43. 行列式的性质

设 $a_{1}$ , $a_{2}$ , $\dots$ , $a_{n}$ , $x$ , $y$ 是 $n + 2$ 个 $n \times 1$ 阵. 设 $A = [a_{1}, a_{2}, \dots, a_{n}]$ . 设 $s$ 是一个数. 行列式有如下性质:

(1) 一个方阵与其转置的行列式相等. 于是, 我不必同时讲行的性质与列的性质, 因为您总可用转置, 译列的性质为行的性质 (或译行的性质为列的性质).

(2) 若一个阵的某一列是二组数的和, 那么此阵的行列式等于二个阵的行列式的和, 而这二个阵, 除这一列以外, 全与原阵的对应的列一样. 用公式写, 就是 $= det [\dots, a_{j - 1}, x + y, a_{j + 1}, \dots] det [\dots, a_{j - 1}, x, a_{j + 1}, \dots] + det [\dots, a_{j - 1}, y, a_{j + 1}, \dots],$ 其中, 未写的列是不变的, 下同.

(3) 以一个数乘阵的一列后得到的阵的行列式等于以此数乘原阵的行列式. 用公式写, 就是 $det [\dots, a_{j - 1}, s x, a_{j + 1}, \dots] = s det [\dots, a_{j - 1}, x, a_{j + 1}, \dots] .$

(4) 若一个阵有二列相同, 则其行列式为 $0$ .

(5) 若交换一个阵的二列, 则其行列式变号.

(6) 若一个阵有二列成比例, 则其行列式为 $0$ . (利用性质 (3) (4) 即知.)

(7) 若加一个阵的一列的倍于另一列, 则其行列式不变. (利用性质 (2) (6) 即知.)

(8) 单位阵的行列式是 $1$ .

(9) 设 $j$ 是不超过 $n$ 的正整数. 则 $det (A) = i = 1 \sum n (- 1)^{i + j} [A]_{i, j} det (A (i ∣ j)) .$

不难看出, (2) (3) 的联合是多线性, 且 (4) 就是交错性.

这些性质是有用的. 我们看几个例.

例 43.1. 设 $n$ 级阵 $A$ 适合: 当 $1 ⩽ j < i ⩽ n$ 时, $[A]_{i, j} = 0$ . 形象地, $A = ⎣ ⎡ [A]_{1, 1} 0 ⋮ 00 [A]_{1, 2} [A]_{2, 2} ⋮ 00 \dots \dots ⋱ \dots \dots [A]_{1, n - 1} [A]_{2, n - 1} ⋮ [A]_{n - 1, n - 1} 0 [A]_{1, n} [A]_{2, n} ⋮ [A]_{n - 1, n} [A]_{n, n} ⎦ ⎤ .$ 我们计算 $det (A)$ .

按列 $1$ 展开, 有 $det (A) = = = = i = 1 \sum n (- 1)^{i + 1} [A]_{i, 1} det (A (i ∣1)) (- 1)^{1 + 1} [A]_{1, 1} det (A (1∣1)) + i = 2 \sum n (- 1)^{i + 1} [A]_{i, 1} det (A (i ∣1)) [A]_{1, 1} det (A (1∣1)) + i = 2 \sum n (- 1)^{i + 1} 0 det (A (i ∣1)) [A]_{1, 1} det (A (1∣1)) .$

不难看出, 当 $1 ⩽ j < i ⩽ n - 1$ 时, 也有 $[A (1∣1)]_{i, j} = 0$ . 于是, 类似地, $det (A (1∣1)) = [A (1∣1)]_{1, 1} det ((A (1∣1)) (1∣1)) = [A]_{2, 2} det (A (1, 2 ∣ 1, 2)) .$ 故 $det (A) = [A]_{1, 1} [A]_{2, 2} det (A (1, 2 ∣ 1, 2)) .$

