笔记内容摘自 猴博士爱讲课@B站 https://www.bilibili.com/video/BV1hs411e7X8?p=4
行列式分为2阶、3阶、4阶……n阶等，其中2阶的计算方法为： ∣ 1 3 2 6 ∣ \begin{vmatrix}1&3\\2&6\end{vmatrix} ∣∣∣∣1236∣∣∣∣
计算方法为对角线相乘求差，即 1 ∗ 6 − 2 ∗ 3 = 0 1*6 - 2*3 = 0 1∗6−2∗3=0 对于3阶及以上的，责需要经过对行列式进行变换后再进行计算，例如： ∣ 1 2 3 2 3 4 4 5 7 ∣ ⟶ R 2 − 2 R 1 = ∣ 1 2 3 2 − 1 ∗ 2 3 − 2 ∗ 2 4 − 3 ∗ 2 4 5 7 ∣ = ∣ 1 2 3 0 − 1 − 2 4 5 7 ∣ ⟶ R 3 − 4 R 1 = ∣ 1 2 3 0 − 1 − 2 4 − 4 ∗ 1 5 − 4 ∗ 2 7 − 4 ∗ 3 ∣ = ∣ 1 2 3 0 − 1 − 2 0 − 3 − 5 ∣ ⟶ R 3 − 3 R 2 = ∣ 1 2 3 0 − 1 − 2 0 − 3 − 3 ∗ ( − 1 ) − 5 − 3 ∗ ( − 2 ) ∣ = ∣ 1 2 3 0 − 1 − 2 0 0 1 ∣ = 1 ∗ ( − 1 ) ∗ 1 = − 1 \begin{vmatrix}1&2&3\\2&3&4\\4&5&7\end{vmatrix} \\ \overset{R_2-2R_1}{\longrightarrow}=\begin{vmatrix}1&2&3\\2-1*2&3-2*2&4-3*2\\4&5&7\end{vmatrix} =\begin{vmatrix}1&2&3\\0&-1&-2\\4&5&7\end{vmatrix} \\ \overset{R_3-4R_1}{\longrightarrow}=\begin{vmatrix}1&2&3\\0&-1&-2\\4-4*1&5-4*2&7-4*3\end{vmatrix} =\begin{vmatrix}1&2&3\\0&-1&-2\\0&-3&-5\end{vmatrix} \\ \overset{R_3-3R_2}{\longrightarrow}=\begin{vmatrix}1&2&3\\0&-1&-2\\0&-3-3*(-1)&-5-3*(-2)\end{vmatrix} =\begin{vmatrix}1&2&3\\0&-1&-2\\0&0&1\end{vmatrix} \\ = 1 * (-1) * 1 \\ =-1 ∣∣∣∣∣∣124235347∣∣∣∣∣∣⟶R2−2R1=∣∣∣∣∣∣12−1∗2423−2∗2534−3∗27∣∣∣∣∣∣=∣∣∣∣∣∣1042−153−27∣∣∣∣∣∣⟶R3−4R1=∣∣∣∣∣∣104−4∗12−15−4∗23−27−4∗3∣∣∣∣∣∣=∣∣∣∣∣∣1002−1−33−2−5∣∣∣∣∣∣⟶R3−3R2=∣∣∣∣∣∣1002−1−3−3∗(−1)3−2−5−3∗(−2)∣∣∣∣∣∣=∣∣∣∣∣∣1002−103−21∣∣∣∣∣∣=1∗(−1)∗1=−1
性质1：某行（列）加上或减去领一行（列）的几倍，行列式不变
上述3阶的计算过程也可以简化为： ∣ 1 2 3 2 3 4 4 5 7 ∣ → R 2 − 2 R 1 → R 3 − 4 R 1 → R 3 − 3 R 2 ∣ 1 2 3 0 − 1 − 2 0 0 1 ∣ = 1 ∗ ( − 1 ) ∗ 1 = − 1 \begin{vmatrix}1&2&3\\2&3&4\\4&5&7\end{vmatrix} \begin{matrix}\xrightarrow{R_2-2R_1} \\ \xrightarrow{R_3-4R_1} \\ \xrightarrow{R_3-3R_2} \end{matrix} \begin{vmatrix}1&2&3\\0&-1&-2\\0&0&1\end{vmatrix} \\ =1*(-1)*1 \\ =-1 ∣∣∣∣∣∣124235347∣∣∣∣∣∣R2−2R1 R3−4R1 R3−3R2 ∣∣∣∣∣∣1002−103−21∣∣∣∣∣∣=1∗(−1)∗1=−1
