사용자:慈居/행렬식

일반

삼각행렬식

${\begin{vmatrix}l_{1,1}&&&&0\\l_{2,1}&l_{2,2}&&&\\l_{3,1}&l_{3,2}&\ddots &&\\\vdots &\vdots &\ddots &\ddots &\\l_{n,1}&l_{n,2}&\ldots &l_{n,n-1}&l_{n,n}\end{vmatrix}}=\prod _{i=1}^{n}l_{i,i}$

${\begin{vmatrix}u_{1,1}&u_{1,2}&u_{1,3}&\ldots &u_{1,n}\\&u_{2,2}&u_{2,3}&\ldots &u_{2,n}\\&&\ddots &\ddots &\vdots \\&&&\ddots &u_{n-1,n}\\0&&&&u_{n,n}\end{vmatrix}}=\prod _{i=1}^{n}u_{i,i}$

각 열의 합이 같은 행렬식

${\begin{vmatrix}1&2&3&4\\2&3&4&1\\3&4&1&2\\4&1&2&3\end{vmatrix}}=10{\begin{vmatrix}1&1&1&1\\2&3&4&1\\3&4&1&2\\4&1&2&3\end{vmatrix}}=10{\begin{vmatrix}1&1&1&1\\0&1&2&-1\\0&1&-2&-1\\0&-3&-2&-1\end{vmatrix}}=10{\begin{vmatrix}1&1&1&1\\0&1&2&-1\\0&0&-4&0\\0&0&0&-4\end{vmatrix}}=10\times (-4)\times (-4)=160$

갈퀴형

${\begin{vmatrix}1+a_{1}&1&1&\cdots &1\\1&1+a_{2}&1&\cdots &1\\1&1&1+a_{3}&\cdots &1\\\vdots &\vdots &\vdots &\ddots &\vdots \\1&1&1&\cdots &1+a_{n}\end{vmatrix}}_{a_{k}\neq 0\ \forall k}={\begin{vmatrix}1+a_{1}&1&1&\cdots &1\\-a_{1}&a_{2}&0&\cdots &0\\-a_{1}&0&a_{3}&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\-a_{1}&0&0&\cdots &a_{n}\end{vmatrix}}={\begin{vmatrix}a_{1}(1+{\frac {1}{a_{1}}}+\cdots +{\frac {1}{a_{n}}})&1&1&\cdots &1\\0&a_{2}&0&\cdots &0\\0&0&a_{3}&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\cdots &a_{n}\end{vmatrix}}=(1+{\frac {1}{a_{1}}}+\cdots +{\frac {1}{a_{n}}})a_{1}a_{2}\cdots a_{n}$

${\begin{vmatrix}x-a&a&\cdots &a\\a&x-a&\cdots &a\\\vdots &\vdots &\ddots &\vdots \\a&a&\cdots &x-a\end{vmatrix}}$

${\begin{vmatrix}x_{1}-a_{1}&x_{2}&\cdots &x_{n}\\x_{1}&x_{2}-a_{2}&\cdots &x_{n}\\\vdots &\vdots &\ddots &\vdots \\x_{1}&x_{2}&\cdots &x_{n}-a_{n}\end{vmatrix}}=(-1)^{n-1}(-1+\sum _{i=1}^{n}{\frac {x_{i}}{a_{i}}})\prod _{i=1}^{n}a_{i}$

가변법

${\begin{vmatrix}a&b&\cdots &b\\b&a&\cdots &b\\\vdots &\vdots &\ddots &\vdots \\b&b&\cdots &a\end{vmatrix}}={\begin{vmatrix}1&b&b&\cdots &b\\0&a&b&\cdots &b\\0&b&a&\cdots &b\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&b&b&\cdots &a\end{vmatrix}}={\begin{vmatrix}1&b&b&\cdots &b\\-1&a-b&0&\cdots &0\\-1&0&a-b&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\-1&0&0&\cdots &a-b\end{vmatrix}}$

더하기

${\begin{vmatrix}1&-4&2\\3&3&2\\302&297&203\end{vmatrix}}={\begin{vmatrix}1&-4&2\\3&3&2\\300&300&200\end{vmatrix}}+{\begin{vmatrix}1&-4&2\\3&3&2\\2&-3&3\end{vmatrix}}=0+{\begin{vmatrix}1&-4&2\\0&-5&1\\0&0&-1\end{vmatrix}}=5$