……

最后, 我们得到 $det (A) = = = [A]_{1, 1} [A]_{2, 2} \dots [A]_{n - 1, n - 1} det (A (1, 2, \dots, n - 1 ∣ 1, 2, \dots, n - 1)) [A]_{1, 1} [A]_{2, 2} \dots [A]_{n - 1, n - 1} [A]_{n, n} [A]_{1, 1} [A]_{2, 2} \dots [A]_{n, n} .$

例 43.2. 设 $n$ 级阵 $A$ 适合: 当 $1 ⩽ i < j ⩽ n$ 时, $[A]_{i, j} = 0$ . 形象地, $A = ⎣ ⎡ [A]_{1, 1} [A]_{2, 1} ⋮ [A]_{n - 1, 1} [A]_{n, 1} 0 [A]_{2, 2} ⋮ [A]_{n - 1, 2} [A]_{n, 2} \dots \dots ⋱ \dots \dots 00 ⋮ [A]_{n - 1, n - 1} [A]_{n, n - 1} 00 ⋮ 0 [A]_{n, n} ⎦ ⎤ .$ 我们计算 $det (A)$ .

不难看出, $A$ 的转置 $A^{T}$ 适合: 当 $1 ⩽ j < i ⩽ n$ 时, $[A^{T}]_{i, j} = [A]_{j, i} = 0$ . 由性质 (1) 与上例的结果, $det (A) = = = det (A^{T}) [A^{T}]_{1, 1} [A^{T}]_{2, 2} \dots [A^{T}]_{n, n} [A]_{1, 1} [A]_{2, 2} \dots [A]_{n, n} .$

例 43.3. 运用性质 (7), 可作出一些 $0$ , 使计算简单. 比如, 设 $A = ⎣ ⎡ 438951276 ⎦ ⎤ .$ 则 $det (A) = = = = = = = = = = det ⎣ ⎡ 438951 2 + 9 7 + 5 6 + 1 ⎦ ⎤ det ⎣ ⎡ 438951 2 + 9 + 4 7 + 5 + 3 6 + 1 + 8 ⎦ ⎤ det ⎣ ⎡ 438951151515 ⎦ ⎤ det ⎣ ⎡ 438 9 - 15 \cdot \frac{1}{15} 5 - 15 \cdot \frac{1}{15} 1 - 15 \cdot \frac{1}{15} 151515 ⎦ ⎤ det ⎣ ⎡ 4 - 15 \cdot \frac{8}{15} 3 - 15 \cdot \frac{8}{15} 8 - 15 \cdot \frac{8}{15} 840151515 ⎦ ⎤ det ⎣ ⎡ - 4 - 5 0 840151515 ⎦ ⎤ det ⎣ ⎡ - 4 + 8 \cdot \frac{5}{4} - 5 + 4 \cdot \frac{5}{4} 0 + 0 \cdot \frac{5}{4} 840151515 ⎦ ⎤ det ⎣ ⎡ 600840151515 ⎦ ⎤ 6 \cdot 4 \cdot 15 360.$ 我们用 $3$ 级阵的行列式的公式验证结果: $det (A) = = = 4 \cdot 5 \cdot 6 + 3 \cdot 1 \cdot 2 + 8 \cdot 9 \cdot 7 - 4 \cdot 1 \cdot 7 - 3 \cdot 9 \cdot 6 - 8 \cdot 5 \cdot 2 120 + 6 + 504 - 28 - 162 - 80 360.$

例 43.4. 设 $A = ⎣ ⎡ 16594310615211714138121 ⎦ ⎤ .$ 则 $det (A) = = = det ⎣ ⎡ 16594 3 - 2 10 - 11 6 - 7 15 - 14 211714138121 ⎦ ⎤ det ⎣ ⎡ 16 - 13 5 - 8 9 - 12 4 - 1 3 - 2 10 - 11 6 - 7 15 - 14 211714138121 ⎦ ⎤ det ⎣ ⎡ 3 - 3 - 3 3 1 - 1 - 1 1 211714138121 ⎦ ⎤ .$ 利用性质 (6) 可知, 最后一个阵的行列式为 $0$ . 所以, $A$ 的行列式也是 $0$ .

