 / 11) d$ and $m = (4 / 11) c + (2 / 11) d$ . This shows that for any choice of $c$ and $d$ there are appropriate $m$ and $n$ . We conclude any member of $S 1$ is a member of $S 2$ because it can be rewritten in this way:

$\begin{pmatrix} c+d \\ -c+d \\ 0 \end{pmatrix} =\begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix}((4/11)c+(2/11)d)+ \begin{pmatrix} -1 \\ -3 \\ 0 \end{pmatrix}((5/11)c-(3/11)d).$

For the other inclusion, $S_2\subseteq S_1$ , we want to do the opposite. We want to show that for any choice of $m$ and $n$ there are appropriate $c$ and $d$ . So fix $m$ and $n$ and solve for $c$ and $d$ :

$\begin{array}{rcl} \begin{array}{*{2}{rc}r} c &+ &d &= &4m-n \\ -c &+ &d &= &m-3n \end{array} &\xrightarrow[]{\rho_1+\rho_2} &\begin{array}{*{2}{rc}r} c &+ &d &= &4m-n \\ & &2d &= &5m-4n \end{array} \end{array}$

shows that $d = (5 / 2) m - 2 n$ and $c = (3 / 2) m + n$ . Thus any vector from $S 2$

$\begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix}m+\begin{pmatrix} -1 \\ -3 \\ 0 \end{pmatrix}n$

is also of the right form for $S 1$

$\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}((3/2)m+n) +\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}((5/2)m-2n).$

Example 4.2

Of course, sometimes sets are not equal. The method of the prior example will help us see the relationship between the two sets. These

$P= \{\begin{pmatrix} x+y \\ 2x \\ y \end{pmatrix}\,\big|\, x,y\in\mathbb{R}\} \quad\text{and}\quad R= \{\begin{pmatrix} m+p \\ n \\ p \end{pmatrix}\,\big|\, m,n,p\in\mathbb{R}\}$

are not equal sets. While $P$ is a subset of $R$ , it is a proper subset of $R$ because $R$ is not a subset of $P$ .

To see that, observe first that given a vector from $P$ we can express it in the form for $R$ — if we fix $x$ and $y$ , we can solve for appropriate $m$ , $n$ , and $p$ :

$\begin{array}{*{3}{rc}r} m & & &+ &p &= &x+y \\ & &n & & &= &2x \\ & & & &p &= &y \end{array}$

shows that that any

$\vec{v}= \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}x+ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}y$

can be expressed as a member of $R$ with $m = x$ , $n = 2 x$ , and $p = y$ :

$\vec{v}= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}x+ \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}2x+ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}y.$

Thus $P\subseteq R$ .

But, for the other direction, the reduction resulting from fixing $m$ , $n$ , and $p$ and looking for $x$ and $y$

$\begin{array}{rcl} \begin{array}{*{2}{rc}r} x &+ &y &= &m+p \\ 2x & & &= &n \\ & &y &= &p \end{array} &\xrightarrow[]{-2\rho_1+\rho_2} &\begin{array}{*{2}{rc}r} x &+ &y &= &m+p \\ & &-2y&= &-2m+n-2p \\ & &y &= &p \end{array} \\ &\xrightarrow[]{(1/2)\rho_2+\rho_3} &\begin{array}{*{2}{rc}r} x &+ &y &= &m+p \\ & &-2y&= &-2m+n-2p \\ & &0 &= &m+(1/2)n \end{array} \end{array}$

shows that the only vectors

$\begin{pmatrix} m+p \\ n \\ p \end{pmatrix}\in R$

representable in the form

$\begin{pmatrix} x+y \\ 2x \\ y \end{pmatrix}$

are those where $0 = m + (1 / 2) n$ . For instance,

$\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$

is in $R$ but not in $P$ .

[edit] Exercises

Problem 1

Decide if the vector is a member of the set.