扩展一个4阶的行列式算一下 ∣ 1 2 3 4 2 3 4 5 4 5 7 8 8 9 12 12 ∣ → R 2 − 2 R 1 → R 3 − 4 R 1 → R 4 − 8 R 1 ∣ 1 2 3 4 0 − 1 − 2 − 3 0 − 3 − 5 − 8 0 − 7 − 12 − 20 ∣ → R 3 − 3 R 2 → R 4 − 7 R 2 ∣ 1 2 3 4 0 − 1 − 2 − 3 0 0 1 1 0 0 2 1 ∣ → R 4 − 2 R 3 ∣ 1 2 3 4 0 − 1 − 2 − 3 0 0 1 1 0 0 0 − 1 ∣ = 1 ∗ ( − 1 ) ∗ 1 ∗ ( − 1 ) = 1 \begin{vmatrix}1&2&3&4\\2&3&4&5\\4&5&7&8\\8&9&12&12\end{vmatrix} \begin{matrix} \xrightarrow{R_2-2R_1} \\ \xrightarrow{R_3-4R_1} \\ \xrightarrow{R_4-8R_1} \end{matrix} \begin{vmatrix}1&2&3&4\\0&-1&-2&-3\\0&-3&-5&-8\\0&-7&-12&-20\end{vmatrix} \begin{matrix} \xrightarrow{R_3-3R_2} \\ \xrightarrow{R_4-7R_2} \end{matrix} \begin{vmatrix}1&2&3&4\\0&-1&-2&-3\\0&0&1&1\\0&0&2&1\end{vmatrix} \begin{matrix} \xrightarrow{R_4-2R_3} \end{matrix} \begin{vmatrix}1&2&3&4\\0&-1&-2&-3\\0&0&1&1\\0&0&0&-1\end{vmatrix} \\ =1*(-1)*1*(-1) \\ =1 ∣∣∣∣∣∣∣∣124823593471245812∣∣∣∣∣∣∣∣R2−2R1 R3−4R1 R4−8R1 ∣∣∣∣∣∣∣∣10002−1−3−73−2−5−124−3−8−20∣∣∣∣∣∣∣∣R3−3R2 R4−7R2 ∣∣∣∣∣∣∣∣10002−1003−2124−311∣∣∣∣∣∣∣∣R4−2R3 ∣∣∣∣∣∣∣∣10002−1003−2104−31−1∣∣∣∣∣∣∣∣=1∗(−1)∗1∗(−1)=1
性质2：某行（列）乘k，等于k乘此行列式
例： 已知 ∣ 1 2 3 4 2 3 4 5 4 5 7 8 8 9 10 12 ∣ = − 1 ，求 ∣ 2 4 6 8 2 3 4 5 4 5 7 8 8 9 10 12 ∣ \text{已知} \begin{vmatrix}1&2&3&4\\2&3&4&5\\4&5&7&8\\8&9&10&12\end{vmatrix}= -1 \text{，求} \begin{vmatrix}2&4&6&8\\2&3&4&5\\4&5&7&8\\8&9&10&12\end{vmatrix} 已知∣∣∣∣∣∣∣∣124823593471045812∣∣∣∣∣∣∣∣=−1，求∣∣∣∣∣∣∣∣224843596471085812∣∣∣∣∣∣∣∣ 观察发现，第一行{2,4,6,8}是{1,2,3,4}的2倍，即 ∣ 2 4 6 8 2 3 4 5 4 5 7 8 8 9 10 12 ∣ = 2 ∗ ∣ 1 2 3 4 2 3 4 5 4 5 7 8 8 9 10 12 ∣ = 2 ∗ ( − 1 ) = − 2 \begin{vmatrix}2&4&6&8\\2&3&4&5\\4&5&7&8\\8&9&10&12\end{vmatrix} =2 *\begin{vmatrix}1&2&3&4\\2&3&4&5\\4&5&7&8\\8&9&10&12\end{vmatrix} =2*(-1) = -2 ∣∣∣∣∣∣∣∣224843596471085812∣∣∣∣∣∣∣∣=2∗∣∣∣∣∣∣∣∣124823593471045812∣∣∣∣∣∣∣∣=2∗(−1)=−2
