## 答延伸寸──關於線性方程的通解表達

http://www.lic.nkfust.edu.tw/ezfiles/5/1005/img/791/982131.pdf

$\begin{bmatrix} 5\\ 0\\ 0 \end{bmatrix}+s\begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix}+t\left[\!\!\begin{array}{r} -3\\ 0\\ 1 \end{array}\!\!\right]$

$\begin{bmatrix} \alpha\\ 5\\ 4 \end{bmatrix}+s\left[\!\!\begin{array}{r} 1\\ \beta\\ -3 \end{array}\!\!\right]+t\begin{bmatrix} 1\\ 2\\ \gamma \end{bmatrix}$

$\mathbf{y}_p=\mathbf{x}_p+s\mathbf{x}_1+t\mathbf{x}_2$

$\mathbf{y}_p-\mathbf{x}_p=s\mathbf{x}_1+t\mathbf{x}_2$

$N(A)=\mathrm{span}\{\mathbf{x}_1,\mathbf{x}_2\}=\mathrm{span}\{\mathbf{y}_1,\mathbf{y}_2\}$

\begin{aligned} \mathbf{y}_1&={s}'\mathbf{x}_1+{t}'\mathbf{x}_2\\ \mathbf{y}_2&={s}''\mathbf{x}_1+{t}''\mathbf{x}_2. \end{aligned}

\begin{aligned} &\left[\!\!\begin{array}{cr} 2&-3\\ 1&0\\ 0&1 \end{array}\!\!\right]\begin{bmatrix} s\\ t \end{bmatrix}=\begin{bmatrix} \alpha-5\\ 5\\ 4 \end{bmatrix}\\ &\left[\!\!\begin{array}{cr} 2&-3\\ 1&0\\ 0&1 \end{array}\!\!\right]\begin{bmatrix} {s}'\\ {t}' \end{bmatrix}=\left[\!\!\begin{array}{r} 1\\ \beta\\ -3 \end{array}\!\!\right]\\ &\left[\!\!\begin{array}{cr} 2&-3\\ 1&0\\ 0&1 \end{array}\!\!\right]\begin{bmatrix} {s}''\\ {t}'' \end{bmatrix}=\begin{bmatrix} 1\\ 2\\ \gamma \end{bmatrix}.\end{aligned}

\begin{aligned} \left[\!\!\begin{array}{crccrc} 2&-3&\vline&\alpha-5&1&1\\ 1&0&\vline&5&\beta&2\\ 0&1&\vline&4&-3&\gamma \end{array}\!\!\right]&\to\left[\!\!\begin{array}{crccrc} 1&0&\vline&5&\beta&2\\ 0&1&\vline&4&-3&\gamma\\ 2&-3&\vline&\alpha-5&1&1 \end{array}\!\!\right]\\ &\to\left[\!\!\begin{array}{cccccc} 1&0&\vline&5&\beta&2\\ 0&1&\vline&4&-3&\gamma\\ 0&0&\vline&\alpha-3&-2\beta-8&3\gamma-3 \end{array}\!\!\right]\end{aligned}

$\left\{\begin{array}{rc} \alpha-3\!&=0 \\ -2\beta-8\!&=0\\ 3\gamma-3\!&=0 \end{array}\right.~~~\Rightarrow~~~\left\{\begin{array}{cl} \alpha\!&=3\\ \beta\!&=-4\\ \gamma\!&=1. \end{array}\right.$

$\mathcal{S}=\mathcal{W}+\mathbf{p}\overset{\underset{\mathrm{def}}{}}{=}\{\mathbf{w}+\mathbf{p}\vert\mathbf{w}\in\mathcal{W}\}$

\begin{aligned} \mathbf{y}_p-\mathbf{x}_p&\in\mathrm{span}\{\mathbf{x}_1,\mathbf{x}_2\}\\ \mathbf{y}_1&\in\mathrm{span}\{\mathbf{x}_1,\mathbf{x}_2\}\\ \mathbf{y}_2&\in\mathrm{span}\{\mathbf{x}_1,\mathbf{x}_2\}. \end{aligned}

\begin{aligned} &(\mathbf{y}_p-\mathbf{x}_p)^T\mathbf{n}=\begin{bmatrix} \alpha-5 &5&4 \end{bmatrix}\left[\!\!\begin{array}{r} 1\\ -2\\ 3 \end{array}\!\!\right]=\alpha-5-10+12=\alpha-3=0\\ &\mathbf{y}_1^T\mathbf{n}=\begin{bmatrix} 1&\beta&-3 \end{bmatrix}\left[\!\!\begin{array}{r} 1\\ -2\\ 3 \end{array}\!\!\right]=1-2\beta-9=-2\beta-8=0\\ &\mathbf{y}_2^T\mathbf{n}=\begin{bmatrix} 1&2&\gamma \end{bmatrix}\left[\!\!\begin{array}{r} 1\\ -2\\ 3 \end{array}\!\!\right]=1-4+3\gamma=3\gamma-3=0.\end{aligned}

### One Response to 答延伸寸──關於線性方程的通解表達

1. 延伸寸 說道：

哇，大師出手，就是不一樣！ 本題所應用的觀念很有價值。一般書本及授課講師著墨不多。