359 lines
5.9 KiB
Markdown
359 lines
5.9 KiB
Markdown
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---
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title: "《线性代数》矩阵乘法与线性变换复合"
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date: 2023-08-06T15:02:43+08:00
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---
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## 矩阵与复合线性变换的关系
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复合线性变换:完成一次变换后再次进行变换,如先旋转再剪切。与一次变换相同,也可通过追踪$\hat{i}$和$\hat{j}$来确定变换后的向量变换
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新矩阵表示了一个单独的作用来完成复合线性变换
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对于一个先旋转后剪切的线性变换,可以用以下方式来进行计算
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选左乘旋转矩阵再左乘剪切矩阵,数值上表示对一个给定向量进行旋转然后剪切
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$$
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\begin{bmatrix}
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1 & 1 \\\\
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0 & 1
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\end{bmatrix}
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(\begin{bmatrix}
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0 & -1 \\\\
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1 & 0
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\end{bmatrix}
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\begin{bmatrix}
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x \\\\
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y
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\end{bmatrix}) =
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\begin{bmatrix}
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1 & -1 \\\\
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1 & 0
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\end{bmatrix}
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\begin{bmatrix}
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x \\\\
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y
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\end{bmatrix}
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$$
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由此可得,对于下列矩阵,需要从右向左读,即先应用右侧矩阵描述的变换再应用左侧矩阵描述的变换
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$$
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\begin{bmatrix}
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1 & 1 \\\\
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0 & 1
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\end{bmatrix}
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\begin{bmatrix}
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0 & -1 \\\\
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1 & 0
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\end{bmatrix} =
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\begin{bmatrix}
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1 & -1 \\\\
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1 & 0
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\end{bmatrix}
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$$
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**两个矩阵相乘有着几何意义,即两个线性变换相继作用**
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## 矩阵相乘计算流程
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$$
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\begin{bmatrix}
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a & b \\\\
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c & d
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\end{bmatrix}
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\begin{bmatrix}
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e & f \\\\
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g & h
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\end{bmatrix} =
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\begin{bmatrix}
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? & ? \\\\
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? & ?
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\end{bmatrix}
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$$
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首先,要得知$\hat{i}$的终点可由第二个矩阵的第一列得知,因此
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$$
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\begin{bmatrix}
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a & b \\\\
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c & d
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\end{bmatrix}
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\begin{bmatrix}
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e \\\\
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g
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\end{bmatrix} =
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e
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\begin{bmatrix}
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a \\\\
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c
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\end{bmatrix} +
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g
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\begin{bmatrix}
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b \\\\
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d
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\end{bmatrix} =
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\begin{bmatrix}
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ae + bg \\\\
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ce + dg
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\end{bmatrix}
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$$
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其次,$\hat{j}$终点在右侧矩阵第二列所表示的位置上
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$$
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\begin{bmatrix}
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a & b \\\\
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c & d
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\end{bmatrix}
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\begin{bmatrix}
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f \\\\
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h
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\end{bmatrix} =
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f
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\begin{bmatrix}
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a \\\\
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c
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\end{bmatrix} +
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h
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\begin{bmatrix}
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b \\\\
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d
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\end{bmatrix} =
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\begin{bmatrix}
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af + bh \\\\
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cf + dh
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\end{bmatrix}
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$$
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可得最终结果为
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$$
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\begin{bmatrix}
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a & b \\\\
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c & d
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\end{bmatrix}
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\begin{bmatrix}
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e & f \\\\
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g & h
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\end{bmatrix} =
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\begin{bmatrix}
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ae + bg & af + bh \\\\
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ce + dg & cd + dh
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\end{bmatrix}
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$$
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### 矩阵相乘顺序
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$$
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M_1M_2 \not ={M_2M_1}
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$$
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矩阵相乘结果受顺序影响
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可由相乘本质是进行多次线性变换得知,改变变换的顺序会导致不同的结果
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![](../../images/%E6%95%B0%E5%AD%A6/%E3%80%8A%E7%BA%BF%E6%80%A7%E4%BB%A3%E6%95%B0%E3%80%8B%E7%9F%A9%E9%98%B5%E4%B9%98%E6%B3%95%E4%B8%8E%E7%BA%BF%E6%80%A7%E5%8F%98%E6%8D%A2%E5%A4%8D%E5%90%88/%E7%9F%A9%E9%98%B5%E7%9B%B8%E4%B9%98%E9%A1%BA%E5%BA%8F%E5%9B%BE%E5%BD%A2%E8%A1%A8%E7%A4%BA.gif)
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对于结合律,本质上没有改变变换的顺序,因而不会导致结果不同
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$$
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A(BC) = (AB)C
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$$
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![](../../images/%E6%95%B0%E5%AD%A6/%E3%80%8A%E7%BA%BF%E6%80%A7%E4%BB%A3%E6%95%B0%E3%80%8B%E7%9F%A9%E9%98%B5%E4%B9%98%E6%B3%95%E4%B8%8E%E7%BA%BF%E6%80%A7%E5%8F%98%E6%8D%A2%E5%A4%8D%E5%90%88/%E7%BB%93%E5%90%88%E5%BE%8B%E5%8F%98%E6%8D%A2%E8%AF%81%E6%98%8E.gif)
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## 三维空间下的线性变换
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三维空间下的线性变换可由二维拓展,都可由基向量表示所有的向量
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三维下需要引入三个基向量,X轴的$\hat{i}$,Y轴的$\hat{j}$,Z轴的$\hat{k}$
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需要得知变换后的向量位置只需要将坐标与矩阵的对应列相乘
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$$
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\begin{bmatrix}
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0 & 1 & 2 \\\\
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3 & 4 & 5 \\\\
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6 & 7 & 8
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\end{bmatrix}
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\begin{bmatrix}
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x \\\\
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y \\\\
