首页工科高等代数线性代数行列式

# 余子式与行列式展开

定理 11：

∣A1,10A2,1A2,2∣=∣A1,1∣∣A2,2∣\begin {vmatrix} A_{1, 1} & 0 \\ A_{2, 1} & A_{2, 2} \end {vmatrix} = |A_{1, 1}| |A_{2, 2}|

例 1.1. 计算：

Δ=∣−10000∗2300∗4500∗∗∗13∗∗∗25∣\Delta = \begin {vmatrix} -1 & 0 & 0 & 0 & 0 \\ * & 2 & 3 & 0 & 0 \\ * & 4 & 5 & 0 & 0 \\ * & * & * & 1 & 3 \\ * & * & * & 2 & 5 \end {vmatrix}

解：

Δ=−1∣2345∣∣1325∣=−2\Delta = -1 \begin {vmatrix} 2 & 3 \\ 4 & 5 \end {vmatrix} \begin {vmatrix} 1 & 3 \\ 2 & 5 \end {vmatrix} = -2

行列式

Δ=∣0⋯a1,j⋯0a2,1⋯a2,j⋯a2,n⋮⋱⋮⋱⋮an,1⋯an,j⋯an,n∣\Delta = \begin {vmatrix} 0 & \cdots & a_{1, j} & \cdots & 0 \\ a_{2, 1} & \cdots & a_{2, j} & \cdots & a_{2, n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n, 1} & \cdots & a_{n, j} & \cdots & a_{n, n} \end {vmatrix}

的第 11 行除了第 jj 列的 a1,ja_{1, j} 以外，其余元都是 00，试将 Δ\Delta 化为 n−1n - 1 阶行列式来计算。

将 Δ\Delta 的第 jj 列依次与它左边的 j−1j - 1 列互换位置，经过 j−1j - 1 次变号变为：

Δ1=∣a1,j0⋯00⋯0a2,ja2,1⋯a2,j−1a2,j+1⋯a2,n⋮⋮⋱⋮⋮⋱⋮an,jan,1⋯an,j−1an,j+1⋯an,n∣=a1,j⋅∣a2,1⋯a2,j−1a2,j+1⋯a2,n⋮⋱⋮⋮⋱⋮an,1⋯an,j−1an,j+1⋯an,n∣=a1,jM1,j\Delta_1 = \begin {vmatrix} a_{1, j} & 0 & \cdots & 0 & 0 & \cdots & 0 \\ a_{2, j} & a_{2, 1} & \cdots & a_{2, j - 1} & a_{2, j + 1} & \cdots & a_{2, n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{n, j} & a_{n, 1} & \cdots & a_{n, j - 1} & a_{n, j + 1} & \cdots & a_{n, n} \end {vmatrix} = a_{1, j} \cdot \begin {vmatrix} a_{2, 1} & \cdots & a_{2, j - 1} & a_{2, j + 1} & \cdots & a_{2, n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{n, 1} & \cdots & a_{n, j - 1} & a_{n, j + 1} & \cdots & a_{n, n} \end {vmatrix} = a_{1, j} M_{1, j}

从而 Δ=(−1)j−1Δ1=a1,j(−1)j−1M1,j\Delta = (-1)^{j - 1} \Delta_1 = a_{1, j} (-1)^{j - 1} M_{1, j}

(−1)j−1M1,j(-1)^{j - 1} M_{1, j} 称为 a1,ja_{1, j} 在 Δ\Delta 中的代数余子式，记作 A1,jA_{1, j}。这样，上式就成为：

Δ=a1,jA1,j\Delta = a_{1, j} A_{1, j}

而对于一般的行列式，可以将第一行 (a1,1,⋯ ,a1,n)(a_{1, 1}, \cdots, a_{1, n}) 拆成 nn 个至少含有 n−1n - 1 个 00 向量的和：

