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The first step is to find dπ. With some algebraic manipulation, left as an exercise, we get
(AZ–1XAT)dπ = – b + µAZ–1e + AZ–1XδD
dπ = (AZ–1XAT)–1(– b + µAZ–1e + AZ–1XδD)
Note that Z–1 = diag{1/z1, 1/z2, … , 1 /zn} and is trivial to compute. Further
multiplications and substitutions give
d z = – δ D + A Td π
dx = Z–1(µe – XZe – Xdz)
(10)