Section02_逆矩阵

Questions

  1. 何为逆矩阵
  2. 逆矩阵是否存在
  3. 逆矩阵如何求

定义

  • 对于 An×n\boldsymbol{A}_{n\times n}，若 ∃ Bn×n\exists\ \boldsymbol{B}_{n\times n}，使得 BA=AB=E\boldsymbol{BA} = \boldsymbol{AB} = \boldsymbol{E}，则称 A\boldsymbol{A} 可逆，B\boldsymbol{B} 为 A\boldsymbol{A} 的逆矩阵，记 A−1=B\boldsymbol{A}^{-1} = \boldsymbol{B}

例题

  1. An×n\boldsymbol{A}_{n\times n}，且 A2+A−3E=0\boldsymbol{A}^{2}+\boldsymbol{A}-3 \boldsymbol{E} = \boldsymbol{0}，求
    1. A−1\boldsymbol{A}^{-1}
    2. (A−E)−1(\boldsymbol{A}-\boldsymbol{E})^{-1} ∵A2+A−3E=0∴A(A+E)=3EAA+E3=E∴A−1=13(A+E)∵A2+A−3E=0∴(A−E)(A+2E)=E∴(A−E)−1=A+2E \begin{array}{ll} \because & \boldsymbol{A}^{2} + \boldsymbol{A} - 3 \boldsymbol{E} = \boldsymbol{0} \\ \therefore & \boldsymbol{A}(\boldsymbol{A} + \boldsymbol{E}) = 3 \boldsymbol{E} \\ & \boldsymbol{A}\frac{\boldsymbol{A} + \boldsymbol{E}}{3} = \boldsymbol{E} \\ \therefore & \boldsymbol{A}^{-1} = \frac{1}{3}(\boldsymbol{A}+ \boldsymbol{E}) \\ \\ \because & \boldsymbol{A}^{2} + \boldsymbol{A} - 3 \boldsymbol{E} = \boldsymbol{0} \\ \therefore & (\boldsymbol{A} - \boldsymbol{E})(\boldsymbol{A} + 2\boldsymbol{E}) = \boldsymbol{E} \\ \therefore & (\boldsymbol{A} - \boldsymbol{E})^{-1} = \boldsymbol{A}+ 2\boldsymbol{E} \\ \end{array}
  2. A≠0,A3=0\boldsymbol{A} \ne \boldsymbol{0}, \boldsymbol{A}^{3} = \boldsymbol{0}，求 (E−A)−1(\boldsymbol{E} - \boldsymbol{A})^{-1} ∵a3−b3=(a−b)(a2+ab+b2)∴E3−A3=(E−A)(A2+AE+E2)∴(E−A)(A2+A+E)=E∴(E−A)−1=A2+A+E \begin{array}{ll} \because & a^{3} - b^{3} = (a-b)(a^{2}+ab +b^{2}) \\ \therefore & \boldsymbol{E}^{3} - \boldsymbol{A}^{3} = (\boldsymbol{E} - \boldsymbol{A})(\boldsymbol{A}^{2} + \boldsymbol{AE} + \boldsymbol{E}^{2}) \\ \therefore & (\boldsymbol{E} - \boldsymbol{A})(\boldsymbol{A}^{2}+ \boldsymbol{A} + \boldsymbol{E}) = \boldsymbol{E} \\ \therefore & (\boldsymbol{E}-\boldsymbol{A})^{-1} = \boldsymbol{A}^{2} + \boldsymbol{A} + \boldsymbol{E} \\ \end{array}

