49. 按前二列展开行列式

本节, 我想证明一个公式. 它在交错性的一个证明与反称性的一个证明里有用.

定理 49.1. 设 $A$ 是 $n$ 级阵 ( $n ⩾ 2$ ). 则 $det (A) = 1 ⩽ i < k ⩽ n \sum det [[A]_{i, 1} [A]_{k, 1} [A]_{i, 2} [A]_{k, 2}] (- 1)^{i + k + 1 + 2} det (A (i, k ∣1, 2)) .$

证. 注意到, 当

d_{2} \neq = d_{1}

时,

[A]_{d_{2}, 2}

是

A (d_{1} ∣1)

的

(d_{2} - ρ (d_{2}, d_{1}), 1)

-元. 从而

= = = = = = = = = det (A) d_{1} = 1 \sum n (- 1)^{d_{1} + 1} [A]_{d_{1}, 1} det (A (d_{1} ∣1)) d_{1} = 1 \sum n (- 1)^{d_{1} + 1} [A]_{d_{1}, 1} 1 ⩽ d_{2} ⩽ n d_{2} \neq = d_{1} \sum (- 1)^{d_{2} - ρ (d_{2}, d_{1}) + 1} [A]_{d_{2}, 2} det (A (d_{1}, d_{2} ∣1, 2)) d_{1} = 1 \sum n 1 ⩽ d_{2} ⩽ n d_{2} \neq = d_{1} \sum (- 1)^{d_{1} + 1} [A]_{d_{1}, 1} (- 1)^{d_{2} - ρ (d_{2}, d_{1}) + 1} [A]_{d_{2}, 2} det (A (d_{1}, d_{2} ∣1, 2)) 1 ⩽ d_{1}, d_{2} ⩽ n d_{1} \neq = d_{2} \sum (- 1)^{d_{1} + 1} [A]_{d_{1}, 1} (- 1)^{d_{2} - ρ (d_{2}, d_{1}) + 1} [A]_{d_{2}, 2} det (A (d_{1}, d_{2} ∣1, 2)) 1 ⩽ d_{1}, d_{2} ⩽ n d_{1} \neq = d_{2} \sum (- 1)^{ρ (d_{1}, d_{2})} [A]_{d_{1}, 1} [A]_{d_{2}, 2} (- 1)^{d_{1} + d_{2} + 1 + 2} det (A (d_{1}, d_{2} ∣1, 2)) + 1 ⩽ d_{1}, d_{2} ⩽ n d_{1} < d_{2} \sum (- 1)^{ρ (d_{1}, d_{2})} [A]_{d_{1}, 1} [A]_{d_{2}, 2} (- 1)^{d_{1} + d_{2} + 1 + 2} det (A (d_{1}, d_{2} ∣1, 2)) + 1 ⩽ d_{1}, d_{2} ⩽ n d_{1} > d_{2} \sum (- 1)^{ρ (d_{1}, d_{2})} [A]_{d_{1}, 1} [A]_{d_{2}, 2} (- 1)^{d_{1} + d_{2} + 1 + 2} det (A (d_{1}, d_{2} ∣1, 2)) + 1 ⩽ i, k ⩽ n i < k \sum (- 1)^{ρ (i, k)} [A]_{i, 1} [A]_{k, 2} (- 1)^{i + k + 1 + 2} det (A (i, k ∣1, 2)) + 1 ⩽ k, i ⩽ n k > i \sum (- 1)^{ρ (k, i)} [A]_{k, 1} [A]_{i, 2} (- 1)^{k + i + 1 + 2} det (A (k, i ∣1, 2)) 1 ⩽ i < k ⩽ n \sum ([A]_{i, 1} [A]_{k, 2} - [A]_{k, 1} [A]_{i, 2}) (- 1)^{i + k + 1 + 2} det (A (i, k ∣1, 2)) 1 ⩽ i < k ⩽ n \sum det [[A]_{i, 1} [A]_{k, 1} [A]_{i, 2} [A]_{k, 2}] (- 1)^{i + k + 1 + 2} det (A (i, k ∣1, 2)) .

证毕.

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49. 按前二列展开行列式