### 线性代数证明题

$$\left|\begin{array}{ccccc} n & s_1 & s_2 & \cdots & s_{n-1} \\ s_1 & s_2 & s_3 & \cdots & s_n \\ \cdots \\ s_{n-1} & s_n & s_{n+1} & \cdots & s_{2 n-2} \end{array}\right|=\prod_{1 < j < i < n}\left(\lambda_i-\lambda_j\right)^2 .$$

$$D_{\mathrm{n}}=\left|\begin{array}{ccccc} 2 n & n & & & 0 \\ n & 2 n & n & & \\ & \ddots & \ddots & \ddots & \\ & & n & 2 n & n \\ 0 & & & n & 2 n \end{array}\right| .$$

$$D=\left|\begin{array}{lll} \boldsymbol{\alpha} & \boldsymbol{\beta} & \boldsymbol{\gamma} \\ \gamma & \boldsymbol{\alpha} & \boldsymbol{\beta} \\ \boldsymbol{\beta} & \gamma & \boldsymbol{\alpha} \end{array}\right|$$

$$F_n=\left|\begin{array}{rrrrrr} 1 & 1 & & & 0 \\ -1 & 1 & 1 & & & \\ & -1 & \ddots & \ddots & & \\ & & \ddots & \ddots & 1 & \\ 0 & & & \ddots & 1 & 1 \\ 0 & & & & -1 & 1 \end{array}\right| .$$

$A_1$ 合同于 $B_1$.， $A_2$ 合同于 $B_2$, 求证:
$$\left[\begin{array}{cc} A_1 & 0 \\ 0 & A_2 \end{array}\right] \text { 合同于 }\left[\begin{array}{cc} B_1 & 0 \\ 0 & B_2 \end{array}\right] \text {. }$$

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