& u_{13} \\ 
0 & u_{22} & u_{23} \\ 
0 & 0 & u_{33}\end{pmatrix}$$This equation is very easy to solve, $$\begin{aligned}
u_{11}=\sqrt{K_{11}} &u_{12}=\frac{\displaystyle{K_{12}}}{\displaystyle\sqrt{K_{11}}}&u_{13}=\frac{K_{13}}{\sqrt{K_{11}}}  \label{utri} \\
u_{22}=\sqrt{K_{22}-\frac{\displaystyle{|K_{12}|^{2}}}{\displaystyle{K_{11}}}}~~ &~~u_{23}=\frac{\displaystyle{K_{23}K_{11}-K_{13}K_{12}^{\ast }}}{\displaystyle\sqrt{K_{22}K_{11}^{2}-K_{11}|K_{12}|^{2}}}~~&~~u_{33}=\sqrt{K_{33}-|u_{23}|^{2}-|u_{13}|^{2}}  \notag\end{aligned}$$The inverse of this upper triangular matrix is also upper triangular, and it is also easily obtained. Obviously we could have chosen to use lower triangular matrices $L$ instead of the upper triangular matrices $U$ and the explicit form of the $L$ would then have been obtained in a similar way in terms of $K$.

This form for the square root of the Kähler matrix is different from the standard form used in the literature [@dreiner; @peddie]. Clearly the standard form is related to our triangular form by an unobservable WBT and therefore the two forms are physically indistinguishable. However it is evident from Eq. (\[utri\]) that from the point of view of calculability it is much simpler to obtain the triangular form than the standard form.

The Kähler corrections to Yukawa couplings
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The form of the Yukawa coupling matrix
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To proceed we need to know the form of the Yukawa couplings coming from the superpotential. A fit to the data using a form for the Yukawa matrices where the 