逆矩阵存在定理

Notes

  1. 行列式性质
    1. ∣A⊺∣=∣A∣\vert \boldsymbol{A}^{\intercal} \vert = \vert \boldsymbol{A} \vert
    2. ∣A−1∣=1∣A∣\vert \boldsymbol{A}^{-1} \vert = \frac{1}{\vert \boldsymbol{A} \vert}
    3. ∣A∗∣=∣A∣n−1\vert \boldsymbol{A}^{*} \vert = \vert \boldsymbol{A} \vert^{n-1}
    4. 对于 An×n\boldsymbol{A}_{n\times n}, ∣kA∣=kn∣A∣\vert k \boldsymbol{A} \vert= k^{n}\vert \boldsymbol{A} \vert
    5. 对于 An×n,Bn×n\boldsymbol{A}_{n\times n}, \boldsymbol{B}_{n\times n}, ∣AB∣=∣A∣∣B∣\vert \boldsymbol{AB} \vert = \vert \boldsymbol{A} \vert \vert \boldsymbol{B} \vert (Laplace 法则)
  2. 逆矩阵的性质
    1. (A−1)−1=A(\boldsymbol{A}^{-1})^{-1} = \boldsymbol{A}
    2. (A⊺)−1=(A−1)⊺(\boldsymbol{A}^{\intercal})^{-1} = (\boldsymbol{A}^{-1})^{\intercal}
    3. (Am)−1=(A−1)m(\boldsymbol{A}^{m})^{-1} = (\boldsymbol{A}^{-1})^{m}
    4. (kA)−1=1kA−1(k≠0)(k\boldsymbol{A})^{-1} = \frac{1}{k}\boldsymbol{A}^{-1}\quad (k\ne 0)
    5. (AB)−1=B−1A−1(\boldsymbol{AB})^{-1} = \boldsymbol{B}^{-1}\boldsymbol{A}^{-1}
    6. Am×m,Bn×n\boldsymbol{A}_{m\times m}, \boldsymbol{B}_{n\times n} 可逆，则 (AB)−1=(A−1B−1)\left( \begin{matrix} \boldsymbol{A} & \\ & \boldsymbol{B} \end{matrix} \right)^{-1} = \left( \begin{matrix} \boldsymbol{A}^{-1} & \\ & \boldsymbol{B}^{-1} \end{matrix} \right)
  • Th An×n\boldsymbol{A}_{n\times n} 可逆 ⇔\Leftrightarrow ∣A∣≠0\vert \boldsymbol{A} \vert\ne 0

    Proof

    ⇒\Rightarrow ∵An×n可逆∴∃ B,AB=E∴∣AB∣=∣E∣∣A∣∣B∣=1≠0∴det⁡(A)≠0 \begin{array}{ll} \because & \boldsymbol{A}_{n\times n} \text{可逆} \\ \therefore & \exists\ \boldsymbol{B}, \boldsymbol{AB} = \boldsymbol{E} \\ \therefore & \vert \boldsymbol{AB} \vert = \vert \boldsymbol{E} \vert \\ & \vert \boldsymbol{A} \vert \vert \boldsymbol{B}\vert = 1 \ne 0 \\ \therefore & \det(\boldsymbol{A})\ne 0 \\ \end{array} ⇐\Leftarrow ∵AA∗=∣A∣E∣A∣≠0∴AA∗∣A∣=E∴A可逆，且A−1=A∗∣A∣ \begin{array}{ll} \because & \boldsymbol{AA}^{*} = \vert \boldsymbol{A} \vert \boldsymbol{E} \\ & \vert\boldsymbol{A}\vert \ne 0 \\ \therefore & \boldsymbol{A}\frac{\boldsymbol{A}^{*}}{\vert \boldsymbol{A} \vert} = \boldsymbol{E} \\ \therefore & \boldsymbol{A}\text{可逆，且} \boldsymbol{A}^{-1} = \frac{\boldsymbol{A}^{*}}{\vert \boldsymbol{A} \vert}\\ \end{array}