(a1,1,⋯ ,a1,n)=(a1,0,⋯ ,0)+(0,a2,⋯ ,0)+⋯+(0,⋯ ,0,a1,n)(a_{1, 1}, \cdots, a_{1, n}) = (a_1, 0, \cdots, 0) + (0, a_2, \cdots, 0) + \cdots + (0, \cdots, 0, a_{1, n})

按照行列式的性质有：

Δ=∣a1,1⋯a1,j⋯a1,na2,1⋯a2,j⋯a2,n⋮⋱⋮⋱⋮an,1⋯an,j⋯an,n∣=∣a1,10⋯0a2,1a2,2⋯a2,n⋮⋮⋱⋮an,1an,2⋯an,n∣+⋯+∣0⋯0a1,j0⋯0a2,1⋯a2,j−1a2,ja2,j+1⋯a2,n⋮⋱⋮⋮⋮⋱⋮an,1⋯an,j−1an,jan,j+1⋯an,n∣+⋯+∣0⋯0a1,na2,1⋯a2,n−1a2,n⋮⋱⋮⋮an,1⋯an,n−1an,n∣=∑j=1na1,jA1,j\Delta = \begin {vmatrix} a_{1, 1} & \cdots & a_{1, j} & \cdots & a_{1, n} \\ a_{2, 1} & \cdots & a_{2, j} & \cdots & a_{2, n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n, 1} & \cdots & a_{n, j} & \cdots & a_{n, n} \end {vmatrix} = \begin {vmatrix} a_{1, 1} & 0 & \cdots & 0 \\ a_{2, 1} & a_{2, 2} & \cdots & a_{2, n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n, 1} & a_{n, 2} & \cdots & a_{n, n} \end {vmatrix} + \cdots + \begin {vmatrix} 0 & \cdots & 0 & a_{1, j} & 0 & \cdots & 0 \\ a_{2, 1} & \cdots & a_{2, j - 1} & a_{2, j} & a_{2, j + 1} & \cdots & a_{2, n} \\ \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n, 1} & \cdots & a_{n, j - 1} & a_{n, j} & a_{n, j + 1} & \cdots & a_{n, n} \end {vmatrix} + \cdots + \begin {vmatrix} 0 & \cdots & 0 & a_{1, n} \\ a_{2, 1} & \cdots & a_{2, n - 1} & a_{2, n} \\ \vdots & \ddots & \vdots & \vdots \\ a_{n, 1} & \cdots & a_{n, n - 1} & a_{n, n} \end {vmatrix} = \sum_{j = 1}^n a_{1, j} A_{1, j}

这就得出了：

引理 11：行列式 Δ\Delta 的值，等于它的第 11 行各元素分别乘以它们的代数余子式所得的乘积之和：

Δ=∑j=1na1,jA1,j\Delta = \sum_{j = 1}^n a_{1, j} A_{1, j}

这称为行列式按第一行展开。

行列式也可按第 ii 行展开：

Δ=∣a1,1⋯a1,j⋯a1,n⋮⋱⋮⋱⋮ai−1,j⋯ai−1,j⋯ai−1,nai,1⋯ai,j⋯ai,nai+1,1⋯ai+1,j⋯ai+1,n⋮⋱⋮⋱⋮an,1⋯an,j⋯an,n∣=(−1)i−1∣ai,1⋯ai,j⋯ai,na1,1⋯a1,j⋯a1,n⋮⋱⋮⋱⋮ai−1,j⋯ai−1,j⋯ai−1,nai+1,1⋯ai+1,j⋯ai+1,n⋮⋱⋮⋱⋮an,1⋯an,j⋯an,n∣=(−1)i−1∑j=1nai,j⋅(−1)1+jMi,j=∑j=1nai,j⋅(−1)i+jMi,j=∑j=1nai,jAi,j\Delta = \begin {vmatrix} a_{1, 1} & \cdots & a_{1, j} & \cdots & a_{1, n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{i - 1, j} & \cdots & a_{i - 1, j} & \cdots & a_{i - 1, n} \\ a_{i, 1} & \cdots & a_{i, j} & \cdots & a_{i, n} \\ a_{i + 1, 1} & \cdots & a_{i + 1, j} & \cdots & a_{i + 1, n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n, 1} & \cdots & a_{n, j} & \cdots & a_{n, n} \end {vmatrix} = (-1)^{i - 1} \begin {vmatrix} a_{i, 1} & \cdots & a_{i, j} & \cdots & a_{i, n} \\ a_{1, 1} & \cdots & a_{1, j} & \cdots & a_{1, n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{i - 1, j} & \cdots & a_{i - 1, j} & \cdots & a_{i - 1, n} \\ a_{i + 1, 1} & \cdots & a_{i + 1, j} & \cdots & a_{i + 1, n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n, 1} & \cdots & a_{n, j} & \cdots & a_{n, n} \end {vmatrix} = (-1)^{i - 1} \sum_{j = 1}^n a_{i, j} \cdot (-1)^{1 + j} M_{i, j} = \sum_{j = 1}^n a_{i, j} \cdot (-1)^{i + j} M_{i, j} = \sum_{j = 1}^n a_{i, j} A_{i, j}

