Quod Erat Demonstrandum

2016/04/29

證 Cramer’s rule

Filed under: NSS,Pure Mathematics — johnmayhk @ 12:04 上午

聽聞代課想向 M2 學生證明 Cramer’s rule,見他手執 n by n 的線性代數材料,並稱內容頗抽象云云,我沒說甚麼。

其實教純數時也略談,技巧不過是在不同的等式,刻意乘上 cofactors 再相加,以致產生 coefficient matrix 的 determinant 云云。

但對 M2 學生來說,估計不是太易,這裡試用 inverse matrix 來解釋,希望明啦~

因為是 M2,考慮 3 by 3 方程組已足夠,見下

\left \{ \begin{array}{ll} a_1x+b_1y+c_1z=d_1\\a_2x+b_2y+c_2z=d_2\\a_3x+b_3y+c_3z=d_3\end{array}\right.

設所謂 coefficient matrix 為

M=\left(\begin{array}{rcl}a_1& b_1& c_1\\a_2& b_2& c_2\\a_3& b_3& c_3\\\end{array}\right)

如果 M^{-1} 存在,那麼

\left(\begin{array}{rcl}x\\y\\z\\\end{array}\right)=M^{-1}\left(\begin{array}{rcl}d_1\\d_2\\d_3\\\end{array}\right) ………. (1)

另外,由於

M^{-1}\left(\begin{array}{rcl}a_1& b_1& c_1\\a_2& b_2& c_2\\a_3& b_3& c_3\\\end{array}\right)=\left(\begin{array}{rcl}1& 0& 0\\0& 1& 0\\0& 0& 1\\\end{array}\right)

所以有

M^{-1}\left(\begin{array}{rcl}b_1\\b_2\\b_3\\\end{array}\right)=\left(\begin{array}{rcl}0\\1\\0\\\end{array}\right) ………. (2)

M^{-1}\left(\begin{array}{rcl}c_1\\c_2\\c_3\\\end{array}\right)=\left(\begin{array}{rcl}0\\0\\1\\\end{array}\right) ………. (3)

由 (1), (2) 及 (3),得

M^{-1}\left(\begin{array}{rcl}d_1& b_1& c_1\\d_2& b_2& c_2\\d_3& b_3& c_3\\\end{array}\right)=\left(\begin{array}{rcl}x& 0& 0\\y& 1& 0\\z& 0& 1\\\end{array}\right)

找其行列式,即

\det(M^{-1})\det\left(\begin{array}{rcl}d_1& b_1& c_1\\d_2& b_2& c_2\\d_3& b_3& c_3\\\end{array}\right)=\det\left(\begin{array}{rcl}x& 0& 0\\y& 1& 0\\z& 0& 1\\\end{array}\right)

故此

x=\displaystyle\frac{\det\left(\begin{array}{rcl}d_1& b_1& c_1\\d_2& b_2& c_2\\d_3& b_3& c_3\\\end{array}\right)}{\det(M)}=\frac{\Delta_x}{\Delta}

同理,得

y=\frac{\Delta_y}{\Delta}z=\frac{\Delta_z}{\Delta}

同學,自行證明吧。

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