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Jan 15, 2023
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Fix math dcases => \left\{ #179

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230 changes: 117 additions & 113 deletions docs/02algorithms/10gaussian-elimination/index.mdx
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Original file line number Diff line number Diff line change
Expand Up @@ -19,13 +19,13 @@ $$
$$
\gdef\customtextcircled#1{\textcircled{\small#1}}
\gdef\formulanumber#1{\,\cdot\cdot\cdot\cdot\cdot\cdot\,\customtextcircled{#1}}
\begin{dcases}
\begin{alignat*}{3.5}
& x_1 - {} & 2 & x_2 + {} & 3 & x_3 = 3 & \formulanumber{1} \\
-& x_1 + {} & 3 & x_2 - {} & 2 & x_3 = 1 & \formulanumber{2} \\
& x_1 - {} & & x_2 + {} & 6 & x_3 = 11 & \formulanumber{3}
\end{alignat*}
\end{dcases}
\left\{
\begin{alignat*}{3.5}
& x_1 - {} & 2 & x_2 + {} & 3 & x_3 = 3 & \formulanumber{1} \\
-& x_1 + {} & 3 & x_2 - {} & 2 & x_3 = 1 & \formulanumber{2} \\
& x_1 - {} & & x_2 + {} & 6 & x_3 = 11 & \formulanumber{3}
\end{alignat*}
\right.
$$

$
Expand All @@ -49,7 +49,7 @@ $$
\gdef\formulanumber#1{\,\cdot\cdot\cdot\cdot\cdot\cdot\,\customtextcircled{#1}}
\begin{align*}
2x_2 + 4x_3 &= 12 \\
\therefore x_2 + 2x_3 &= 6 \formulanumber{5}
\therefore x_2 + 2x_3 &= 6 \formulanumber{5}
\end{align*}
$$

Expand All @@ -59,20 +59,24 @@ $
\customtextcircled{4},\customtextcircled{5}$ より、

$$
\begin{dcases}
x_2 = 2 \\
x_3 = 2
\end{dcases}
\left\{
\begin{align*}
x_2 &= 2 \\
x_3 &= 2
\end{align*}
\right.
$$

よって、

$$
\begin{dcases}
x_1 = 1 \\
x_2 = 2 \\
x_3 = 2
\end{dcases}
\left\{
\begin{align*}
x_1 &= 1 \\
x_2 &= 2 \\
x_3 &= 2
\end{align*}
\right.
$$

しかし、このようなアルゴリズムでプログラムを作るのは、難しそうです。
Expand All @@ -88,15 +92,15 @@ Gauss の消去法は、前進消去と後退代入の二段階から成りま
次のように $n$ 個の未知数 $x_1, x_2, x_3, \dots , x_n$ に対して、$m$ 個の方程式を考えます。

$$
\begin{dcases}
\begin{alignat*}{5}
a_{1, 1} & x_1 + {} & a_{1, 2} & x_2 + {} & a_{1, 3} & x_3 + \dots + {} & a_{1, n} & x_n = c_1 \\
a_{2, 1} & x_1 + {} & a_{2, 2} & x_2 + {} & a_{2, 3} & x_3 + \dots + {} & a_{2, n} & x_n = c_2 \\
a_{3, 1} & x_1 + {} & a_{3, 2} & x_2 + {} & a_{3, 3} & x_3 + \dots + {} & a_{3, n} & x_n = c_3 \\
& \cdots\cdot\cdot \\
a_{m, 1} & x_1 + {} & a_{m, 2} & x_2 + {} & a_{m, 3} & x_3 + \dots + {} & a_{m, n} & x_n = c_m
\end{alignat*}
\end{dcases}
\left\{
\begin{alignat*}{5}
a_{1, 1} & x_1 + {} & a_{1, 2} & x_2 + {} & a_{1, 3} & x_3 + \dots + {} & a_{1, n} & x_n = c_1 \\
a_{2, 1} & x_1 + {} & a_{2, 2} & x_2 + {} & a_{2, 3} & x_3 + \dots + {} & a_{2, n} & x_n = c_2 \\
a_{3, 1} & x_1 + {} & a_{3, 2} & x_2 + {} & a_{3, 3} & x_3 + \dots + {} & a_{3, n} & x_n = c_3 \\
& \cdots\cdot\cdot \\
a_{m, 1} & x_1 + {} & a_{m, 2} & x_2 + {} & a_{m, 3} & x_3 + \dots + {} & a_{m, n} & x_n = c_m
\end{alignat*}
\right.
$$