其中 Mi,jM_{i, j} 称为 ai,ja_{i, j} 的余子式，Ai,j=(−1)i+jMi,jA_{i, j} = (-1)^{i + j} M_{i, j} 称为 ai,ja_{i, j} 的代数余子式，同样可以得到行列式按列展开的公式。

记 M=A(i1i2⋯irk1k2⋯kr)M = A \begin {pmatrix} i_1 & i_2 & \cdots & i_r \\ k_1 & k_2 & \cdots & k_r \end {pmatrix} 是 AA 中第 i1,i2,⋯ ,iri_1, i_2, \cdots, i_r 行和第 k1,k2,⋯ ,krk_1, k_2, \cdots, k_r 列交叉处的元组成的子式，则 A(ir+1⋯inkr+1⋯kn)A \begin {pmatrix} i_{r + 1} & \cdots & i_n \\ k_{r + 1} & \cdots & k_n \end {pmatrix} 是在 AA 中将 MM 所在的 rr 行和 rr 列全部删去剩下的元按原来的顺序排成的子式，称为 MM 的余子式，MM 的余子式与 (−1)i1+i2+⋯+ir+k1+k2+⋯+kr(-1)^{i_1 + i_2 + \cdots + i_r + k_1 + k_2 + \cdots + k_r} 的乘积称为 MM 的代数余子式，由此我们得到：

定理 22（拉普拉斯展开定理）：设 ∣A∣|A| 是 nn 阶行列式，对任意正整数 r<nr < n，任意取定 rr 个指标 i1<i2<⋯<ir≤ni_1 < i_2 < \cdots < i_r \le n，则 ∣A∣|A| 的值等于它的第 i1,i2,⋯ ,iri_1, i_2, \cdots, i_r 行（或列）元组成的所有的 rr 阶子式分别于它们的代数余子式的乘积之和。

# 习题

  1. 计算 nn 阶行列式：

    (1) ∣ab0⋯000ab⋯00⋮⋮⋮⋱⋮⋮000⋯abb00⋯0a∣\begin {vmatrix} a & b & 0 & \cdots & 0 & 0 \\ 0 & a & b & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a & b \\ b & 0 & 0 & \cdots & 0 & a \end {vmatrix}

    (2) ∣x1a1a2⋯an−2an−1a1x2a2⋯an−2an−1a1a2x3⋯an−2an−1⋮⋮⋮⋱⋮⋮a1a2a3⋯an−1xn∣\begin {vmatrix} x_1 & a_1 & a_2 & \cdots & a_{n - 2} & a_{n - 1} \\ a_1 & x_2 & a_2 & \cdots & a_{n - 2} & a_{n - 1} \\ a_1 & a_2 & x_3 & \cdots & a_{n - 2} & a_{n - 1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ a_1 & a_2 & a_3 & \cdots & a_{n - 1} & x_n \end {vmatrix}

