The equations may be written as
Carbon: ( R Q ) a + c = 2 Hydrogen: − 3 b + 1.7 c + 2 d = 6 Oxygen: ( 2 ( R Q ) − 2 ) a + 0.4 c + d = 1 Nitrogen: b − 0.15 c = 0 \begin{matrix}
\text{Carbon: }& (RQ)a + c = 2 \\
\text{Hydrogen: }& -3b + 1.7c + 2d = 6\\
\text{Oxygen: }& (2(RQ)-2)a + 0.4c + d = 1\\
\text{Nitrogen: }& b - 0.15c = 0
\end{matrix} Carbon: Hydrogen: Oxygen: Nitrogen: ( RQ ) a + c = 2 − 3 b + 1.7 c + 2 d = 6 ( 2 ( RQ ) − 2 ) a + 0.4 c + d = 1 b − 0.15 c = 0
Or substituting RQ = 0.8 in matrix form
[ 0.8 0 1 0 0 − 3 1.7 2 − 0.4 0 0.4 1 0 1 − 0.15 0 ] [ a b c d ] = [ 2 6 1 0 ] \begin{bmatrix}
0.8 & 0 & 1 & 0 \\
0 & -3 & 1.7 & 2 \\
-0.4 & 0 & 0.4 & 1 \\
0 & 1 & -0.15 & 0
\end{bmatrix}
\begin{bmatrix}
a \\
b \\
c \\
d
\end{bmatrix} =
\begin{bmatrix}
2 \\
6 \\
1 \\
0
\end{bmatrix} ⎣ ⎡ 0.8 0 − 0.4 0 0 − 3 0 1 1 1.7 0.4 − 0.15 0 2 1 0 ⎦ ⎤ ⎣ ⎡ a b c d ⎦ ⎤ = ⎣ ⎡ 2 6 1 0 ⎦ ⎤
Adding 1/2 of the first row to the third one:
[ 0.8 0 1 0 0 − 3 1.7 2 0 0 0.9 1 0 1 − 0.15 0 ] [ a b c d ] = [ 2 6 2 0 ] \begin{bmatrix}
0.8 & 0 & 1 & 0 \\
0 & -3 & 1.7 & 2 \\
0 & 0 & 0.9 & 1 \\
0 & 1 & -0.15 & 0
\end{bmatrix}
\begin{bmatrix}
a \\
b \\
c \\
d
\end{bmatrix} =
\begin{bmatrix}
2 \\
6 \\
2 \\
0
\end{bmatrix} ⎣ ⎡ 0.8 0 0 0 0 − 3 0 1 1 1.7 0.9 − 0.15 0 2 1 0 ⎦ ⎤ ⎣ ⎡ a b c d ⎦ ⎤ = ⎣ ⎡ 2 6 2 0 ⎦ ⎤
Add 1/3 of the second row to the fourth:
[ 0.8 0 1 0 0 − 3 1.7 2 0 0 0.9 1 0 0 0.417 0.667 ] [ a b c d ] = [ 2 6 2 2 ] \begin{bmatrix}
0.8 & 0 & 1 & 0 \\
0 & -3 & 1.7 & 2\\
0 & 0 & 0.9 & 1 \\
0 & 0 & 0.417 & 0.667
\end{bmatrix}
\begin{bmatrix}
a \\
b \\
c \\
d
\end{bmatrix} =
\begin{bmatrix}
2 \\
6 \\
2 \\
2
\end{bmatrix} ⎣ ⎡ 0.8 0 0 0 0 − 3 0 0 1 1.7 0.9 0.417 0 2 1 0.667 ⎦ ⎤ ⎣ ⎡ a b c d ⎦ ⎤ = ⎣ ⎡ 2 6 2 2 ⎦ ⎤
Subtract 4.17/9 of the third row from the fourth:
[ 0.8 0 1 0 0 − 3 1.7 2 0 0 0.9 1 0 0 0 0.204 ] [ a b c d ] = [ 2 6 2 1.07 ] \begin{bmatrix}
0.8 & 0 & 1 & 0 \\
0 & -3 & 1.7 & 2\\
0 & 0 & 0.9 & 1 \\
0 & 0 & 0 & 0.204
\end{bmatrix}
\begin{bmatrix}
a \\
b \\
c \\
d
\end{bmatrix} =
\begin{bmatrix}
2 \\
6 \\
2 \\
1.07
\end{bmatrix} ⎣ ⎡ 0.8 0 0 0 0 − 3 0 0 1 1.7 0.9 0 0 2 1 0.204 ⎦ ⎤ ⎣ ⎡ a b c d ⎦ ⎤ = ⎣ ⎡ 2 6 2 1.07 ⎦ ⎤
Therefor
d = 1.07 / 0.204 c = ( 2 − d ) / 0.9 b = ( 6 − 1.7 c − 2 d ) / ( − 3 ) a = ( 2 − c ) / 0.8 d = 1.07 / 0.204 \\
c = (2 - d) / 0.9 \\
b = (6 - 1.7 c - 2 d) / (-3) \\
a = (2 - c) / 0.8 d = 1.07/0.204 c = ( 2 − d ) /0.9 b = ( 6 − 1.7 c − 2 d ) / ( − 3 ) a = ( 2 − c ) /0.8
[ a b c d ] = [ 7.0 − 0.55 − 3.6 5.24 ] \begin{bmatrix}
a \\
b \\
c \\
d
\end{bmatrix} =
\begin{bmatrix}
7.0 \\
-0.55 \\
-3.6 \\
5.24
\end{bmatrix} ⎣ ⎡ a b c d ⎦ ⎤ = ⎣ ⎡ 7.0 − 0.55 − 3.6 5.24 ⎦ ⎤
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