$\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ , $\{\begin{pmatrix} 1 \\ 2 \end{pmatrix}k\,\big|\, k\in\mathbb{R}\}$
$\begin{pmatrix} -3 \\ 3 \end{pmatrix}$ , $\{\begin{pmatrix} 1 \\ -1 \end{pmatrix}k\,\big|\, k\in\mathbb{R}\}$
$\begin{pmatrix} -3 \\ 3 \\ 4 \end{pmatrix}$ , $\{\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}k\,\big|\, k\in\mathbb{R}\}$
$\begin{pmatrix} -3 \\ 3 \\ 4 \end{pmatrix}$ , $\{\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}k+\begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix}m \,\big|\, k,m\in\mathbb{R}\}$
$\begin{pmatrix} 1 \\ 4 \\ 14 \end{pmatrix}$ , $\{\begin{pmatrix} 2 \\ 2 \\ 5 \end{pmatrix}k+\begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}m \,\big|\, k,m\in\mathbb{R}\}$
$\begin{pmatrix} 1 \\ 4 \\ 6 \end{pmatrix}$ , $\{\begin{pmatrix} 2 \\ 2 \\ 5 \end{pmatrix}k+\begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}m \,\big|\, k,m\in\mathbb{R}\}$

Answer

No.
Yes.
No.
Yes.
Yes; use Gauss' method to get $k = 4$ and $m = - 3$ .
No; use Gauss' method to conclude that there is no solution.

Problem 2

Produce two descriptions of this set that are different than this one.

$\{\begin{pmatrix} 2 \\ -5 \end{pmatrix}k\,\big|\, k\in\mathbb{R}\}$

Answer

One easy thing to do is to double and triple the vector:

$\{\begin{pmatrix} 4 \\ -10 \end{pmatrix}k\,\big|\, k\in\mathbb{R}\} \quad\text{and}\quad \{\begin{pmatrix} 6 \\ -55 \end{pmatrix}k\,\big|\, k\in\mathbb{R}\}.$

This exercise is recommended for all readers.

Problem 3

Show that the three descriptions given at the start of this subsection all describe the same set.

Answer

Instead of showing all three equalities, we can show that the first equals the second, and that the second equals the third. Both equalities are easy, using the methods of this subsection.

This exercise is recommended for all readers.

Problem 4

Show that these sets are equal

$\{\begin{pmatrix} 1 \\ 4 \\ 1 \\ 1 \end{pmatrix} +\begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}z\,\big|\, z\in\mathbb{R} \} \quad\text{and}\quad \{\begin{pmatrix} 0 \\ 4 \\ 2 \\ 1 \end{pmatrix} +\begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}k\,\big|\, k\in\mathbb{R} \},$

and that both describe the solution set of this system.

$\begin{array}{*{4}{rc}r} x &- &y &+ &z &+ &w &= &-1 \\ & &y & & &- &w &= &3 \\ x & & &+ &z &+ &2w &= &4 \end{array}$

Answer

That system reduces like this:

$\begin{array}{rcl} &\xrightarrow[]{-\rho_1+\rho_2} &\begin{array}{*{4}{rc}r} x &- &y &+ &z &+ &w &= &-1 \\ & &y & & &- &w &= &3 \\ & &y & & &+ &w &= &5 \end{array} \\ &\xrightarrow[]{-\rho_2+\rho_3} &\begin{array}{*{4}{rc}r} x &- &y &+ &z &+ &w &= &-1 \\ & &y & & &- &w &= &3 \\ & & & & & &2w &= &2 \end{array} \end{array}$

showing that $w = 1$ , $y = 4$ and $x = 2 - z$ .

This exercise is recommended for all readers.

Problem 5

Decide if the sets are equal.

$\{\begin{pmatrix} 1 \\ 2 \end{pmatrix} +\begin{pmatrix} 0 \\ 3 \end{pmatrix}t \,\big|\, t\in\mathbb{R}\}$ and $\{\begin{pmatrix} 1 \\ 8 \end{pmatrix} +\begin{pmatrix} 0 \\ -1 \end{pmatrix}s \,\big|\, s\in\mathbb{R}\}$
$\{\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}t +\begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix}s \,\big|\, t,s\in\mathbb{R}\}$ and $\{\begin{pmatrix} 4 \\ 7 \\ 7 \end{pmatrix}m +\begin{pmatrix} -4 \\ -2 \\ -10 \end{pmatrix}n \,\big|\, m,n\in\mathbb{R}\}$
$\{\begin{pmatrix} 1 \\ 2 \end{pmatrix}t \,\big|\, t\in\mathbb{R}\}$ and $\{\begin{pmatrix} 2 \\ 4 \end{pmatrix}m +\begin{pmatrix} 4 \\ 8 \end{pmatrix}n \,\big|\, m,n\in\mathbb{R}\}$
$\{\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}s +\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}t \,\big|\, s,t\in\mathbb{R}\}$ and $\{\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}m +\begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix}n \,\big|\, m,n\in\mathbb{R}\}$
$\{\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}t +\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}s \,\big|\, t,s\in\mathbb{R}\}$ and $\{\begin{pmatrix} 3 \\ 7 \\ 7 \end{pmatrix}t +\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}s \,\big|\, t,s\in\mathbb{R}\}$

