# Linear Algebra/Comparing Set Descriptions/Solutions
## Solutions
### 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:
    $\{{\begin{pmatrix}4\\-10\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}\quad {\text{and}}\quad \{{\begin{pmatrix}6\\-15\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}.$
    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.

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