The Inverse of a Partitioned Matrix - Chalmers

1 The Inverse of a Partitioned MatrixHerman J. BierensJuly 21, 2013 Consider a pairA,Bofn nmatrices, Partitioned asA= A11A12A21A22!,B= B11B12B21B22!,whereA11andB11arek kmatrices. Suppose thatAis nonsingular andB=A this note it will be shown how to derive theBij s in terms oftheAij s, given thatdet(A11)6=0anddet(A22)6=0.(1)The latter conditions are sufficientforthenonsingularityofA. However, ingeneral they are not necessary conditions. For example, consider the caseA= A11A12A21A22!= 0110!.On the other hand, ifAis positive definite then the conditions (1) are nec-essary as 1thenAB= A11A12A21A22! B11B12B21B22!(2)= A11B11+A12B21A11B12+A12B22A21B11+A22B21A 21B12+A22B22!= IkOk,n kOn k,kIn k!,whereasusualIdenotes the unit Matrix andOa zero Matrix , with sizesindicated by the subscripts solve (2), we need to solve four Matrix equations.

2 A11B11+A12B21=Ik(3)A11B12+A12B22=Ok,n k(4)A21B11+A22B21=On k,k(5)A21B12+A22B22=In k(6)Itfollowsfrom(4)and(5)thatB12= A 111A12B22,(7)B21= A 122A21B11,(8)so that (3) and (6) become A11 A12A 122A21 B11=Ik A22 A21A 111A12 B22=In kHenceB11= A11 A12A 122A21 1B22= A22 A21A 111A12 1 Substituting these solutions in (7) and (8) it follows thatB12= A 111A12 A22 A21A 111A12 1B21= A 122A21 A11 A12A 122A21 1 Thus,A 1= A11 A12A 122A21 1 A 111A12 A22 A21A 111A12 1 A 122A21 A11 A12A 122A21 1 A22 A21A 111A12 1 Moreover, 1=InimpliesA 1A=In,we also haveA 1= A11 A12A 122A21 1 A11 A12A 122A21 1A12A 122 A22 A21A 111A12 1A21A 111 A22 A21A 111A12 1 2