工程数学（线性代数、复变函数）

<div> <h2>分块矩阵的逆矩阵求法</h2> <p>分块矩阵的逆矩阵求法是一种将大矩阵拆分成较小的子矩阵，然后利用这些子矩阵的性质来求解原矩阵的方法。</p> <p>假设有一个可逆的分块矩阵：</p> <div> <span>A = </span><span>\(\begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}\)</span> </div> <p>其中 \(A_{11}, A_{12}, A_{21}, A_{22}\) 都是方阵。则其逆矩阵为：</p> <div> <span>\(A^{-1} = \begin{bmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{bmatrix}\)</span> </div> <p>其中：</p> <ul> <li>\(B_{11} = (A_{11} - A_{12}A_{22}^{-1}A_{21})^{-1}\)</li> <li>\(B_{12} = -A_{11}^{-1}A_{12}(A_{22} - A_{21}A_{11}^{-1}A_{12})^{-1}\)</li> <li>\(B_{21} = -A_{22}^{-1}A_{21}(A_{11} - A_{12}A_{22}^{-1}A_{21})^{-1}\)</li> <li>\(B_{22} = (A_{22} - A_{21}A_{11}^{-1}A_{12})^{-1}\)</li> </ul> <p>以上公式需要满足 \(A_{11}, A_{22}\) 以及 \(A_{22} - A_{21}A_{11}^{-1}A_{12}\) 和 \(A_{11} - A_{12}A_{22}^{-1}A_{21}\) 都是可逆矩阵。</p> <h3>例题</h3> <p>设有一个分块矩阵：</p> <div> <span>A = </span><span>\(\begin{bmatrix} 2 & 0 & 1 & 0 \\ 0 & 3 & 0 & 1 \\ 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 2 \end{bmatrix}\)</span> </div> <p>将其分为：</p> <div> <span>\(A_{11} = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}\)</span>, <span>\(A_{12} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)</span>, <span>\(A_{21} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)</span>, <span>\(A_{22} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\)</span> </div> <p>计算逆矩阵 \(A^{-1}\)：</p> <div> <span>\(A_{11}^{-1} = \begin{bmatrix} 0.5 & 0 \\ 0 & 1/3 \end{bmatrix}\)</span>, <span>\(A_{22}^{-1} = \begin{bmatrix} 0.5 & 0 \\ 0 & 0.5 \end{bmatrix}\)</span> </div> <div> <span>\(A_{22} - A_{21}A_{11}^{-1}A_{12} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0.5 & 0 \\ 0 & 1/3 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1.5 & 0 \\ 0 & 1.5 \end{bmatrix}\)</span> </div> <div> <span>\((A_{22} - A_{21}A_{11}^{-1}A_{12})^{-1} = \begin{bmatrix} 2/3 & 0 \\ 0 & 2/3 \end{bmatrix}\)</span> </div> <div> <span>\(A_{11} - A_{12}A_{22}^{-1}A_{21} = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0.5 & 0 \\ 0 & 0.5 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1.5 & 0 \\ 0 & 2.5 \end{bmatrix}\)</span> </div> <div> <span>\((A_{11} - A_{12}A_{22}^{-1}A_{21})^{-1} = \begin{bmatrix} 2/3 & 0 \\ 0 & 2/5 \end{bmatrix}\)</span> </div> <div> <span>\(B_{11} = (A_{11} - A_{12}A_{22}^{-1}A_{21})^{-1} = \begin{bmatrix} 2/3 & 0 \\ 0 & 2/5 \end{bmatrix}\)</span> </div> <div> <span>\(B_{12} = -A_{11}^{-1}A_{12}(A_{22} - A_{21}A_{11}^{-1}A_{12})^{-1} = -\begin{bmatrix} 0.5 & 0 \\ 0 & 1/3 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2/3 & 0 \\ 0 & 2/3 \end{bmatrix} = \begin{bmatrix} -1/3 & 0 \\ 0 & -2/9 \end{bmatrix}\)</span> </div> <div> <span>\(B_{21} = -A_{22}^{-1}A_{21}(A_{11} - A_{12}A_{22}^{-1}A_{21})^{-1} = -\begin{bmatrix} 0.5 & 0 \\ 0 & 0.5 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2/3 & 0 \\ 0 & 2/5 \end{bmatrix} = \begin{bmatrix} -1/3 & 0 \\ 0 & -1/5 \end{bmatrix}\)</span> </div> <div> <span>\(B_{22} = (A_{22} - A_{21}A_{11}^{-1}A_{12})^{-1} = \begin{bmatrix} 2/3 & 0 \\ 0 & 2/3 \end{bmatrix}\)</span> </div> <p>因此，</p> <div> <span>\(A^{-1} = \begin{bmatrix} 2/3 & 0 & -1/3 & 0 \\ 0 & 2/5 & 0 & -2/9 \\ -1/3 & 0 & 2/3 & 0 \\ 0 & -1/5 & 0 & 2/3 \end{bmatrix}\)</span> </div> </div>

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