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Work locally in $z$ where $u\ne 0$, and take the matrix entries literally as they appear in the original PDF. At fixed $z$, abbreviateThen $A=D{\lambda }^2+\left(\begin{array}{cc}0& U\\ V& 0\end{array}\right)\lambda +\left(\begin{array}{cc}d& P\\ R& -d\end{array}\right)$. DefineThe literal coefficients giveIn general neither $h=z/2$ nor $\theta =0$ is valid for the printed system.

The normalized formal fundamental matrix iswith a chosen logarithm branch. In particular the diagonal normal form has ${\Lambda }_1=0$, ${\Lambda }_2=hD$, ${\Lambda }_3=\theta D$. Its first two prefactor coefficients areThese formulas include the diagonal terms required by the derivative of the prefactor. Setting every diagonal prefactor coefficient to zero while retaining only the displayed exponent would generally fail the differential equation.

Here is the coefficient calculation and the full recursion. Write $A_1=\left(\begin{array}{cc}0& U\\ V& 0\end{array}\right)$, $A_2=\left(\begin{array}{cc}d& P\\ R& -d\end{array}\right)$ and $G_0=I$, with negative-index $G_j=0$. Substitution into the gauge transformation of a linear differential system gives, for $m\ge 1$,At $m=1$, the commutator determines $(G_1{)}_{12}=-U/2$, $(G_1{)}_{21}=V/2$ and the absence of a linear diagonal term. The diagonal part at $m=2$ gives $h=d+UV/2$. The next diagonal part gives $\theta =(UR+VP)/2$. At $m=4$ diagonal solvability gives $-\rho =PR/2-UV(d+h)/4$; the next order fixes the two diagonal entries of $G_2$ above. The off-diagonal entries are fixed at each stage because $[D,X{]}_{12}=2X_{12}$ and $[D,X{]}_{21}=-2X_{21}$. Later diagonal solvability fixes the remaining normalization coefficients. Thus this is a complete formal solution, not a convergence assertion. It is the formal cubic-exponential two-by-two system expansion. The printed hint's second prefactor term must have denominator ${\lambda }^2$, and the remainder after the $1/\lambda$ term of the diagonal expansion must start at $1/{\lambda }^2$.

Now impose an isomonodromic deformation in the full irregular sense: its Stokes matrices and formal exponent are fixed in $z$, with the leading formal normalization fixed. Sectorial matrices satisfy $Y_{j+1}=Y_j{C}_j$ with $(C_j{)}_z=0$, so $B_j=(Y_j{)}_z{Y}_j^{-1}$ is the same on overlaps. Since the spectral system has no finite singularities, its fundamental matrix and inverse are entire in $\lambda$, making $B$ entire. Differentiating the formal expression, using ${\theta }_z=0$, givesBut $[G_1,D]=A_1$. The sectorial cover makes this a global growth bound, so Liouville's theorem applied to $B-h_z(\lambda D+A_1)$ gives the literal consequenceAn arbitrary parameter dependence of $u,w$ need not be isomonodromic; this analytic argument presupposes the stated constancy of generalized monodromy data.

The requested matrix would require $h_z=1/2$ from its diagonal and upper entry, and then its lower entry would require $(z-2)w=0$. On an open deformation domain this forces $w\equiv 0$. Thus that requested expression is not the general deformation matrix for the printed coefficients.

There is a consistent nearby correction: replace the lower entry of $A_1$ by $-2w/u$. Then $V=-2w/u$, so $h=z/2$ and $\theta =0$, and the same derivation proves exactlyFor this corrected system use the same prefactor formulas with $V=-2w/u$: $\rho =-w(w+z)/2-w(u^{\prime }/u{)}^2$ and both diagonal entries of $G_2$ are ${\rho }^2/2+w/4$. Compatibility determines the allowed $u,w$ in the next solution. This correction is stated explicitly rather than silently substituting it for the original PDF.