例题

  1. A=(11−1121102)\boldsymbol{A} = \left( \begin{matrix} 1 & 1 & -1 \\ 1 & 2 & 1 \\ 1 & 0 & 2 \end{matrix} \right)，则 A\boldsymbol{A} 是否可逆? 若可逆，求 A−1\boldsymbol{A}^{-1} det⁡(A)=5≠0∴A 可逆∵A∗=(4−1−2−2313−21)⊺=(4−23−13−2−211)∴A−1=A∗∣A∣=15(4−23−13−2−211) \begin{array}{ll} & \det(\boldsymbol{A}) = 5 \ne 0 \\ \therefore & \boldsymbol{A}\text{ 可逆} \\ \because & \boldsymbol{A}^{*} = \left( \begin{matrix} 4 & -1 & -2 \\ -2 & 3 & 1 \\ 3 & -2 & 1 \end{matrix} \right)^{\intercal} = \left( \begin{matrix} 4 & -2 & 3 \\ -1 & 3 & -2 \\ -2 & 1 & 1 \end{matrix} \right)\\ \therefore & \boldsymbol{A}^{-1} = \frac{\boldsymbol{A}^{*}}{\vert \boldsymbol{A} \vert} = \frac{1}{5} \left( \begin{matrix} 4 & -2 & 3 \\ -1 & 3 & -2 \\ -2 & 1 & 1 \end{matrix} \right)\\ \end{array}
    • Note 若 A\boldsymbol{A} 可逆，A−1=1∣A∣A∗⇔∣A∣A−1=A∗\boldsymbol{A}^{-1} = \frac{1}{\vert \boldsymbol{A} \vert}\boldsymbol{A}^{*} \Leftrightarrow \vert \boldsymbol{A} \vert\boldsymbol{A}^{-1} = \boldsymbol{A}^{*}
  2. A,B\boldsymbol{A}, \boldsymbol{B} 为 nn 阶可逆矩阵，则 (AB)∗(\boldsymbol{AB})^{*} 为? (AB)∗=∣AB∣(AB)−1=∣A∣∣B∣B−1A−1=B∗A∗ \begin{array}{ll} & (\boldsymbol{AB})^{*} = \vert \boldsymbol{AB} \vert (\boldsymbol{AB})^{-1} \\ & = \vert \boldsymbol{A} \vert \vert \boldsymbol{B} \vert \boldsymbol{B}^{-1} \boldsymbol{A}^{-1} \\ & = \boldsymbol{B}^{*} \boldsymbol{A}^{*} \end{array}
  3. Am×m,Bn×n\boldsymbol{A}_{m\times m}, \boldsymbol{B}_{n\times n}可逆，求 (AB)∗\left( \begin{matrix} \boldsymbol{A} & \\ & \boldsymbol{B} \end{matrix} \right)^{*} (AB)∗=∣AB∣(AB)−1=∣A∣∣B∣(A−1B−1)=(∣B∣A∗∣A∣B∗) \begin{array}{ll} & \left( \begin{matrix} \boldsymbol{A} & \\ & \boldsymbol{B} \end{matrix} \right)^{*} = \left\vert \begin{matrix} \boldsymbol{A} & \\ & \boldsymbol{B} \end{matrix} \right\vert \left( \begin{matrix} \boldsymbol{A} & \\ & \boldsymbol{B} \end{matrix} \right)^{-1} \\ & = \vert \boldsymbol{A} \vert \vert \boldsymbol{B} \vert \left( \begin{matrix} \boldsymbol{A}^{-1} & \\ & \boldsymbol{B}^{-1} \end{matrix} \right) \\ & = \left( \begin{matrix} \vert \boldsymbol{B} \vert\boldsymbol{A}^{*} & \\ & \vert \boldsymbol{A} \vert \boldsymbol{B}^{*} \end{matrix} \right) \end{array} \\

A−1\boldsymbol{A}^{-1} 的求法

法一 伴随矩阵法

A−1=A∗det⁡(A) \boldsymbol{A}^{-1} = \frac{\boldsymbol{A}^{*}}{\det(\boldsymbol{A})}

法二 初等变换法

基础知识

  1. 方程组的三种同解变形
    1. 对调两个方程
    2. 一方程乘以不为零的数
    3. 一方程乘以 kk 后加到另一方程上
  2. 矩阵的三种初等行变化
    1. 对调两行
    2. 一行乘以不为零的数
    3. 一行乘以 kk 加到另一行上