我们当然也可用 $4$ 级阵的行列式的公式验证结果. 不过, 这是复杂的. 首先, 根据定义, $= det (A) + (- 1)^{1 + 1} [A]_{1, 1} det (A (1∣1)) + (- 1)^{2 + 1} [A]_{2, 1} det (A (2∣1)) + (- 1)^{3 + 1} [A]_{3, 1} det (A (3∣1)) + (- 1)^{4 + 1} [A]_{4, 1} det (A (4∣1)) .$ 可算出 $det (A (1∣1)) = det ⎣ ⎡ 10615117148121 ⎦ ⎤ = 136, det (A (2∣1)) = det ⎣ ⎡ 3615271413121 ⎦ ⎤ = - 408, det (A (3∣1)) = det ⎣ ⎡ 31015211141381 ⎦ ⎤ = - 408, det (A (4∣1)) = det ⎣ ⎡ 3106211713812 ⎦ ⎤ = 136.$ 所以 $det (A) = 16 \cdot 136 - 5 \cdot (- 408) + 9 \cdot (- 408) - 4 \cdot 136 = 0.$

例 43.5. 设 $x$ , $y$ 是数. 作 $n$ 级阵 $A$ 如下: $[A]_{i, j} = {x, y, i = j; i \neq = j .$ 形象地, $A = ⎣ ⎡ x y ⋮ y y y x ⋮ y y \dots \dots \dots \dots y y ⋮ x y y y ⋮ y x ⎦ ⎤ .$ 我们计算 $det (A)$ .

我们加 $A$ 的列 $1$ , $2$ , $\dots$ , $n - 1$ 于列 $n$ , 得 $n$ 级阵 $A_{1} = ⎣ ⎡ x y ⋮ y y y x ⋮ y y \dots \dots \dots \dots y y ⋮ x y x + (n - 1) y x + (n - 1) y ⋮ x + (n - 1) y x + (n - 1) y ⎦ ⎤ .$ 注意到, $A_{1}$ 的前 $n - 1$ 列与 $A$ 的前 $n - 1$ 列一样, 但 $A_{1}$ 的列 $n$ 的元全为 $x + (n - 1) y$ . 用 $n - 1$ 次性质 (7), 有 $det (A) = det (A_{1})$ .

作 $n$ 级阵 $A_{2} = ⎣ ⎡ x y ⋮ y y y x ⋮ y y \dots \dots \dots \dots y y ⋮ x y 11 ⋮ 11 ⎦ ⎤ .$ 注意到, $A_{2}$ 的前 $n - 1$ 列与 $A_{1}$ 的前 $n - 1$ 列一样, 但 $A_{2}$ 的列 $n$ 的元全为 $1$ . 用性质 (3), 有 $det (A) = det (A_{1}) = (x + (n - 1) y) det (A_{2})$ .

我们加 $A_{2}$ 的列 $n$ 的 $- y$ 倍于列 $1$ , 列 $n$ 的 $- y$ 倍于列 $2$ , ……, 列 $n$ 的 $- y$ 倍于列 $n - 1$ , 得 $n$ 级阵 $A_{3} = ⎣ ⎡ x - y 0 ⋮ 00 0 x - y ⋮ 00 \dots \dots \dots \dots 00 ⋮ x - y 0 11 ⋮ 11 ⎦ ⎤ .$ 用 $n - 1$ 次性质 (7), 有 $det (A) = (x + (n - 1) y) det (A_{2}) = (x + (n - 1) y) det (A_{3})$ . 注意到, $1 ⩽ j < i ⩽ n$ 时, 有 $[A_{3}]_{i, j} = 0$ . 故 $det (A_{3}) = (x - y)^{n - 1} \cdot 1 = (x - y)^{n - 1} .$ 故 $det (A) = (x + (n - 1) y) det (A_{3}) = (x + (n - 1) y) (x - y)^{n - 1} .$

