Question: Partitioning large square matrices can sometimes make their inverses easier to compute, particularly if the blocks have a nice form. Verify by block multiplication that

Where P = (A - BD-1 C)-1,
Q = -PBD-1, R = -D-1CP, and S = D-1 + D-1CPBD-1
A B P Q %D
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ABA B A C D C D A B P Q C D RS 3 CP P AP ... View full answer
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