Answer

For each item, we call the first set $S 1$ and the other $S 2$ .

They are equal. To see that $S_1\subseteq S_2$ , we must show that any element of the first set is in the second, that is, for any vector of the form

$\vec{v}=\begin{pmatrix} 1 \\ 2 \end{pmatrix} +\begin{pmatrix} 0 \\ 3 \end{pmatrix}t$

there is an appropriate $s$ such that

$\vec{v}=\begin{pmatrix} 1 \\ 8 \end{pmatrix} +\begin{pmatrix} 0 \\ -1 \end{pmatrix}s.$

Restated, given $t$ we must find $s$ so that this holds.

$\begin{array}{*{2}{rc}r} 1 &+ &0s &= &1+0t \\ 8 &- &1s &= &2+3t \end{array}$

That system reduces to

$\begin{array}{*{1}{rc}r} 1 &= &1 \\ s &= &6-3t \end{array}$

That is,

$\begin{pmatrix} 1 \\ 2 \end{pmatrix} +\begin{pmatrix} 0 \\ 3 \end{pmatrix}t =\begin{pmatrix} 1 \\ 8 \end{pmatrix} +\begin{pmatrix} 0 \\ -1 \end{pmatrix}(6-3t)$

and so any vector in the form for $S 1$ can be stated in the form needed for inclusion in $S 2$ . For $S_2\subseteq S_1$ , we look for $t$ so that these equations hold.

$\begin{array}{*{2}{rc}r} 1 &+ &0t &= &1+0s \\ 2 &+ &3t &= &8-1s \end{array}$

Rewrite that as

$\begin{array}{*{1}{rc}r} 1 &= &1 \\ t &= &2-(1/3)s \end{array}$

and so

$\begin{pmatrix} 1 \\ 8 \end{pmatrix} +\begin{pmatrix} 0 \\ -1 \end{pmatrix}s =\begin{pmatrix} 1 \\ 2 \end{pmatrix} +\begin{pmatrix} 0 \\ 3 \end{pmatrix}(2-(1/3)s).$
These two are equal. To show that $S_1\subseteq S_2$ , we check that for any $t, s$ we can find an appropriate $m, n$ so that these hold.

$\begin{array}{*{2}{rc}r} 4m &- &4n &= &1t+2s \\ 7m &- &2n &= &3t+1s \\ 7m &- &10n &= &1t+5s \end{array}$

Use Gauss' method

$\begin{array}{rcl} \left(\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 7 &-2 &3t+1s \\ 7 &-10 &1t+5s \end{array}\right) &\xrightarrow[(-7/4)\rho_1+\rho_3]{(-7/4)\rho_1+\rho_2} &\left(\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 0 &5 &(5/4)t-(10/4)s \\ 0 &-3 &-(3/4)t+(6/4)s \end{array}\right) \\ &\xrightarrow[]{(3/5)\rho_2+\rho_3} &\left(\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 0 &5 &(5/4)t-(10/4)s \\ 0 &0 &0 \end{array}\right) \end{array}$

to conclude that

$\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}t +\begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix}s =\begin{pmatrix} 4 \\ 7 \\ 7 \end{pmatrix}((1/2)t) +\begin{pmatrix} -4 \\ -2 \\ -10 \end{pmatrix}((1/4)t-(1/2)s)$

and so $S_1\subseteq S_2$ . For $S_2\subseteq S_1$ , solve

$\begin{array}{*{2}{rc}r} 1t &+ &2s &= &4m-4n \\ 3t &+ &1s &= &7m-2n \\ 1t &+ &5s &= &7m-10n \end{array}$