この方程式系に対して、以下を行ったものは元の方程式系と同値です。
Expand Down Expand Up @@ -188,18 +192,18 @@ $$
これから、作られる方程式系は次のようになりこれははじめの方程式系と同値です。

$$
\begin{dcases}
\begin{alignat*}{4}
& x_{j_1} + \dots + {} & b_{1, j_2} & x_{j_2} + \dots + {} & b_{1, j_3} & x_{j_3} + \dots + {} & b_{1, n} & x_n = d_1 \\
0 & x_{j_1} + \dots + {} & & x_{j_2} + \dots + {} & b_{2, j_3} & x_{j_3} + \dots + {} & b_{2, n} & x_n = d_2 \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & & x_{j_3} + \dots + {} & b_{3, n} & x_n = d_3 \\
& \dots \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & & x_n = d_l \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_{l + 1} \\
& \dots \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_m \\
\end{alignat*}
\end{dcases}
\left\{
\begin{alignat*}{4}
& x_{j_1} + \dots + {} & b_{1, j_2} & x_{j_2} + \dots + {} & b_{1, j_3} & x_{j_3} + \dots + {} & b_{1, n} & x_n = d_1 \\
0 & x_{j_1} + \dots + {} & & x_{j_2} + \dots + {} & b_{2, j_3} & x_{j_3} + \dots + {} & b_{2, n} & x_n = d_2 \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & & x_{j_3} + \dots + {} & b_{3, n} & x_n = d_3 \\
& \dots \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & & x_n = d_l \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_{l + 1} \\
& \dots \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_m \\
\end{alignat*}
\right.
$$

これで $x_n$ から順番に求めていくことで連立方程式を解くことができます。
Expand All @@ -208,13 +212,13 @@ $$
Gauss の消去法を次の方程式系について行うと、以下のようになります。

$$
\begin{dcases}
\begin{alignat*}{3.5}
& x_1 - {} & 2 & x_2 + {} & 3 & x_3 = 3 \\
-& x_1 + {} & 3 & x_2 - {} & 2 & x_3 = 1 \\
& x_1 - {} & & x_2 + {} & 6 & x_3 = 11
\end{alignat*}
\end{dcases}
\left\{
\begin{alignat*}{3.5}
& x_1 - {} & 2 & x_2 + {} & 3 & x_3 = 3 \\
-& x_1 + {} & 3 & x_2 - {} & 2 & x_3 = 1 \\
& x_1 - {} & & x_2 + {} & 6 & x_3 = 11
\end{alignat*}
\right.
$$

まずは、前進消去を行います。
Expand Down Expand Up @@ -268,37 +272,37 @@ $$
次に、連立方程式に戻して後退代入を行っていきます。

$$
\begin{dcases}
\begin{alignat*}{3}
x_1 - 2 & x_2 + {} & 3 & x_3 & {} = {} & 3 \\
& x_2 + {} & & x_3 & {} = {} & 4 \\
& & & x_3 & {} = {} & 2
\end{alignat*}
\end{dcases} \\
\therefore
\begin{dcases}
\begin{alignat*}{3}
x_1 - 2 & x_2 & & & {} = {} & 3 - 3\times 2 = -3 \\
& x_2 & & & {} = {} & 4 - 2 = 2 \\
& & & x_3 & {} = {} & 2
\end{alignat*}
\end{dcases} \\
\left\{
\begin{alignat*}{3}
x_1 - 2 & x_2 + {} & 3 & x_3 & {} = {} & 3 \\
& x_2 + {} & & x_3 & {} = {} & 4 \\
& & & x_3 & {} = {} & 2
\end{alignat*}
\right. \\
\therefore
\begin{dcases}
\begin{alignat*}{3}
x_1 & & & & {} = {} & -3 + 2\times 2 = 1 \\
& x_2 & & & {} = {} & 2 \\
\left\{
\begin{alignat*}{3}
x_1 - 2 & x_2 & & & {} = {} & 3 - 3\times 2 = -3 \\
& x_2 & & & {} = {} & 4 - 2 = 2 \\
& & & x_3 & {} = {} & 2
\end{alignat*}
\end{dcases} \\
\end{alignat*}
\right. \\
\therefore
\left\{
\begin{alignat*}{3}
x_1 & & & & {} = {} & -3 + 2\times 2 = 1 \\
& x_2 & & & {} = {} & 2 \\
& & & x_3 & {} = {} & 2
\end{alignat*}
\right. \\
\therefore
\begin{dcases}
\begin{align*}
x_1 &= 1\\
x_2 &= 2\\
x_3 &= 2
\end{align*}
\end{dcases}
\left\{
\begin{align*}
x_1 &= 1\\
x_2 &= 2\\
x_3 &= 2
\end{align*}
\right.
$$