∣ 2 4 6 8 2 3 4 5 12 15 21 24 8 9 10 12 ∣ = 2 ∗ 3 ∗ ∣ 1 2 3 4 2 3 4 5 4 5 7 8 8 9 10 12 ∣ = 2 ∗ 3 ∗ ( − 1 ) = − 6 \begin{vmatrix}2&4&6&8\\2&3&4&5\\12&15&21&24\\8&9&10&12\end{vmatrix} =2*3*\begin{vmatrix}1&2&3&4\\2&3&4&5\\4&5&7&8\\8&9&10&12\end{vmatrix} =2*3*(-1)=-6 ∣∣∣∣∣∣∣∣2212843159642110852412∣∣∣∣∣∣∣∣=2∗3∗∣∣∣∣∣∣∣∣124823593471045812∣∣∣∣∣∣∣∣=2∗3∗(−1)=−6
性质3：互换两行（列），行列式变号
例1： ∣ 2 3 4 5 1 2 3 4 4 5 7 8 8 9 10 12 ∣ → R 1 < − > R 2 − 1 ∗ ∣ 1 2 3 4 2 3 4 5 4 5 7 8 8 9 10 12 ∣ = − 1 ∗ ( − 1 ) = 1 \begin{vmatrix}2&3&4&5\\1&2&3&4\\4&5&7&8\\8&9&10&12\end{vmatrix} \begin{matrix}\xrightarrow{R_1<->R_2}\end{matrix} -1*\begin{vmatrix}1&2&3&4\\2&3&4&5\\4&5&7&8\\8&9&10&12\end{vmatrix} = -1*(-1) = 1 ∣∣∣∣∣∣∣∣214832594371054812∣∣∣∣∣∣∣∣R1<−>R2 −1∗∣∣∣∣∣∣∣∣124823593471045812∣∣∣∣∣∣∣∣=−1∗(−1)=1 例2： ∣ 0 0 0 3 0 0 3 2 1 2 3 4 0 5 2 4 ∣ → R 1 < − > R 4 − 1 ∗ ∣ 0 5 2 4 0 0 3 2 1 2 3 4 0 0 0 3 ∣ → R 2 < − > R 3 − 1 ∗ ( − 1 ) ∣ 0 5 2 4 1 2 3 4 0 0 3 2 0 0 0 3 ∣ → R 1 < − > R 2 − 1 ∗ ( − 1 ) ∗ ( − 1 ) ∣ 1 2 3 4 0 5 2 4 0 0 3 2 0 0 0 3 ∣ = − 1 ∗ ( − 1 ) ∗ ( − 1 ) ∗ 1 ∗ 5 ∗ 3 ∗ 3 = − 45 \begin{vmatrix}0&0&0&3\\0&0&3&2\\1&2&3&4\\0&5&2&4\end{vmatrix} \begin{matrix} \xrightarrow{R_1<->R_4} \end{matrix} -1*\begin{vmatrix}0&5&2&4\\0&0&3&2\\1&2&3&4\\0&0&0&3\end{vmatrix} \begin{matrix} \xrightarrow{R_2<->R_3} \end{matrix} -1*(-1)\begin{vmatrix}0&5&2&4\\1&2&3&4\\0&0&3&2\\0&0&0&3\end{vmatrix} \\ \begin{matrix} \xrightarrow{R_1<->R_2} \end{matrix} -1*(-1)*(-1)\begin{vmatrix}1&2&3&4\\0&5&2&4\\0&0&3&2\\0&0&0&3\end{vmatrix} \\ =-1*(-1)*(-1)*1*5*3*3= -45 ∣∣∣∣∣∣∣∣0010002503323244∣∣∣∣∣∣∣∣R1<−>R4 −1∗∣∣∣∣∣∣∣∣0010502023304243∣∣∣∣∣∣∣∣R2<−>R3 −1∗(−1)∣∣∣∣∣∣∣∣0100520023304423∣∣∣∣∣∣∣∣R1<−>R2 −1∗(−1)∗(−1)∣∣∣∣∣∣∣∣1000250032304423∣∣∣∣∣∣∣∣=−1∗(−1)∗(−1)∗1∗5∗3∗3=−45
