forward substitution backward substitution the first equation can be solved efficiently via forward substitutionand the second via backward substitutionas illustrated in the following example example (solving linear equations with an lu decompositionlet plu be the same as in example we wish to solve ax [ ]firstsolving / - / givesy and / / / by forward substitution next - / / gives / - / / and ( / )/ - / so [- ]/ woodbury identity lu (or more generally pludecompositions can also be applied to block matrices starting point is the following lu decomposition for general matrix# / bc/ which holds as long as this can be seen by simply writing out the matrix product the block matrix generalization for matrices rnxn rnxk rkxn rkxk is #in - onxk ( : ca- okxn ik provided that is invertible (againwrite out the block matrix productherewe use the notation pxq to denote the matrix of zeros we can further rewrite this as##in onxk onxk in - sca- ik okxn ca- okxn ik thusinverting both sideswe obtain #- #- #- in - onxk in onxk - okxn ik okxn ca- ca- ik inversion of the above block matrices gives (again write out##in onxk in - - - onxk - okxn ik okxn ( ca- )- -ca- ik ( assuming that is invertiblewe could also perform block ul (as opposed to ludecomposition# bd- in onxk ( okxn - ik