方阵行列式
1 ) 简单的方阵行列式
- 行列式是数学的一个函数，可以看做是几何空间中，一个线性变换对”面积”或”体积”的影响
- 方阵行列式，n阶方阵A的行列式表示为 ∣ A ∣ |A| ∣A∣ 或者 det(A)
- 1×1的方阵，其行列式等于该元素本身. A = ( a 11 ) ∣ A ∣ = a 11 A = (a_{11}) \ \ \ |A|= a_{11} A=(a11) ∣A∣=a11
- 2×2的方阵, 其行列式用主对角线元素乘积减去次对角线元素的乘积
- 可见，行列式是一个数值，可正，可负
A = ( a 11 a 12 a 21 a 22 ) A = \left ( \begin{array}{cccc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right ) A=(a11a21a12a22)
∣ A ∣ = a 11 ∗ a 22 − a 12 ∗ a 21 |A| = a_{11} * a_{22} – a_{12} * a_{21} ∣A∣=a11∗a22−a12∗a21
2 ) n阶方阵行列式
- n阶方阵A的行列式计算规则为：主对角线元素乘积和减去次对角线元素乘积和。
- 设 r i r_i ri为第i个主对角线的积， I i I_i Ii为第i个次对角线的积。 0 ≤ i ≤ n 0 \leq i \leq n 0≤i≤n
- A = { a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ a m 1 a m 2 ⋯ a m n } A = \left \{\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1n} \\a_{21} & a_{22} & \cdots & a_{2n} \\\cdots & \cdots & \cdots & \cdots \\a_{m1} & a_{m2} & \cdots & a_{mn}\end{array} \right \} A=⎩⎪⎪⎨⎪⎪⎧a11a21⋯am1a12a22⋯am2⋯⋯⋯⋯a1na2n⋯amn⎭⎪⎪⎬⎪⎪⎫
- r i = Π k = 1 i a k ( n + k − i ) ∗ Π k = i + 1 n a k ( k − i ) r_i = \Pi_{k=1}^i a_{k(n+k-i)} * \Pi_{k=i+1}^n a_{k(k-i)} ri=Πk=1iak(n+k−i)∗Πk=i+1nak(k−i)
- l i = Π k = 1 i a k ( i − k + 1 ) ∗ Π k = i + 1 n a k ( n − k + i + 1 ) l_i = \Pi_{k=1}^i a_{k(i-k+1)} * \Pi_{k=i+1}^n a_{k(n-k+i+1)} li=Πk=1iak(i−k+1)∗Πk=i+1nak(n−k+i+1)
- ∣ A ∣ = ∑ i = 1 n r i − ∑ i = 1 n l i |A| = \sum_{i=1}^n r_i – \sum_{i=1}^n l_i ∣A∣=∑i=1nri−∑i=1nli
3 ) 3阶方阵行列式 演示
- 根据方阵行列式的计算规则可以得到三阶方阵A的行列式为
- A = { a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 } A =\left \{\begin{array}{cccc}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{array} \right \} A=⎩⎨⎧a11a21a31a12a22a32a13a23a33⎭⎬⎫
- r i = Π k = 1 i a k ( 3 − i + k ) ∗ Π k = i + 1 3 a k ( k − i ) r_i = \Pi_{k=1}^i a_{k(3-i+k)} * \Pi_{k=i+1}^3 a_{k(k-i)} ri=Πk=1iak(3−i+k)∗Πk=i+13ak(k−i)