∣ x a ⋯ a a x ⋯ a ⋮ ⋮ ⋱ ⋮ a a ⋯ x ∣ = ( x − a ) ( n − 1 ) [ x + ( n − 1 ) a ] \begin{vmatrix}x&a&\cdots&a\\a&x&\cdots&a\\\vdots&\vdots&\ddots&\vdots\\a&a&\cdots&x\end{vmatrix} =(x-a)^{(n-1)}[x+(n-1)a] ∣∣∣∣∣∣∣∣∣xa⋮aax⋮a⋯⋯⋱⋯aa⋮x∣∣∣∣∣∣∣∣∣=(x−a)(n−1)[x+(n−1)a]
例1：请计算如下行列式的值 ∣ 2 3 3 3 3 2 3 3 3 3 2 3 3 3 3 2 ∣ 即 x = 2 a = 3 n = 4 , 代 入 以 上 公 式 可 得 ∣ 2 3 3 3 3 2 3 3 3 3 2 3 3 3 3 2 ∣ = ( 2 − 3 ) ( 4 − 1 ) [ 2 + ( 4 − 1 ) ∗ 3 ] = − 11 \begin{vmatrix} 2&3&3&3\\3&2&3&3\\3&3&2&3\\3&3&3&2 \end{vmatrix} \\即 x=2 ~\ ~\ ~\ a=3 ~\ ~\ ~\ n =4,代入以上公式可得 \\ \begin{vmatrix} 2&3&3&3\\3&2&3&3\\3&3&2&3\\3&3&3&2 \end{vmatrix} =(2-3)^{(4-1)}[2+(4-1)*3] =-11 ∣∣∣∣∣∣∣∣2333323333233332∣∣∣∣∣∣∣∣即x=2 a=3 n=4,代入以上公式可得∣∣∣∣∣∣∣∣2333323333233332∣∣∣∣∣∣∣∣=(2−3)(4−1)[2+(4−1)∗3]=−11
∣ 1 1 ⋯ 1 x 1 x 2 ⋯ x n x 1 2 x 2 2 ⋯ x n 2 ⋮ ⋮ ⋱ ⋮ x 1 n − 1 x 2 n − 1 ⋯ x n n − 1 ∣ = ( x n − x n − 1 ) ( x n − x n − 2 ) ( x n − x n − 3 ) ⋯ ⋯ ( x n − x 1 ) ∗ ( x n − 1 − x n − 2 ) ( x n − 1 − x n − 3 ) ⋯ ⋯ ( x n − 1 − x 1 ) ∗ ⋯ ⋯ ∗ ( x 2 − x 1 ) \begin{vmatrix} 1&1&\cdots&1\\x_1&x_2&\cdots&x_n\\x_1^2&x_2^2&\cdots&x_n^2\\\vdots&\vdots&\ddots&\vdots\\x_1^{n-1}&x_2^{n-1}&\cdots&x_n^{n-1} \end{vmatrix} \\=(x_n-x_{n-1})(x_n-x_{n-2})(x_n-x_{n-3})\cdots\cdots(x_n-x_1) \\*(x_{n-1}-x_{n-2})(x_{n-1}-x_{n-3})\cdots\cdots(x_{n-1}-x_1) \\*\cdots\cdots \\*(x_2-x_1) ∣∣∣∣∣∣∣∣∣∣∣1x1x12⋮x1n−11x2x22⋮x2n−1⋯⋯⋯⋱⋯1xnxn2⋮xnn−1∣∣∣∣∣∣∣∣∣∣∣=(xn−xn−1)(xn−xn−2)(xn−xn−3)⋯⋯(xn−x1)∗(xn−1−xn−2)(xn−1−xn−3)⋯⋯(xn−1−x1)∗⋯⋯∗(x2−x1)
例2：请计算如下行列式的值 ∣ 1 1 1 1 3 4 5 6 3 2 4 2 5 2 6 2 3 3 4 3 5 3 6 3 ∣ 即 x 1 = 3 x 2 = 4 x 3 = 5 x 4 = 6 n = 4 ， 代 入 以 上 公 式 可 得 ( x 4 − x 3 ) ( x 4 − x 2 ) ( x 4 − x 1 ) ( x 3 − x 2 ) ( x 3 − x 1 ) ( x 2 − x 1 ) = ( 6 − 5 ) ∗ ( 6 − 4 ) ∗ ( 6 − 3 ) ∗ ( 5 − 4 ) ∗ ( 5 − 3 ) ∗ ( 4 − 3 ) = 12 \begin{vmatrix} 1&1&1&1\\3&4&5&6\\3^2&4^2&5^2&6^2\\3^3&4^3&5^3&6^3 \end{vmatrix} \\即 x_1=3 ~\ ~\ ~\ x_2=4 ~\ ~\ ~\ x_3=5 ~\ ~\ ~\ x_4=6 ~\ ~\ ~\ n=4，代入以上公式可得 \\(x_4-x_3)(x_4-x_2)(x_4-x_1)(x_3-x_2)(x_3-x_1)(x_2-x_1) \\=(6-5)*(6-4)*(6-3)*(5-4)*(5-3)*(4-3) \\=12 ∣∣∣∣∣∣∣∣133233144243155253166263∣∣∣∣∣∣∣∣即x1=3 x2=4 x3=5 x4=6 n=4，代入以上公式可得(x4−x3)(x4−x2)(x4−x1)(x3−x2)(x3−x1)(x2−x1)=(6−5)∗(6−4)∗(6−3)∗(5−4)∗(5−3)∗(4−3)=12