例 43.6. 设 $x_{1}$ , $x_{2}$ , $\dots$ , $x_{n}$ 是数. 作 $n$ 级阵 $V (x_{1}, x_{2}, \dots, x_{n})$ 如下: $[V (x_{1}, x_{2}, \dots, x_{n})]_{i, j} = x_{i}^{j - 1} .$ 形象地, $V (x_{1}, x_{2}, \dots, x_{n}) = ⎣ ⎡ 111 ⋮ 11 x_{1} x_{2} x_{3} ⋮ x_{n - 1} x_{n} x_{1}^{2} x_{2}^{2} x_{3}^{2} ⋮ x_{n - 1}^{2} x_{n}^{2} \dots \dots \dots \dots \dots x_{1}^{n - 2} x_{2}^{n - 2} x_{3}^{n - 2} ⋮ x_{n - 1}^{n - 2} x_{n}^{n - 2} x_{1}^{n - 1} x_{2}^{n - 1} x_{3}^{n - 1} ⋮ x_{n - 1}^{n - 1} x_{n}^{n - 1} ⎦ ⎤ .$ 我们计算 $det (V (x_{1}, x_{2}, \dots, x_{n}))$ .

我们加 $V (x_{1}, x_{2}, \dots, x_{n})$ 的列 $n - 1$ 的 $- x_{n}$ 倍于列 $n$ , 列 $n - 2$ 的 $- x_{n}$ 倍于列 $n - 1$ , ……, 列 $1$ 的 $- x_{n}$ 倍于列 $2$ , 得 $n$ 级阵 $= = A ⎣ ⎡ 111 ⋮ 11 x_{1} - x_{n} x_{2} - x_{n} x_{3} - x_{n} ⋮ x_{n - 1} - x_{n} 0 x_{1}^{2} - x_{1} x_{n} x_{2}^{2} - x_{2} x_{n} x_{3}^{2} - x_{3} x_{n} ⋮ x_{n - 1}^{2} - x_{n - 1} x_{n} 0 \dots \dots \dots \dots \dots x_{1}^{n - 2} - x_{1}^{n - 3} x_{n} x_{2}^{n - 2} - x_{2}^{n - 3} x_{n} x_{3}^{n - 2} - x_{3}^{n - 3} x_{n} ⋮ x_{n - 1}^{n - 2} - x_{n - 1}^{n - 3} x_{n} 0 x_{1}^{n - 1} - x_{1}^{n - 2} x_{n} x_{2}^{n - 1} - x_{2}^{n - 2} x_{n} x_{3}^{n - 1} - x_{3}^{n - 2} x_{n} ⋮ x_{n - 1}^{n - 1} - x_{n - 1}^{n - 2} x_{n} 0 ⎦ ⎤ ⎣ ⎡ 111 ⋮ 11 x_{1} - x_{n} x_{2} - x_{n} x_{3} - x_{n} ⋮ x_{n - 1} - x_{n} 0 x_{1} (x_{1} - x_{n}) x_{2} (x_{2} - x_{n}) x_{3} (x_{3} - x_{n}) ⋮ x_{n - 1} (x_{n - 1} - x_{n}) 0 \dots \dots \dots \dots \dots x_{1}^{n - 3} (x_{1} - x_{n}) x_{2}^{n - 3} (x_{2} - x_{n}) x_{3}^{n - 3} (x_{3} - x_{n}) ⋮ x_{n - 1}^{n - 3} (x_{n - 1} - x_{n}) 0 x_{1}^{n - 2} (x_{1} - x_{n}) x_{2}^{n - 2} (x_{2} - x_{n}) x_{3}^{n - 2} (x_{3} - x_{n}) ⋮ x_{n - 1}^{n - 2} (x_{n - 1} - x_{n}) 0 ⎦ ⎤;$ 具体地, $[A]_{i, j} = {1, x_{i}^{j - 1} - x_{i}^{j - 2} x_{n} = x_{i}^{j - 2} (x_{i} - x_{n}), j = 1; j > 2.$ 由性质 (7), 有 $det (V (x_{1}, x_{2}, \dots, x_{n})) = det (A) .$

