- 方程组如下:
$$
\begin{cases}
a_{00}*x+a_{01}*y+a_{01}*z=b_0 \\
a_{10}*x+a_{11}*y+a_{11}*z=b_0 \\
a_{20}*x+a_{21}*y+a_{21}*z=b_0 \\
\end{cases}
$$ - 矩阵形式为:
A =
\begin{vmatrix}
a_{00}&a_{01}&a_{02}\\
a_{10}&a_{11}&a_{12}\\
a_{20}&a_{21}&a_{22}\\
\end{vmatrix}
X=
\begin{vmatrix}
x\\
y\\
z\\
\end{vmatrix}
B=
\begin{vmatrix}
b_0\\
b_1\\
b_2\\
\end{vmatrix} 计算det值
det(A)=$$a_{00}*a_{11}*a_{22}+a_{01}*a_{12}*a_{20}+a_{02}*a_{10}*a_{21}-a_{02}*a_{11}*a_{20}-a_{01}*a_{10}*a_{22}-a_{00}*a_{12}*a{21}$$
基于余因子扩展计算矩阵行列式的方法,结果为:
det(A)=$$a_{00}*(a_{11}*a_{22}-a_{12}*a_{21})-a_{01}*(a_{10}*a_{22}-a_{12}*a_{20})+a_{02}*(a_{10}*a_{21}-a_{11}*a_{20})$$
(提出某一行的所有元素。元素分别乘以元素不在的行和列组成的矩阵的行列式,加减号由元素所在的行列数和决定,奇数减,偶数加。这的行列从0开始计数)克莱姆法表示为一组求解x,y,z的商:
对于每个变量(x,y,z),都有一个用于求解它的商,其中分子分别是用列向量B替换矩阵A的第一列、第二列和第三列得到的矩阵的行列式,分母都为det(A):
令:det(tx)=
$$\begin{vmatrix}
b_0&a_{01}&a_{02}\\
b_1&a_{11}&a_{12}\\
b_2&a_{21}&a_{22}\\
\end{vmatrix}$$
则:
x=det(tx)/det(A)
令:det(ty)=
$$\begin{vmatrix}
a_{00}&b_{0}&a_{02}\\
a_{10}&b_{1}&a_{12}\\
a_{20}&b_{2}&a_{22}\\
\end{vmatrix}$$
则:
y=det(ty)/det(A)
令:det(tz)=
$$\begin{vmatrix}
a_{00}&a_{01}&b_{0}\\
a_{10}&a_{11}&b_{1}\\
a_{20}&a_{21}&b_{2}\\
\end{vmatrix}$$
则:
z=det(tz)/det(A)