${\begin{vmatrix}\end{vmatrix}}=1$

${\begin{vmatrix}a_{1}+b_{1}\end{vmatrix}}=a_{1}+b_{1}$

${\begin{vmatrix}a_{1}+b_{1}&a_{1}+b_{2}\\a_{2}+b_{1}&a_{2}+b_{2}\end{vmatrix}}=-(a_{1}-a_{2})(b_{1}-b_{2})$

${\begin{vmatrix}a_{1}+b_{1}&a_{1}+b_{2}&a_{1}+b_{3}\\a_{2}+b_{1}&a_{2}+b_{2}&a_{2}+b_{3}\\a_{3}+b_{1}&a_{3}+b_{2}&a_{3}+b_{3}\end{vmatrix}}=0$

$D_{n}={\begin{vmatrix}a&b&\cdots &b&b\\c&a&\cdots &b&b\\\vdots &\vdots &\ddots &\vdots &\vdots \\c&c&\cdots &a&b\\c&c&\cdots &c&a\\\end{vmatrix}}_{n\times n,\ b\neq c}$

$D_{n}={\begin{vmatrix}a-b&0&\cdots &0&0\\c&a&\cdots &b&b\\\vdots &\vdots &\ddots &\vdots &\vdots \\c&c&\cdots &a&b\\c&c&\cdots &c&a\\\end{vmatrix}}+{\begin{vmatrix}b&b&\cdots &b&b\\c&a&\cdots &b&b\\\vdots &\vdots &\ddots &\vdots &\vdots \\c&c&\cdots &a&b\\c&c&\cdots &c&a\\\end{vmatrix}}=(a-b)D_{n-1}+b{\begin{vmatrix}1&1&\cdots &1&1\\c&a&\cdots &b&b\\\vdots &\vdots &\ddots &\vdots &\vdots \\c&c&\cdots &a&b\\c&c&\cdots &c&a\\\end{vmatrix}}=(a-b)D_{n-1}+b{\begin{vmatrix}1&1&\cdots &1&1\\0&a-c&\cdots &b-c&b-c\\\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&\cdots &a-c&b-c\\0&0&\cdots &0&a-c\end{vmatrix}}=(a-b)D_{n-1}+b(a-c)^{n-1}$

$D_{n}=(a-b)D_{n-1}+b(a-c)^{n-1}$

$D_{n}=(a-c)D_{n-1}+c(a-b)^{n-1}$

$\therefore D_{n}={\frac {b(a-c)^{n}-c(a-b)^{n}}{b-c}}$

라플라스 전개, 소행렬식 M_ij, 여인자 C_ij

• $\det A=\sum _{j=1}^{n}a_{ij}C_{ij}$, 여기서 $C_{ij}=(-1)^{i+j}M_{ij}$
• $\det A(t)=\det A+t\sum _{i,j=1}^{n}C_{ij}$, 여기서 $A(t)_{ij}=A_{ij}+t$

${\begin{vmatrix}a&b&0&\cdots &0\\0&a&b&\cdots &0\\0&0&a&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\b&0&0&\cdots &a\end{vmatrix}}_{n\times n}=a{\begin{vmatrix}a&b&\cdots &0\\0&a&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &a\end{vmatrix}}_{(n-1)\times (n-1)}+(-1)^{n+1}b{\begin{vmatrix}b&a&\cdots &0\\0&b&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &b\end{vmatrix}}{(n-1)\times (n-1)}=a^{n}+(-1)^{n+1}b^{n}$

${\begin{vmatrix}1&2&2&\cdots &2\\2&2&2&\cdots &2\\2&2&3&\cdots &2\\\vdots &\vdots &\vdots &\ddots &\vdots \\2&2&2&\cdots &n\\\end{vmatrix}}=-2(n-2)!$

${\begin{vmatrix}1+x_{1}y_{1}&1+x_{1}y_{2}&\cdots &1+x_{1}y_{n}\\1+x_{2}y_{1}&1+x_{2}y_{2}&\cdots &1+x_{2}y_{n}\\\vdots &\vdots &\ddots &\vdots \\1+x_{n}y_{1}&1+x_{n}y_{2}&\cdots &1+x_{n}y_{n}\\\end{vmatrix}}={\begin{cases}0,&n\geq 3\\(x_{1}-x_{2})(y_{1}-y_{2}),&n=2\\1+x_{1}y_{1},&n=1\\1,&n=0\end{cases}}$

