x 1 C 3 H 8 + x 2 O 2 → x 3 C O 2 + x 4 H 2 O x_1C_3H_8+x_2O_2\rightarrow x_3CO_2+x_4H_2O x1C3H8+x2O2→x3CO2+x4H2O
四种物质的成分的列向量 C 3 H 8 : [ 3 8 0 ] O 2 : [ 0 0 2 ] C O 2 : [ 1 0 2 ] H 2 O : [ 0 2 1 ] C_3H_8:\begin{bmatrix}3 \\ 8 \\ 0\end{bmatrix}O_2:\begin{bmatrix}0 \\ 0\\ 2\end{bmatrix}CO_2:\begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix}H_2O:\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix} C3H8:⎣⎡380⎦⎤O2:⎣⎡002⎦⎤CO2:⎣⎡102⎦⎤H2O:⎣⎡021⎦⎤配平方程为 x 1 [ 3 8 0 ] + x 2 [ 0 0 2 ] = x 3 [ 1 0 2 ] + x 4 [ 0 2 1 ] x_1\begin{bmatrix}3 \\ 8 \\ 0\end{bmatrix}+x_2\begin{bmatrix}0 \\ 0\\ 2\end{bmatrix}=x_3\begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix}+x_4\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix} x1⎣⎡380⎦⎤+x2⎣⎡002⎦⎤=x3⎣⎡102⎦⎤+x4⎣⎡021⎦⎤移项得到 A = [ 3 0 − 1 0 8 0 0 − 2 0 2 − 2 − 1 ] b = [ 0 0 0 ] A=\begin{bmatrix}3&0&-1&0 \\ 8&0&0&-2 \\ 0&2&-2&-1\end{bmatrix}b=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix} A=⎣⎡380002−10−20−2−1⎦⎤b=⎣⎡000⎦⎤利用matlab求解rref(A) 令 x 4 = 4 x_4=4 x4=4，得 { x 1 = 1 x 2 = 5 x 3 = 3 x 4 = 4 \left\{\begin{matrix} x_1 = 1\\ x_2=5\\ x_3=3\\ x_4=4\end{matrix}\right. ⎩⎪⎪⎨⎪⎪⎧x1=1x2=5x3=3x4=4 C 3 H 8 + 5 O 2 → 3 C O 2 + 4 H 2 O C_3H_8+5O_2\rightarrow 3CO_2+4H_2O C3H8+5O2→3CO2+4H2O