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How to solve this type of problem: We've got a block 2x2 matrix : $$A=\begin{bmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\\\end{bmatrix}$$ If matrices $A$ and $A_{22}$ are invertible, show that a matrix $B = A_{11} - A_{12}A_{22}^{-1}A_{21}$is also invertible.

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Observe that for a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ the claim follows from $(ad - bc)/d = a - bd^{-1}c$.

Let vertical bars indicate a determinant. Then \begin{align} |A_{22}|^{-1}|A| &= \begin{vmatrix} I & 0 \\ 0 & A_{22}^{-1}\end{vmatrix}\begin{vmatrix} A_{11} & A_{12} \\ A_{21}& A_{22}\end{vmatrix} = \begin{vmatrix} A_{11} & A_{12} \\ A_{22}^{-1}A_{21} & I \end{vmatrix}\\ & = \begin{vmatrix} I & A_{12} \\ 0 & I \end{vmatrix}\ \begin{vmatrix} A_{11} - A_{12}A_{22}^{-1}A_{21} & 0 \\ A_{22}^{-1}A_{21} & I \end{vmatrix} = |A_{11} - A_{12}A_{22}^{-1}A_{21}|, \end{align} from which the claim follows.

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