with Gaussian reduction

$\begin{array}{rcl} \left(\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 3 &1 &7m-2n \\ 1 &5 &7m-10n \end{array}\right) &\xrightarrow[-\rho_1+\rho_3]{-3\rho_1+\rho_2} &\left(\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 0 &-5 &-5m+10n\\ 0 &3 &3m-6n \end{array}\right) \\ &\xrightarrow[]{(3/5)\rho_2+\rho_3} &\left(\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 0 &-5 &-5m+10n\\ 0 &0 &0 \end{array}\right) \end{array}$

to get

$\begin{pmatrix} 4 \\ 7 \\ 7 \end{pmatrix}m +\begin{pmatrix} -4 \\ -2 \\ -10 \end{pmatrix}n =\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}(2m) +\begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix}(m-2n)$

and so any member of $S 2$ can be expressed in the form needed for $S 1$ .
These sets are equal. To prove that $S_1\subseteq S_2$ , we must be able to solve

$\begin{array}{*{2}{rc}r} 2m &+ &4n &= &1t \\ 4m &+ &8n &= &2t \end{array}$

for $m$ and $n$ in terms of $t$ . Apply Gaussian reduction

$\left(\begin{array}{*{2}{c}|c} 2 &4 &1t \\ 4 &8 &2t \end{array}\right) \xrightarrow[]{-2\rho_1+\rho_2} \left(\begin{array}{*{2}{c}|c} 2 &4 &1t \\ 0 &0 &0 \end{array}\right)$

to conclude that any pair $m, n$ where $2 m + 4 n = t$ will do. For instance,

$\begin{pmatrix} 1 \\ 2 \end{pmatrix}t =\begin{pmatrix} 2 \\ 4 \end{pmatrix}((1/2)t) +\begin{pmatrix} 4 \\ 8 \end{pmatrix}(0)$

or

$\begin{pmatrix} 1 \\ 2 \end{pmatrix}t =\begin{pmatrix} 2 \\ 4 \end{pmatrix}((-3/2)t) +\begin{pmatrix} 4 \\ 8 \end{pmatrix}(t).$

Thus $S_1\subseteq S_2$ . For $S_2\subseteq S_1$ , we solve

$\begin{array}{*{2}{rc}r} 1t &= &2m+4n \\ 2t &= &4m+8n \end{array}$

with Gauss' method

$\begin{array}{rcl} \left(\begin{array}{*{1}{c}|c} 1 &2m+4n \\ 2 &4m+8n \end{array}\right) &\xrightarrow[]{-2\rho_1+\rho_2} &\left(\begin{array}{*{1}{c}|c} 1 &2m+4n \\ 0 &0 \end{array}\right) \end{array}$

to deduce that any vector in $S 2$ is also in $S 1$ .

$\begin{pmatrix} 2 \\ 4 \end{pmatrix}m +\begin{pmatrix} 4 \\ 8 \end{pmatrix}n =\begin{pmatrix} 1 \\ 2 \end{pmatrix}(2m+4n)\in S_1.$
Neither set is a subset of the other. For $S_1\subseteq S_2$ to hold we must be able to solve

$\begin{array}{*{2}{rc}r} -1m &+ &0n &= &1s-1t \\ 1m &+ &1n &= &0s+1t \\ 1m &+ &3n &= &2s+0t \end{array}$

for $m$ and $n$ in terms of $t$ and $s$ . Gauss' method

$\begin{array}{rcl} \left(\begin{array}{*{2}{c}|c} -1 &0 &1s-1t \\ 1 &1 &0s+1t \\ 1 &3 &2s+0t \end{array}\right) &\xrightarrow[\rho_1+\rho_3]{\rho_1+\rho_2} &\left(\begin{array}{*{2}{c}|c} -1 &0 &1s-1t \\ 0 &1 &1s+0t \\ 0 &3 &3s-1t \end{array}\right) \\ &\xrightarrow[]{-3\rho_2+\rho_3} &\left(\begin{array}{*{2}{c}|c} -1 &0 &1s-1t \\ 0 &1 &1s+0t \\ 0 &3 &0s-1t \end{array}\right) \end{array}$

shows that we can only find an appropriate pair $m, n$ when $t = 0$ . That is,

$\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$

has no expression of the form

$\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}m+\begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix}n.$

