Answer
$2 . N a_{3} P O_{4}+3 . B a\left(N O_{3}\right)_{2} \rightarrow B a\left(P O_{4}\right)_{2}+6 . N a N O_{3}$
Work Step by Step
The following vectors list the numbers of atoms of
sodium (Na), phosphorus (P), oxygen (O) barium (Ba), and nitrogen(N)
\[
x_{1} \cdot\left[\begin{array}{c}
3 \\
1 \\
4 \\
0 \\
0
\end{array}\right]+x_{2} \cdot\left[\begin{array}{c}
0 \\
0 \\
6 \\
1 \\
2
\end{array}\right]=x_{3} \cdot\left[\begin{array}{c}
0 \\
2 \\
8 \\
3 \\
0
\end{array}\right]+x_{4} \cdot\left[\begin{array}{c}
1 \\
0 \\
3 \\
0 \\
1
\end{array}\right]
\]
The coefficients in the equation
\[
x_{1} \cdot N a_{3} P O_{4}+x_{2} \cdot B a\left(N O_{3}\right)_{2} \rightarrow x_{3} \cdot B a\left(P O_{4}\right)_{2}+x_{4} \cdot N a N O_{3}
\]
Satisfy left side equation
\[
\left[\begin{array}{ccccc}
3 & 0 & 0 & -1 & 0 \\
1 & 0 & -2 & 0 & 0 \\
4 & 6 & -8 & -3 & 0 \\
0 & 1 & -3 & 0 & 0 \\
0 & 2 & 0 & -1 & 0
\end{array}\right]
\]Move the right terms to the left side (changing the sign of each entry in the third and fourth vectors) and
row reduce the augmented matrix of the homogeneous
system:
4
\[
\left[\begin{array}{ccccc}
1 & 0 & -2 & 0 & 0 \\
3 & 0 & 0 & -1 & 0 \\
4 & 6 & -8 & -3 & 0 \\
0 & 1 & -3 & 0 & 0 \\
0 & 2 & 0 & -1 & 0
\end{array}\right]
\]
Swapping row 1 and row 3
5
\[
\left[\begin{array}{ccccc}
1 & 0 & -2 & 0 & 0 \\
0 & 0 & 6 & -1 & 0 \\
0 & 6 & 0 & -3 & 0 \\
0 & 1 & -3 & 0 & 0 \\
0 & 2 & 0 & -1 & 0
\end{array}\right]
\]
6
\[
\left[\begin{array}{ccccc}
1 & 0 & -2 & 0 & 0 \\
0 & 1 & -3 & 0 & 0 \\
0 & 6 & 0 & -3 & 0 \\
0 & 0 & 6 & -1 & 0 \\
0 & 2 & 0 & -1 & 0
\end{array}\right]
\]
\[
\left[\begin{array}{ccccc}
1 & 0 & -2 & 0 & 0 \\
0 & 1 & -3 & 0 & 0 \\
0 & 0 & 18 & -3 & 0 \\
0 & 0 & 6 & -1 & 0 \\
0 & 0 & 6 & -1 & 0
\end{array}\right]
\]
Taking 18 out as common from row 3 and
\[
R_{5}=R_{5}-R_{4}
\]
The general solution is
\[
\begin{array}{l}
x_{1}=(1 / 3) x_{4} \\
x_{2}=(1 / 2) x_{4} \\
x_{3}=(1 / 6) x_{4}
\end{array}
\]
with $x_{4}$ free. Take $x_{4}=6 .$ Then $x_{1}=2, x_{2}=3$ and $x_{3}=1 .$ The balanced equation is
$2 . N a_{3} P O_{4}+3 . B a\left(N O_{3}\right)_{2} \rightarrow B a\left(P O_{4}\right)_{2}+6 . N a N O_{3}$
