## 矩陣和之行列式 (上)

$A=[a_{ij}]$$B=[b_{ij}]$$n\times n$ 階矩陣。矩陣乘積 $AB$ 的行列式存在一個優美的公式

$\det(AB)=(\det A)(\det B)$

\begin{aligned} \det(A+B)&=\begin{vmatrix} a_{11}+b_{11}&a_{12}+b_{12}\\ a_{21}+b_{21}&a_{22}+b_{22} \end{vmatrix}\\ &=\begin{vmatrix} a_{11}&a_{12}\\ a_{21}+b_{21}&a_{22}+b_{22} \end{vmatrix}+\begin{vmatrix} b_{11}&b_{12}\\ a_{21}+b_{21}&a_{22}+b_{22} \end{vmatrix}\\ &=\begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{vmatrix}+\begin{vmatrix} a_{11}&a_{12}\\ b_{21}&b_{22} \end{vmatrix}+\begin{vmatrix} b_{11}&b_{12}\\ a_{21}&a_{22} \end{vmatrix}+\begin{vmatrix} b_{11}&b_{12}\\ b_{21}&b_{22} \end{vmatrix}\end{aligned}

$\det\left(A+UV^T\right)=\det\left(I_m+V^TA^{-1}U\right)(\det A)$

$m=1$ 時，$U=\mathbf{u}$$V=\mathbf{v}$$n$ 維行向量，上式可表示成

$\det\left(A+\mathbf{u}\mathbf{v}^T\right)=\left(1+\mathbf{v}^TA^{-1}\mathbf{u}\right)(\det A)$

$A(\mathrm{adj}A)=(\det A)I$

$\mathrm{adj}A=(\det A)A^{-1}$

$\det\left(A+\mathbf{u}\mathbf{v}^T\right)=\det A+\mathbf{v}^T(\mathrm{adj}A)\mathbf{u}$

$\begin{bmatrix} A^{-1}&0\\ V^TA^{-1}&I_m \end{bmatrix}\begin{bmatrix} A&U\\ -V^T&I_m \end{bmatrix}=\begin{bmatrix} I_n&A^{-1}U\\ 0&I_m+V^TA^{-1}U \end{bmatrix}$

$\begin{vmatrix} A^{-1}&0\\ V^TA^{-1}&I_m \end{vmatrix}\cdot\begin{vmatrix} A&U\\ -V^T&I_m \end{vmatrix}=\begin{vmatrix} I_n&A^{-1}U\\ 0&I_m+V^TA^{-1}U \end{vmatrix}$

$\begin{vmatrix} A^{-1}&0\\ V^TA^{-1}&I_m \end{vmatrix}=\det(A^{-1})=(\det A)^{-1}$

$\begin{vmatrix} I_n&A^{-1}U\\ 0&I_m+V^TA^{-1}U \end{vmatrix}=\det\left(I_m+V^TA^{-1}U\right)$

$\begin{vmatrix} A&U\\ -V^T&I_m \end{vmatrix}=\det\left(I_m+V^TA^{-1}U\right)(\det A)$

$\begin{bmatrix} A&U\\ -V^T&I_m \end{bmatrix}\begin{bmatrix} I_n&0\\ V^T&I_m \end{bmatrix}=\begin{bmatrix} A+UV^T&U\\ 0&I_m \end{bmatrix}$

$\begin{vmatrix} I_n&0\\ V^T&I_m \end{vmatrix}=1$

$\begin{vmatrix} A+UV^T&U\\ 0&I_m \end{vmatrix}=\det\left(A+UV^T\right)$

\begin{aligned} \det\left(A+UV^T\right)&=\begin{vmatrix} A^{-1}&U\\ -V^T&I_m \end{vmatrix}=\det\left(I_m+V^TA^{-1}U\right)(\det A)\end{aligned}

\begin{aligned} \det\left(I_n+UV^T\right)&=\det\left(I_m+V^TI_n^{-1}U\right)(\det I_n)\\ &=\det\left(I_m+V^TU\right)\end{aligned}

$\det A=\begin{vmatrix} 2&1&\cdots&1\\ 1&2&\cdots&1\\ \vdots&\vdots&\ddots&\vdots\\ 1&1&\cdots&2 \end{vmatrix}$

$U=V=\begin{bmatrix} 1\\ 1\\ \vdots\\ 1 \end{bmatrix}$，套用 Sylvester 行列式定理，立得 $\det A=1+V^TU=n+1$