- l i = Π k = 1 i a k ( i − k + 1 ) ∗ Π k = i + 1 3 a k ( 4 − k + i ) l_i = \Pi_{k=1}^i a_{k(i-k+1)} * \Pi_{k=i+1}^3 a_{k(4-k+i)} li=Πk=1iak(i−k+1)∗Πk=i+13ak(4−k+i)
- ∣ A ∣ = ∑ i = 1 3 r i − ∑ i = 1 3 l i = a 13 a 21 a 32 + a 12 a 23 a 31 + a 11 a 22 a 33 − a 11 a 23 a 32 − a 12 a 21 a 33 − a 13 a 22 a 31 |A| = \sum_{i=1}^3 r_i – \sum_{i=1}^3 l_i = a_{13}a_{21}a_{32} + a_{12}a_{23}a_{31} + a_{11}a_{22}a_{33} – a_{11}a_{23}a_{32} – a_{12}a_{21}a_{33} – a_{13}a_{22}a_{31} ∣A∣=∑i=13ri−∑i=13li=a13a21a32+a12a23a31+a11a22a33−a11a23a32−a12a21a33−a13a22a31
代数余子式
1 ）概念
- 在一个n阶的行列式A中，把元素 a i j ( i , j = 1 , 2 , 3 , . . . , n ) a_{ij}(i,j=1,2,3,…,n) aij(i,j=1,2,3,...,n)所在的行和列划去后, 剩下的 ( n − 1 ) 2 (n-1)^2 (n−1)2个元素按照原来的次序组成的一个n-1阶行列式 M i j M_{ij} Mij, 称为元素 a i j a_{ij} aij的余子式。
- M i j M_{ij} Mij带上符号 ( − 1 ) i + j (-1)^{i+j} (−1)i+j称为 a i j a_{ij} aij的代数余子式，记为： A i j = ( − 1 ) i + j M i j A_{ij} = (-1)^{i+j}M_{ij} Aij=(−1)i+jMij
- 可见，余子式，代数余子式都是一个数值
2 ）利用代数余子式求方阵A的行列式|A|
- ∀ 1 ≤ i ≤ n , ∣ A ∣ = ∑ j = 1 n a i j ⋅ ( − 1 ) i + j M i j \forall 1 \leq i \leq n, |A| = \sum_{j=1}^n a_{ij} · (-1)^{i+j} M_{ij} ∀1≤i≤n,∣A∣=∑j=1naij⋅(−1)i+jMij
- 简写为： ∣ A ∣ = ∑ j = 1 n a i j ⋅ A i j |A| = \sum_{j=1}^n a_{ij} · A_{ij} ∣A∣=∑j=1naij⋅Aij
- ∀ 1 ≤ j ≤ n , ∣ A ∣ = ∑ i = 1 n a i j ⋅ ( − 1 ) i + j M i j \forall 1 \leq j \leq n, |A| = \sum_{i=1}^n a_{ij} · (-1)^{i+j} M_{ij} ∀1≤j≤n,∣A∣=∑i=1naij⋅(−1)i+jMij
- 简写为： ∣ A ∣ = ∑ i = 1 n a i j ⋅ A i j |A| = \sum_{i=1}^n a_{ij} · A_{ij} ∣A∣=∑i=1naij⋅Aij
- A = { a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 } A = \left \{\begin{array}{cccc}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{array} \right \} A=⎩⎨⎧a11a21a31a12a22a32a13a23a33⎭⎬⎫
- 使用原始的方阵行列式计算
- ∣ A ∣ = a 13 a 21 a 32 + a 12 a 23 a 31 + a 11 a 22 a 33 − a 11 a 23 a 32 − a 12 a 21 a 33 − a 13 a 22 a 31 |A| = a_{13}a_{21}a_{32} + a_{12}a_{23}a_{31} + a_{11}a_{22}a_{33} – a_{11}a_{23}a_{32} – a_{12}a_{21}a_{33} – a_{13}a_{22}a_{31} ∣A∣=a13a21a32+a12a23a31+a11a22a33−a11a23a32−a12a21a33−a13a22a31
