Markdown矩阵编辑

矩阵

$\begin{vmatrix}
x & y & z & 1\\
x_{1} & y_{1} & z_{1} & 1\\
x_{2} & y_{2} & z_{2} & 1\\
x_{3} & y_{3} & z_{3} & 1
\end{vmatrix}$=0

∣ x y z 1 x 1 y 1 z 1 1 x 2 y 2 z 2 1 x 3 y 3 z 3 1 ∣ \begin{vmatrix} x & y & z & 1\\ x_{1} & y_{1} & z_{1} & 1\\ x_{2} & y_{2} & z_{2} & 1\\ x_{3} & y_{3} & z_{3} & 1 \end{vmatrix} xx1x2x3yy1y2y3zz1z2z31111=0

矩阵方程

$$\left[\begin{matrix}
A_1 & B_1 & C_1\\
A_1 & B_1 & C_1\\
A_1 & B_1 & C_1
\end{matrix}\right]
\left[\begin{matrix}
x\\
y\\
z
\end{matrix}\right]
+
\left[\begin{matrix}
D_1\\
D_2\\
D_3
\end{matrix}\right]
=0
$$

[ A 1 B 1 C 1 A 1 B 1 C 1 A 1 B 1 C 1 ] [ x y z ] + [ D 1 D 2 D 3 ] = 0 \left[\begin{matrix} A_1 & B_1 & C_1\\ A_1 & B_1 & C_1\\ A_1 & B_1 & C_1 \end{matrix}\right] \left[\begin{matrix} x\\ y\\ z \end{matrix}\right] + \left[\begin{matrix} D_1\\ D_2\\ D_3 \end{matrix}\right] =0 A1A1A1B1B1B1C1C1C1xyz+D1D2D3=0

矩阵转置

$$\left[\begin{matrix}
x_0\\
y_0\\
z_0
\end{matrix}\right]=-
\left[\begin{matrix}
A_1 & B_1 & C_1\\
A_1 & B_1 & C_1\\
A_1 & B_1 & C_1
\end{matrix}\right]^\mathrm{-1}
\left[\begin{matrix}
D_1\\
D_2\\
D_3
\end{matrix}\right]
$$

[ x 0 y 0 z 0 ] = − [ A 1 B 1 C 1 A 1 B 1 C 1 A 1 B 1 C 1 ] − 1 [ D 1 D 2 D 3 ] \left[\begin{matrix} x_0\\ y_0\\ z_0 \end{matrix}\right]=- \left[\begin{matrix} A_1 & B_1 & C_1\\ A_1 & B_1 & C_1\\ A_1 & B_1 & C_1 \end{matrix}\right]^\mathrm{-1} \left[\begin{matrix} D_1\\ D_2\\ D_3 \end{matrix}\right] x0y0z0=A1A1A1B1B1B1C1C1C11D1D2D3

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转载自blog.csdn.net/qq_52671517/article/details/124578261