      Notes

      1. 将行变为列，则称为矩阵的初等列变化
      2. 初等行变化和初等列变化合称为初等变化
  3. 三个初等矩阵
    1. Eij≜(1⋱0⋯1⋱1⋯0⋱1)\boldsymbol{E}_{ij} \triangleq \left( \begin{smallmatrix} 1 \\ & \ddots \\ & & 0 & \cdots & 1 \\ & & & \ddots & \\ & & 1 &\cdots & 0 \\ & & & & & \ddots \\ & & & & & & 1 \\ \end{smallmatrix} \right) (将 E\boldsymbol{E} 的 i,ji,j 行(或列)对调)

      Notes

      • EijA\boldsymbol{E}_{ij}\boldsymbol{A}: 对调 AA 的 i,ji,j 行
      • AEij\boldsymbol{A}\boldsymbol{E}_{ij}: 对调 AA 的 i,ji,j 列
      • det⁡(Eij)=−1≠0\det(\boldsymbol{E}_{ij}) = -1 \ne 0
      • Eij−1=Eij\boldsymbol{E}_{ij}^{-1} = \boldsymbol{E}_{ij}
    2. Ei(c)≜(1⋱1c1⋱1)\boldsymbol{E}_{i}(c) \triangleq \left( \begin{smallmatrix} 1 \\ & \ddots \\ & & 1 \\ & & & c \\ & & & & 1 \\ & & & & & \ddots \\ & & & & & & 1 \\ \end{smallmatrix} \right) (将 E\boldsymbol{E} 的第 ii 行(或列)乘以 cc)

      Notes

      • Ei(c)A\boldsymbol{E}_{i}(c)\boldsymbol{A}: 使 A\boldsymbol{A} 的 ii 行乘以 cc
      • AEi(c)\boldsymbol{A}\boldsymbol{E}_{i}(c): 使 A\boldsymbol{A} 的 ii 列乘以 cc
      • det⁡(Ei(c))=c≠0\det(\boldsymbol{E}_{i}(c)) = c \ne 0
      • Ei(c)−1=Ei(1c)\boldsymbol{E}_{i}(c)^{-1} = \boldsymbol{E}_{i}(\frac{1}{c})
    3. Eij(k)=(1⋱1⋯k⋱⋮1⋱1)\boldsymbol{E}_{ij}(k) = \left( \begin{smallmatrix} 1 \\ & \ddots \\ & & 1 & \cdots & k \\ & & & \ddots & \vdots \\ & & & & 1 \\ & & & & & \ddots \\ & & & & & & 1 \\ \end{smallmatrix} \right) (将 EE 的第 jj 行乘以 kk 后加到 ii 行上)

      Notes

      • Eij(k)A\boldsymbol{E}_{ij}(k)\boldsymbol{A}: 使 A\boldsymbol{A} 的 jj 行乘以 kk 加到 ii 行上
      • AEij(k)\boldsymbol{A}\boldsymbol{E}_{ij}(k): 使 A\boldsymbol{A} 的 ii 列乘以 kk 加到 jj 列上
      • det⁡(Eij(k))=1\det(\boldsymbol{E}_{ij}(k)) = 1
      • Eij(k)−1=Eij(−k)\boldsymbol{E}_{ij}(k)^{-1} = E_{ij}(-k)

方程组的三种同解变形⇕矩阵的三种初等行变化⇕矩阵左乘Eij,Ei(c),Eij(k) \begin{array}{c} \color{#D0104C}\text{方程组的三种同解变形} \\ \Updownarrow \\ \color{#D0104C}\text{矩阵的三种初等行变化} \\ \Updownarrow \\ \color{#D0104C}\text{矩阵左乘}\boldsymbol{E}_{ij}, \boldsymbol{E}_{i}(c), \boldsymbol{E}_{ij}(k) \\ \end{array}

