## 矩陣和之行列式 (下)

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

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

$A+\mathbf{u}\mathbf{v}^T=\mathrm{diag}(1-n,2-n,3-n,\ldots,-1,0)+nE$

\begin{aligned} \det A&=(1-n)\cdots(-1)0=0\end{aligned}

$\det(\mathrm{diag}(1-n,2-n,\ldots,-1,0)+nE)=(-1)^{n-1}n!$

$\begin{vmatrix} 1&1&1&\cdots&1&0\\ 1&1&1&\cdots&0&1\\ 1&1&1&\cdots&1&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 1&0&1&\cdots&1&1\\ 0&1&1&\cdots&1&1 \end{vmatrix}$

$A+\mathbf{u}\mathbf{v}^T=-R+E$

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

\begin{aligned} \mathrm{adj}R&=(\det R)R^{-1}=(-1)^tR\end{aligned}

\begin{aligned} \mathrm{adj}(-R)&=\det(-R)(-R)^{-1}\\ &=(-1)^{n+1}(\det R)R^{-1}\\ &=(-1)^{n+1}(-1)^{t}R\\ &=(-1)^{n+t+1}R\end{aligned}

\begin{aligned} \det(-R+E)&=\det(-R)+\mathbf{e}^T\mathrm{adj}(-R)\mathbf{e}\\ &=(-1)^{n+t}+(-1)^{n+t+1}\mathbf{e}^TR\mathbf{e}\\ &=(-1)^{n+t}+(-1)^{n+t+1}n\\ &=(-1)^{n+t}(1-n)\end{aligned}

$\det(-R+E)=(-1)^{n(n+1)/2}(1-n)$

$\begin{vmatrix} 1&2&3&\cdots&n-1&n+1\\ 2&4&6&\cdots&2(n-1)+1&2n\\ 3&6&9&\cdots&3(n-1)&3n\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ n-1&2(n-1)+1&3(n-1)&\cdots&(n-1)^2&n(n-1)\\ n+1&2n&3n&\cdots&n(n-1)&n^2 \end{vmatrix}$

$A+\mathbf{u}\mathbf{v}^T=R+P$

\begin{aligned} \det(R+P)&=\det R+\mathbf{u}^T(\mathrm{adj}R)\mathbf{v}\\ &=(-1)^t+(-1)^{t}\mathbf{u}^TR\mathbf{u}\\ &=\displaystyle(-1)^t\left(1+\sum_{k=1}^nk(n-k+1)\right)\end{aligned}

$\displaystyle\sum_{k=1}^nk(n-k+1)=\frac{1}{6}n(n+1)(n+2)$

$\det(R+P)=\displaystyle(-1)^t\left(1+\frac{1}{6}n(n+1)(n+2)\right)$

$f(x)=\begin{vmatrix} x+1&x^2&x^3&\cdots&x^{n-1}&x^n\\ x^2&x^3+1&x^4&\cdots&x^n&x^{n+1}\\ x^3&x^4&x^5+1&\cdots&x^{n+1}&x^{n+2}\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ x^{n-1}&x^{n}&x^{n+1}&\cdots&x^{2n-3}+1&x^{2n-2}\\ x^n&x^{n+1}&x^{n+2}&\cdots&x^{2n-2}&x^{2n-1}+1 \end{vmatrix}$

$A+\mathbf{u}\mathbf{v}^T=I+B$

\begin{aligned} f(x)&=1+x\mathbf{u}^T\mathbf{u}=1+x\displaystyle\sum_{k=0}^{n-1}x^{2k}=1+\sum_{k=0}^{n-1}x^{2k+1}\end{aligned}

Marvin Marcus, Determinants of Sums, College Mathematics Journal, 21, 130-135, 1990.