    (3) ∣α+βαβ0⋯001α+βαβ⋯0001α+β⋯00⋮⋮⋮⋱⋮⋮000⋯α+βαβ000⋯1α+β∣\begin {vmatrix} \alpha + \beta & \alpha \beta & 0 & \cdots & 0 & 0 \\ 1 & \alpha + \beta & \alpha \beta & \cdots & 0 & 0 \\ 0 & 1 & \alpha + \beta & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \alpha + \beta & \alpha \beta \\ 0 & 0 & 0 & \cdots & 1 & \alpha + \beta \end {vmatrix}

    (4) ∣ab0⋯00cab⋯000ca⋯00⋮⋮⋮⋱⋮⋮000⋯ab000⋯ca∣\begin {vmatrix} a & b & 0 & \cdots & 0 & 0 \\ c & a & b & \cdots &0 & 0 \\ 0 & c & a & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a & b \\ 0 & 0 & 0 & \cdots & c & a \end {vmatrix}

    (5) 对角元都为 00，主对角线上方元素都为 11，下方元素都为 −1-1 的偶数阶行列式：

    ∣01⋯1−10⋯1⋮⋮⋱⋮−1−1⋯0∣\begin {vmatrix} 0 & 1 & \cdots & 1 \\ -1 & 0 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ -1 & -1 & \cdots & 0 \end {vmatrix}

    (1) 解：

    ∣ab0⋯000ab⋯00⋮⋮⋮⋱⋮⋮000⋯abb00⋯0a∣=−ba(1)+(2),⋯ ,−ba(n−1)+(n)∣a00⋯000a0⋯00⋮⋮⋮⋱⋮⋮000⋯a0b−b2ab3a2⋯(−1)nbn−1an−2(−1)n+1bnan−1∣=(−1)n+1bn \begin {vmatrix} a & b & 0 & \cdots & 0 & 0 \\ 0 & a & b & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a & b \\ b & 0 & 0 & \cdots & 0 & a \end {vmatrix} \xlongequal {- \frac b a (1) + (2), \cdots, - \frac b a (n - 1) + (n)} \begin {vmatrix} a & 0 & 0 & \cdots & 0 & 0 \\ 0 & a & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a & 0 \\ b & - \frac {b^2} a & \frac {b^3} {a^2} & \cdots & (-1)^{n} \frac {b^{n - 1}} {a^{n - 2}} & (-1)^{n + 1} \frac {b^n} {a^{n - 1}} \end {vmatrix} = (-1)^{n + 1} b^n

    (2) 解：

    Δ=∣x1−a1a1−x20⋯000x2−a2a2−x3⋯0000x3−a3⋯00⋮⋮⋮⋱⋮⋮000⋯xn−1−an−1an−1−xna1a2a3⋯an−1xn∣=∣x1−a100⋯00x2−a20⋯000x3−a3⋯0⋮⋮⋮⋱⋮a1a2+x2−a1x1−a1a1a3+x3−a2x2−a2a2+(x3−a2)(x2−a1)(x2−a2)(x1−a1)a1⋯xn+∑i=1n−1(∏i=1n−1xj+1−ajxj−ajai)∣=[xn+∑i=1n−1(∏i=1n−1xj+1−ajxj−ajai)]∏i=1n−1(xi−ai) \Delta = \begin {vmatrix} x_1 - a_1 & a_1 - x_2 & 0 & \cdots & 0 & 0 \\ 0 & x_2 - a_2 & a_2 - x_3 & \cdots & 0 & 0 \\ 0 & 0 & x_3 - a_3 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x_{n - 1} - a_{n - 1} & a_{n - 1} - x_n \\ a_1 & a_2 & a_3 & \cdots & a_{n - 1} & x_n \end {vmatrix} = \begin {vmatrix} x_1 - a_1 & 0 & 0 & \cdots & 0 \\ 0 & x_2 - a_2 & 0 & \cdots & 0 \\ 0 & 0 & x_3 - a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_1 & a_2 + \frac {x_2 - a_1} {x_1 - a_1} a_1 & a_3 + \frac {x_3 - a_2} {x_2 - a_2} a_2 + \frac {(x_3 - a_2)(x_2 - a_1)} {(x_2 - a_2)(x_1 - a_1)} a_1 & \cdots & x_n + \sum\limits_{i = 1}^{n - 1} \left( \prod\limits_{i = 1}^{n - 1} \frac {x_{j + 1} - a_j} {x_j - a_j} a_i \right) \end {vmatrix} = \left[ x_n + \sum\limits_{i = 1}^{n - 1} \left( \prod\limits_{i = 1}^{n - 1} \frac {x_{j + 1} - a_j} {x_j - a_j} a_i \right) \right] \prod_{i = 1}^{n - 1} (x_i - a_i)