①两行（列）相同或者成比例时，行列式为0
②某行（列）为两项相加相减时，行列式可拆解成两个行列式相加减
例3：请计算如下行列式的值 已知 ∣ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ∣ = 1 ， 试 求 ∣ a 1 + c 1 b 1 a 1 + b 1 a 2 + c 2 b 2 a 2 + b 2 a 3 + c 3 b 3 a 3 + b 3 ∣ ∣ a 1 + c 1 b 1 a 1 + b 1 a 2 + c 2 b 2 a 2 + b 2 a 3 + c 3 b 3 a 3 + b 3 ∣ = ∣ a 1 b 1 a 1 + b 1 a 2 b 2 a 2 + b 2 a 3 b 3 a 3 + b 3 ∣ + ∣ c 1 b 1 a 1 + b 1 c 2 b 2 a 2 + b 2 c 3 b 3 a 3 + b 3 ∣ = ∣ a 1 b 1 a 1 a 2 b 2 a 2 a 3 b 3 a 3 ∣ + ∣ a 1 b 1 b 1 a 2 b 2 b 2 a 3 b 3 b 3 ∣ + ∣ c 1 b 1 a 1 c 2 b 2 a 2 c 3 b 3 a 3 ∣ + ∣ c 1 b 1 b 1 c 2 b 2 b 2 c 3 b 3 b 3 ∣ = 0 + 0 + ∣ c 1 b 1 a 1 c 2 b 2 a 2 c 3 b 3 a 3 ∣ + 0 = − 1 ∗ ∣ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ∣ = − 1 ∗ 1 = − 1 \text{已知} \begin{vmatrix} a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3 \end{vmatrix} =1，试求 \begin{vmatrix} a_1+c_1&b_1&a_1+b_1\\a_2+c_2&b_2&a_2+b_2\\a_3+c_3&b_3&a_3+b_3 \end{vmatrix} \\ \begin{vmatrix} a_1+c_1&b_1&a_1+b_1\\a_2+c_2&b_2&a_2+b_2\\a_3+c_3&b_3&a_3+b_3 \end{vmatrix} =\begin{vmatrix} a_1&b_1&a_1+b_1\\a_2&b_2&a_2+b_2\\a_3&b_3&a_3+b_3 \end{vmatrix}+ \begin{vmatrix} c_1&b_1&a_1+b_1\\c_2&b_2&a_2+b_2\\c_3&b_3&a_3+b_3 \end{vmatrix} \\ =\begin{vmatrix} a_1&b_1&a_1\\a_2&b_2&a_2\\a_3&b_3&a_3 \end{vmatrix}+ \begin{vmatrix} a_1&b_1&b_1\\a_2&b_2&b_2\\a_3&b_3&b_3 \end{vmatrix}+ \begin{vmatrix} c_1&b_1&a_1\\c_2&b_2&a_2\\c_3&b_3&a_3 \end{vmatrix}+ \begin{vmatrix} c_1&b_1&b_1\\c_2&b_2&b_2\\c_3&b_3&b_3 \end{vmatrix} \\ =0+0+\begin{vmatrix} c_1&b_1&a_1\\c_2&b_2&a_2\\c_3&b_3&a_3 \end{vmatrix}+0 =-1*\begin{vmatrix} a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3 \end{vmatrix}=-1*1=-1 已知∣∣∣∣∣∣a1a2a3b1b2b3c1c2c3∣∣∣∣∣∣=1，试求∣∣∣∣∣∣a1+c1a2+c2a3+c3b1b2b3a1+b1a2+b2a3+b3∣∣∣∣∣∣∣∣∣∣∣∣a1+c1a2+c2a3+c3b1b2b3a1+b1a2+b2a3+b3∣∣∣∣∣∣=∣∣∣∣∣∣a1a2a3b1b2b3a1+b1a2+b2a3+b3∣∣∣∣∣∣+∣∣∣∣∣∣c1c2c3b1b2b3a1+b1a2+b2a3+b3∣∣∣∣∣∣=∣∣∣∣∣∣a1a2a3b1b2b3a1a2a3∣∣∣∣∣∣+∣∣∣∣∣∣a1a2a3b1b2b3b1b2b3∣∣∣∣∣∣+∣∣∣∣∣∣c1c2c3b1b2b3a1a2a3∣∣∣∣∣∣+∣∣∣∣∣∣c1c2c3b1b2b3b1b2b3∣∣∣∣∣∣=0+0+∣∣∣∣∣∣c1c2c3b1b2b3a1a2a3∣∣∣∣∣∣+0=−1∗∣∣∣∣∣∣a1a2a3b1b2b3c1c2c3∣∣∣∣∣∣=−1∗1=−1
例4： 试 求 ∣ 1 2 3 5 6 7 9 10 11 ∣ 中 a 23 的 余 子 式 ， a 12 的 代 数 余 子 式 试求 \begin{vmatrix} 1&2&3\\5&6&7\\9&10&11 \end{vmatrix} 中 a_{23}的余子式，a_{12}的代数余子式 试求∣∣∣∣∣∣15926103711∣∣∣∣∣∣中a23的余子式，a12的代数余子式
余子式M： M 23 = ∣ 1 2 9 10 ∣ = − 8 M_{23}= \begin{vmatrix} 1&2\\9&10 \end{vmatrix} =-8 M23=∣∣∣∣19210∣∣∣∣=−8 代数余子式A： A 12 = ( − 1 ) 1 + 2 M 12 = ( − 1 ) 3 ∗ ∣ 5 7 9 11 ∣ = − 1 ∗ ( 5 ∗ 11 − 9 ∗ 7 ) = − 1 ∗ ( − 8 ) = 8 A_{12}=(-1)^{1+2}M_{12} \\ =(-1)^3*\begin{vmatrix}5&7\\9&11 \end{vmatrix} \\ =-1*(5*11-9*7)=-1*(-8)=8 A12=(−1)1+2M12=(−1)3∗∣∣∣∣59711∣∣∣∣=−1∗(5∗11−9∗7)=−1∗(−8)=8
D = a i 1 A i 1 + a i 2 A i 2 + ⋯ ⋯ + a i n A i n ( 第 i 行 ) D = a 1 j A 1 j + a 2 j A 2 j + ⋯ ⋯ + a n j A n j ( 第 j 列 ) D = a_{i1}A_{i1}+a_{i2}A_{i2}+\cdots\cdots+a_{in}A_{in}(第i行) \\ D = a_{1j}A_{1j}+a_{2j}A_{2j}+\cdots\cdots+a_{nj}A_{nj}(第j列) D=ai1Ai1+ai2Ai2+⋯⋯+ainAin(第i行)D=a1jA1j+a2jA2j+⋯⋯+anjAnj(第j列)
例5： ∣ 1 2 3 5 6 7 9 10 11 ∣ = a 11 A 11 + a 12 A 12 + a 13 A 13 = a 11 ( − 1 ) 1 + 1 M 11 + a 12 ( − 1 ) 1 + 2 M 12 + a 13 ( − 1 ) 1 + 3 M 13 = 1 ∗ ( − 1 ) 2 ∗ ∣ 6 7 10 11 ∣ + 2 ∗ ( − 1 ) 3 ∗ ∣ 5 9 7 11 ∣ + 3 ∗ ( − 1 ) 4 ∗ ∣ 5 6 9 10 ∣ = − 3 + 2 ∗ ( − 1 ) ∗ ( − 8 ) + 3 ∗ ( − 4 ) = 1 \begin{vmatrix} 1&2&3\\5&6&7\\9&10&11 \end{vmatrix} =a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13} \\ =a_{11}(-1)^{1+1}M_{11}+a_{12}(-1)^{1+2}M_{12}+a_{13}(-1)^{1+3}M_{13} \\=1*(-1)^2*\begin{vmatrix}6&7\\10&11\end{vmatrix}+2*(-1)^3*\begin{vmatrix}5&9\\7&11\end{vmatrix}+3*(-1)^4*\begin{vmatrix}5&6\\9&10\end{vmatrix} \\ =-3+2*(-1)*(-8)+3*(-4)=1 ∣∣∣∣∣∣15926103711∣∣∣∣∣∣=a11A11+a12A12+a13A13=a11(−1)1+1M11+a12(−1)1+2M12+a13(−1)1+3M13=1∗(−1)2∗∣∣∣∣610711∣∣∣∣+2∗(−1)3∗∣∣∣∣57911∣∣∣∣+3∗(−1)4∗∣∣∣∣59610∣∣∣∣=−3+2∗(−1)∗(−8)+3∗(−4)=1
∣ 1 2 3 5 6 7 9 10 11 ∣ = a 12 ∗ A 12 + a 22 ∗ A 22 + a 32 ∗ A 32 = a 12 ∗ ( − 1 ) 1 + 2 M 12 + a 22 ∗ ( − 1 ) 2 + 2 M 22 + a 32 ∗ ( − 1 ) 3 + 2 M 32 = 2 ∗ ( − 1 ) 3 ∣ 5 7 9 11 ∣ + 6 ∗ ( − 1 ) 4 ∣ 1 3 9 11 ∣ + 10 ∗ ( − 1 ) 5 ∣ 1 5 3 7 ∣ = 2 ∗ ( − 1 ) ∗ ( − 8 ) + 6 ∗ ( − 1 ) 4 ∗ ( − 16 ) + 10 ∗ ( − 1 ) 5 ∗ ( − 8 ) = 16 − 96 + 80 = 0 \begin{vmatrix} 1&2&3\\5&6&7\\9&10&11 \end{vmatrix} =a_{12}*A_{12}+a_{22}*A_{22}+a_{32}*A_{32} \\ =a_{12}*(-1)^{1+2}M_{12}+a_{22}*(-1)^{2+2}M_{22}+a_{32}*(-1)^{3+2}M_{32} \\ =2*(-1)^3\begin{vmatrix}5&7\\9&11\end{vmatrix}+6*(-1)^4\begin{vmatrix}1&3\\9&11\end{vmatrix}+10*(-1)^5\begin{vmatrix}1&5\\3&7\end{vmatrix} \\ =2*(-1)*(-8)+6*(-1)^4*(-16)+10*(-1)^5*(-8) \\ =16-96+80=0 ∣∣∣∣∣∣15926103711∣∣∣∣∣∣=a12∗A12+a22∗A22+a32∗A32=a12∗(−1)1+2M12+a22∗(−1)2+2M22+a32∗(−1)3+2M32=2∗(−1)3∣∣∣∣59711∣∣∣∣+6∗(−1)4∣∣∣∣19311∣∣∣∣+10∗(−1)5∣∣∣∣1357∣∣∣∣=2∗(−1)∗(−8)+6∗(−1)4∗(−16)+10∗(−1)5∗(−8)=16−96+80=0
已 知 D = ∣ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ∣ ， 试 求 ① 3 A 11 + 4 A 12 + 5 A 13 + 6 A 14 ② 3 A 11 + 4 A 21 + 5 A 31 + 6 A 41 ③ 3 M 11 + 4 M 21 + 5 M 31 + 6 M 41 已知 D = \begin{vmatrix} 1 & 2 & 3 & 4\\ 5 & 6 & 7 & 8\\ 9 & 10 & 11 & 12\\ 13 & 14 & 15 & 16 \end{vmatrix} ，试求 ① 3A_{11}+4A_{12}+5A_{13}+6A_{14} ~\ ~\ ~\ ② 3A_{11}+4A_{21}+5A_{31}+6A_{41} ~\ ~\ ~\ ③3M_{11}+4M_{21}+5M_{31}+6M_{41} 已知D=∣∣∣∣∣∣∣∣15913261014371115481216∣∣∣∣∣∣∣∣，试求①3A11+4A12+5A13+6A14 ②3A11+4A21+5A31+6A41 ③3M11+4M21+5M31+6M41
