Problem Set 3: Row Reduction, Vectors, and Span

When potassium permanganate ($\mathrm{KMnO_4}$) is mixed with hydrochloric acid ($\mathrm{HCl}$), the result is potassium chloride ($\mathrm{KCl}$), manganese chloride ($\mathrm{MnCl_2}$), water ($\mathrm{H_2O}$), and chlorine gas ($\mathrm{Cl_2}$): \[ x_1\,\mathrm{KMnO_4} + x_2\,\mathrm{HCl} \longrightarrow x_3\,\mathrm{KCl} + x_4\,\mathrm{MnCl_2} + x_5\,\mathrm{H_2O} + x_6\,\mathrm{Cl_2}. \] This is a standard way of producing chlorine gas in a lab. Do not try it at home.

1. Before we set out trying to solve for the $x_i$, how many solutions should we expect to find? Explain.
2. Write down the system of linear equations that $x_1, \dots, x_6$ must satisfy in order for this reaction to be balanced. Use row reduction to find the general solution to your linear system. You may use the Row Reducer to help you calculate.
3. A chemist wants the balanced reaction with the smallest possible positive whole number coefficients. Help them find this solution.
4. (Challenge) Carbon monoxide, carbon dioxide, and hydrogen gas can react to produce methane and water: \[ x_1\,\mathrm{CO} + x_2\,\mathrm{CO_2} + x_3\,\mathrm{H_2} \longrightarrow x_4\,\mathrm{CH_4} + x_5\,\mathrm{H_2O}. \] Find the general solution to the system of equations for this reaction. You should find two free variables. Explain what this means chemically: show that every balanced version of this reaction is built out of two simpler reactions, and find those two reactions.

Given a triangle with vertices $p, q, r$, the medial triangle is the triangle whose vertices are the midpoints of the three sides. We label the vertices of the medial triangle $p', q', r'$ so that $p'$ is the midpoint of the side opposite $p$, $q'$ is the midpoint of the side opposite $q$, and $r'$ is the midpoint of the side opposite $r$. See the picture below for an example: the solid dots are the vertices of the triangle and the hollow dots are the vertices of the medial triangle.

1. Let $p = (-2, 1)$, $q = (4, -1)$, and $r = (2, 5)$. Find the vertices $p', q', r'$ of the medial triangle. Note that these are not the same points as in the picture.
2. Now suppose you are only told the vertices of the medial triangle: $p' = (1, 4)$, $q' = (-2, 1)$, and $r' = (3, 0)$. Find the vertices $p, q, r$ of the original triangle.
3. Now let $p, q, r$ be any triangle and let $p', q', r'$ be the vertices of its medial triangle. Find formulas for $p, q, r$ in terms of $p', q', r'$.

Recall that the span of a collection of vectors $v_1, \dots, v_k$ is the set \[ \sp\{v_1,\ldots, v_k\} := \{c_1 v_1 + \cdots + c_k v_k : c_i \in \RR\} \] of all their linear combinations. For each of the following, determine whether $b$ is in $\sp\{v_1, \dots, v_k\}$. If it is, write $b$ explicitly as a linear combination of the $v_i$. If it is not, explain how you know.

1. \( \blue{ v_1 = \begin{bmatrix*}[r] 1 \\ 2 \\ -1 \end{bmatrix*},\quad v_2 = \begin{bmatrix*}[r] 2 \\ -1 \\ 3 \end{bmatrix*},\quad b = \begin{bmatrix} 4 \\ 3 \\ 1 \end{bmatrix} } \)
2. \( \blue{ v_1 = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix},\quad v_2 = \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix},\quad b = \begin{bmatrix} 1 \\ 4 \\ 0 \end{bmatrix} } \)
3. \( \blue{ v_1 = \begin{bmatrix} 1 \\ 0 \\ 2 \\ 1 \end{bmatrix},\quad v_2 = \begin{bmatrix*}[r] 0 \\ 1 \\ 1 \\ -1 \end{bmatrix*},\quad v_3 = \begin{bmatrix} 1 \\ 1 \\ 3 \\ 0 \end{bmatrix},\quad b = \begin{bmatrix} 2 \\ 1 \\ 5 \\ 1 \end{bmatrix} } \)
4. \( \blue{ v_1 = \begin{bmatrix} 1 \\ 0 \\ 2 \\ 1 \end{bmatrix},\quad v_2 = \begin{bmatrix*}[r] 0 \\ 1 \\ 1 \\ -1 \end{bmatrix*},\quad v_3 = \begin{bmatrix} 1 \\ 1 \\ 3 \\ 0 \end{bmatrix},\quad b = \begin{bmatrix} 2 \\ 1 \\ 5 \\ 2 \end{bmatrix} } \)

The Row Reducer is happy to help you with arithmetic, but it only works when the matrix is filled with actual numbers. In this problem some of the entries are an unknown constant $c$, and that means you get to do the row reduction by hand.

Consider the system \[ \blue{ \begin{alignedat}{4} x &{}-{}& 2y &{}+{}& 3z &=&{} 1 \\ 2x &{}-{}& 3y &{}+{}& 4z &=&{} 3 \\ 3x &{}-{}& 5y &{}+{}& cz &=&{}\,\, 5 \end{alignedat} } \] where $c$ is a real number. For which values of $c$ does the system have no solutions? Exactly one solution? Infinitely many solutions? You do not need to find the exact solutions.