Partitioning large square matrices can sometimes make their...

Partitioning large square matrices can sometimes make their inverses easier to compute, particularly if the blocks have a nice form. Using block multiplication, compute the inverse of the given matrix, partitioned as shown. (Assume that all inverses exist as needed. Recall that "O" represents the 0 matrix.) IB CI

$\mathrm{A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3} \ldots \mathrm{A}_{\mathrm{n}}$ and $\mathrm{B}_{1}, \mathrm{~B}_{2}, \mathrm{~B}_{3} \ldots . \mathrm{B}_{\mathrm{n}}$ are non-singular square matrices order $\mathrm{n}$ such that $\mathrm{A}_{1} \mathrm{~B}_{1}=\mathrm{I}_{\mathrm{n}}, \mathrm{A}_{2} \mathrm{~B}_{2}=\mathrm{In}, \mathrm{A}_{3} \mathrm{~B}_{3}=\mathrm{In}-\mathrm{A}_{\mathrm{n}} \mathrm{B}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}}$, then $\left(\mathrm{A}_{1} \mathrm{~A}_{2} \mathrm{~A}_{3}-\mathrm{A}_{\mathrm{n}}\right)^{-1}=$ (1) $\mathrm{B}_{1} \mathrm{~B}_{2} \mathrm{~B}_{3}-\ldots \mathrm{B}_{\mathrm{n}}$ (2) $\mathrm{B}_{1}^{-1} \mathrm{~B}_{2}^{-1} \mathrm{~B}_{3}^{-1}-\mathrm{B}_{\mathrm{n}}^{-1}$ (3) $\mathrm{B}_{\mathrm{n}} \mathrm{B}_{\mathrm{n}-1} \mathrm{~B}_{\mathrm{n}-2}-\ldots_{1}$ (4) $\mathrm{B}_{\mathrm{n}}{ }^{-1} \mathrm{~B}_{\mathrm{n}-1}{ }^{-1} \mathrm{~B}_{\mathrm{n}-2}{ }^{-1}-\mathrm{B}_{1}^{-1}$

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