按行 $n$ 展开, 有 $det (V (x_{1}, x_{2}, \dots, x_{n})) = det (A) = (- 1)^{n + 1} det (B),$ 其中 $B = A (n ∣1)$ , 且 $[B]_{i, j} = x_{i}^{j - 1} (x_{i} - x_{n})$ ; 形象地, $B = ⎣ ⎡ x_{1} - x_{n} x_{2} - x_{n} x_{3} - x_{n} ⋮ x_{n - 1} - x_{n} x_{1} (x_{1} - x_{n}) x_{2} (x_{2} - x_{n}) x_{3} (x_{3} - x_{n}) ⋮ x_{n - 1} (x_{n - 1} - x_{n}) \dots \dots \dots \dots x_{1}^{n - 3} (x_{1} - x_{n}) x_{2}^{n - 3} (x_{2} - x_{n}) x_{3}^{n - 3} (x_{3} - x_{n}) ⋮ x_{n - 1}^{n - 3} (x_{n - 1} - x_{n}) x_{1}^{n - 2} (x_{1} - x_{n}) x_{2}^{n - 2} (x_{2} - x_{n}) x_{3}^{n - 2} (x_{3} - x_{n}) ⋮ x_{n - 1}^{n - 2} (x_{n - 1} - x_{n}) ⎦ ⎤ .$

用 $n - 1$ 次性质 (3) (关于行), 有 $det (B) = = (x_{1} - x_{n}) (x_{2} - x_{n}) \dots (x_{n - 1} - x_{n}) det (V (x_{1}, x_{2}, \dots, x_{n - 1})) (- 1)^{n - 1} (x_{n} - x_{1}) (x_{n} - x_{2}) \dots (x_{n} - x_{n - 1}) det (V (x_{1}, x_{2}, \dots, x_{n - 1})) .$ 故 $= = = det (V (x_{1}, x_{2}, \dots, x_{n})) (- 1)^{n + 1} det (B) (- 1)^{n + 1} (- 1)^{n - 1} (x_{n} - x_{1}) (x_{n} - x_{2}) \dots (x_{n} - x_{n - 1}) det (V (x_{1}, x_{2}, \dots, x_{n - 1})) (x_{n} - x_{1}) (x_{n} - x_{2}) \dots (x_{n} - x_{n - 1}) det (V (x_{1}, x_{2}, \dots, x_{n - 1})) .$

类似地, $= det (V (x_{1}, x_{2}, \dots, x_{n - 1})) (x_{n - 1} - x_{1}) (x_{n - 1} - x_{2}) \dots (x_{n - 1} - x_{n - 2}) det (V (x_{1}, x_{2}, \dots, x_{n - 2})) .$ 故 $det (V (x_{1}, x_{2}, \dots, x_{n})) = \cdot (x_{n} - x_{1}) (x_{n} - x_{2}) \dots (x_{n} - x_{n - 1}) \cdot (x_{n - 1} - x_{1}) (x_{n - 1} - x_{2}) \dots (x_{n - 1} - x_{n - 2}) \cdot det (V (x_{1}, x_{2}, \dots, x_{n - 2})) .$

