2.2.7 矩阵乘法和行列式的计算

设 $A, B$ 是两个 $n$ 阶矩阵,则 $∣ AB ∣ = ∣ A ∣ ∣ B ∣$ . 这个结论可以用来简化某些行列式的计算. 方法是将要计算的行列式通过矩阵乘法化为两个容易计算的行列式之积, 再分别计算出两个行列式, 将结果求积就可以了. 下面几个例子告诉我们如何来灵活地应用这种方法.

例 2.48 若 $n \geq 3$ ,求证下列矩阵 $A$ 的行列式值等于零:

A = 1 + x_{1} y_{1} 1 + x_{2} y_{1} ⋮ 1 + x_{n} y_{1} 1 + x_{1} y_{2} 1 + x_{2} y_{2} ⋮ 1 + x_{n} y_{2} \dots \dots \dots 1 + x_{1} y_{n} 1 + x_{2} y_{n} ⋮ 1 + x_{n} y_{n} .

证明从下列分解即可得到结论:

A = 111 ⋮ 1 x_{1} x_{2} x_{3} ⋮ x_{n} 000 ⋮ 0 \dots \dots \dots \dots 000 ⋮ 0 1 y_{1} 0 ⋮ 0 1 y_{2} 0 ⋮ 0 1 y_{3} 0 ⋮ 0 \dots \dots \dots \dots 1 y_{n} 0 ⋮ 0 .

例 2.49 计算下列 $n + 1$ 阶矩阵的行列式的值:

A = (a_{0} + b_{0})^{n} (a_{1} + b_{0})^{n} ⋮ (a_{n} + b_{0})^{n} (a_{0} + b_{1})^{n} (a_{1} + b_{1})^{n} ⋮ (a_{n} + b_{1})^{n} \dots \dots \dots (a_{0} + b_{n})^{n} (a_{1} + b_{n})^{n} ⋮ (a_{n} + b_{n})^{n}

证明将 $A$ 分解为

A = 11 ⋮ 1 C_{n}^{1} a_{0} C_{n}^{1} a_{1} ⋮ C_{n}^{1} a_{n} C_{n}^{2} a_{0}^{2} C_{n}^{2} a_{1}^{2} ⋮ C_{n}^{2} a_{n}^{2} \dots \dots \dots C_{n}^{n} a_{0}^{n} C_{n}^{n} a_{1}^{n} ⋮ C_{n}^{n} a_{n}^{n} b_{0}^{n} b_{0}^{n - 1} ⋮ 1 b_{1}^{n} b_{1}^{n - 1} ⋮ 1 b_{2}^{n} b_{2}^{n - 1} ⋮ 1 \dots \dots \dots b_{n}^{n} b_{n}^{n - 1} ⋮ 1,

于是

∣ A ∣ = C_{n}^{1} C_{n}^{2} \dots C_{n}^{n} 0 \leq i < j \leq n \prod (a_{j} - a_{i}) (b_{i} - b_{j}) .

例 2.50 设 $s_{k} = x_{1}^{k} + x_{2}^{k} + \dots + x_{n}^{k} (k \geq 1), s_{0} = n$ ,

S = s_{0} s_{1} s_{2} ⋮ s_{n - 1} s_{1} s_{2} s_{3} ⋮ s_{n} s_{2} s_{3} s_{4} ⋮ s_{n + 1} \dots \dots \dots \dots s_{n - 1} s_{n} s_{n + 1} ⋮ s_{2 n - 2}

求 $∣ S ∣$ 的值并证明若 $x_{i}$ 是实数,则 $∣ S ∣ \geq 0$ .

证明设

V = 1 x_{1} x_{1}^{2} ⋮ x_{1}^{n - 1} 1 x_{2} x_{2}^{2} ⋮ x_{2}^{n - 1} 1 x_{3} x_{3}^{2} ⋮ x_{3}^{n - 1} \dots \dots \dots \dots 1 x_{n} x_{n}^{2} ⋮ x_{n}^{n - 1}

则 $S = V V^{'}$ ,因此

∣ S ∣ = ∣ V ∣^{2} = 1 \leq i < j \leq n \prod (x_{j} - x_{i})^{2} \geq 0.

例 2.51 计算下列矩阵 $A$ 的行列式的值:

A = x y z w y - x - w z - z - w x y w - z y - x

解注意到

A A^{'} = x y z w y - x - w z - z - w x y w - z y - x x y - z w y - x - w - z z - w x y w z y - x = u 000 0 u 00 00 u 0 000 u,

其中 $u = x^{2} + y^{2} + z^{2} + w^{2}$ ,因此

∣ A ∣^{2} = (x^{2} + y^{2} + z^{2} + w^{2})^{4} .