${\begin{vmatrix}1+t&t&\cdots &t\\t&2+t&\cdots &t\\\vdots &\vdots &\ddots &\vdots \\t&t&\cdots &n+t\end{vmatrix}}=n!(1+t\sum _{i=1}^{n}{\frac {1}{i}})$

귀납

$D_{n}={\begin{vmatrix}a+b&ab&0&\cdots &0\\1&a+b&ab&\cdots &0\\0&1&a+b&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\cdots &a+b\end{vmatrix}}_{n\times n,\ a\neq b}$

$D_{n}=(a+b)D_{n-1}-abD_{n-2}$

$D_{n}-aD_{n-1}=b(D_{n-1}-aD_{n-2})$

$D_{n}-aD_{n-1}=b^{n-1}(D_{1}-D_{0})=b^{n}$

$D_{n}-bD_{n-1}=a^{n}$

$\therefore D_{n}={\frac {\begin{vmatrix}b^{n}&-a\\a^{n}&-b\end{vmatrix}}{\begin{vmatrix}1&-a\\1&-b\end{vmatrix}}}={\frac {a^{n+1}-b^{n+1}}{a-b}}=\sum _{k=0}^{n}a^{n-k}b^{k}$

${\begin{vmatrix}x&0&\cdots &0&a_{0}\\-1&x&\cdots &0&a_{1}\\\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&\cdots &x&a_{n-2}\\0&0&\cdots &-1&a_{n-1}\end{vmatrix}}=x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}$

방데르몽드 행렬식

${\begin{vmatrix}1&1&\cdots &1&1\\a_{1}&a_{2}&\cdots &a_{n-1}&a_{n}\\\vdots &\vdots &\ddots &\vdots &\vdots \\a_{1}^{n-2}&a_{2}^{n-2}&\cdots &a_{n-1}^{n-2}&a_{n}^{n-2}\\a_{1}^{n-1}&a_{2}^{n-1}&\cdots &a_{n-1}^{n-1}&a_{n}^{n-1}\end{vmatrix}}=\prod _{1\leq i<j\leq n}(a_{j}-a_{i})$

$D={\begin{vmatrix}1&1&\cdots &1&1\\x_{1}&x_{2}&\cdots &x_{n-1}&x_{n}\\\vdots &\vdots &\ddots &\vdots &\vdots \\x_{1}^{n-2}&x_{2}^{n-2}&\cdots &x_{n-1}^{n-2}&x_{n}^{n-2}\\x_{1}^{n}&x_{2}^{n}&\cdots &x_{n-1}^{n}&x_{n}^{n}\end{vmatrix}}$

$D(y):={\begin{vmatrix}1&1&\cdots &1&1\\x_{1}&x_{2}&\cdots &x_{n}&y\\\vdots &\vdots &\ddots &\vdots &\vdots \\x_{1}^{n-1}&x_{2}^{n-1}&\cdots &x_{n}^{n-1}&y^{n-1}\\x_{1}^{n}&x_{2}^{n}&\cdots &x_{n}^{n}&y^{n}\end{vmatrix}}=\prod _{i=1}^{n}(y-x_{i})\prod _{1\leq i<j\leq n}(x_{j}-x_{i})=t_{0}+t_{1}y+\cdots +t_{n-1}y^{n-1}+t_{n}y^{n}$

$D=t_{n-1}=-\sum _{i=1}^{n}x_{i}\prod _{1\leq i<j\leq n}(x_{j}-x_{i})$

치환

$\sum _{j_{1}j_{2}\cdots j_{n}\in S_{n}}{\begin{vmatrix}a_{1j_{1}}&a_{1j_{2}}&\cdots &a_{1j_{n}}\\a_{2j_{1}}&a_{2j_{2}}&\cdots &a_{nj_{n}}\\\vdots &\vdots &\ddots &\vdots \\a_{nj_{1}}&a_{nj_{2}}&\cdots &a_{nj_{n}}\end{vmatrix}}=0$