……

最后, 我们有 $det (V (x_{1}, x_{2}, \dots, x_{n})) = = = = \cdot (x_{n} - x_{1}) (x_{n} - x_{2}) \dots (x_{n} - x_{n - 1}) \cdot (x_{n - 1} - x_{1}) (x_{n - 1} - x_{2}) \dots (x_{n - 1} - x_{n - 2}) \cdot \dots\dots\dots\dots\dots\dots\dots\dots \cdot (x_{3} - x_{1}) (x_{3} - x_{2}) \cdot (x_{2} - x_{1}) \cdot det (V (x_{1})) \cdot (x_{n} - x_{1}) (x_{n} - x_{2}) \dots (x_{n} - x_{n - 1}) \cdot (x_{n - 1} - x_{1}) (x_{n - 1} - x_{2}) \dots (x_{n - 1} - x_{n - 2}) \cdot \dots\dots\dots\dots\dots\dots\dots\dots \cdot (x_{3} - x_{1}) (x_{3} - x_{2}) \cdot (x_{2} - x_{1}) \cdot 1 2 ⩽ v ⩽ n \prod 1 ⩽ u ⩽ v - 1 \prod (x_{v} - x_{u}) 1 ⩽ u < v ⩽ n \prod (x_{v} - x_{u}) .$

不难看出, 若 $V (x_{1}, x_{2}, \dots, x_{n})$ 的行列式非零, 则 $x_{1}$ , $x_{2}$ , $\dots$ , $x_{n}$ 互不相同; 反过来, 若 $x_{1}$ , $x_{2}$ , $\dots$ , $x_{n}$ 互不相同, 则 $V (x_{1}, x_{2}, \dots, x_{n})$ 的行列式非零.

例 43.7. 设 $a_{1}$ , $a_{2}$ , $\dots$ , $a_{n}$ 是 $n$ 个数. 作 $n$ 级阵 $C (a_{1}, a_{2}, \dots, a_{n})$ 如下: $[C (a_{1}, a_{2}, \dots, a_{n})]_{i, j} = ⎩ ⎨ ⎧ - a_{n - i + 1}, 1, 0, j = n; 2 ⩽ i = j + 1 ⩽ n; 其他 .$ 形象地, $C (a_{1}, a_{2}, \dots, a_{n}) = ⎣ ⎡ 010 ⋮ 00 001 ⋮ 00 \dots \dots \dots \dots \dots 000 ⋮ 10 000 ⋮ 01 - a_{n} - a_{n - 1} - a_{n - 2} ⋮ - a_{2} - a_{1} ⎦ ⎤ .$ 设 $x$ 是一个数. 记 $P (x; a_{1}, a_{2}, \dots, a_{n}) = x I_{n} - C (a_{1}, a_{2}, \dots, a_{n})$ . 形象地, $P (x; a_{1}, a_{2}, \dots, a_{n}) = ⎣ ⎡ x - 1 0 ⋮ 00 0 x - 1 ⋮ 00 \dots \dots \dots \dots \dots 000 ⋮ - 1 0 000 ⋮ x - 1 a_{n} a_{n - 1} a_{n - 2} ⋮ a_{2} x + a_{1} ⎦ ⎤ .$ 我们计算 $det (P (x; a_{1}, a_{2}, \dots, a_{n}))$ .