    (3) 解：

    Δ=∣α+β00⋯01α2+αβ+β2α+β0⋯001α3+α2β+αβ2+β3α2+αβ+β2⋯00⋮⋮⋮⋱⋮000⋯∑i=0nαiβn−i∑i=0n−1αiβn−1−i∣=∑i=0nαiβn−i \Delta = \begin {vmatrix} \alpha + \beta & 0 & 0 & \cdots & 0 \\ 1 & \frac {\alpha^2 + \alpha \beta + \beta^2} {\alpha + \beta} & 0 & \cdots & 0 \\ 0 & 1 & \frac {\alpha^3 + \alpha^2 \beta + \alpha \beta^2 + \beta^3} {\alpha^2 + \alpha \beta + \beta^2} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \frac {\sum\limits_{i = 0}^n \alpha^i \beta^{n - i}} {\sum\limits_{i = 0}^{n - 1} \alpha^i \beta^{n - 1 - i}} \end {vmatrix} = \sum\limits_{i = 0}^n \alpha^i \beta^{n - i}

    (4) 解：

    Δ=cn∣acbc0⋯001acbc⋯0001ac⋯00⋮⋮⋮⋱⋮⋮000⋯acbc000⋯1ac∣ \Delta = c^n \begin {vmatrix} \frac a c & \frac b c & 0 & \cdots & 0 & 0 \\ 1 & \frac a c & \frac b c & \cdots & 0 & 0 \\ 0 & 1 & \frac a c & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \frac a c & \frac b c \\ 0 & 0 & 0 & \cdots & 1 & \frac a c \end {vmatrix}

    令 ac=α+β,bc=αβ\dfrac a c = \alpha + \beta, \dfrac b c = \alpha \beta，可解得 α=a+a2−2bc2c,β=a−a2−2bc2c\alpha = \dfrac {a + \sqrt {a^2 - 2bc}} {2c}, \beta = \dfrac {a - \sqrt {a^2 - 2bc}} {2c}
    当 c=0c = 0 时，显然 Δ=an\Delta = a^n；
    当 c≠0c \not = 0 时，由 (3)(3) 可知：

    Δ=∑i=0n(a+a2−2bc)i(a−a2−2bc)n−i2n \Delta = \frac {\sum\limits_{i = 0}^n (a + \sqrt {a^2 - 2bc})^i (a - \sqrt {a^2 - 2bc})^{n - i}} {2^n}

    (5) 解：

    Δn=∣01⋯0−10⋯0⋮⋮⋱⋮−1−1⋯1∣+∣01⋯1−10⋯1⋮⋮⋱⋮−1−1⋯−1∣=Δn−1+∣−10⋯0−2−1⋯0⋮⋮⋱⋮−1−1⋯−1∣=Δn−1+(−1)n=Δn−2 \Delta_n = \begin {vmatrix} 0 & 1 & \cdots & 0 \\ -1 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ -1 & -1 & \cdots & 1 \end {vmatrix} + \begin {vmatrix} 0 & 1 & \cdots & 1 \\ -1 & 0 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ -1 & -1 & \cdots & -1 \end {vmatrix} = \Delta_{n - 1} + \begin {vmatrix} -1 & 0 & \cdots & 0 \\ -2 & -1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ -1 & -1 & \cdots & -1 \end {vmatrix} = \Delta_{n - 1} + (-1)^n = \Delta_{n - 2}