Gauss の消去法を使えば、このようにシステマティックに連立方程式を解けます。
Expand Down Expand Up @@ -330,18 +334,18 @@ $$
これから、作られる方程式系は次のようになりこれは元の方程式系と同値です。

$$
\begin{dcases}
\begin{alignat*}{4}
& x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_1 \\
0 & x_{j_1} + \dots + {} & & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_2 \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & & x_{j_3} + \dots + {} & 0 & x_n = d_3 \\
& \dots \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & & x_n = d_l \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_{l + 1} \\
& \dots \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_m \\
\end{alignat*}
\end{dcases}
\left\{
\begin{alignat*}{4}
& x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_1 \\
0 & x_{j_1} + \dots + {} & & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_2 \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & & x_{j_3} + \dots + {} & 0 & x_n = d_3 \\
& \dots \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & & x_n = d_l \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_{l + 1} \\
& \dots \\
0 & x_{j_1} + \dots + {} & 0 & x_{j_2} + \dots + {} & 0 & x_{j_3} + \dots + {} & 0 & x_n = d_m \\
\end{alignat*}
\right.
$$

これで連立方程式を解くことができました。
Expand All @@ -350,13 +354,13 @@ $$
これを次の方程式系について行うと、以下のようになります。

$$
\begin{dcases}
\begin{alignat*}{3.5}
& x_1 - {} & 2 & x_2 + {} & 3 & x_3 = 3 \\
-& x_1 + {} & 3 & x_2 - {} & 2 & x_3 = 1 \\
& x_1 - {} & & x_2 + {} & 6 & x_3 = 11
\end{alignat*}
\end{dcases}
\left\{
\begin{alignat*}{3.5}
& x_1 - {} & 2 & x_2 + {} & 3 & x_3 = 3 \\
-& x_1 + {} & 3 & x_2 - {} & 2 & x_3 = 1 \\
& x_1 - {} & & x_2 + {} & 6 & x_3 = 11
\end{alignat*}
\right.
$$

$$
Expand Down Expand Up @@ -430,21 +434,21 @@ $$
$$

$$
\begin{dcases}
\begin{alignat*}{4}
& x_1 + {} & 0 & x_2 + {} & 0 & x_3 & & = 1 \\
0 & x_1 + {} & & x_2 + {} & 0 & x_3 & & = 2 \\
0 & x_1 + {} & 0 & x_2 + {} & & x_3 & & = 2
\end{alignat*}
\end{dcases} \\
\left\{
\begin{alignat*}{4}
& x_1 + {} & 0 & x_2 + {} & 0 & x_3 & & = 1 \\
0 & x_1 + {} & & x_2 + {} & 0 & x_3 & & = 2 \\
0 & x_1 + {} & 0 & x_2 + {} & & x_3 & & = 2
\end{alignat*}
\right. \\
\therefore
\begin{dcases}
\begin{align*}
x_1 &= 1 \\
x_2 &= 2 \\
x_3 &= 2
\end{align*}
\end{dcases}
\left\{
\begin{align*}
x_1 &= 1 \\
x_2 &= 2 \\
x_3 &= 2
\end{align*}
\right.
$$

:::
Expand Down Expand Up @@ -473,13 +477,13 @@ $x_n$ は $d_n$ になります。求まった $x_n$ をそれよりも上の式
次のような連立方程式を解こうとすると、次のようなエラーが出てしまいます。

$$
\begin{dcases}
\begin{alignat*}{3.5}
& & -2 & x_2 + {} & 3 & x_3 = 2 \\
-& x_1 + {} & 3 & x_2 - {} & 2 & x_3 = 1 \\
& x_1 - {} & & x_2 + {} & 6 & x_3 = 11
\end{alignat*}
\end{dcases}
\left\{
\begin{alignat*}{3.5}
& & -2 & x_2 + {} & 3 & x_3 = 2 \\
-& x_1 + {} & 3 & x_2 - {} & 2 & x_3 = 1 \\
& x_1 - {} & & x_2 + {} & 6 & x_3 = 11
\end{alignat*}
\right.
$$

<ViewSource path="/gaussian-elimination/gaussian_elimination_revised.ipynb" />
Expand Down