逆矩陣的列和

$\displaystyle (-1,3,-7)\times(-2,1,-5)=\left(\left|\!\!\begin{array}{cr} 3&-7\\ 1&-5 \end{array}\!\!\right|,\left|\!\!\begin{array}{cc} -7&-1\\ -5&-2 \end{array}\!\!\right|,\left|\!\!\begin{array}{cc} -1&3\\ -2&1 \end{array}\!\!\right|\right)=(-8,9,5)$

$\displaystyle A=\left[\!\!\begin{array}{crr} 0&3&-5\\ 1&0&2\\ 3&-1&7 \end{array}\!\!\right]$

$\displaystyle A^{-1}=\left[\!\!\begin{array}{rrr} 1&-8&3\\ -\frac{1}{2}&\frac{15}{2}&-\frac{5}{2}\\[0.3em] -\frac{1}{2}&\frac{9}{2}&-\frac{3}{2} \end{array}\!\!\right]$

$\displaystyle \left[\!\!\begin{array}{r} -4\\ \frac{9}{2}\\[0.3em] \frac{5}{2} \end{array}\!\!\right]$

$\displaystyle D=\left[\!\!\begin{array}{crr} 1&-1&0\\ 0&1&-1\\ 0&0&1 \end{array}\!\!\right]$

$\displaystyle D^{-1}=\begin{bmatrix} 1&1&1\\ 0&1&1\\ 0&0&1 \end{bmatrix}$

\displaystyle\begin{aligned} DA&=\begin{bmatrix} a_{11}-a_{21}&a_{12}-a_{22}&a_{13}-a_{23}\\ a_{21}-a_{31}&a_{22}-a_{32}&a_{23}-a_{33}\\ a_{31}&a_{32}&a_{33} \end{bmatrix}\\ BD^{-1}&=\begin{bmatrix} b_{11}&b_{11}+b_{12}&b_{11}+b_{12}+b_{13}\\ b_{21}&b_{21}+b_{22}&b_{21}+b_{22}+b_{23}\\ b_{31}&b_{31}+b_{32}&b_{31}+b_{32}+b_{33} \end{bmatrix}. \end{aligned}

$\displaystyle \begin{bmatrix} a_{11}-a_{21}&a_{12}-a_{22}&a_{13}-a_{23}\end{bmatrix} =\begin{bmatrix} a_{11}&a_{12}&a_{13} \end{bmatrix}-\begin{bmatrix} a_{21}&a_{22}&a_{23} \end{bmatrix}$

$\displaystyle \begin{bmatrix} a_{21}-a_{31}&a_{22}-a_{32}&a_{23}-a_{33}\end{bmatrix} =\begin{bmatrix} a_{21}&a_{22}&a_{23} \end{bmatrix}-\begin{bmatrix} a_{31}&a_{32}&a_{33} \end{bmatrix}$

[1] 維基百科：跳躍性思維。

\displaystyle\begin{aligned} 3b-5c+d&=0\\ a+2c+d&=0\\ 3a-b+7c+d&=0,\end{aligned}

$\displaystyle \left[\!\!\begin{array}{crr} 0&3&-5\\ 1&0&2\\ 3&-1&7 \end{array}\!\!\right]\begin{bmatrix} a\\ b\\ c\end{bmatrix}=-d\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$

$\displaystyle \begin{bmatrix} a\\ b\\ c\end{bmatrix}=(-d)\left[\!\!\begin{array}{crr} 0&3&-5\\ 1&0&2\\ 3&-1&7 \end{array}\!\!\right]^{-1}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$

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5 Responses to 逆矩陣的列和

1. yuyumagic424 says:

我想到的是
平面方程式$ax+by+cz=d$
三個點代入都合方程式，
所以寫出：
$\begin{bmatrix} ~0 & 3 & -5~\\ ~1 & 0 & 2~\\ ~3 & -1 & 7~ \end{bmatrix}\cdot \begin{bmatrix} ~a~\\ ~b~\\ ~c~ \end{bmatrix} =\begin{bmatrix} ~d~\\ ~d~\\ ~d~ \end{bmatrix}$
這便有
$\begin{bmatrix} ~a~\\ ~b~\\ ~c~ \end{bmatrix} = \begin{bmatrix} ~0 & 3 & -5~\\ ~1 & 0 & 2~\\ ~3 & -1 & 7~ \end{bmatrix}^{-1} \cdot \begin{bmatrix} ~d~\\ ~d~\\ ~d~ \end{bmatrix}$

• ccjou says:

謝謝分享，這個直覺推論比我原來寫的簡單多了。

我將你的回應編輯過以正確顯示，LaTeX語法是
\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}

• phenix says:

好妙啊，高数在学平面方程的时候根本就没往这方面想😅

2. suehang says:

哎,LaTeX,我永远的痛!

3. xiaokeshen says:

Thanks a lot.
後記 should be ax+by+cz instead of
ax+by+cy. It should be a typo.