# Linear Algebra/Comparing Set Descriptions/Solutions

## Solutions

Problem 1

Decide if the vector is a member of the set.

1. ${\displaystyle {\begin{pmatrix}2\\3\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}1\\2\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}$
2. ${\displaystyle {\begin{pmatrix}-3\\3\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}1\\-1\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}$
3. ${\displaystyle {\begin{pmatrix}-3\\3\\4\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}1\\-1\\2\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}$
4. ${\displaystyle {\begin{pmatrix}-3\\3\\4\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}1\\-1\\2\end{pmatrix}}k+{\begin{pmatrix}0\\0\\2\end{pmatrix}}m\,{\big |}\,k,m\in \mathbb {R} \}}$
5. ${\displaystyle {\begin{pmatrix}1\\4\\14\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}2\\2\\5\end{pmatrix}}k+{\begin{pmatrix}-1\\0\\2\end{pmatrix}}m\,{\big |}\,k,m\in \mathbb {R} \}}$
6. ${\displaystyle {\begin{pmatrix}1\\4\\6\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}2\\2\\5\end{pmatrix}}k+{\begin{pmatrix}-1\\0\\2\end{pmatrix}}m\,{\big |}\,k,m\in \mathbb {R} \}}$
1. No.
2. Yes.
3. No.
4. Yes.
5. Yes; use Gauss' method to get ${\displaystyle k=4}$ and ${\displaystyle m=-3}$.
6. 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.

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

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

${\displaystyle \{{\begin{pmatrix}4\\-10\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}\quad {\text{and}}\quad \{{\begin{pmatrix}6\\-15\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}.}$
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.

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

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

and that both describe the solution set of this system.

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

That system reduces like this:

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

showing that ${\displaystyle w=1}$, ${\displaystyle y=4}$ and ${\displaystyle x=2-z}$.

This exercise is recommended for all readers.
Problem 5

Decide if the sets are equal.

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

For each item, we call the first set ${\displaystyle S_{1}}$ and the other ${\displaystyle S_{2}}$.

