问题 2016A04

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问题 2016A04

设下列矩阵 M 是可逆阵, 试求其逆阵 $M^{- 1}$ :
$M = a_{1}^{2} a_{2} a_{1} + 1 ⋮ a_{n} a_{1} + 1 a_{1} a_{2} + 1 a_{2}^{2} ⋮ a_{n} a_{2} + 1 \dots \dots \dots a_{1} a_{n} + 1 a_{2} a_{n} + 1 ⋮ a_{n}^{2} .$

解答

令 $A = - I_{n}, C = I_{2}, B^{'} = D = (a_{1} 1 a_{2} 1 \dots \dots a_{n} 1)$ ，则 $M = A + B C D$ 。由行列式的降阶公式可知 (参考例 2.67):

∣ M ∣ = ∣ C ∣∣ A ∣∣ C^{- 1} + D A^{- 1} B ∣ = (- 1)^{n} I_{2} - (\sum_{i = 1}^{n} a_{i}^{2} \sum_{i = 1}^{n} a_{i} \sum_{i = 1}^{n} a_{i} n)

= (- 1)^{n} ((1 - n) (1 - i = 1 \sum n a_{i}^{2}) - (i = 1 \sum n a_{i})^{2}) \neq = 0.

记 $c = (1 - n) (1 - \sum_{i = 1}^{n} a_{i}^{2}) - (\sum_{i = 1}^{n} a_{i})^{2} \neq = 0$ , 再由例 2.25 (Sherman-Morrison-Woodbury 公式) 可得

M^{- 1} = A^{- 1} - A^{- 1} B (C^{- 1} + D A^{- 1} B)^{- 1} D A^{- 1}

= - I_{n} - \frac{1}{c} a_{1} a_{2} ⋮ a_{n} 11 ⋮ 1 (1 - n \sum_{i = 1}^{n} a_{i} \sum_{i = 1}^{n} a_{i} 1 - \sum_{i = 1}^{n} a_{i}^{2}) (a_{1} 1 a_{2} 1 \dots \dots a_{n} 1)

= - I_{n} - \frac{1}{c} (a_{i} a_{j} (1 - n) + (a_{i} + a_{j}) i = 1 \sum n a_{i} - i = 1 \sum n a_{i}^{2} + 1)_{n \times n} .

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问题 2016A04

问题 2016A04

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被以下题目直接调用

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