Having shown that $S 1$ is not a subset of $S 2$ , we know $S_1\neq S_2$ so, strictly speaking, we need not go further. But we shall also show that $S 2$ is not a subset of $S 1$ . For $S_2\subseteq S_1$ to hold, we must be able to solve

$\begin{array}{*{2}{rc}r} 1s &- &1t &= &-1m+0n \\ 0s &+ &1t &= &1m+1n \\ 2s &+ &0t &= &1m+3n \end{array}$

for $s$ and $t$ . Apply row reduction

$\begin{array}{rcl} \left(\begin{array}{*{2}{c}|c} 1 &-1 &-1m+0n \\ 0 &1 &1m+1n \\ 2 &0 &1m+3n \end{array}\right) &\xrightarrow[]{-2\rho_1+\rho_3} &\left(\begin{array}{*{2}{c}|c} 1 &-1 &-1m+0n \\ 0 &1 &1m+1n \\ 0 &2 &3m+3n \end{array}\right) \\ &\xrightarrow[]{-2\rho_2+\rho_3} &\left(\begin{array}{*{2}{c}|c} 1 &-1 &-1m+0n \\ 0 &1 &1m+1n \\ 0 &0 &1m+1n \end{array}\right) \end{array}$

to deduce that the only vectors from $S 2$ that are also in $S 1$ are of the form

$\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}m +\begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix}(-m).$

For instance,

$\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$

is in $S 2$ but not in $S 1$ .
These sets are equal. First we change the parameters:

$S_2=\{\begin{pmatrix} 3 \\ 7 \\ 7 \end{pmatrix}m +\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}n \,\big|\, m,n\in\mathbb{R}\}.$

Now, to show that $S_1\subseteq S_2$ , we solve

$\begin{array}{*{2}{rc}r} 3m &+ &1n &= &1t+2s \\ 7m &+ &3n &= &3t+4s \\ 7m &+ &1n &= &1t+6s \end{array}$

with Gauss' method

$\begin{array}{rcl} \left(\begin{array}{*{2}{c}|c} 3 &1 &1t+2s \\ 7 &3 &3t+4s \\ 7 &1 &1t+6s \end{array}\right) &\xrightarrow[(-7/3)\rho_1+\rho_3]{(-7/3)\rho_1+\rho_2} &\left(\begin{array}{*{2}{c}|c} 3 &1 &1t+2s \\ 0 &2/3 &(2/3)t-(2/3)s \\ 0 &-4/3 &(-4/3)t+(4/3)s \end{array}\right) \\ &\xrightarrow[]{2\rho_2+\rho_3} &\left(\begin{array}{*{2}{c}|c} 3 &1 &1t+2s \\ 0 &2/3 &(2/3)t-(2/3)s \\ 0 &0 &0 \end{array}\right) \end{array}$

to get that

$\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}t +\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}s =\begin{pmatrix} 3 \\ 7 \\ 7 \end{pmatrix}(s) +\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}(t-s)$

and so $S_1\subseteq S_2$ . The proof that $S_2\subseteq S_1$ involves solving

$\begin{array}{*{2}{rc}r} 1t &+ &2s &= &3m+1n \\ 3t &+ &4s &= &7m+3n \\ 1t &+ &6s &= &7m+1n \end{array}$

with Gaussian reduction

$\begin{array}{rcl} \left(\begin{array}{*{2}{c}|c} 1 &2 &3m+1n \\ 3 &4 &7m+3n \\ 1 &6 &7m+1n \end{array}\right) &\xrightarrow[-\rho_1+\rho_3]{-3\rho_1+\rho_2} &\left(\begin{array}{*{2}{c}|c} 1 &2 &3m+1n \\ 0 &-2 &-2m \\ 0 &4 &4m \end{array}\right) \\ &\xrightarrow[]{2\rho_2+\rho_3} &\left(\begin{array}{*{2}{c}|c} 1 &2 &3m+1n \\ 0 &-2 &-2m \\ 0 &0 &0 \end{array}\right) \end{array}$

to conclude

$\begin{pmatrix} 3 \\ 7 \\ 7 \end{pmatrix}m +\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}n =\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}(m+n) +\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}(m)$

and so any vector in $S 2$ is also in $S 1$ .