一 引例
求解二元一次方程組
{ a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 \begin{cases} a_{11}x_1+a_{12}x_2=b_1\\ a_{21}x_1+a_{22}x_2=b_2\\ \end{cases} {a11?x1?+a12?x2?=b1?a21?x1?+a22?x2?=b2??
解: 1 × a 21 ? 2 × a 11 ? x 2 = a 11 b 2 ? a 21 b 1 a 11 a 22 ? a 12 a 21 x 1 = a 22 b 1 ? a 12 b 2 a 11 a 22 ? a 12 a 21 解:\\ 1\times a_{21}-2\times a_{11}\Rightarrow\\ x_2=\frac{a_{11}b_2-a_{21}b_1}{a_{11}a_{22}-a_{12}a_{21}}\\ x_1=\frac{a_{22}b_1-a_{12}b_2}{a_{11}a_{22}-a_{12}a_{21}}\\ 解:1×a21??2×a11??x2?=a11?a22??a12?a21?a11?b2??a21?b1??x1?=a11?a22??a12?a21?a22?b1??a12?b2??
a 11 a 12 a 21 a 22 \begin{matrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{matrix} a11?a21??a12?a22?? 兩行兩列數表。
定義:表達式 a 11 a 22 ? a 12 a 21 = Δ ∣ a 11 a 12 a 21 a 22 ∣ a_{11}a_{22}-a_{12}a_{21} \overset{\Delta}{=} \begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\\end{vmatrix} a11?a22??a12?a21?=Δ ?a11?a21??a12?a22?? ?為數表所確定的二階行列式。
注:
- ∣ a 11 a 12 a 21 a 22 ∣ \begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\\end{vmatrix} ?a11?a21??a12?a22?? ?, a i j , i = 1 , 2 , j = 1 , 2 a_{ij},i=1,2,j=1,2 aij?,i=1,2,j=1,2為行列式的元素。i為元素所在的行,j位元素所在的列。
二 計算
a 11 a 22 ? a 12 a 21 = ∣ a 11 a 12 a 21 a 22 ∣ = D a_{11}a_{22}-a_{12}a_{21}= \begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\\ \end{vmatrix} =D a11?a22??a12?a21?= ?a11?a21??a12?a22?? ?=D
b 1 a 22 ? b 2 a 12 = ∣ b 1 a 12 b 2 a 22 ∣ = D 1 b_1a_{22}-b_2a_{12}= \begin{vmatrix} b_1&a_{12}\\ b_2&a_{22}\\ \end{vmatrix} =D_1 b1?a22??b2?a12?= ?b1?b2??a12?a22?? ?=D1?
a 21 b 1 ? a 11 b 2 = ∣ a 21 a 11 b 2 b 1 ∣ = D 2 a_{21}b_1-a_{11}b_2= \begin{vmatrix} a_{21}&a_{11}\\ b_2&b_1\\ \end{vmatrix} =D_2 a21?b1??a11?b2?= ?a21?b2??a11?b1?? ?=D2?
一種二元一次方程組的解為 x 1 = D 1 D , x 2 = D 2 D x_1=\frac{D_1}{D},x_2=\frac{D_2}{D} x1?=DD1??,x2?=DD2??
例 { 3 x 1 ? 2 x 2 = 12 2 x 1 + x 2 = 1 \begin{cases}3x_1-2x_2=12\\2x_1+x_2=1\end{cases} {3x1??2x2?=122x1?+x2?=1?解 x 1 , x 2 x_1,x_2 x1?,x2?
x 1 = 12 ? ( ? 2 ) 3 ? ( ? 4 ) = 2 x 2 = 3 ? 24 7 = ? 3 x_1=\frac{12-(-2)}{3-(-4)}=2\\ x_2=\frac{3-24}{7}=-3 x1?=3?(?4)12?(?2)?=2x2?=73?24?=?3
三 二階行列式的幾何意義
a 11 a 22 ? a 12 a 21 = ∣ a 11 a 12 a 21 a 22 ∣ 令 ∣ a ? ∣ = l , ∣ b ? ∣ = m , 則平行四邊形的面積: S = a ? × b ? = l ? m ? sin ? ( β ? α ) = l ? m ( sin ? β cos ? α ? cos ? β sin ? α ) = l cos ? α ? m sin ? β ? l sin ? α ? m cos ? β = a 11 a 22 ? a 12 a 21 a_{11}a_{22}-a_{12}a_{21}= \begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\\ \end{vmatrix}\\ 令\vert\vec a\vert=l,\vert\vec b\vert=m,則 平行四邊形的面積:\\ S=\vec a\times\vec b=l\cdot m\cdot\sin(\beta-\alpha)\\ =l\cdot m(\sin\beta\cos\alpha-\cos\beta\sin\alpha)\\ =l\cos\alpha\cdot m\sin\beta-l\sin\alpha\cdot m\cos\beta\\ =a_{11}a_{22}-a_{12}a_{21} a11?a22??a12?a21?= ?a11?a21??a12?a22?? ?令∣a∣=l,∣b∣=m,則平行四邊形的面積:S=a×b=l?m?sin(β?α)=l?m(sinβcosα?cosβsinα)=lcosα?msinβ?lsinα?mcosβ=a11?a22??a12?a21?
上述二階行列式幾何意義:以行向量(第一行為向量和以第二行為向量)為鄰邊所構成平行四邊形的面積。
三階行列式放在后面N階行列式講解。
結語
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參考:
[1]同濟六版《線性代數》全程教學視頻[CP/OL].2020-02-07.p2.