Answer
\begin{array}{l}
x_{1}=\frac{x_{6}}{6} \quad x_{2}=\frac{14 x_{6}}{99} \quad x_{3}=\frac{x_{6}}{18} \\
x_{4}=\frac{7 x_{6}}{99} \quad x_{5}=\frac{28 x_{6}}{99} \quad x_{6} \text { is free }
\end{array}
Work Step by Step
Equation involves five types of atoms. So we need vector in $\mathbb{R}^{5}$ that lists the number of atoms in molecule, for example:
\[
v=\left[\begin{array}{cc}
\text { no. of } \mathrm{Pb} \\
\text { no. of } \mathrm{N} \\
\text { no. of } \mathrm{Cr} \\
\text { no. of } \mathrm{Mn} \\
\text { no. of } \mathrm{O}
\end{array}\right.
\]
So we have:
\[
\begin{array}{c}
\mathrm{PbN}_{6}=\left[\begin{array}{c}
1 \\
6 \\
0 \\
0 \\
0
\end{array}\right] & \mathrm{CrMn}_{2} \mathrm{O}_{8}=\left[\begin{array}{c}
0 \\
0 \\
1 \\
2 \\
8
\end{array}\right] \quad \mathrm{Pb}_{3} \mathrm{O}_{4}=\left[\begin{array}{c}
3 \\
0 \\
0 \\
0 \\
4
\end{array}\right] \\
\mathrm{Cr}_{2} \mathrm{O}_{3}=\left[\begin{array}{c}
0 \\
0 \\
2 \\
3
\end{array}\right] \quad \mathrm{MnO}_{2}=\left[\begin{array}{c}
0 \\
0 \\
0 \\
1 \\
2
\end{array}\right] \quad \mathrm{NO}=\left[\begin{array}{c}
0 \\
1 \\
0 \\
0 \\
1
\end{array}\right]
\end{array}
\]
Let $x_{1}, \ldots x_{6}$ be the coefficients in the equation. They must satisfy:
\[
x_{1}\left[\begin{array}{l}
1 \\
6 \\
0 \\
0 \\
0
\end{array}\right]+x_{2}\left[\begin{array}{l}
0 \\
0 \\
1 \\
2 \\
8
\end{array}\right]=x_{3}\left[\begin{array}{l}
3 \\
0 \\
0 \\
0 \\
4
\end{array}\right]+x_{4}\left[\begin{array}{l}
0 \\
0 \\
2 \\
0 \\
3
\end{array}\right]+x_{5}\left[\begin{array}{l}
0 \\
0 \\
0 \\
1 \\
2
\end{array}\right]+x_{6}\left[\begin{array}{l}
0 \\
1 \\
0 \\
0 \\
1
\end{array}\right]
\]
\[
x_{1}\left[\begin{array}{l}
1 \\
6 \\
0 \\
0 \\
0
\end{array}\right]+x_{2}\left[\begin{array}{l}
0 \\
0 \\
1 \\
2 \\
8
\end{array}\right]+x_{3}\left[\begin{array}{c}
-3 \\
0 \\
0 \\
0 \\
-4
\end{array}\right]+x_{4}\left[\begin{array}{c}
0 \\
0 \\
-2 \\
0 \\
3
\end{array}\right]+x_{5}\left[\begin{array}{c}
0 \\
0 \\
0 \\
-1 \\
-2
\end{array}\right]+x_{6}\left[\begin{array}{c}
0 \\
-1 \\
0 \\
0 \\
-1
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0 \\
0 \\
0
\end{array}\right]
\]
2
The augmented matrix is:
\[
\left[\begin{array}{ccccccc}
1 & 0 & -3 & 0 & 0 & 0 & 0 \\
6 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & 0 & -2 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 & -1 & 0 & 0 \\
0 & 8 & 4 & 3 & -2 & -1 & 0
\end{array}\right]
\]
Using MATLAB we can find its reduced echelon form:
\[
\left[\begin{array}{ccccccc}
1 & 0 & -3 & 0 & 0 & 0 & 0 \\
6 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & 0 & -2 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 & -1 & 0 & 0 \\
0 & 8 & 4 & 3 & -2 & -1 & 0
\end{array}\right] \sim\left[\begin{array}{ccccccc}
1 & 0 & 0 & 0 & 0 & -1 / 6 & 0 \\
0 & 1 & 0 & 0 & 0 & -14 / 99 & 0 \\
0 & 0 & 1 & 0 & 0 & -1 / 18 & 0 \\
0 & 0 & 0 & 1 & 0 & -7 / 99 & 0 \\
0 & 0 & 0 & 0 & 1 & -28 / 99 & 0
\end{array}\right]
\]
The general solution is:
\[
\begin{aligned}
x_{1}=\frac{x_{6}}{6} & x_{2}=\frac{14 x_{6}}{99} \quad x_{3}=\frac{x_{6}}{18} \\
x_{4}=\frac{7 x_{6}}{99} \quad x_{5}=\frac{28 x_{6}}{99} & x_{6} \text { is free }
\end{aligned}
\]