1. They are equal. To see that ${\displaystyle S_{1}\subseteq S_{2}}$, we must show that any element of the first set is in the second, that is, for any vector of the form
${\displaystyle {\vec {v}}={\begin{pmatrix}1\\2\end{pmatrix}}+{\begin{pmatrix}0\\3\end{pmatrix}}t}$
there is an appropriate ${\displaystyle s}$ such that
${\displaystyle {\vec {v}}={\begin{pmatrix}1\\8\end{pmatrix}}+{\begin{pmatrix}0\\-1\end{pmatrix}}s.}$
Restated, given ${\displaystyle t}$ we must find ${\displaystyle s}$ so that this holds.
${\displaystyle {\begin{array}{*{2}{rc}r}1&+&0s&=&1+0t\\8&-&1s&=&2+3t\end{array}}}$
That system reduces to
${\displaystyle {\begin{array}{*{1}{rc}r}1&=&1\\s&=&6-3t\end{array}}}$
That is,
${\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}+{\begin{pmatrix}0\\3\end{pmatrix}}t={\begin{pmatrix}1\\8\end{pmatrix}}+{\begin{pmatrix}0\\-1\end{pmatrix}}(6-3t)}$
and so any vector in the form for ${\displaystyle S_{1}}$ can be stated in the form needed for inclusion in ${\displaystyle S_{2}}$. For ${\displaystyle S_{2}\subseteq S_{1}}$, we look for ${\displaystyle t}$ so that these equations hold.
${\displaystyle {\begin{array}{*{2}{rc}r}1&+&0t&=&1+0s\\2&+&3t&=&8-1s\end{array}}}$
Rewrite that as
${\displaystyle {\begin{array}{*{1}{rc}r}1&=&1\\t&=&2-(1/3)s\end{array}}}$
and so
${\displaystyle {\begin{pmatrix}1\\8\end{pmatrix}}+{\begin{pmatrix}0\\-1\end{pmatrix}}s={\begin{pmatrix}1\\2\end{pmatrix}}+{\begin{pmatrix}0\\3\end{pmatrix}}(2-(1/3)s).}$
2. These two are equal. To show that ${\displaystyle S_{1}\subseteq S_{2}}$, we check that for any ${\displaystyle t,s}$ we can find an appropriate ${\displaystyle m,n}$ so that these hold.
${\displaystyle {\begin{array}{*{2}{rc}r}4m&-&4n&=&1t+2s\\7m&-&2n&=&3t+1s\\7m&-&10n&=&1t+5s\end{array}}}$
Use Gauss' method
${\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}4&-4&1t+2s\\7&-2&3t+1s\\7&-10&1t+5s\end{array}}\right)&{\xrightarrow[{(-7/4)\rho _{1}+\rho _{3}}]{(-7/4)\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}4&-4&1t+2s\\0&5&(5/4)t-(10/4)s\\0&-3&-(3/4)t+(6/4)s\end{array}}\right)\\&{\xrightarrow[{}]{(3/5)\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}4&-4&1t+2s\\0&5&(5/4)t-(10/4)s\\0&0&0\end{array}}\right)\end{array}}}$
to conclude that
${\displaystyle {\begin{pmatrix}1\\3\\1\end{pmatrix}}t+{\begin{pmatrix}2\\1\\5\end{pmatrix}}s={\begin{pmatrix}4\\7\\7\end{pmatrix}}((1/2)t)+{\begin{pmatrix}-4\\-2\\-10\end{pmatrix}}((1/4)t-(1/2)s)}$
and so ${\displaystyle S_{1}\subseteq S_{2}}$. For ${\displaystyle S_{2}\subseteq S_{1}}$, solve
${\displaystyle {\begin{array}{*{2}{rc}r}1t&+&2s&=&4m-4n\\3t&+&1s&=&7m-2n\\1t&+&5s&=&7m-10n\end{array}}}$
with Gaussian reduction
${\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}1&2&4m-4n\\3&1&7m-2n\\1&5&7m-10n\end{array}}\right)&{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-3\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}1&2&4m-4n\\0&-5&-5m+10n\\0&3&3m-6n\end{array}}\right)\\&{\xrightarrow[{}]{(3/5)\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&2&4m-4n\\0&-5&-5m+10n\\0&0&0\end{array}}\right)\end{array}}}$
to get
${\displaystyle {\begin{pmatrix}4\\7\\7\end{pmatrix}}m+{\begin{pmatrix}-4\\-2\\-10\end{pmatrix}}n={\begin{pmatrix}1\\3\\1\end{pmatrix}}(2m)+{\begin{pmatrix}2\\1\\5\end{pmatrix}}(m-2n)}$
and so any member of ${\displaystyle S_{2}}$ can be expressed in the form needed for ${\displaystyle S_{1}}$.
3. These sets are equal. To prove that ${\displaystyle S_{1}\subseteq S_{2}}$, we must be able to solve
${\displaystyle {\begin{array}{*{2}{rc}r}2m&+&4n&=&1t\\4m&+&8n&=&2t\end{array}}}$
for ${\displaystyle m}$ and ${\displaystyle n}$ in terms of ${\displaystyle t}$. Apply Gaussian reduction
${\displaystyle \left({\begin{array}{*{2}{c}|c}2&4&1t\\4&8&2t\end{array}}\right){\xrightarrow[{}]{-2\rho _{1}+\rho _{2}}}\left({\begin{array}{*{2}{c}|c}2&4&1t\\0&0&0\end{array}}\right)}$
to conclude that any pair ${\displaystyle m,n}$ where ${\displaystyle 2m+4n=t}$ will do. For instance,
${\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}t={\begin{pmatrix}2\\4\end{pmatrix}}((1/2)t)+{\begin{pmatrix}4\\8\end{pmatrix}}(0)}$
or
${\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}t={\begin{pmatrix}2\\4\end{pmatrix}}((-3/2)t)+{\begin{pmatrix}4\\8\end{pmatrix}}(t).}$
Thus ${\displaystyle S_{1}\subseteq S_{2}}$. For ${\displaystyle S_{2}\subseteq S_{1}}$, we solve
${\displaystyle {\begin{array}{*{2}{rc}r}1t&=&2m+4n\\2t&=&4m+8n\end{array}}}$
with Gauss' method
${\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{1}{c}|c}1&2m+4n\\2&4m+8n\end{array}}\right)&{\xrightarrow[{}]{-2\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{1}{c}|c}1&2m+4n\\0&0\end{array}}\right)\end{array}}}$