对于A，直接找到对应的位置，将系数与对应的项进行替换即可：
3 A 11 + 4 A 12 + 5 A 13 + 6 A 14 = ∣ 3 4 5 6 5 6 7 8 9 10 11 12 13 14 15 16 ∣ 3 A 11 + 4 A 21 + 5 A 31 + 6 A 41 = ∣ 3 2 3 4 4 6 7 8 5 10 11 12 6 14 15 16 ∣ 3A_{11}+4A_{12}+5A_{13}+6A_{14} \\ =\begin{vmatrix} 3 & 4 & 5 & 6 \\ 5 & 6 & 7 & 8\\ 9 & 10 & 11 & 12\\ 13 & 14 & 15 & 16 \end{vmatrix} \\ 3A_{11}+4A_{21}+5A_{31}+6A_{41} \\ =\begin{vmatrix} 3 & 2 & 3 & 4 \\ 4 & 6 & 7 & 8\\ 5 & 10 & 11 & 12\\ 6 & 14 & 15 & 16 \end{vmatrix} 3A11+4A12+5A13+6A14=∣∣∣∣∣∣∣∣35913461014571115681216∣∣∣∣∣∣∣∣3A11+4A21+5A31+6A41=∣∣∣∣∣∣∣∣3456261014371115481216∣∣∣∣∣∣∣∣
对于M，则先通过M与A的对应关系，即 A i j = ( − 1 ) i + j M i j A_{ij}=(-1)^{i+j}M_{ij} Aij=(−1)i+jMij
3 M 11 + 4 M 21 + 5 M 31 + 6 M 41 A 11 = ( − 1 ) 1 + 1 M 11 → A 11 = M 11 A 21 = ( − 1 ) 2 + 1 M 21 → A 21 = − M 21 A 31 = ( − 1 ) 3 + 1 M 31 → A 31 = M 31 A 41 = ( − 1 ) 4 + 1 M 241 → A 41 = − M 41 ∴ 3 M 11 + 4 M 21 + 5 M 31 + 6 M 41 = 3 A 11 − 4 A 21 + 5 A 31 − 6 M 41 = ∣ 3 2 3 4 − 4 6 7 8 5 10 11 12 − 6 14 15 16 ∣ 3M_{11}+4M_{21}+5M_{31}+6M_{41} \\ \begin{matrix} A_{11}=(-1)^{1+1}M_{11} \xrightarrow{} A_{11}=M_{11} \end{matrix} \\ \begin{matrix} A_{21}=(-1)^{2+1}M_{21} \xrightarrow{} A_{21}=-M_{21} \end{matrix} \\ \begin{matrix} A_{31}=(-1)^{3+1}M_{31} \xrightarrow{} A_{31}=M_{31} \end{matrix} \\ \begin{matrix} A_{41}=(-1)^{4+1}M_{241} \xrightarrow{} A_{41}=-M_{41} \end{matrix} \\ \therefore ~\ ~\ ~\ 3M_{11}+4M_{21}+5M_{31}+6M_{41} \\= 3A_{11}-4A_{21}+5A_{31}-6M_{41} \\ =\begin{vmatrix} 3 & 2 & 3 & 4 \\ -4 & 6 & 7 & 8\\ 5 & 10 & 11 & 12\\ -6 & 14 & 15 & 16 \end{vmatrix} 3M11+4M21+5M31+6M41A11=(−1)1+1M11 A11=M11A21=(−1)2+1M21 A21=−M21A31=(−1)3+1M31 A31=M31A41=(−1)4+1M241 A41=−M41∴ 3M11+4M21+5M31+6M41=3A11−4A21+5A31−6M41=∣∣∣∣∣∣∣∣3−45−6261014371115481216∣∣∣∣∣∣∣∣
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