筆記內容摘自 猴博士愛講課@B站 https://www.bilibili.com/video/BV1hs411e7X8?p=4
行列式
行列式的計算
行列式分爲2階、3階、4階……n
階等,其中2階的計算方法爲:
∣∣∣∣1236∣∣∣∣
計算方法爲對角線相乘求差,即
1∗6−2∗3=0
對於3階及以上的,責需要經過對行列式進行變換後再進行計算,例如:
∣∣∣∣∣∣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
性質1:某行(列)加上或減去領一行(列)的幾倍,行列式不變
上述3階的計算過程也可以簡化爲:
∣∣∣∣∣∣124235347∣∣∣∣∣∣R2−2R1R3−4R1R3−3R2∣∣∣∣∣∣1002−103−21∣∣∣∣∣∣=1∗(−1)∗1=−1
擴展一個4階的行列式算一下
∣∣∣∣∣∣∣∣124823593471245812∣∣∣∣∣∣∣∣R2−2R1R3−4R1R4−8R1∣∣∣∣∣∣∣∣10002−1−3−73−2−5−124−3−8−20∣∣∣∣∣∣∣∣R3−3R2R4−7R2∣∣∣∣∣∣∣∣10002−1003−2124−311∣∣∣∣∣∣∣∣R4−2R3∣∣∣∣∣∣∣∣10002−1003−2104−31−1∣∣∣∣∣∣∣∣=1∗(−1)∗1∗(−1)=1
性質2:某行(列)乘k
,等於k乘此行列式
例:
已知∣∣∣∣∣∣∣∣124823593471045812∣∣∣∣∣∣∣∣=−1,求∣∣∣∣∣∣∣∣224843596471085812∣∣∣∣∣∣∣∣
觀察發現,第一行{2,4,6,8}是{1,2,3,4}的2倍,即
∣∣∣∣∣∣∣∣224843596471085812∣∣∣∣∣∣∣∣=2∗∣∣∣∣∣∣∣∣124823593471045812∣∣∣∣∣∣∣∣=2∗(−1)=−2
∣∣∣∣∣∣∣∣2212843159642110852412∣∣∣∣∣∣∣∣=2∗3∗∣∣∣∣∣∣∣∣124823593471045812∣∣∣∣∣∣∣∣=2∗3∗(−1)=−6
性質3:互換兩行(列),行列式變號
例1:
∣∣∣∣∣∣∣∣214832594371054812∣∣∣∣∣∣∣∣R1<−>R2−1∗∣∣∣∣∣∣∣∣124823593471045812∣∣∣∣∣∣∣∣=−1∗(−1)=1
例2:
∣∣∣∣∣∣∣∣0010002503323244∣∣∣∣∣∣∣∣R1<−>R4−1∗∣∣∣∣∣∣∣∣0010502023304243∣∣∣∣∣∣∣∣R2<−>R3−1∗(−1)∣∣∣∣∣∣∣∣0100520023304423∣∣∣∣∣∣∣∣R1<−>R2−1∗(−1)∗(−1)∣∣∣∣∣∣∣∣1000250032304423∣∣∣∣∣∣∣∣=−1∗(−1)∗(−1)∗1∗5∗3∗3=−45
常用的幾種行列式運算題型
1/7 對稱的N行N列計算
∣∣∣∣∣∣∣∣∣xa⋮aax⋮a⋯⋯⋱⋯aa⋮x∣∣∣∣∣∣∣∣∣=(x−a)(n−1)[x+(n−1)a]
例1:請計算如下行列式的值
∣∣∣∣∣∣∣∣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
2/7 指數遞增行列式求解
∣∣∣∣∣∣∣∣∣∣∣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:請計算如下行列式的值
∣∣∣∣∣∣∣∣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
3/7 特殊行列式計算
①兩行(列)相同或者成比例時,行列式爲0
②某行(列)爲兩項相加相減時,行列式可拆解成兩個行列式相加減
例3:請計算如下行列式的值
已知∣∣∣∣∣∣a1a2a3b1b2b3c1c2c3∣∣∣∣∣∣=1,試求∣∣∣∣∣∣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
4/7 求餘子式、代數餘子式
例4:
試求∣∣∣∣∣∣15926103711∣∣∣∣∣∣中a23的餘子式,a12的代數餘子式
餘子式M:
M23=∣∣∣∣19210∣∣∣∣=−8
代數餘子式A:
A12=(−1)1+2M12=(−1)3∗∣∣∣∣59711∣∣∣∣=−1∗(5∗11−9∗7)=−1∗(−8)=8
5/7 任意行/列計算行列式
D=ai1Ai1+ai2Ai2+⋯⋯+ainAin(第i行)D=a1jA1j+a2jA2j+⋯⋯+anjAnj(第j列)
例5:
∣∣∣∣∣∣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
∣∣∣∣∣∣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
6/7 多個A或M相加減
已知D=∣∣∣∣∣∣∣∣15913261014371115481216∣∣∣∣∣∣∣∣,試求①3A11+4A12+5A13+6A14 ②3A11+4A21+5A31+6A41 ③3M11+4M21+5M31+6M41
對於A,直接找到對應的位置,將係數與對應的項進行替換即可:
3A11+4A12+5A13+6A14=∣∣∣∣∣∣∣∣35913461014571115681216∣∣∣∣∣∣∣∣3A11+4A21+5A31+6A41=∣∣∣∣∣∣∣∣3456261014371115481216∣∣∣∣∣∣∣∣
對於M,則先通過M與A的對應關係,即
Aij=(−1)i+jMij
3M11+4M21+5M31+6M41A11=(−1)1+1M11A11=M11A21=(−1)2+1M21A21=−M21A31=(−1)3+1M31A31=M31A41=(−1)4+1M241A41=−M41∴ 3M11+4M21+5M31+6M41=3A11−4A21+5A31−6M41=∣∣∣∣∣∣∣∣3−45−6261014371115481216∣∣∣∣∣∣∣∣
7/7 給一方程組,判讀其解的情況
方程組 |
D≠0 |
D=0 |
其次 |
只有一組零解 |
有零解與非零解 |
非其次 |
只有一組非零解 |
有多個解或無解 |
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