注意到, 当 $1 < j < n$ 时, $[P (x; a_{1}, a_{2}, \dots, a_{n})]_{1, j} = 0$ . 于是, 按行 $1$ 展开, 有 $= det (P (x; a_{1}, a_{2}, \dots, a_{n})) + (- 1)^{1 + 1} [P (x; a_{1}, a_{2}, \dots, a_{n})]_{1, 1} det ((P (x; a_{1}, a_{2}, \dots, a_{n})) (1∣1)) + (- 1)^{1 + n} [P (x; a_{1}, a_{2}, \dots, a_{n})]_{1, n} det ((P (x; a_{1}, a_{2}, \dots, a_{n})) (1∣ n)) .$ 不难写出, $[P (x; a_{1}, a_{2}, \dots, a_{n})]_{1, 1} = x, [P (x; a_{1}, a_{2}, \dots, a_{n})]_{1, n} = a_{n} .$ 不难写出, $(P (x; a_{1}, a_{2}, \dots, a_{n})) (1∣1) = = ⎣ ⎡ x - 1 ⋮ 00 \dots \dots \dots \dots 00 ⋮ - 1 0 00 ⋮ x - 1 a_{n - 1} a_{n - 2} ⋮ a_{2} x + a_{1} ⎦ ⎤ P (x; a_{1}, a_{2}, \dots, a_{n - 1}) .$ 故 $det ((P (x; a_{1}, a_{2}, \dots, a_{n})) (1∣1)) = det (P (x; a_{1}, a_{2}, \dots, a_{n - 1})) .$ 不难写出, $(P (x; a_{1}, a_{2}, \dots, a_{n})) (1∣ n) = ⎣ ⎡ - 1 0 ⋮ 00 x - 1 ⋮ 00 \dots \dots \dots \dots 00 ⋮ - 1 0 00 ⋮ x - 1 ⎦ ⎤;$ 具体地, $[(P (x; a_{1}, a_{2}, \dots, a_{n})) (1∣ n)]_{i, j} = ⎩ ⎨ ⎧ - 1, x, 0, 1 ⩽ i = j ⩽ n - 1; 1 ⩽ i = j - 1 ⩽ n - 2; 其他 .$ 故 $1 ⩽ j < i ⩽ n - 1$ 时, 有 $[(P (x; a_{1}, a_{2}, \dots, a_{n})) (1∣ n)]_{i, j} = 0$ . 则 $det ((P (x; a_{1}, a_{2}, \dots, a_{n})) (1∣ n)) = (- 1)^{n - 1} .$

综上, $= = = det (P (x; a_{1}, a_{2}, \dots, a_{n})) + (- 1)^{1 + 1} [P (x; a_{1}, a_{2}, \dots, a_{n})]_{1, 1} det ((P (x; a_{1}, a_{2}, \dots, a_{n})) (1∣1)) + (- 1)^{1 + n} [P (x; a_{1}, a_{2}, \dots, a_{n})]_{1, n} det ((P (x; a_{1}, a_{2}, \dots, a_{n})) (1∣ n)) x det (P (x; a_{1}, a_{2}, \dots, a_{n - 1})) + (- 1)^{1 + n} a_{n} (- 1)^{n - 1} x det (P (x; a_{1}, a_{2}, \dots, a_{n - 1})) + a_{n} .$

类似地, $det (P (x; a_{1}, a_{2}, \dots, a_{n - 1})) = x det (P (x; a_{1}, a_{2}, \dots, a_{n - 2})) + a_{n - 1} .$ 故 $det (P (x; a_{1}, a_{2}, \dots, a_{n})) = = = x det (P (x; a_{1}, a_{2}, \dots, a_{n - 1})) + a_{n} x (x det (P (x; a_{1}, a_{2}, \dots, a_{n - 2})) + a_{n - 1}) + a_{n} x^{2} det (P (x; a_{1}, a_{2}, \dots, a_{n - 2})) + a_{n - 1} x + a_{n} .$

……

最后, 我们有 $det (P (x; a_{1}, a_{2}, \dots, a_{n})) = = = = = = x det (P (x; a_{1}, a_{2}, \dots, a_{n - 1})) + a_{n} x^{2} det (P (x; a_{1}, a_{2}, \dots, a_{n - 2})) + a_{n - 1} x + a_{n} \dots\dots\dots\dots x^{n - 1} det (P (x; a_{1})) + a_{2} x^{n - 2} + \dots + a_{n - 1} x + a_{n} x^{n - 1} (x + a_{1}) + a_{2} x^{n - 2} + \dots + a_{n - 1} x + a_{n} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n - 1} x + a_{n} .$

虽然我写了多的例, 我并不想在本书具体地介绍如何计算一个阵的行列式; 那或许不是我的事. 我的目标只是带您入门行列式; 这些例的目的是使您更好地理解行列式的性质. 若您想知道计算一个阵的行列式的更多的方法, 您可以见一些线性代数教材, 或者找相关的文献.

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43. 行列式的性质