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z
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\end{bmatrix} =
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x
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\begin{bmatrix}
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0 \\\\
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3 \\\\
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6
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\end{bmatrix}+
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y
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\begin{bmatrix}
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1 \\\\
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4 \\\\
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7
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\end{bmatrix} +
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z
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\begin{bmatrix}
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2 \\\\
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5 \\\\
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8
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\end{bmatrix} =
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\begin{bmatrix}
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0+y+2z \\\\
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3x+4y+5z \\\\
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6x+7y+8z
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\end{bmatrix}
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$$
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对于两个矩阵相乘也是类似的,第二个矩阵的三个列分别对应三个基向量的位置
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$$
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\begin{bmatrix}
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0 & -2 & 2 \\\\
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5 & 1 & 5 \\\\
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1 & 4 & -1
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\end{bmatrix}
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\begin{bmatrix}
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0 & 1 & 2 \\\\
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3 & 4 & 5 \\\\
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6 & 7 & 8
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\end{bmatrix}
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$$
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$\hat{i}$的终点可由第二个矩阵的第一列得知
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$$
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\begin{bmatrix}
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0 & -2 & 2 \\\\
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5 & 1 & 5 \\\\
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1 & 4 & -1
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\end{bmatrix}
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\begin{bmatrix}
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0 \\\\
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3 \\\\
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6
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\end{bmatrix} =
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0
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\begin{bmatrix}
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0 \\\\
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5 \\\\
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1
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\end{bmatrix} +
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3
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\begin{bmatrix}
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-2 \\\\
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1 \\\\
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4
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\end{bmatrix} +
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6
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\begin{bmatrix}
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2 \\\\
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5 \\\\
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-1
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\end{bmatrix} =
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\begin{bmatrix}
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0-6+12 \\\\
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0+3+30 \\\\
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0+12-6
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\end{bmatrix}
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$$
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$\hat{j}$的终点可由第二个矩阵的第二列得知
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$$
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\begin{bmatrix}
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0 & -2 & 2 \\\\
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5 & 1 & 5 \\\\
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1 & 4 & -1
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\end{bmatrix}
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\begin{bmatrix}
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1 \\\\
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4 \\\\
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7
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\end{bmatrix} =
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1
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\begin{bmatrix}
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0 \\\\
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5 \\\\
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1
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\end{bmatrix} +
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4
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\begin{bmatrix}
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-2 \\\\
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1 \\\\
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4
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\end{bmatrix} +
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7
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\begin{bmatrix}
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2 \\\\
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5 \\\\
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-1
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\end{bmatrix} =
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\begin{bmatrix}
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0-8+14 \\\\
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5+4+35 \\\\
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1+16-7
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\end{bmatrix}
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$$
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$\hat{k}$的终点可由第二个矩阵的第三列得知
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$$
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\begin{bmatrix}
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0 & -2 & 2 \\\\
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5 & 1 & 5 \\\\
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1 & 4 & -1
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\end{bmatrix}
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\begin{bmatrix}
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2 \\\\
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5 \\\\
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8
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\end{bmatrix} =
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2
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\begin{bmatrix}
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0 \\\\
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5 \\\\
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1
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\end{bmatrix} +
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5
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\begin{bmatrix}
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-2 \\\\
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1 \\\\
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4
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\end{bmatrix} +
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8
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\begin{bmatrix}
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2 \\\\
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5 \\\\
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-1
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\end{bmatrix} =
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\begin{bmatrix}
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0-10+16 \\\\
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10+5+40 \\\\
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2+20-8
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\end{bmatrix}
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$$
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所以最终结果为
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$$
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\begin{bmatrix}
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0 & -2 & 2 \\\\
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5 & 1 & 5 \\\\
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1 & 4 & -1
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\end{bmatrix}
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\begin{bmatrix}
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0 & 1 & 2 \\\\
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3 & 4 & 5 \\\\
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6 & 7 & 8
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\end{bmatrix} =
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\begin{bmatrix}
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0-6+12 & 0-8+14 & 0-10+16 \\\\
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0+3+30 & 5+4+35 & 10+5+40 \\\\
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0+12-6 & 1+16-7 & 2+20-8
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\end{bmatrix}
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$$
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