    由于是偶数阶行列式，且 ∣01−10∣=1\begin {vmatrix} 0 & 1 \\ -1 & 0 \end {vmatrix} = 1，因此该行列式值为 11。

  2. 求 44 阶行列式 D4=∣304022220−70053−22∣D_4 = \begin {vmatrix} 3 & 0 & 4 & 0 \\ 2 & 2 & 2 & 2 \\ 0 & -7 & 0 & 0 \\ 5 & 3 & -2 & 2 \end {vmatrix} 中第四行各元素余子式之和。

    解：55 的余子式为：

    ∣040222−700∣=−7∣4022∣=−56 \begin {vmatrix} 0 & 4 & 0 \\ 2 & 2 & 2 \\ -7 & 0 & 0 \end {vmatrix} = -7 \begin {vmatrix} 4 & 0 \\ 2 & 2 \end {vmatrix} = -56

    33 的余子式为：

    ∣340222000∣=0 \begin {vmatrix} 3 & 4 & 0 \\ 2 & 2 & 2 \\ 0 & 0 & 0 \end {vmatrix} = 0

    −2-2 的余子式为：

    ∣3002220−70∣=−7∣0322∣=42 \begin {vmatrix} 3 & 0 & 0 \\ 2 & 2 & 2 \\ 0 & -7 & 0 \end {vmatrix} = -7 \begin {vmatrix} 0 & 3 \\ 2 & 2 \end {vmatrix} = 42

    22 的余子式为：

    ∣3042220−70∣=−7∣4322∣=−14 \begin {vmatrix} 3 & 0 & 4 \\ 2 & 2 & 2 \\ 0 & -7 & 0 \end {vmatrix} = -7 \begin {vmatrix} 4 & 3 \\ 2 & 2 \end {vmatrix} = -14

    则余子式之和为 −28-28

  3. 证明 nn 阶行列式：

    ∣cos⁡θ10⋯0012cos⁡θ1⋯00012cos⁡θ⋯00⋮⋮⋮⋱⋮⋮000⋯2cos⁡θ1000⋯12cos⁡θ∣=cos⁡nθ\begin {vmatrix} \cos \theta & 1 & 0 & \cdots & 0 & 0 \\ 1 & 2 \cos \theta & 1 & \cdots & 0 & 0 \\ 0 & 1 & 2 \cos \theta & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 2 \cos \theta & 1 \\ 0 & 0 & 0 & \cdots & 1 & 2 \cos \theta \end {vmatrix} = \cos n \theta

    证明：

    Δ=∣cos⁡θ00⋯001cos⁡2θcos⁡θ0⋯0001cos⁡3θcos⁡2θ⋯00⋮⋮⋮⋱⋮⋮000⋯cos⁡(n−1)θcos⁡(n−2)θ0000⋯1cos⁡nθcos⁡(n−1)θ∣=cos⁡nθ \Delta = \begin {vmatrix} \cos \theta & 0 & 0 & \cdots & 0 & 0 \\ 1 & \frac {\cos 2 \theta} {\cos \theta} & 0 & \cdots & 0 & 0 \\ 0 & 1 & \frac {\cos 3 \theta} {\cos 2 \theta} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \frac {\cos (n - 1) \theta} {\cos (n - 2) \theta} & 0 \\ 0 & 0 & 0 & \cdots & 1 & \frac {\cos n \theta} {\cos (n - 1) \theta} \end {vmatrix} = \cos n \theta

  4. 求 D4=∣1111123x149x21827x3∣D_4 = \begin {vmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & x \\ 1 & 4 & 9 & x^2 \\ 1 & 8 & 27 & x^3 \end {vmatrix} 展开式中的二次项 x2x^2 的系数。

    解：容易发现 D4D_4 实际上是个范德蒙德行列式，因此 D4=2(x−1)(x−2)(x−3)D_4 = 2 (x - 1) (x - 2) (x - 3)，因此二次项的系数为 −12-12。