  1. 两个定理
    • Th1 A\boldsymbol{A} 可逆 ⇔\Leftrightarrow A\boldsymbol{A} 为若干初等矩阵之积
    • Th2
      • A→rowE⇔∃\boldsymbol{A}\xrightarrow{\text{row}} \boldsymbol{E} \Leftrightarrow \exists 可逆阵 P\boldsymbol{P}，使得 PA=E\boldsymbol{PA} = \boldsymbol{E}
      • A→columnE⇔∃\boldsymbol{A}\xrightarrow{\text{column}} \boldsymbol{E} \Leftrightarrow \exists 可逆阵 Q\boldsymbol{Q}，使得 AQ=E\boldsymbol{AQ} = \boldsymbol{E}
        • Q2 An×n,r(A)=r<n\boldsymbol{A}_{n\times n}, r(\boldsymbol{A}) = r<n A→row(Er000)\boldsymbol{A}\xrightarrow{\text{row}} \left( \begin{smallmatrix} \boldsymbol{E}_{r} & 0 \\ 0 & 0 \end{smallmatrix} \right) 或 ∃ 可逆阵P, 使得PA=(Er000)\exists\ \text{可逆阵}\boldsymbol{P}\text{, 使得}\boldsymbol{PA}= \left( \begin{smallmatrix} \boldsymbol{E}_{r} & 0 \\ 0 & 0 \end{smallmatrix} \right)? ✗
        • Q3 An×n,r(A)=r<n\boldsymbol{A}_{n\times n}, r(\boldsymbol{A}) = r<n A→columnrow(Er000)\boldsymbol{A}\xrightarrow[\text{column}]{\text{row}} \left( \begin{smallmatrix} \boldsymbol{E}_{r} & 0 \\ 0 & 0 \end{smallmatrix} \right) 或 ∃ 可逆阵P,Q, 使得PAQ=(Er000)\exists\ \text{可逆阵}\boldsymbol{P}, \boldsymbol{Q}\text{, 使得}\boldsymbol{PAQ}= \left( \begin{smallmatrix} \boldsymbol{E}_{r} & 0 \\ 0 & 0 \end{smallmatrix} \right)? ✓
  2. 初等变换法求逆矩阵
  3. Th An×n\boldsymbol{A}_{n\times n} 可逆，则 (AE)→row(EA−1)(AE)→column(EA−1) \begin{array}{c} \left( \begin{array}{c:c} \boldsymbol{A} & \boldsymbol{E} \\ \end{array} \right)\xrightarrow{\text{row}} \left( \begin{array}{c:c} \boldsymbol{E} & \boldsymbol{A}^{-1} \end{array} \right) \\\\ \left( \begin{array}{c} \boldsymbol{A} \\ \hdashline\boldsymbol{E} \\ \end{array} \right)\xrightarrow{\text{column}} \left( \begin{array}{c} \boldsymbol{E} \\ \hdashline \boldsymbol{A}^{-1} \\ \end{array} \right) \end{array}

例题

  1. A3×3\boldsymbol{A}_{3\times 3}，A\boldsymbol{A} 第一行 33 倍加到第二行，第二三列对调成E\boldsymbol{E}，则 A=?\boldsymbol{A} = ? ∵(100310001)A(100001010)=E∴A=(100310001)−1(100001010)−1=(100−310001)(100001010)=(100−301010) \begin{array}{ll} \because & \left( \begin{smallmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix} \right) \boldsymbol{A} \left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix} \right) = \boldsymbol{E}\\ \therefore & \boldsymbol{A} = \left( \begin{smallmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix} \right)^{-1} \left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix} \right)^{-1} \\ & = \left( \begin{smallmatrix} 1 & 0 & 0 \\ -3 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix} \right) \left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix} \right) = \left( \begin{smallmatrix} 1 & 0 & 0 \\ -3 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix} \right) \end{array}
  2. A=[1101211−11]\boldsymbol{A}= \left[ \begin{smallmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 1 & -1 & 1 \end{smallmatrix} \right]，则 A\boldsymbol{A} 是否可逆，若可逆，求 A−1\boldsymbol{A}^{-1}

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