- 使用代数余子式计算
- ∣ A ∣ = a 11 A 11 + a 12 A 12 + a 13 A 13 = . . . |A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = … ∣A∣=a11A11+a12A12+a13A13=... 注：此处省略计算和上面结果一样
- 可见，使用代数余子式可进行降维运算，在此例中是三维降到二维
- 使用原始的方阵行列式计算
3 ) 行列式计算例子
- 例：计算: A = ∣ 3 1 4 − 1 2 − 5 1 3 0 ∣ A =\left |\begin{array}{cccc}3 & 1 & 4 \\-1 & 2 & -5 \\1 & 3 & 0\end{array} \right | A=∣∣∣∣∣∣3−111234−50∣∣∣∣∣∣ 的 ∣ A ∣ |A| ∣A∣
- 按第3行展开，比较快，因为有个0，不用计算最后一项了
- ∣ A ∣ = 1 × ∣ 1 4 2 − 5 ∣ + 3 × ( − ∣ 3 4 − 1 − 5 ∣ ) = − 13 + 3 × 11 = 20 |A| = 1 × \left | \begin{array}{cccc} 1 & 4 \\ 2 & -5 \end{array} \right | + 3 × ( – \left | \begin{array}{cccc} 3 & 4 \\ -1 & -5 \end{array} \right |) = -13 + 3 × 11 = 20 ∣A∣=1×∣∣∣∣124−5∣∣∣∣+3×(−∣∣∣∣3−14−5∣∣∣∣)=−13+3×11=20
伴随矩阵
- 对于n阶方阵的任意元素 a i j a_{ij} aij都有各自的代数余子式 A i j = ( − 1 ) i + j M i j A_{ij}=(-1)^{i+j} M_{ij} Aij=(−1)i+jMij, 将所有的代数余子式按次序进行排列，可以得到一个n阶的方阵 A ∗ A^* A∗，那么 A ∗ A^* A∗称为矩阵A的伴随矩阵。
- 注意： A i j A_{ij} Aij位于 A ∗ A^* A∗的第j行第i列，这里的’按次序’就是进行了转置(位置调换,行变列,列变行)， 如下所示：
- A ∗ = { A 11 A 21 ⋯ A n 1 A 12 A 22 ⋯ A n 2 ⋯ ⋯ ⋯ ⋯ A 1 n A 2 n ⋯ A n n } A^* =\left \{\begin{array}{cccc}A_{11} & A_{21} & \cdots & A_{n1} \\A_{12} & A_{22} & \cdots & A_{n2} \\\cdots & \cdots & \cdots & \cdots \\A_{1n} & A_{2n} & \cdots & A_{nn}\end{array} \right \} A∗=⎩⎪⎪⎨⎪⎪⎧A11A12⋯A1nA21A22⋯A2n⋯⋯⋯⋯An1An2⋯Ann⎭⎪⎪⎬⎪⎪⎫
- A ⋅ A ∗ = ∣ A ∣ ⋅ E = { ∣ A ∣ 0 ⋯ 0 0 ∣ A ∣ ⋯ 0 ⋯ ⋯ ⋯ ⋯ 0 0 ⋯ ∣ A ∣ } A · A^* = |A| · E = \left \{\begin{array}{cccc}|A| & 0 & \cdots & 0 \\ 0 & |A| & \cdots & 0 \\\cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & |A| \end{array} \right \} A⋅A∗=∣A∣⋅E=⎩⎪⎪⎨⎪⎪⎧∣A∣0⋯00∣A∣⋯0⋯⋯⋯⋯00⋯∣A∣⎭⎪⎪⎬⎪⎪⎫
- 从这里可以看出转置的优点了，方便相乘运算： a 11 A 11 + a 12 A 12 + . . . a_{11} A_{11} + a_{12}A_{12} + … a11A11+a12A12+...