在矩阵 $A$ 中令 $x = 1, y = z = w = 0$ ,显然 $∣ A ∣ = 1$ ,故

∣ A ∣ = (x^{2} + y^{2} + z^{2} + w^{2})^{2} .

A = a_{1} a_{n} a_{n - 1} ⋮ a_{2} a_{2} a_{1} a_{n} ⋮ a_{3} a_{3} a_{2} a_{1} ⋮ a_{4} \dots \dots \dots \dots a_{n} a_{n - 1} a_{n - 2} ⋮ a_{1}

解作多项式 $f (x) = a_{1} + a_{2} x + a_{3} x^{2} + \dots + a_{n} x^{n - 1}$ ,令 $ε_{1}, ε_{2}, \dots, ε_{n}$ 是 1 的所有 $n$ 次方根. 又令

V = 1 ε_{1} ε_{1}^{2} ⋮ ε_{1}^{n - 1} 1 ε_{2} ε_{2}^{2} ⋮ ε_{2}^{n - 1} 1 ε_{3} ε_{3}^{2} ⋮ ε_{3}^{n - 1} \dots \dots \dots \dots 1 ε_{n} ε_{n}^{2} ⋮ ε_{n}^{n - 1}

则

AV = f (ε_{1}) ε_{1} f (ε_{1}) ε_{1}^{2} f (ε_{1}) ⋮ ε_{1}^{n - 1} f (ε_{1}) f (ε_{2}) ε_{2} f (ε_{2}) ε_{2}^{2} f (ε_{2}) ⋮ ε_{2}^{n - 1} f (ε_{2}) f (ε_{3}) ε_{3} f (ε_{3}) ε_{3}^{2} f (ε_{3}) ⋮ ε_{3}^{n - 1} f (ε_{3}) \dots \dots \dots \dots f (ε_{n}) ε_{n} f (ε_{n}) ε_{n}^{2} f (ε_{n}) ⋮ ε_{n}^{n - 1} f (ε_{n}) .

因此

∣ A ∣ ∣ V ∣ = ∣ AV ∣ = f (ε_{1}) f (ε_{2}) \dots f (ε_{n}) ∣ V ∣ .

因为 $ε_{i}$ 互不相同,所以 $∣ V ∣ \neq = 0$ ,从而

∣ A ∣ = f (ε_{1}) f (ε_{2}) \dots f (ε_{n}) .

A = cos θ cos n θ cos (n - 1) θ ⋮ cos 2 θ cos 2 θ cos θ cos n θ ⋮ cos 3 θ cos 3 θ cos 2 θ cos θ ⋮ cos 4 θ \dots \dots \dots \dots cos n θ cos (n - 1) θ cos (n - 2) θ ⋮ cos θ .

解由上面的结论可知

∣ A ∣ = f (ε_{1}) f (ε_{2}) \dots f (ε_{n})

其中 $f (x) = cos θ + x cos 2 θ + \dots + x^{n - 1} cos n θ$ . 令

g (x) = sin θ + x sin 2 θ + \dots + x^{n - 1} sin n θ,

则

f (x) + i g (x) = (cos θ + i sin θ) + x (cos θ + i sin θ)^{2} + \dots + x^{n - 1} (cos θ + i sin θ)^{n} .

用等比级数求和再比较实部, 得

f (x) = \frac{( x cos θ - 1 ) [ cos ( n + 1 ) θ - cos θ ] + x ^{n + 1} sin θ [ sin ( n + 1 ) θ - sin θ ]}{x ^{2} + 1 - 2 x cos θ} .

设 $ε_{1}, ε_{2}, \dots, ε_{n}$ 是 1 的所有 $n$ 次方根,对任意的 $ε_{i}$ ,经计算并化简,得

f (ε_{i}) = \frac{[ cos θ - cos ( n + 1 ) θ ] - ε _{i} ( 1 - cos n θ )}{[ ( cos θ + i sin θ ) - ε _{i} ] [ ( cos θ - i sin θ ) - ε _{i} ]} .

注意到对任意的 $a, b$ ,有 $a^{n} - b^{n} = (a - ε_{1} b) (a - ε_{2} b) \dots (a - ε_{n} b)$ ,因此

∣ A ∣ = i = 1 \prod n f (ε_{i}) = \frac{[ cos θ - cos ( n + 1 ) θ ] ^{n} - ( 1 - cos n θ ) ^{n}}{[ ( cos n θ + i sin n θ ) - 1 ] [ ( cos n θ - i sin n θ ) - 1 ]}

= \frac{[ cos θ - cos ( n + 1 ) θ ] ^{n} - ( 1 - cos n θ ) ^{n}}{2 ( 1 - cos n θ )}

= 2^{n - 2} sin^{n - 2} \frac{n θ}{2} (sin^{n} \frac{( n + 2 ) θ}{2} - sin^{n} \frac{n θ}{2}) . □

Youliang Zhong