to deduce that any vector in ${\displaystyle S_{2}}$ is also in ${\displaystyle S_{1}}$.
${\displaystyle {\begin{pmatrix}2\\4\end{pmatrix}}m+{\begin{pmatrix}4\\8\end{pmatrix}}n={\begin{pmatrix}1\\2\end{pmatrix}}(2m+4n)\in S_{1}.}$
4. Neither set is a subset of the other. For ${\displaystyle S_{1}\subseteq S_{2}}$ to hold we must be able to solve
${\displaystyle {\begin{array}{*{2}{rc}r}-1m&+&0n&=&1s-1t\\1m&+&1n&=&0s+1t\\1m&+&3n&=&2s+0t\end{array}}}$
for ${\displaystyle m}$ and ${\displaystyle n}$ in terms of ${\displaystyle t}$ and ${\displaystyle s}$. Gauss' method
${\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}-1&0&1s-1t\\1&1&0s+1t\\1&3&2s+0t\end{array}}\right)&{\xrightarrow[{\rho _{1}+\rho _{3}}]{\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}-1&0&1s-1t\\0&1&1s+0t\\0&3&3s-1t\end{array}}\right)\\&{\xrightarrow[{}]{-3\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}-1&0&1s-1t\\0&1&1s+0t\\0&3&0s-1t\end{array}}\right)\end{array}}}$
shows that we can only find an appropriate pair ${\displaystyle m,n}$ when ${\displaystyle t=0}$. That is,
${\displaystyle {\begin{pmatrix}-1\\1\\0\end{pmatrix}}}$
has no expression of the form
${\displaystyle {\begin{pmatrix}-1\\1\\1\end{pmatrix}}m+{\begin{pmatrix}0\\1\\3\end{pmatrix}}n.}$
Having shown that ${\displaystyle S_{1}}$ is not a subset of ${\displaystyle S_{2}}$, we know ${\displaystyle S_{1}\neq S_{2}}$ so, strictly speaking, we need not go further. But we shall also show that ${\displaystyle S_{2}}$ is not a subset of ${\displaystyle S_{1}}$. For ${\displaystyle S_{2}\subseteq S_{1}}$ to hold, we must be able to solve
${\displaystyle {\begin{array}{*{2}{rc}r}1s&-&1t&=&-1m+0n\\0s&+&1t&=&1m+1n\\2s&+&0t&=&1m+3n\end{array}}}$
for ${\displaystyle s}$ and ${\displaystyle t}$. Apply row reduction
${\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}1&-1&-1m+0n\\0&1&1m+1n\\2&0&1m+3n\end{array}}\right)&{\xrightarrow[{}]{-2\rho _{1}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&-1&-1m+0n\\0&1&1m+1n\\0&2&3m+3n\end{array}}\right)\\&{\xrightarrow[{}]{-2\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&-1&-1m+0n\\0&1&1m+1n\\0&0&1m+1n\end{array}}\right)\end{array}}}$
to deduce that the only vectors from ${\displaystyle S_{2}}$ that are also in ${\displaystyle S_{1}}$ are of the form
${\displaystyle {\begin{pmatrix}-1\\1\\1\end{pmatrix}}m+{\begin{pmatrix}0\\1\\3\end{pmatrix}}(-m).}$
For instance,
${\displaystyle {\begin{pmatrix}-1\\1\\1\end{pmatrix}}}$
is in ${\displaystyle S_{2}}$ but not in ${\displaystyle S_{1}}$.
5. These sets are equal. First we change the parameters:
${\displaystyle S_{2}=\{{\begin{pmatrix}3\\7\\7\end{pmatrix}}m+{\begin{pmatrix}1\\3\\1\end{pmatrix}}n\,{\big |}\,m,n\in \mathbb {R} \}.}$
Now, to show that ${\displaystyle S_{1}\subseteq S_{2}}$, we solve
${\displaystyle {\begin{array}{*{2}{rc}r}3m&+&1n&=&1t+2s\\7m&+&3n&=&3t+4s\\7m&+&1n&=&1t+6s\end{array}}}$
with Gauss' method
${\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}3&1&1t+2s\\7&3&3t+4s\\7&1&1t+6s\end{array}}\right)&{\xrightarrow[{(-7/3)\rho _{1}+\rho _{3}}]{(-7/3)\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}3&1&1t+2s\\0&2/3&(2/3)t-(2/3)s\\0&-4/3&(-4/3)t+(4/3)s\end{array}}\right)\\&{\xrightarrow[{}]{2\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}3&1&1t+2s\\0&2/3&(2/3)t-(2/3)s\\0&0&0\end{array}}\right)\end{array}}}$
to get that
${\displaystyle {\begin{pmatrix}1\\3\\1\end{pmatrix}}t+{\begin{pmatrix}2\\4\\6\end{pmatrix}}s={\begin{pmatrix}3\\7\\7\end{pmatrix}}(s)+{\begin{pmatrix}1\\3\\1\end{pmatrix}}(t-s)}$
and so ${\displaystyle S_{1}\subseteq S_{2}}$. The proof that ${\displaystyle S_{2}\subseteq S_{1}}$ involves solving
${\displaystyle {\begin{array}{*{2}{rc}r}1t&+&2s&=&3m+1n\\3t&+&4s&=&7m+3n\\1t&+&6s&=&7m+1n\end{array}}}$
with Gaussian reduction
${\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}1&2&3m+1n\\3&4&7m+3n\\1&6&7m+1n\end{array}}\right)&{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-3\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}1&2&3m+1n\\0&-2&-2m\\0&4&4m\end{array}}\right)\\&{\xrightarrow[{}]{2\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&2&3m+1n\\0&-2&-2m\\0&0&0\end{array}}\right)\end{array}}}$
to conclude
${\displaystyle {\begin{pmatrix}3\\7\\7\end{pmatrix}}m+{\begin{pmatrix}1\\3\\1\end{pmatrix}}n={\begin{pmatrix}1\\3\\1\end{pmatrix}}(m+n)+{\begin{pmatrix}2\\4\\6\end{pmatrix}}(m)}$
and so any vector in ${\displaystyle S_{2}}$ is also in ${\displaystyle S_{1}}$.