- 这里的E是单位矩阵(主对角线全是1，其余位置全是0的矩阵)
方阵的逆
- 设A是数域上的一个n阶方阵，若在相同的数域上存在另一个n阶方阵B, 使得 A B = B A = E AB=BA=E AB=BA=E, 那么称B为A的逆矩阵，而A被称为可逆矩阵或非奇异矩阵
- 如果A不存在逆矩阵，那么A称为奇异矩阵。A的逆矩阵记为： A − 1 A^{-1} A−1
性质
- 如果矩阵A是可逆的，那么矩阵A的逆矩阵是唯一的
- A的逆矩阵的逆矩阵还是A, 记为 ( A − 1 ) − 1 = A (A^{-1})^{-1} = A (A−1)−1=A
- 可逆矩阵A的转置矩阵 A T A^T AT也可逆，并且 ( A T ) − 1 = ( A − 1 ) T (A^T)^{-1} = (A^{-1})^T (AT)−1=(A−1)T
- 若矩阵A可逆，则矩阵A满足消去律，即：AB=AC => B=C
- 矩阵A可逆的充要条件是行列式 ∣ A ∣ ≠ 0 |A| \neq 0 ∣A∣=0
运算规律
- A ⋅ A ∗ = ∣ A ∣ ⋅ E ⇒ A ⋅ A ∗ ∣ A ∣ = E ⇒ A − 1 = A ∗ ∣ A ∣ A · A^{*} = |A| · E \Rightarrow A · \frac{A^*}{|A|} = E \Rightarrow A^{-1} = \frac{A^*}{|A|} A⋅A∗=∣A∣⋅E⇒A⋅∣A∣A∗=E⇒A−1=∣A∣A∗
- ∀ 1 ≤ i ≤ n , ∣ A ∣ = ∑ j = 1 n a i j ⋅ ( − 1 ) i + j M i j \forall 1 \leq i \leq n, \ \ \ |A| = \sum_{j=1}^n a_{ij} · (-1)^{i+j} M_{ij} ∀1≤i≤n, ∣A∣=∑j=1naij⋅(−1)i+jMij
- A = { a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ a n 1 a n 2 ⋯ a n n } A =\left \{\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1n} \\a_{21} & a_{22} & \cdots & a_{2n} \\\cdots & \cdots & \cdots & \cdots \\a_{n1} & a_{n2} & \cdots & a_{nn}\end{array} \right \} A=⎩⎪⎪⎨⎪⎪⎧a11a21⋯an1a12a22⋯an2⋯⋯⋯⋯a1na2n⋯ann⎭⎪⎪⎬⎪⎪⎫
- A ∗ = { A 11 A 21 ⋯ A n 1 A 12 A 22 ⋯ A n 2 ⋯ ⋯ ⋯ ⋯ A 1 n A 2 n ⋯ A n n } A^* =\left \{\begin{array}{cccc}A_{11} & A_{21} & \cdots & A_{n1} \\A_{12} & A_{22} & \cdots & A_{n2} \\\cdots & \cdots & \cdots & \cdots \\A_{1n} & A_{2n} & \cdots & A_{nn}\end{array} \right \} A∗=⎩⎪⎪⎨⎪⎪⎧A11A12⋯A1nA21A22⋯A2n⋯⋯⋯⋯An1An2⋯Ann⎭⎪⎪⎬⎪⎪⎫
- A ⋅ A ∗ = { ∣ A ∣ 0 ⋯ 0 0 ∣ A ∣ ⋯ 0 ⋯ ⋯ ⋯ ⋯ 0 0 ⋯ ∣ A ∣ } = ∣ A ∣ E ⇒ A − 1 = 1 ∣ A ∣ A ∗ A · A^* =\left \{\begin{array}{cccc}|A| & 0 & \cdots & 0 \\0 & |A| & \cdots & 0 \\\cdots & \cdots & \cdots & \cdots \\0 & 0 & \cdots & |A|\end{array} \right \} =|A|E \Rightarrow A^{-1} = \frac{1}{|A|} A^* A⋅A∗=⎩⎪⎪⎨⎪⎪⎧∣A∣0⋯00∣A∣⋯0⋯⋯⋯⋯00⋯∣A∣⎭⎪⎪⎬⎪⎪⎫=∣A∣E⇒A−1=∣A∣1A∗
总结：
