Recall the formula for the inverse of a 2×2 block matrix:
[ABCD]−1=[A−1+A−1BS−1CA−1−A−1BS−1−S−1CA−1S−1]S=D−CA−1BNow consider a 3×3 block matrix
X=[EFGHJKLMN]Apply the 2×2 block inverse formula, plugging in: ˜A=E, ˜B=[FG], ˜C=[HL], and ˜D=[JKMN]:
X−1=[E[FG][HL][JKMN]]−1=[E−1+E−1[FG]Z−1[HL]E−1−E−1[FG]Z−1−Z−1[HL]E−1Z−1]Z=[JKMN]−[HL]E−1[FG]The factor Z, in turn, is another 2×2 block matrix.
Z=[JKMN]−[HL]E−1[FG]=[JKMN]−[HE−1FHE−1GLE−1FLE−1G]=[J−HE−1FK−HE−1GM−LE−1FN−LE−1G]Again apply again the 2×2 block inverse formula to get Z−1, defining:
A=J−HE−1FB=K−HE−1GC=M−LE−1FD=N−LE−1GZ−1=[ABCD]−1=[A−1+A−1BS−1CA−1−A−1BS−1−S−1CA−1S−1]S=D−CA−1BFurther expanding the forumula for X−1 in terms of this is tedious and somewhat unsatisfying. Define
U=G−FA−1BV=L−CA−1Hthen expand and simplify:
E−1+E−1[FG]Z−1[HL]E−1=E−1{I+[FG][A−1+A−1BS−1CA−1−A−1BS−1−S−1CA−1S−1][HL]E−1}=E−1{I+[FG][[A−1+A−1BS−1CA−1]H−A−1BS−1L−S−1CA−1H+S−1L]E−1}=E−1{I+{F[(A−1+A−1BS−1CA−1)H−A−1BS−1L]+G[−S−1CA−1H+S−1L]}E−1}=E−1{I+{FA−1[H+BS−1(CA−1H−L)]−GS−1[CA−1H−L]}E−1}=E−1{I+{FA−1H+[FA−1B−G]S−1[CA−1H−L]}E−1}=E−1+E−1{FA−1H+US−1V}E−1−E−1[FG]Z−1=−E−1[FG][A−1+A−1BS−1CA−1−A−1BS−1−S−1CA−1S−1]=−E−1[FA−1+FA−1BS−1CA−1−GS−1CA−1−FA−1BS−1+GS−1]=[−E−1{F+(FA−1B−G)S−1C}A−1E−1[FA−1B−G]S−1]=[−E−1[F−US−1C]A−1−E−1US−1]−Z−1[HL]E−1=−[A−1+A−1BS−1CA−1−A−1BS−1−S−1CA−1S−1][HL]E−1=[−A−1H−A−1BS−1CA−1H+A−1BS−1LS−1CA−1H−S−1L]E−1=[−A−1[H+BS−1(CA−1H−L)]E−1S−1(CA−1H−L)E−1]=[−A−1[H−BS−1V]E−1−S−1VE−1]This gives the expanded formula:
X−1=[E−1+E−1[FA−1H+US−1V]E−1−E−1[F−US−1C]A−1−E−1US−1−A−1[H−BS−1V]E−1A−1+A−1BS−1CA−1−A−1BS−1−S−1VE−1−S−1CA−1S−1]
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