- A可逆 ⇒ A − 1 \Rightarrow A^{-1} ⇒A−1 可逆，且 ( A − 1 ) − 1 = A (A^{-1})^{-1} = A (A−1)−1=A
- A可逆， k ≠ 0 ⇒ k A k \neq 0 \Rightarrow kA k=0⇒kA 可逆，且 ( k A ) − 1 = 1 k A − 1 (kA)^{-1} = \frac{1}{k} A^{-1} (kA)−1=k1A−1
- A,B 同阶可逆 ⇒ \Rightarrow ⇒ AB 可逆，且 ( A B ) − 1 = B − 1 A − 1 (AB)^{-1} = B^{-1} A^{-1} (AB)−1=B−1A−1
- A可逆 ⇒ A T \Rightarrow A^T ⇒AT 可逆，且 ( A T ) − 1 = ( A − 1 ) T (A^T)^{-1} = (A^{-1})^T (AT)−1=(A−1)T
- A可逆 ⇒ ∣ A − 1 ∣ = 1 ∣ A ∣ \Rightarrow |A^{-1}| = \frac{1}{|A|} ⇒∣A−1∣=∣A∣1
例子
- 求方阵 A = ( 1 2 3 2 2 1 3 4 3 ) A =\left (\begin{array}{cccc}1 & 2 & 3 \\2 & 2 & 1 \\3 & 4 & 3 \end{array} \right ) A=⎝⎛123224313⎠⎞的逆矩阵
- 分析
- 因为 ∣ A ∣ = ∣ 1 2 3 2 2 1 3 4 3 ∣ = 2 ≠ 0 |A| =\left |\begin{array}{cccc}1 & 2 & 3 \\2 & 2 & 1 \\3 & 4 & 3\end{array} \right | = 2 \neq 0 ∣A∣=∣∣∣∣∣∣123224313∣∣∣∣∣∣=2=0
- 所以 A − 1 A^{-1} A−1 存在
- A 11 = ∣ 2 1 4 3 ∣ = 2 A_{11} =\left |\begin{array}{cccc}2 & 1 \\4 & 3\end{array} \right | = 2 A11=∣∣∣∣2413∣∣∣∣=2
- A 12 = ∣ 2 1 3 3 ∣ = − 3 A_{12} = \left | \begin{array}{cccc} 2 & 1 \\ 3 & 3 \end{array} \right | = -3 A12=∣∣∣∣2313∣∣∣∣=−3
- 同理可得： A 13 = 2 , A 21 = 6 , A 22 = − 6 , A 23 = 2 , A 31 = − 4 , A 32 = 5 , A 33 = − 2 A_{13} = 2, A_{21} = 6, A_{22} = -6, A_{23} = 2, A_{31} = -4, A_{32} = 5, A_{33} = -2 A13=2,A21=6,A22=−6,A23=2,A31=−4,A32=5,A33=−2
- 得： A ∗ = ( 2 6 − 4 − 3 − 6 5 2 2 − 2 ) A^* =\left (\begin{array}{cccc}2 & 6 & -4 \\-3 & -6 & 5 \\2 & 2 & -2\end{array} \right ) A∗=⎝⎛2−326−62−45−2⎠⎞
- A − 1 = 1 ∣ A ∣ A ∗ = 1 2 ( 2 6 − 4 − 3 − 6 5 2 2 − 2 ) = ( 1 3 − 2 − 3 2 − 3 5 2 1 1 − 1 ) A^{-1} = \frac{1}{|A|} A^* = \frac{1}{2}\left (\begin{array}{cccc}2 & 6 & -4 \\-3 & -6 & 5 \\2 & 2 & -2\end{array} \right ) = \left (\begin{array}{cccc}1 & 3 & -2 \\-\frac{3}{2} & -3 & \frac{5}{2} \\1 & 1 & -1\end{array} \right ) A−1=∣A∣1A∗=21⎝⎛2−326−62−45−2⎠⎞=⎝⎛1−2313−31−225−1⎠⎞
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